From 6446c86686dc2274a43bfd01bff79c54fd9d215f Mon Sep 17 00:00:00 2001 From: Lukas Werner Date: Sun, 5 Jun 2022 22:01:42 -0700 Subject: [PATCH] save progress --- 1/main.py | 11 + 10/main.go | 34 + 10/main.py | 12 + 11/main.py | 83 + 2/main.py | 20 + 3/main.py | 24 + 4/main.py | 27 + 5/main.py | 22 + 6/main.py | 6 + 7/main.py | 13 + 8/main.py | 43 + 9/main.go | 22 + 9/main.py | 22 + project_euler_problems.txt | 12798 +++++++++++++++++++++++++++++++++++ 14 files changed, 13137 insertions(+) create mode 100644 1/main.py create mode 100644 10/main.go create mode 100644 10/main.py create mode 100644 11/main.py create mode 100644 2/main.py create mode 100644 3/main.py create mode 100644 4/main.py create mode 100644 5/main.py create mode 100644 6/main.py create mode 100644 7/main.py create mode 100644 8/main.py create mode 100644 9/main.go create mode 100644 9/main.py create mode 100644 project_euler_problems.txt diff --git a/1/main.py b/1/main.py new file mode 100644 index 0000000..0473d6d --- /dev/null +++ b/1/main.py @@ -0,0 +1,11 @@ + + +threes = [] +fives = [] + +for i in range(1000): + if i % 3 == 0: + threes.append(i) + elif i % 5 == 0: + fives.append(i) +print(sum(threes) + sum(fives)) diff --git a/10/main.go b/10/main.go new file mode 100644 index 0000000..8604a38 --- /dev/null +++ b/10/main.go @@ -0,0 +1,34 @@ +package main + +import "fmt" + +func main() { + primes := []int{2, 3} + + for i := 3; i < 2_050_000; i++ { + primeFlag := false + for _, j := range primes { + if i%j == 0 { + primeFlag = true + break + } + } + if !primeFlag { + primes = append(primes, i) + } + } + fmt.Println(primes[len(primes)-1]) + + ok_primes := []int{} + // find all the primes that are below 2 mil + for i := 0; primes[i] < 2_000_000; i++ { + ok_primes = append(ok_primes, primes[i]) + } + fmt.Println(ok_primes[len(ok_primes)-1]) + // find the sum + total := 0 + for _, v := range ok_primes { + total += v + } + fmt.Println(total) +} diff --git a/10/main.py b/10/main.py new file mode 100644 index 0000000..0b4eaf2 --- /dev/null +++ b/10/main.py @@ -0,0 +1,12 @@ +primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] + +for i in range(100, 105_000): + if i % 1000 == 0: + print(f"status {i} len({len(primes)})", end="\r") + for j in primes: + if i % j == 0: + break + else: + primes.append(i) + + diff --git a/11/main.py b/11/main.py new file mode 100644 index 0000000..7f0a1ab --- /dev/null +++ b/11/main.py @@ -0,0 +1,83 @@ +matrix = [ + [8, 2, 22, 97, 38, 15, 0, 40, 0, 75, 4, 5, 7, 78, 52, 12, 50, 77, 91, 8], + [49, 49, 99, 40, 17, 81, 18, 57, 60, 87, 17, 40, 98, 43, 69, 48, 4, 56, 62, 0], + [81, 49, 31, 73, 55, 79, 14, 29, 93, 71, 40, 67, 53, 88, 30, 3, 49, 13, 36, 65], + [52, 70, 95, 23, 4, 60, 11, 42, 69, 24, 68, 56, 1, 32, 56, 71, 37, 2, 36, 91], + [22, 31, 16, 71, 51, 67, 63, 89, 41, 92, 36, 54, 22, 40, 40, 28, 66, 33, 13, 80], + [24, 47, 32, 60, 99, 3, 45, 2, 44, 75, 33, 53, 78, 36, 84, 20, 35, 17, 12, 50], + [32, 98, 81, 28, 64, 23, 67, 10, 26, 38, 40, 67, 59, 54, 70, 66, 18, 38, 64, 70], + [67, 26, 20, 68, 2, 62, 12, 20, 95, 63, 94, 39, 63, 8, 40, 91, 66, 49, 94, 21], + [24, 55, 58, 5, 66, 73, 99, 26, 97, 17, 78, 78, 96, 83, 14, 88, 34, 89, 63, 72], + [21, 36, 23, 9, 75, 0, 76, 44, 20, 45, 35, 14, 0, 61, 33, 97, 34, 31, 33, 95], + [78, 17, 53, 28, 22, 75, 31, 67, 15, 94, 3, 80, 4, 62, 16, 14, 9, 53, 56, 92], + [16, 39, 5, 42, 96, 35, 31, 47, 55, 58, 88, 24, 0, 17, 54, 24, 36, 29, 85, 57], + [86, 56, 0, 48, 35, 71, 89, 7, 5, 44, 44, 37, 44, 60, 21, 58, 51, 54, 17, 58], + [19, 80, 81, 68, 5, 94, 47, 69, 28, 73, 92, 13, 86, 52, 17, 77, 4, 89, 55, 40], + [4, 52, 8, 83, 97, 35, 99, 16, 7, 97, 57, 32, 16, 26, 26, 79, 33, 27, 98, 66], + [88, 36, 68, 87, 57, 62, 20, 72, 3, 46, 33, 67, 46, 55, 12, 32, 63, 93, 53, 69], + [4, 42, 16, 73, 38, 25, 39, 11, 24, 94, 72, 18, 8, 46, 29, 32, 40, 62, 76, 36], + [20, 69, 36, 41, 72, 30, 23, 88, 34, 62, 99, 69, 82, 67, 59, 85, 74, 4, 36, 16], + [20, 73, 35, 29, 78, 31, 90, 1, 74, 31, 49, 71, 48, 86, 81, 16, 23, 57, 5, 54], + [1, 70, 54, 71, 83, 51, 54, 69, 16, 92, 33, 48, 61, 43, 52, 1, 89, 19, 67, 48] + ] + +def down_diagonal(sx, sy, ln): + results = [] + for i in range(ln): + results.append(matrix[sx+i][sy+i]) + return results + +def up_diagonal(sx, sy, ln): + results = [] + for i in range(ln): + results.append(matrix[sx-i][sy+i]) + return results + +def down_col(sx, sy, ln): + results = [] + for i in range(ln): + results.append(matrix[sx-i][sy]) + return results + +def left_row(sx, sy, ln): + results = [] + for i in range(ln): + results.append(matrix[sx][sy+i]) + return results + +def product(l): + result = 1 + for i in l: + result *= i + return result + +dd_max = 0 +for i in range(17): + for j in range(17): + prod = product(down_diagonal(i,j,4)) + if prod > dd_max: + dd_max = prod + +ud_max = 0 +for i in range(3, 20): + for j in range(0, 17): + prod = product(up_diagonal(i, j, 4)) + if prod > ud_max: + ud_max = prod + +col_max = 0 +for i in range(0, 17): + for j in range(20): + prod = product(down_col(i, j, 4)) + if prod > col_max: + col_max = prod + +row_max = 0 +for i in range(20): + for j in range(0, 17): + prod = product(left_row(i, j, 4)) + if prod > row_max: + row_max = prod + +print(max([dd_max, ud_max, col_max, row_max])) + diff --git a/2/main.py b/2/main.py new file mode 100644 index 0000000..3c80129 --- /dev/null +++ b/2/main.py @@ -0,0 +1,20 @@ + + +fibs = [1] + +a = 1 +b = 1 + +while a <= 4_000_000 or b <= 4_000_000: + num = a + b + fibs.append(num) + a = b + b = num +print(fibs) + +even_fibs = [] +for i in fibs: + if i % 2 == 0: + even_fibs.append(i) + +print(sum(even_fibs)) diff --git a/3/main.py b/3/main.py new file mode 100644 index 0000000..c74218d --- /dev/null +++ b/3/main.py @@ -0,0 +1,24 @@ +num = 600_851_475_143 +#num = 13195 +primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] + + +for i in range(100, 40000): + for j in primes: + if i % j == 0: + break + else: + primes.append(i) + +print("Got Primes") + + +working_primes = [] +for i in primes: + dividend = num / i + #if float(dividend).is_integer(): + # working_primes.append(i) + if str(float(dividend)).split(".")[1] == "0": + working_primes.append(i) + +print(working_primes) diff --git a/4/main.py b/4/main.py new file mode 100644 index 0000000..73bb37b --- /dev/null +++ b/4/main.py @@ -0,0 +1,27 @@ +three_digit_nums = [i for i in range(100, 999)] + + +def is_palindrome(n): + norm = str(n) + even = len(norm) % 2 == 0 + if even: + half = norm[:int(len(norm)/2)] + other_half = [c for c in norm[int(len(norm)/2):]] + other_half.reverse() + for i, c in enumerate(half): + if c != other_half[i]: + return False + else: + return True + else: + return False + +products = [] + +for i in three_digit_nums: + for j in three_digit_nums: + product = i*j + if is_palindrome(product): + products.append(product) + +print(max(products)) diff --git a/5/main.py b/5/main.py new file mode 100644 index 0000000..f0f2eb1 --- /dev/null +++ b/5/main.py @@ -0,0 +1,22 @@ +num = 2520 + +factors = [i for i in range(1, 20+1)] + + +def perf_num(num): + for f in factors: + if num % f == 0: + pass + else: + return False + else: + return True + +i = 1 +while True: + if i % 1000 == 0: + print("status", i, end='\r') + if perf_num(i): + print("found num", i) + break + i += 1 diff --git a/6/main.py b/6/main.py new file mode 100644 index 0000000..30fea4b --- /dev/null +++ b/6/main.py @@ -0,0 +1,6 @@ +nums = [i for i in range(1, 100+1)] + +square_of_sums = sum(nums)**2 +sum_of_squares = sum([i**2 for i in nums]) + +print(square_of_sums-sum_of_squares) diff --git a/7/main.py b/7/main.py new file mode 100644 index 0000000..4e10c16 --- /dev/null +++ b/7/main.py @@ -0,0 +1,13 @@ +primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97] + +for i in range(100, 105_000): + if i % 1000 == 0: + print(f"status {i} len({len(primes)})", end="\r") + for j in primes: + if i % j == 0: + break + else: + primes.append(i) + +print() +print(primes[10_000]) diff --git a/8/main.py b/8/main.py new file mode 100644 index 0000000..9c7c603 --- /dev/null +++ b/8/main.py @@ -0,0 +1,43 @@ +big_num = """73167176531330624919225119674426574742355349194934 +96983520312774506326239578318016984801869478851843 +85861560789112949495459501737958331952853208805511 +12540698747158523863050715693290963295227443043557 +66896648950445244523161731856403098711121722383113 +62229893423380308135336276614282806444486645238749 +30358907296290491560440772390713810515859307960866 +70172427121883998797908792274921901699720888093776 +65727333001053367881220235421809751254540594752243 +52584907711670556013604839586446706324415722155397 +53697817977846174064955149290862569321978468622482 +83972241375657056057490261407972968652414535100474 +82166370484403199890008895243450658541227588666881 +16427171479924442928230863465674813919123162824586 +17866458359124566529476545682848912883142607690042 +24219022671055626321111109370544217506941658960408 +07198403850962455444362981230987879927244284909188 +84580156166097919133875499200524063689912560717606 +05886116467109405077541002256983155200055935729725 +71636269561882670428252483600823257530420752963450 +""" +big_num = big_num.replace("\n", "") + + +window = 13 + +biggest_product = 1 +start = 0 +end = window + +while start+window < len(big_num): + part = big_num[start:end] + digits = [int(c) for c in part] + total = 1 + for i in digits: + total *= i + if total > biggest_product: + biggest_product = total + start += 1 + end += 1 +print(biggest_product) + + diff --git a/9/main.go b/9/main.go new file mode 100644 index 0000000..42b2607 --- /dev/null +++ b/9/main.go @@ -0,0 +1,22 @@ +package main + +import "fmt" + +func main() { + for a := 1; a < 1000; a++ { + for b := 1; b < 1000; b++ { + for c := 1; c < 1000; c++ { + if c%100 == 0 { + fmt.Printf("iter %v %v %v\r", a, b, c) + } + if is_pythag(a, b, c) && a+b+c == 1000 { + fmt.Printf("\n\n winner {a:%v b:%v c:%v} sum: %v\n", a, b, c, a*b*c) + } + } + } + } +} + +func is_pythag(a int, b int, c int) bool { + return (a < b && b < c) && (a*a+b*b == c*c) +} diff --git a/9/main.py b/9/main.py new file mode 100644 index 0000000..924b6f9 --- /dev/null +++ b/9/main.py @@ -0,0 +1,22 @@ +a = 1 +b = 1 +c = 1 + +def is_pythag(a, b, c): + return a < b < c and a**2 + b**2 == c**2 + + +for a in range(1, 1000): + for b in range(1, 1000): + for c in range(1, 1000): + + if c % 100 == 0: + print(f"iter {a} {b} {c}", end="\r") + + + + + if is_pythag(a, b, c) and a+b+c==1000: + print(f"a {a} b {b} c {c}") + else: + continue diff --git a/project_euler_problems.txt b/project_euler_problems.txt new file mode 100644 index 0000000..c59f46d --- /dev/null +++ b/project_euler_problems.txt @@ -0,0 +1,12798 @@ + +Project Euler is protected under +Attribution-Non-Commercial-Share Alike 2.0 UK: England & Wales + +^sup and [sub] are used for subscripts/superscripts. +Many symbols are utf8 + +Solutions are hashed with md5sum +echo -n 'myanswer' | md5sum +some systems, like macos, use the command 'md5' instead of 'md5sum' + + + + + + +Problem 1 +========= + + + If we list all the natural numbers below 10 that are multiples of 3 or 5, + we get 3, 5, 6 and 9. The sum of these multiples is 23. + + Find the sum of all the multiples of 3 or 5 below 1000. + + + Answer: e1edf9d1967ca96767dcc2b2d6df69f4 + + +Problem 2 +========= + + + Each new term in the Fibonacci sequence is generated by adding the + previous two terms. By starting with 1 and 2, the first 10 terms will be: + + 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... + + By considering the terms in the Fibonacci sequence whose values do not + exceed four million, find the sum of the even-valued terms. + + + Answer: 4194eb91842c8e7e6df099ca73c38f28 + + +Problem 3 +========= + + + The prime factors of 13195 are 5, 7, 13 and 29. + + What is the largest prime factor of the number 600851475143 ? + + + Answer: 94c4dd41f9dddce696557d3717d98d82 + + +Problem 4 +========= + + + A palindromic number reads the same both ways. The largest palindrome made + from the product of two 2-digit numbers is 9009 = 91 × 99. + + Find the largest palindrome made from the product of two 3-digit numbers. + + + Answer: d4cfc27d16ea72a96b83d9bdef6ce2ec + + +Problem 5 +========= + + + 2520 is the smallest number that can be divided by each of the numbers + from 1 to 10 without any remainder. + + What is the smallest positive number that is evenly divisible by all of + the numbers from 1 to 20? + + + Answer: bc0d0a22a7a46212135ed0ba77d22f3a + + +Problem 6 +========= + + + The sum of the squares of the first ten natural numbers is, + + 1^2 + 2^2 + ... + 10^2 = 385 + + The square of the sum of the first ten natural numbers is, + + (1 + 2 + ... + 10)^2 = 55^2 = 3025 + + Hence the difference between the sum of the squares of the first ten + natural numbers and the square of the sum is 3025 − 385 = 2640. + + Find the difference between the sum of the squares of the first one + hundred natural numbers and the square of the sum. + + + Answer: 867380888952c39a131fe1d832246ecc + + +Problem 7 +========= + + + By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see + that the 6th prime is 13. + + What is the 10 001st prime number? + + + Answer: 8c32ab09ec0210af60d392e9b2009560 + + +Problem 8 +========= + + + The four adjacent digits in the 1000-digit number that have the greatest + product are 9 × 9 × 8 × 9 = 5832. + + 73167176531330624919225119674426574742355349194934 + 96983520312774506326239578318016984801869478851843 + 85861560789112949495459501737958331952853208805511 + 12540698747158523863050715693290963295227443043557 + 66896648950445244523161731856403098711121722383113 + 62229893423380308135336276614282806444486645238749 + 30358907296290491560440772390713810515859307960866 + 70172427121883998797908792274921901699720888093776 + 65727333001053367881220235421809751254540594752243 + 52584907711670556013604839586446706324415722155397 + 53697817977846174064955149290862569321978468622482 + 83972241375657056057490261407972968652414535100474 + 82166370484403199890008895243450658541227588666881 + 16427171479924442928230863465674813919123162824586 + 17866458359124566529476545682848912883142607690042 + 24219022671055626321111109370544217506941658960408 + 07198403850962455444362981230987879927244284909188 + 84580156166097919133875499200524063689912560717606 + 05886116467109405077541002256983155200055935729725 + 71636269561882670428252483600823257530420752963450 + + Find the thirteen adjacent digits in the 1000-digit number that have the + greatest product. What is the value of this product? + + + Answer: 0f53ea7949d32ef24f9186207600403c + + +Problem 9 +========= + + + A Pythagorean triplet is a set of three natural numbers, a < b < c, for + which, + + a^2 + b^2 = c^2 + + For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2. + + There exists exactly one Pythagorean triplet for which a + b + c = 1000. + Find the product abc. + + + Answer: 24eaa9820350012ff678de47cb85b639 + + +Problem 10 +========== + + + The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17. + + Find the sum of all the primes below two million. + + + Answer: d915b2a9ac8749a6b837404815f1ae25 + + +Problem 11 +========== + + + In the 20×20 grid below, four numbers along a diagonal line have been + marked in red. + + 08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08 + 49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00 + 81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65 + 52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91 + 22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80 + 24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50 + 32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70 + 67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21 + 24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72 + 21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95 + 78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92 + 16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57 + 86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58 + 19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40 + 04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66 + 88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69 + 04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36 + 20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16 + 20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54 + 01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48 + + The product of these numbers is 26 × 63 × 78 × 14 = 1788696. + + What is the greatest product of four adjacent numbers in the same + direction (up, down, left, right, or diagonally) in the 20×20 grid? + + + Answer: 678f5d2e1eaa42f04fa53411b4f441ac + + +Problem 12 +========== + + + The sequence of triangle numbers is generated by adding the natural + numbers. So the 7^th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = + 28. The first ten terms would be: + + 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... + + Let us list the factors of the first seven triangle numbers: + +  1: 1 +  3: 1,3 +  6: 1,2,3,6 + 10: 1,2,5,10 + 15: 1,3,5,15 + 21: 1,3,7,21 + 28: 1,2,4,7,14,28 + + We can see that 28 is the first triangle number to have over five + divisors. + + What is the value of the first triangle number to have over five hundred + divisors? + + + Answer: 8091de7d285989bbfa9a2f9f3bdcc7c0 + + +Problem 13 +========== + + + Work out the first ten digits of the sum of the following one-hundred + 50-digit numbers. + + 37107287533902102798797998220837590246510135740250 + 46376937677490009712648124896970078050417018260538 + 74324986199524741059474233309513058123726617309629 + 91942213363574161572522430563301811072406154908250 + 23067588207539346171171980310421047513778063246676 + 89261670696623633820136378418383684178734361726757 + 28112879812849979408065481931592621691275889832738 + 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71329612474782464538636993009049310363619763878039 + 62184073572399794223406235393808339651327408011116 + 66627891981488087797941876876144230030984490851411 + 60661826293682836764744779239180335110989069790714 + 85786944089552990653640447425576083659976645795096 + 66024396409905389607120198219976047599490197230297 + 64913982680032973156037120041377903785566085089252 + 16730939319872750275468906903707539413042652315011 + 94809377245048795150954100921645863754710598436791 + 78639167021187492431995700641917969777599028300699 + 15368713711936614952811305876380278410754449733078 + 40789923115535562561142322423255033685442488917353 + 44889911501440648020369068063960672322193204149535 + 41503128880339536053299340368006977710650566631954 + 81234880673210146739058568557934581403627822703280 + 82616570773948327592232845941706525094512325230608 + 22918802058777319719839450180888072429661980811197 + 77158542502016545090413245809786882778948721859617 + 72107838435069186155435662884062257473692284509516 + 20849603980134001723930671666823555245252804609722 + 53503534226472524250874054075591789781264330331690 + + Answer: 361113f19fd302adc31268f8283a4f2d + + +Problem 14 +========== + + + The following iterative sequence is defined for the set of positive + integers: + + n → n/2 (n is even) + n → 3n + 1 (n is odd) + + Using the rule above and starting with 13, we generate the following + sequence: + + 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1 + + It can be seen that this sequence (starting at 13 and finishing at 1) + contains 10 terms. Although it has not been proved yet (Collatz Problem), + it is thought that all starting numbers finish at 1. + + Which starting number, under one million, produces the longest chain? + + NOTE: Once the chain starts the terms are allowed to go above one million. + + + Answer: 5052c3765262bb2c6be537abd60b305e + + +Problem 15 +========== + + + Starting in the top left corner of a 2×2 grid, and only being able to move + to the right and down, there are exactly 6 routes to the bottom right + corner. + + How many such routes are there through a 20×20 grid? + + + p_015.gif + Answer: 928f3957168ac592c4215dcd04e0b678 + + +Problem 16 +========== + + + 2^15 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26. + + What is the sum of the digits of the number 2^1000? + + + Answer: 6a5889bb0190d0211a991f47bb19a777 + + +Problem 17 +========== + + + If the numbers 1 to 5 are written out in words: one, two, three, four, + five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total. + + If all the numbers from 1 to 1000 (one thousand) inclusive were written + out in words, how many letters would be used? + + NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and + forty-two) contains 23 letters and 115 (one hundred and fifteen) contains + 20 letters. The use of "and" when writing out numbers is in compliance + with British usage. + + + Answer: 6a979d4a9cf85135408529edc8a133d0 + + +Problem 18 +========== + + + By starting at the top of the triangle below and moving to adjacent + numbers on the row below, the maximum total from top to bottom is 23. + + 3 + 7 4 + 2 4 6 + 8 5 9 3 + + That is, 3 + 7 + 4 + 9 = 23. + + Find the maximum total from top to bottom of the triangle below: + + 75 + 95 64 + 17 47 82 + 18 35 87 10 + 20 04 82 47 65 + 19 01 23 75 03 34 + 88 02 77 73 07 63 67 + 99 65 04 28 06 16 70 92 + 41 41 26 56 83 40 80 70 33 + 41 48 72 33 47 32 37 16 94 29 + 53 71 44 65 25 43 91 52 97 51 14 + 70 11 33 28 77 73 17 78 39 68 17 57 + 91 71 52 38 17 14 91 43 58 50 27 29 48 + 63 66 04 68 89 53 67 30 73 16 69 87 40 31 + 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23 + + NOTE: As there are only 16384 routes, it is possible to solve this problem + by trying every route. However, [1]Problem 67, is the same challenge with + a triangle containing one-hundred rows; it cannot be solved by brute + force, and requires a clever method! ;o) + + + Visible links + 1. problem=67 + Answer: 708f3cf8100d5e71834b1db77dfa15d6 + + +Problem 19 +========== + + + You are given the following information, but you may prefer to do some + research for yourself. + + • 1 Jan 1900 was a Monday. + • Thirty days has September, + April, June and November. + All the rest have thirty-one, + Saving February alone, + Which has twenty-eight, rain or shine. + And on leap years, twenty-nine. + • A leap year occurs on any year evenly divisible by 4, but not on a + century unless it is divisible by 400. + + How many Sundays fell on the first of the month during the twentieth + century (1 Jan 1901 to 31 Dec 2000)? + + + Answer: a4a042cf4fd6bfb47701cbc8a1653ada + + +Problem 20 +========== + + + n! means n × (n − 1) × ... × 3 × 2 × 1 + + For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800, + and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = + 27. + + Find the sum of the digits in the number 100! + + + Answer: 443cb001c138b2561a0d90720d6ce111 + + +Problem 21 +========== + + + Let d(n) be defined as the sum of proper divisors of n (numbers less than + n which divide evenly into n). + If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair + and each of a and b are called amicable numbers. + + For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, + 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, + 2, 4, 71 and 142; so d(284) = 220. + + Evaluate the sum of all the amicable numbers under 10000. + + + Answer: 51e04cd4e55e7e415bf24de9e1b0f3ff + + +Problem 22 +========== + + + Using [1]names.txt, a 46K text file containing over five-thousand first + names, begin by sorting it into alphabetical order. Then working out the + alphabetical value for each name, multiply this value by its alphabetical + position in the list to obtain a name score. + + For example, when the list is sorted into alphabetical order, COLIN, which + is worth 3 + 15 + 12 + 9 + 14 = 53, is the 938th name in the list. So, + COLIN would obtain a score of 938 × 53 = 49714. + + What is the total of all the name scores in the file? + + + Visible links + 1. names.txt + Answer: f2c9c91cb025746f781fa4db8be3983f + + +Problem 23 +========== + + + A perfect number is a number for which the sum of its proper divisors is + exactly equal to the number. For example, the sum of the proper divisors + of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect + number. + + A number n is called deficient if the sum of its proper divisors is less + than n and it is called abundant if this sum exceeds n. + + As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the + smallest number that can be written as the sum of two abundant numbers is + 24. By mathematical analysis, it can be shown that all integers greater + than 28123 can be written as the sum of two abundant numbers. However, + this upper limit cannot be reduced any further by analysis even though it + is known that the greatest number that cannot be expressed as the sum of + two abundant numbers is less than this limit. + + Find the sum of all the positive integers which cannot be written as the + sum of two abundant numbers. + + + Answer: 2c8258c0604152962f7787571511cf28 + + +Problem 24 +========== + + + A permutation is an ordered arrangement of objects. For example, 3124 is + one possible permutation of the digits 1, 2, 3 and 4. If all of the + permutations are listed numerically or alphabetically, we call it + lexicographic order. The lexicographic permutations of 0, 1 and 2 are: + + 012   021   102   120   201   210 + + What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, + 4, 5, 6, 7, 8 and 9? + + + Answer: 7f155b45cb3f0a6e518d59ec348bff84 + + +Problem 25 +========== + + + The Fibonacci sequence is defined by the recurrence relation: + + F[n] = F[n−1] + F[n−2], where F[1] = 1 and F[2] = 1. + + Hence the first 12 terms will be: + + F[1] = 1 + F[2] = 1 + F[3] = 2 + F[4] = 3 + F[5] = 5 + F[6] = 8 + F[7] = 13 + F[8] = 21 + F[9] = 34 + F[10] = 55 + F[11] = 89 + F[12] = 144 + + The 12th term, F[12], is the first term to contain three digits. + + What is the first term in the Fibonacci sequence to contain 1000 digits? + + + Answer: a376802c0811f1b9088828288eb0d3f0 + + +Problem 26 +========== + + + A unit fraction contains 1 in the numerator. The decimal representation of + the unit fractions with denominators 2 to 10 are given: + + 1/2 =  0.5 + 1/3 =  0.(3) + 1/4 =  0.25 + 1/5 =  0.2 + 1/6 =  0.1(6) + 1/7 =  0.(142857) + 1/8 =  0.125 + 1/9 =  0.(1) + 1/10 =  0.1 + + Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can + be seen that 1/7 has a 6-digit recurring cycle. + + Find the value of d < 1000 for which ^1/[d] contains the longest recurring + cycle in its decimal fraction part. + + + Answer: 6aab1270668d8cac7cef2566a1c5f569 + + +Problem 27 +========== + + + Euler discovered the remarkable quadratic formula: + + n² + n + 41 + + It turns out that the formula will produce 40 primes for the consecutive + values n = 0 to 39. However, when n = 40, 40^2 + 40 + 41 = 40(40 + 1) + 41 + is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly + divisible by 41. + + The incredible formula  n² − 79n + 1601 was discovered, which produces 80 + primes for the consecutive values n = 0 to 79. The product of the + coefficients, −79 and 1601, is −126479. + + Considering quadratics of the form: + + n² + an + b, where |a| < 1000 and |b| < 1000 + + where |n| is the modulus/absolute value of n + e.g. |11| = 11 and |−4| = 4 + + Find the product of the coefficients, a and b, for the quadratic + expression that produces the maximum number of primes for consecutive + values of n, starting with n = 0. + + + Answer: 69d9e3218fd7abb6ff453ea96505183d + + +Problem 28 +========== + + + Starting with the number 1 and moving to the right in a clockwise + direction a 5 by 5 spiral is formed as follows: + + 21 22 23 24 25 + 20  7  8  9 10 + 19  6  1  2 11 + 18  5  4  3 12 + 17 16 15 14 13 + + It can be verified that the sum of the numbers on the diagonals is 101. + + What is the sum of the numbers on the diagonals in a 1001 by 1001 spiral + formed in the same way? + + + Answer: 0d53425bd7c5bf9919df3718c8e49fa6 + + +Problem 29 +========== + + + Consider all integer combinations of a^b for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5: + + 2^2=4, 2^3=8, 2^4=16, 2^5=32 + 3^2=9, 3^3=27, 3^4=81, 3^5=243 + 4^2=16, 4^3=64, 4^4=256, 4^5=1024 + 5^2=25, 5^3=125, 5^4=625, 5^5=3125 + + If they are then placed in numerical order, with any repeats removed, we + get the following sequence of 15 distinct terms: + + 4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125 + + How many distinct terms are in the sequence generated by a^b for 2 ≤ a ≤ + 100 and 2 ≤ b ≤ 100? + + + Answer: 6f0ca67289d79eb35d19decbc0a08453 + + +Problem 30 +========== + + + Surprisingly there are only three numbers that can be written as the sum + of fourth powers of their digits: + + 1634 = 1^4 + 6^4 + 3^4 + 4^4 + 8208 = 8^4 + 2^4 + 0^4 + 8^4 + 9474 = 9^4 + 4^4 + 7^4 + 4^4 + + As 1 = 1^4 is not a sum it is not included. + + The sum of these numbers is 1634 + 8208 + 9474 = 19316. + + Find the sum of all the numbers that can be written as the sum of fifth + powers of their digits. + + + Answer: 27a1779a8a8c323a307ac8a70bc4489d + + +Problem 31 +========== + + + In England the currency is made up of pound, £, and pence, p, and there + are eight coins in general circulation: + + 1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p). + + It is possible to make £2 in the following way: + + 1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p + + How many different ways can £2 be made using any number of coins? + + + Answer: 142dfe4a33d624d2b830a9257e96726d + + +Problem 32 +========== + + + We shall say that an n-digit number is pandigital if it makes use of all + the digits 1 to n exactly once; for example, the 5-digit number, 15234, is + 1 through 5 pandigital. + + The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing + multiplicand, multiplier, and product is 1 through 9 pandigital. + + Find the sum of all products whose multiplicand/multiplier/product + identity can be written as a 1 through 9 pandigital. + + HINT: Some products can be obtained in more than one way so be sure to + only include it once in your sum. + + Answer: 100f6e37d0b0564490a2ee27eff0660d + + +Problem 33 +========== + + + The fraction 49/98 is a curious fraction, as an inexperienced + mathematician in attempting to simplify it may incorrectly believe that + 49/98 = 4/8, which is correct, is obtained by cancelling the 9s. + + We shall consider fractions like, 30/50 = 3/5, to be trivial + examples. + + There are exactly four non-trivial examples of this type of fraction, less + than one in value, and containing two digits in the numerator and + denominator. + + If the product of these four fractions is given in its lowest common + terms, find the value of the denominator. + + + Answer: f899139df5e1059396431415e770c6dd + + +Problem 34 +========== + + + 145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145. + + Find the sum of all numbers which are equal to the sum of the factorial of + their digits. + + Note: as 1! = 1 and 2! = 2 are not sums they are not included. + + + Answer: 60803ea798a0c0dfb7f36397d8d4d772 + + +Problem 35 +========== + + + The number, 197, is called a circular prime because all rotations of the + digits: 197, 971, and 719, are themselves prime. + + There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, + 71, 73, 79, and 97. + + How many circular primes are there below one million? + + + Answer: b53b3a3d6ab90ce0268229151c9bde11 + + +Problem 36 +========== + + + The decimal number, 585 = 1001001001[2] (binary), is palindromic in both + bases. + + Find the sum of all numbers, less than one million, which are palindromic + in base 10 and base 2. + + (Please note that the palindromic number, in either base, may not include + leading zeros.) + + + Answer: 0e175dc2f28833885f62e7345addff03 + + +Problem 37 +========== + + + The number 3797 has an interesting property. Being prime itself, it is + possible to continuously remove digits from left to right, and remain + prime at each stage: 3797, 797, 97, and 7. Similarly we can work from + right to left: 3797, 379, 37, and 3. + + Find the sum of the only eleven primes that are both truncatable from left + to right and right to left. + + NOTE: 2, 3, 5, and 7 are not considered to be truncatable primes. + + + Answer: cace46c61b00de1b60874936a093981d + + +Problem 38 +========== + + + Take the number 192 and multiply it by each of 1, 2, and 3: + + 192 × 1 = 192 + 192 × 2 = 384 + 192 × 3 = 576 + + By concatenating each product we get the 1 to 9 pandigital, 192384576. We + will call 192384576 the concatenated product of 192 and (1,2,3) + + The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, + and 5, giving the pandigital, 918273645, which is the concatenated product + of 9 and (1,2,3,4,5). + + What is the largest 1 to 9 pandigital 9-digit number that can be formed as + the concatenated product of an integer with (1,2, ... , n) where n > 1? + + + Answer: f2a29ede8dc9fae7926dc7a4357ac25e + + +Problem 39 +========== + + + If p is the perimeter of a right angle triangle with integral length + sides, {a,b,c}, there are exactly three solutions for p = 120. + + {20,48,52}, {24,45,51}, {30,40,50} + + For which value of p ≤ 1000, is the number of solutions maximised? + + + Answer: fa83a11a198d5a7f0bf77a1987bcd006 + + +Problem 40 +========== + + + An irrational decimal fraction is created by concatenating the positive + integers: + + 0.123456789101112131415161718192021... + + It can be seen that the 12^th digit of the fractional part is 1. + + If d[n] represents the n^th digit of the fractional part, find the value + of the following expression. + + d[1] × d[10] × d[100] × d[1000] × d[10000] × d[100000] × d[1000000] + + + Answer: 6f3ef77ac0e3619e98159e9b6febf557 + + +Problem 41 +========== + + + We shall say that an n-digit number is pandigital if it makes use of all + the digits 1 to n exactly once. For example, 2143 is a 4-digit pandigital + and is also prime. + + What is the largest n-digit pandigital prime that exists? + + + Answer: d0a1bd6ab4229b2d0754be8923431404 + + +Problem 42 +========== + + + The n^th term of the sequence of triangle numbers is given by, t[n] = + ½n(n+1); so the first ten triangle numbers are: + + 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... + + By converting each letter in a word to a number corresponding to its + alphabetical position and adding these values we form a word value. For + example, the word value for SKY is 19 + 11 + 25 = 55 = t[10]. If the word + value is a triangle number then we shall call the word a triangle word. + + Using [1]words.txt, a 16K text file containing nearly two-thousand common + English words, how many are triangle words? + + + Visible links + 1. words.txt + Answer: 82aa4b0af34c2313a562076992e50aa3 + + +Problem 43 +========== + + + The number, 1406357289, is a 0 to 9 pandigital number because it is made + up of each of the digits 0 to 9 in some order, but it also has a rather + interesting sub-string divisibility property. + + Let d[1] be the 1^st digit, d[2] be the 2^nd digit, and so on. In this + way, we note the following: + + • d[2]d[3]d[4]=406 is divisible by 2 + • d[3]d[4]d[5]=063 is divisible by 3 + • d[4]d[5]d[6]=635 is divisible by 5 + • d[5]d[6]d[7]=357 is divisible by 7 + • d[6]d[7]d[8]=572 is divisible by 11 + • d[7]d[8]d[9]=728 is divisible by 13 + • d[8]d[9]d[10]=289 is divisible by 17 + + Find the sum of all 0 to 9 pandigital numbers with this property. + + + Answer: 115253b7721af0fdff25cd391dfc70cf + + +Problem 44 +========== + + + Pentagonal numbers are generated by the formula, P[n]=n(3n−1)/2. The first + ten pentagonal numbers are: + + 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ... + + It can be seen that P[4] + P[7] = 22 + 70 = 92 = P[8]. However, their + difference, 70 − 22 = 48, is not pentagonal. + + Find the pair of pentagonal numbers, P[j] and P[k], for which their sum + and difference are pentagonal and D = |P[k] − P[j]| is minimised; what is + the value of D? + + + Answer: 2c2556cb85621309ca647465ffa62370 + + +Problem 45 +========== + + + Triangle, pentagonal, and hexagonal numbers are generated by the following + formulae: + + Triangle   T[n]=n(n+1)/2   1, 3, 6, 10, 15, ... + Pentagonal   P[n]=n(3n−1)/2   1, 5, 12, 22, 35, ... + Hexagonal   H[n]=n(2n−1)   1, 6, 15, 28, 45, ... + + It can be verified that T[285] = P[165] = H[143] = 40755. + + Find the next triangle number that is also pentagonal and hexagonal. + + + Answer: 30dfe3e3b286add9d12e493ca7be63fc + + +Problem 46 +========== + + + It was proposed by Christian Goldbach that every odd composite number can + be written as the sum of a prime and twice a square. + + 9 = 7 + 2×1^2 + 15 = 7 + 2×2^2 + 21 = 3 + 2×3^2 + 25 = 7 + 2×3^2 + 27 = 19 + 2×2^2 + 33 = 31 + 2×1^2 + + It turns out that the conjecture was false. + + What is the smallest odd composite that cannot be written as the sum of a + prime and twice a square? + + + Answer: 89abe98de6071178edb1b28901a8f459 + + +Problem 47 +========== + + + The first two consecutive numbers to have two distinct prime factors are: + + 14 = 2 × 7 + 15 = 3 × 5 + + The first three consecutive numbers to have three distinct prime factors + are: + + 644 = 2² × 7 × 23 + 645 = 3 × 5 × 43 + 646 = 2 × 17 × 19. + + Find the first four consecutive integers to have four distinct prime + factors. What is the first of these numbers? + + + Answer: 748f517ecdc29106e2738f88aa7530f4 + + +Problem 48 +========== + + + The series, 1^1 + 2^2 + 3^3 + ... + 10^10 = 10405071317. + + Find the last ten digits of the series, 1^1 + 2^2 + 3^3 + ... + 1000^1000. + + + Answer: 0829124724747ae1c65da8cae5263346 + + +Problem 49 +========== + + + The arithmetic sequence, 1487, 4817, 8147, in which each of the terms + increases by 3330, is unusual in two ways: (i) each of the three terms are + prime, and, (ii) each of the 4-digit numbers are permutations of one + another. + + There are no arithmetic sequences made up of three 1-, 2-, or 3-digit + primes, exhibiting this property, but there is one other 4-digit + increasing sequence. + + What 12-digit number do you form by concatenating the three terms in this + sequence? + + + Answer: 0b99933d3e2a9addccbb663d46cbb592 + + +Problem 50 +========== + + + The prime 41, can be written as the sum of six consecutive primes: + + 41 = 2 + 3 + 5 + 7 + 11 + 13 + + This is the longest sum of consecutive primes that adds to a prime below + one-hundred. + + The longest sum of consecutive primes below one-thousand that adds to a + prime, contains 21 terms, and is equal to 953. + + Which prime, below one-million, can be written as the sum of the most + consecutive primes? + + + Answer: 73229bab6c5dc1c7cf7a4fa123caf6bc + + +Problem 51 +========== + + + By replacing the 1^st digit of the 2-digit number *3, it turns out that + six of the nine possible values: 13, 23, 43, 53, 73, and 83, are all + prime. + + By replacing the 3^rd and 4^th digits of 56**3 with the same digit, this + 5-digit number is the first example having seven primes among the ten + generated numbers, yielding the family: 56003, 56113, 56333, 56443, 56663, + 56773, and 56993. Consequently 56003, being the first member of this + family, is the smallest prime with this property. + + Find the smallest prime which, by replacing part of the number (not + necessarily adjacent digits) with the same digit, is part of an eight + prime value family. + + + Answer: e2a8daa5eb919905dadd795593084c22 + + +Problem 52 +========== + + + It can be seen that the number, 125874, and its double, 251748, contain + exactly the same digits, but in a different order. + + Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, + contain the same digits. + + + Answer: a420384997c8a1a93d5a84046117c2aa + + +Problem 53 +========== + + + There are exactly ten ways of selecting three from five, 12345: + + 123, 124, 125, 134, 135, 145, 234, 235, 245, and 345 + + In combinatorics, we use the notation, ^5C[3] = 10. + + In general, + + ^nC[r] = n! ,where r ≤ n, n! = n×(n−1)×...×3×2×1, and 0! = 1. + r!(n−r)! + + It is not until n = 23, that a value exceeds one-million: ^23C[10] = + 1144066. + + How many, not necessarily distinct, values of  ^nC[r], for 1 ≤ n ≤ 100, + are greater than one-million? + + + Answer: e3b21256183cf7c2c7a66be163579d37 + + +Problem 54 +========== + + + In the card game poker, a hand consists of five cards and are ranked, from + lowest to highest, in the following way: + + • High Card: Highest value card. + • One Pair: Two cards of the same value. + • Two Pairs: Two different pairs. + • Three of a Kind: Three cards of the same value. + • Straight: All cards are consecutive values. + • Flush: All cards of the same suit. + • Full House: Three of a kind and a pair. + • Four of a Kind: Four cards of the same value. + • Straight Flush: All cards are consecutive values of same suit. + • Royal Flush: Ten, Jack, Queen, King, Ace, in same suit. + + The cards are valued in the order: + 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace. + + If two players have the same ranked hands then the rank made up of the + highest value wins; for example, a pair of eights beats a pair of fives + (see example 1 below). But if two ranks tie, for example, both players + have a pair of queens, then highest cards in each hand are compared (see + example 4 below); if the highest cards tie then the next highest cards are + compared, and so on. + + Consider the following five hands dealt to two players: + + Hand   Player 1   Player 2   Winner + 1   5H 5C 6S 7S KD   2C 3S 8S 8D TD   Player 2 + Pair of Fives Pair of Eights + 2   5D 8C 9S JS AC   2C 5C 7D 8S QH   Player 1 + Highest card Ace Highest card Queen + 3   2D 9C AS AH AC   3D 6D 7D TD QD   Player 2 + Three Aces Flush with Diamonds + 4D 6S 9H QH QC 3D 6D 7H QD QS + 4   Pair of Queens   Pair of Queens   Player 1 + Highest card Nine Highest card Seven + 2H 2D 4C 4D 4S 3C 3D 3S 9S 9D + 5   Full House   Full House   Player 1 + With Three Fours with Three Threes + + The file, [1]poker.txt, contains one-thousand random hands dealt to two + players. Each line of the file contains ten cards (separated by a single + space): the first five are Player 1's cards and the last five are Player + 2's cards. You can assume that all hands are valid (no invalid characters + or repeated cards), each player's hand is in no specific order, and in + each hand there is a clear winner. + + How many hands does Player 1 win? + + + Visible links + 1. poker.txt + Answer: 142949df56ea8ae0be8b5306971900a4 + + +Problem 55 +========== + + + If we take 47, reverse and add, 47 + 74 = 121, which is palindromic. + + Not all numbers produce palindromes so quickly. For example, + + 349 + 943 = 1292, + 1292 + 2921 = 4213 + 4213 + 3124 = 7337 + + That is, 349 took three iterations to arrive at a palindrome. + + Although no one has proved it yet, it is thought that some numbers, like + 196, never produce a palindrome. A number that never forms a palindrome + through the reverse and add process is called a Lychrel number. Due to the + theoretical nature of these numbers, and for the purpose of this problem, + we shall assume that a number is Lychrel until proven otherwise. In + addition you are given that for every number below ten-thousand, it will + either (i) become a palindrome in less than fifty iterations, or, (ii) no + one, with all the computing power that exists, has managed so far to map + it to a palindrome. In fact, 10677 is the first number to be shown to + require over fifty iterations before producing a palindrome: + 4668731596684224866951378664 (53 iterations, 28-digits). + + Surprisingly, there are palindromic numbers that are themselves Lychrel + numbers; the first example is 4994. + + How many Lychrel numbers are there below ten-thousand? + + NOTE: Wording was modified slightly on 24 April 2007 to emphasise the + theoretical nature of Lychrel numbers. + + + Answer: 077e29b11be80ab57e1a2ecabb7da330 + + +Problem 56 +========== + + + A googol (10^100) is a massive number: one followed by one-hundred zeros; + 100^100 is almost unimaginably large: one followed by two-hundred zeros. + Despite their size, the sum of the digits in each number is only 1. + + Considering natural numbers of the form, a^b, where a, b < 100, what is + the maximum digital sum? + + + Answer: c22abfa379f38b5b0411bc11fa9bf92f + + +Problem 57 +========== + + + It is possible to show that the square root of two can be expressed as an + infinite continued fraction. + + √ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213... + + By expanding this for the first four iterations, we get: + + 1 + 1/2 = 3/2 = 1.5 + 1 + 1/(2 + 1/2) = 7/5 = 1.4 + 1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666... + 1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379... + + The next three expansions are 99/70, 239/169, and 577/408, but the eighth + expansion, 1393/985, is the first example where the number of digits in + the numerator exceeds the number of digits in the denominator. + + In the first one-thousand expansions, how many fractions contain a + numerator with more digits than denominator? + + + Answer: b3e3e393c77e35a4a3f3cbd1e429b5dc + + +Problem 58 +========== + + + Starting with 1 and spiralling anticlockwise in the following way, a + square spiral with side length 7 is formed. + + 37 36 35 34 33 32 31 + 38 17 16 15 14 13 30 + 39 18  5  4  3 12 29 + 40 19  6  1  2 11 28 + 41 20  7  8  9 10 27 + 42 21 22 23 24 25 26 + 43 44 45 46 47 48 49 + + It is interesting to note that the odd squares lie along the bottom right + diagonal, but what is more interesting is that 8 out of the 13 numbers + lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%. + + If one complete new layer is wrapped around the spiral above, a square + spiral with side length 9 will be formed. If this process is continued, + what is the side length of the square spiral for which the ratio of primes + along both diagonals first falls below 10%? + + + Answer: b62fc92a2561538525c89be63f36bf7b + + +Problem 59 +========== + + + Each character on a computer is assigned a unique code and the preferred + standard is ASCII (American Standard Code for Information Interchange). + For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107. + + A modern encryption method is to take a text file, convert the bytes to + ASCII, then XOR each byte with a given value, taken from a secret key. The + advantage with the XOR function is that using the same encryption key on + the cipher text, restores the plain text; for example, 65 XOR 42 = 107, + then 107 XOR 42 = 65. + + For unbreakable encryption, the key is the same length as the plain text + message, and the key is made up of random bytes. The user would keep the + encrypted message and the encryption key in different locations, and + without both "halves", it is impossible to decrypt the message. + + Unfortunately, this method is impractical for most users, so the modified + method is to use a password as a key. If the password is shorter than the + message, which is likely, the key is repeated cyclically throughout the + message. The balance for this method is using a sufficiently long password + key for security, but short enough to be memorable. + + Your task has been made easy, as the encryption key consists of three + lower case characters. Using [1]cipher1.txt, a file containing the + encrypted ASCII codes, and the knowledge that the plain text must contain + common English words, decrypt the message and find the sum of the ASCII + values in the original text. + + + Visible links + 1. cipher1.txt + Answer: 68f891fe214e2bfa07c998ad5d0a390f + + +Problem 60 +========== + + + The primes 3, 7, 109, and 673, are quite remarkable. By taking any two + primes and concatenating them in any order the result will always be + prime. For example, taking 7 and 109, both 7109 and 1097 are prime. The + sum of these four primes, 792, represents the lowest sum for a set of four + primes with this property. + + Find the lowest sum for a set of five primes for which any two primes + concatenate to produce another prime. + + + Answer: a4b5a70ca8cf24d0eb4330748d1e72e5 + + +Problem 61 +========== + + + Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers + are all figurate (polygonal) numbers and are generated by the following + formulae: + + Triangle   P[3,n]=n(n+1)/2   1, 3, 6, 10, 15, ... + Square   P[4,n]=n^2   1, 4, 9, 16, 25, ... + Pentagonal   P[5,n]=n(3n−1)/2   1, 5, 12, 22, 35, ... + Hexagonal   P[6,n]=n(2n−1)   1, 6, 15, 28, 45, ... + Heptagonal   P[7,n]=n(5n−3)/2   1, 7, 18, 34, 55, ... + Octagonal   P[8,n]=n(3n−2)   1, 8, 21, 40, 65, ... + + The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three + interesting properties. + +  1. The set is cyclic, in that the last two digits of each number is the + first two digits of the next number (including the last number with + the first). +  2. Each polygonal type: triangle (P[3,127]=8128), square (P[4,91]=8281), + and pentagonal (P[5,44]=2882), is represented by a different number in + the set. +  3. This is the only set of 4-digit numbers with this property. + + Find the sum of the only ordered set of six cyclic 4-digit numbers for + which each polygonal type: triangle, square, pentagonal, hexagonal, + heptagonal, and octagonal, is represented by a different number in the + set. + + + Answer: caec17d84884addeec35c3610645ab63 + + +Problem 62 +========== + + + The cube, 41063625 (345^3), can be permuted to produce two other cubes: + 56623104 (384^3) and 66430125 (405^3). In fact, 41063625 is the smallest + cube which has exactly three permutations of its digits which are also + cube. + + Find the smallest cube for which exactly five permutations of its digits + are cube. + + + Answer: 8f46b522b5401b8b6df99a7410eea44b + + +Problem 63 +========== + + + The 5-digit number, 16807=7^5, is also a fifth power. Similarly, the + 9-digit number, 134217728=8^9, is a ninth power. + + How many n-digit positive integers exist which are also an nth power? + + + Answer: f457c545a9ded88f18ecee47145a72c0 + + +Problem 64 +========== + + + All square roots are periodic when written as continued fractions and can + be written in the form: + + √N = a[0] + 1 +   a[1] + 1 +     a[2] + 1 +       a[3] + ... + + For example, let us consider √23: + + √23 = 4 + √23 — 4 = 4 +  1  = 4 +  1 +   1   1 +  √23 – 3 + √23—4 7 + + If we continue we would get the following expansion: + + √23 = 4 + 1 +   1 + 1 +     3 + 1 +       1 + 1 +         8 + ... + + The process can be summarised as follows: + + a[0] = 4,   1  =  √23+4  = 1 +  √23—3 + √23—4 7 7 + a[1] = 1,   7  =  7(√23+3)  = 3 +  √23—3 + √23—3 14 2 + a[2] = 3,   2  =  2(√23+3)  = 1 +  √23—4 + √23—3 14 7 + a[3] = 1,   7  =  7(√23+4)  = 8 +  √23—4 + √23—4 7 + a[4] = 8,   1  =  √23+4  = 1 +  √23—3 + √23—4 7 7 + a[5] = 1,   7  =  7(√23+3)  = 3 +  √23—3 + √23—3 14 2 + a[6] = 3,   2  =  2(√23+3)  = 1 +  √23—4 + √23—3 14 7 + a[7] = 1,   7  =  7(√23+4)  = 8 +  √23—4 + √23—4 7 + + It can be seen that the sequence is repeating. For conciseness, we use the + notation √23 = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats + indefinitely. + + The first ten continued fraction representations of (irrational) square + roots are: + + √2=[1;(2)], period=1 + √3=[1;(1,2)], period=2 + √5=[2;(4)], period=1 + √6=[2;(2,4)], period=2 + √7=[2;(1,1,1,4)], period=4 + √8=[2;(1,4)], period=2 + √10=[3;(6)], period=1 + √11=[3;(3,6)], period=2 + √12= [3;(2,6)], period=2 + √13=[3;(1,1,1,1,6)], period=5 + + Exactly four continued fractions, for N ≤ 13, have an odd period. + + How many continued fractions for N ≤ 10000 have an odd period? + + + Answer: dc960c46c38bd16e953d97cdeefdbc68 + + +Problem 65 +========== + + + The square root of 2 can be written as an infinite continued fraction. + + √2 = 1 + 1 +   2 + 1 +     2 + 1 +       2 + 1 +         2 + ... + + The infinite continued fraction can be written, √2 = [1;(2)], (2) + indicates that 2 repeats ad infinitum. In a similar way, √23 = + [4;(1,3,1,8)]. + + It turns out that the sequence of partial values of continued fractions + for square roots provide the best rational approximations. Let us consider + the convergents for √2. + + 1 + 1 = 3/2 +   2   + + 1 + 1 = 7/5 +   2 + 1 +     2   + + 1 + 1 = 17/12 +   2 + 1   +     2 + 1   +       2   + + 1 + 1 = 41/29 +   2 + 1 +     2 + 1   +       2 + 1   +         2   + + Hence the sequence of the first ten convergents for √2 are: + + 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, + ... + + What is most surprising is that the important mathematical constant, + e = [2; 1,2,1, 1,4,1, 1,6,1 , ... , 1,2k,1, ...]. + + The first ten terms in the sequence of convergents for e are: + + 2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ... + + The sum of digits in the numerator of the 10^th convergent is 1+4+5+7=17. + + Find the sum of digits in the numerator of the 100^th convergent of the + continued fraction for e. + + + Answer: 7a614fd06c325499f1680b9896beedeb + + +Problem 66 +========== + + + Consider quadratic Diophantine equations of the form: + + x^2 – Dy^2 = 1 + + For example, when D=13, the minimal solution in x is 649^2 – 13×180^2 = 1. + + It can be assumed that there are no solutions in positive integers when D + is square. + + By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the + following: + + 3^2 – 2×2^2 = 1 + 2^2 – 3×1^2 = 1 + 9^2 – 5×4^2 = 1 + 5^2 – 6×2^2 = 1 + 8^2 – 7×3^2 = 1 + + Hence, by considering minimal solutions in x for D ≤ 7, the largest x is + obtained when D=5. + + Find the value of D ≤ 1000 in minimal solutions of x for which the largest + value of x is obtained. + + + Answer: 3a066bda8c96b9478bb0512f0a43028c + + +Problem 67 +========== + + + By starting at the top of the triangle below and moving to adjacent + numbers on the row below, the maximum total from top to bottom is 23. + + 3 + 7 4 + 2 4 6 + 8 5 9 3 + + That is, 3 + 7 + 4 + 9 = 23. + + Find the maximum total from top to bottom in [1]triangle.txt, a 15K text + file containing a triangle with one-hundred rows. + + NOTE: This is a much more difficult version of [2]Problem 18. It is not + possible to try every route to solve this problem, as there are 2^99 + altogether! If you could check one trillion (10^12) routes every second it + would take over twenty billion years to check them all. There is an + efficient algorithm to solve it. ;o) + + + Visible links + 1. triangle.txt + 2. problem=18 + Answer: 9d702ffd99ad9c70ac37e506facc8c38 + + +Problem 68 +========== + + + Consider the following "magic" 3-gon ring, filled with the numbers 1 to 6, + and each line adding to nine. + + Working clockwise, and starting from the group of three with the + numerically lowest external node (4,3,2 in this example), each solution + can be described uniquely. For example, the above solution can be + described by the set: 4,3,2; 6,2,1; 5,1,3. + + It is possible to complete the ring with four different totals: 9, 10, 11, + and 12. There are eight solutions in total. + + Total Solution Set + 9 4,2,3; 5,3,1; 6,1,2 + 9 4,3,2; 6,2,1; 5,1,3 + 10 2,3,5; 4,5,1; 6,1,3 + 10 2,5,3; 6,3,1; 4,1,5 + 11 1,4,6; 3,6,2; 5,2,4 + 11 1,6,4; 5,4,2; 3,2,6 + 12 1,5,6; 2,6,4; 3,4,5 + 12 1,6,5; 3,5,4; 2,4,6 + + By concatenating each group it is possible to form 9-digit strings; the + maximum string for a 3-gon ring is 432621513. + + Using the numbers 1 to 10, and depending on arrangements, it is possible + to form 16- and 17-digit strings. What is the maximum 16-digit string for + a "magic" 5-gon ring? + + + p_068_1.gif + p_068_2.gif + Answer: 26227442c6fed0292a528ac3790175be + + +Problem 69 +========== + + + Euler's Totient function, φ(n) [sometimes called the phi function], is + used to determine the number of numbers less than n which are relatively + prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine + and relatively prime to nine, φ(9)=6. + + ┌────┬──────────────────┬──────┬───────────┐ + │ n │ Relatively Prime │ φ(n) │ n/φ(n) │ + ├────┼──────────────────┼──────┼───────────┤ + │ 2 │ 1 │ 1 │ 2 │ + ├────┼──────────────────┼──────┼───────────┤ + │ 3 │ 1,2 │ 2 │ 1.5 │ + ├────┼──────────────────┼──────┼───────────┤ + │ 4 │ 1,3 │ 2 │ 2 │ + ├────┼──────────────────┼──────┼───────────┤ + │ 5 │ 1,2,3,4 │ 4 │ 1.25 │ + ├────┼──────────────────┼──────┼───────────┤ + │ 6 │ 1,5 │ 2 │ 3 │ + ├────┼──────────────────┼──────┼───────────┤ + │ 7 │ 1,2,3,4,5,6 │ 6 │ 1.1666... │ + ├────┼──────────────────┼──────┼───────────┤ + │ 8 │ 1,3,5,7 │ 4 │ 2 │ + ├────┼──────────────────┼──────┼───────────┤ + │ 9 │ 1,2,4,5,7,8 │ 6 │ 1.5 │ + ├────┼──────────────────┼──────┼───────────┤ + │ 10 │ 1,3,7,9 │ 4 │ 2.5 │ + └────┴──────────────────┴──────┴───────────┘ + + It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10. + + Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum. + + + Answer: bf08b01ead83cbd62a9839ca1cf35ada + + +Problem 70 +========== + + + Euler's Totient function, φ(n) [sometimes called the phi function], is + used to determine the number of positive numbers less than or equal to n + which are relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are + all less than nine and relatively prime to nine, φ(9)=6. + The number 1 is considered to be relatively prime to every positive + number, so φ(1)=1. + + Interestingly, φ(87109)=79180, and it can be seen that 87109 is a + permutation of 79180. + + Find the value of n, 1 < n < 10^7, for which φ(n) is a permutation of n + and the ratio n/φ(n) produces a minimum. + + + Answer: 1884dde67ced589082c8b7043abce181 + + +Problem 71 +========== + + + Consider the fraction, n/d, where n and d are positive integers. If n 519432^525806 would be much more + difficult, as both numbers contain over three million digits. + + Using [1]base_exp.txt, a 22K text file containing one thousand lines with + a base/exponent pair on each line, determine which line number has the + greatest numerical value. + + NOTE: The first two lines in the file represent the numbers in the example + given above. + + + Visible links + 1. base_exp.txt + Answer: 1ecfb463472ec9115b10c292ef8bc986 + + +Problem 100 +=========== + + + If a box contains twenty-one coloured discs, composed of fifteen blue + discs and six red discs, and two discs were taken at random, it can be + seen that the probability of taking two blue discs, P(BB) = + (15/21)×(14/20) = 1/2. + + The next such arrangement, for which there is exactly 50% chance of taking + two blue discs at random, is a box containing eighty-five blue discs and + thirty-five red discs. + + By finding the first arrangement to contain over 10^12 = 1,000,000,000,000 + discs in total, determine the number of blue discs that the box would + contain. + + + Answer: 21156e3acc4ca35b7a318c541a0648d5 + + +Problem 101 +=========== + + + If we are presented with the first k terms of a sequence it is impossible + to say with certainty the value of the next term, as there are infinitely + many polynomial functions that can model the sequence. + + As an example, let us consider the sequence of cube numbers. This is + defined by the generating function, + u[n] = n^3: 1, 8, 27, 64, 125, 216, ... + + Suppose we were only given the first two terms of this sequence. Working + on the principle that "simple is best" we should assume a linear + relationship and predict the next term to be 15 (common difference 7). + Even if we were presented with the first three terms, by the same + principle of simplicity, a quadratic relationship should be assumed. + + We shall define OP(k, n) to be the n^th term of the optimum polynomial + generating function for the first k terms of a sequence. It should be + clear that OP(k, n) will accurately generate the terms of the sequence for + n ≤ k, and potentially the first incorrect term (FIT) will be OP(k, k+1); + in which case we shall call it a bad OP (BOP). + + As a basis, if we were only given the first term of sequence, it would be + most sensible to assume constancy; that is, for n ≥ 2, OP(1, n) = u[1]. + + Hence we obtain the following OPs for the cubic sequence: + + OP(1, n) = 1 1, 1, 1, 1, ... + OP(2, n) = 7n−6 1, 8, 15, ... + OP(3, n) = 6n^2−11n+6      1, 8, 27, 58, ... + OP(4, n) = n^3 1, 8, 27, 64, 125, ... + + Clearly no BOPs exist for k ≥ 4. + + By considering the sum of FITs generated by the BOPs (indicated in red + above), we obtain 1 + 15 + 58 = 74. + + Consider the following tenth degree polynomial generating function: + + u[n] = 1 − n + n^2 − n^3 + n^4 − n^5 + n^6 − n^7 + n^8 − n^9 + n^10 + + Find the sum of FITs for the BOPs. + + + Answer: d382b0cc25e82446da83d3a792e1cd27 + + +Problem 102 +=========== + + + Three distinct points are plotted at random on a Cartesian plane, for + which -1000 ≤ x, y ≤ 1000, such that a triangle is formed. + + Consider the following two triangles: + + A(-340,495), B(-153,-910), C(835,-947) + + X(-175,41), Y(-421,-714), Z(574,-645) + + It can be verified that triangle ABC contains the origin, whereas triangle + XYZ does not. + + Using [1]triangles.txt, a 27K text file containing the co-ordinates of one + thousand "random" triangles, find the number of triangles for which the + interior contains the origin. + + NOTE: The first two examples in the file represent the triangles in the + example given above. + + + Visible links + 1. triangles.txt + Answer: 74db120f0a8e5646ef5a30154e9f6deb + + +Problem 103 +=========== + + + Let S(A) represent the sum of elements in set A of size n. We shall call + it a special sum set if for any two non-empty disjoint subsets, B and C, + the following properties are true: + + i. S(B) ≠ S(C); that is, sums of subsets cannot be equal. + ii. If B contains more elements than C then S(B) > S(C). + + If S(A) is minimised for a given n, we shall call it an optimum special + sum set. The first five optimum special sum sets are given below. + + n = 1: {1} + n = 2: {1, 2} + n = 3: {2, 3, 4} + n = 4: {3, 5, 6, 7} + n = 5: {6, 9, 11, 12, 13} + + It seems that for a given optimum set, A = {a[1], a[2], ... , a[n]}, the + next optimum set is of the form B = {b, a[1]+b, a[2]+b, ... ,a[n]+b}, + where b is the "middle" element on the previous row. + + By applying this "rule" we would expect the optimum set for n = 6 to be A + = {11, 17, 20, 22, 23, 24}, with S(A) = 117. However, this is not the + optimum set, as we have merely applied an algorithm to provide a near + optimum set. The optimum set for n = 6 is A = {11, 18, 19, 20, 22, 25}, + with S(A) = 115 and corresponding set string: 111819202225. + + Given that A is an optimum special sum set for n = 7, find its set string. + + NOTE: This problem is related to [1]Problem 105 and [2]Problem 106. + + + Visible links + 1. problem=105 + 2. problem=106 + Answer: af8c238336c2a79bb81a24b3fef3330d + + +Problem 104 +=========== + + + The Fibonacci sequence is defined by the recurrence relation: + + F[n] = F[n−1] + F[n−2], where F[1] = 1 and F[2] = 1. + + It turns out that F[541], which contains 113 digits, is the first + Fibonacci number for which the last nine digits are 1-9 pandigital + (contain all the digits 1 to 9, but not necessarily in order). And + F[2749], which contains 575 digits, is the first Fibonacci number for + which the first nine digits are 1-9 pandigital. + + Given that F[k] is the first Fibonacci number for which the first nine + digits AND the last nine digits are 1-9 pandigital, find k. + + + Answer: c8771ddd4df191098d70a8e94dd1cde7 + + +Problem 105 +=========== + + + Let S(A) represent the sum of elements in set A of size n. We shall call + it a special sum set if for any two non-empty disjoint subsets, B and C, + the following properties are true: + + i. S(B) ≠ S(C); that is, sums of subsets cannot be equal. + ii. If B contains more elements than C then S(B) > S(C). + + For example, {81, 88, 75, 42, 87, 84, 86, 65} is not a special sum set + because 65 + 87 + 88 = 75 + 81 + 84, whereas {157, 150, 164, 119, 79, 159, + 161, 139, 158} satisfies both rules for all possible subset pair + combinations and S(A) = 1286. + + Using [1]sets.txt (right click and "Save Link/Target As..."), a 4K text + file with one-hundred sets containing seven to twelve elements (the two + examples given above are the first two sets in the file), identify all the + special sum sets, A[1], A[2], ..., A[k], and find the value of S(A[1]) + + S(A[2]) + ... + S(A[k]). + + NOTE: This problem is related to [2]Problem 103 and [3]Problem 106. + + + Visible links + 1. sets.txt + 2. problem=103 + 3. problem=106 + Answer: c87d30e494eff438fe37b4c810167da0 + + +Problem 106 +=========== + + + Let S(A) represent the sum of elements in set A of size n. We shall call + it a special sum set if for any two non-empty disjoint subsets, B and C, + the following properties are true: + + i. S(B) ≠ S(C); that is, sums of subsets cannot be equal. + ii. If B contains more elements than C then S(B) > S(C). + + For this problem we shall assume that a given set contains n strictly + increasing elements and it already satisfies the second rule. + + Surprisingly, out of the 25 possible subset pairs that can be obtained + from a set for which n = 4, only 1 of these pairs need to be tested for + equality (first rule). Similarly, when n = 7, only 70 out of the 966 + subset pairs need to be tested. + + For n = 12, how many of the 261625 subset pairs that can be obtained need + to be tested for equality? + + NOTE: This problem is related to [1]Problem 103 and [2]Problem 105. + + + Visible links + 1. problem=103 + 2. problem=105 + Answer: c8fd9e36fdeb06bcc93a0732c667b6d8 + + +Problem 107 +=========== + + + The following undirected network consists of seven vertices and twelve + edges with a total weight of 243. + + The same network can be represented by the matrix below. + + ┌──────┬────┬────┬────┬────┬────┬────┬────┐ + │      │ A │ B │ C │ D │ E │ F │ G │ + ├──────┼────┼────┼────┼────┼────┼────┼────┤ + │ A │ - │ 16 │ 12 │ 21 │ - │ - │ - │ + ├──────┼────┼────┼────┼────┼────┼────┼────┤ + │ B │ 16 │ - │ - │ 17 │ 20 │ - │ - │ + ├──────┼────┼────┼────┼────┼────┼────┼────┤ + │ C │ 12 │ - │ - │ 28 │ - │ 31 │ - │ + ├──────┼────┼────┼────┼────┼────┼────┼────┤ + │ D │ 21 │ 17 │ 28 │ - │ 18 │ 19 │ 23 │ + ├──────┼────┼────┼────┼────┼────┼────┼────┤ + │ E │ - │ 20 │ - │ 18 │ - │ - │ 11 │ + ├──────┼────┼────┼────┼────┼────┼────┼────┤ + │ F │ - │ - │ 31 │ 19 │ - │ - │ 27 │ + ├──────┼────┼────┼────┼────┼────┼────┼────┤ + │ G │ - │ - │ - │ 23 │ 11 │ 27 │ - │ + └──────┴────┴────┴────┴────┴────┴────┴────┘ + + However, it is possible to optimise the network by removing some edges and + still ensure that all points on the network remain connected. The network + which achieves the maximum saving is shown below. It has a weight of 93, + representing a saving of 243 − 93 = 150 from the original network. + + Using [1]network.txt, a 6K text file containing a network with forty + vertices, and given in matrix form, find the maximum saving which can be + achieved by removing redundant edges whilst ensuring that the network + remains connected. + + + Visible links + 1. network.txt + p_107_1.gif + p_107_2.gif + Answer: b0db1202ec966e7855ca23626eb285b8 + + +Problem 108 +=========== + + + In the following equation x, y, and n are positive integers. + + 1 1 1 + ─ + ─ = ─ + x y n + + For n = 4 there are exactly three distinct solutions: + + 1 1 1 + ─ + ─ = ─ + 5 20 4 + 1 1 1 + ─ + ─ = ─ + 6 12 4 + 1 1 1 + ─ + ─ = ─ + 8 8 4 + + What is the least value of n for which the number of distinct solutions + exceeds one-thousand? + + NOTE: This problem is an easier version of [1]Problem 110; it is strongly + advised that you solve this one first. + + + Visible links + 1. problem=110 + Answer: 765ba18edd2844db2db95fba25d2f3e7 + + +Problem 109 +=========== + + + In the game of darts a player throws three darts at a target board which + is split into twenty equal sized sections numbered one to twenty. + + The score of a dart is determined by the number of the region that the + dart lands in. A dart landing outside the red/green outer ring scores + zero. The black and cream regions inside this ring represent single + scores. However, the red/green outer ring and middle ring score double and + treble scores respectively. + + At the centre of the board are two concentric circles called the bull + region, or bulls-eye. The outer bull is worth 25 points and the inner bull + is a double, worth 50 points. + + There are many variations of rules but in the most popular game the + players will begin with a score 301 or 501 and the first player to reduce + their running total to zero is a winner. However, it is normal to play a + "doubles out" system, which means that the player must land a double + (including the double bulls-eye at the centre of the board) on their final + dart to win; any other dart that would reduce their running total to one + or lower means the score for that set of three darts is "bust". + + When a player is able to finish on their current score it is called a + "checkout" and the highest checkout is 170: T20 T20 D25 (two treble 20s + and double bull). + + There are exactly eleven distinct ways to checkout on a score of 6: + + ┌──┬──┬──┐ + │D3│  │  │ + ├──┼──┼──┤ + │D1│D2│  │ + ├──┼──┼──┤ + │S2│D2│  │ + ├──┼──┼──┤ + │D2│D1│  │ + ├──┼──┼──┤ + │S4│D1│  │ + ├──┼──┼──┤ + │S1│S1│D2│ + ├──┼──┼──┤ + │S1│T1│D1│ + ├──┼──┼──┤ + │S1│S3│D1│ + ├──┼──┼──┤ + │D1│D1│D1│ + ├──┼──┼──┤ + │D1│S2│D1│ + ├──┼──┼──┤ + │S2│S2│D1│ + └──┴──┴──┘ + + Note that D1 D2 is considered different to D2 D1 as they finish on + different doubles. However, the combination S1 T1 D1 is considered the + same as T1 S1 D1. + + In addition we shall not include misses in considering combinations; for + example, D3 is the same as 0 D3 and 0 0 D3. + + Incredibly there are 42336 distinct ways of checking out in total. + + How many distinct ways can a player checkout with a score less than 100? + + + p_109.gif + Answer: e6aebd5be1ba81557dbcc5f6f57bbe5c + + +Problem 110 +=========== + + + In the following equation x, y, and n are positive integers. + + 1 1 1 + ─ + ─ = ─ + x y n + + It can be verified that when n = 1260 there are 113 distinct solutions and + this is the least value of n for which the total number of distinct + solutions exceeds one hundred. + + What is the least value of n for which the number of distinct solutions + exceeds four million? + + NOTE: This problem is a much more difficult version of [1]Problem 108 and + as it is well beyond the limitations of a brute force approach it requires + a clever implementation. + + + Visible links + 1. problem=108 + Answer: 591a7a92f10322866e6a02f3b2386a1c + + +Problem 111 +=========== + + + Considering 4-digit primes containing repeated digits it is clear that + they cannot all be the same: 1111 is divisible by 11, 2222 is divisible by + 22, and so on. But there are nine 4-digit primes containing three ones: + + 1117, 1151, 1171, 1181, 1511, 1811, 2111, 4111, 8111 + + We shall say that M(n, d) represents the maximum number of repeated digits + for an n-digit prime where d is the repeated digit, N(n, d) represents the + number of such primes, and S(n, d) represents the sum of these primes. + + So M(4, 1) = 3 is the maximum number of repeated digits for a 4-digit + prime where one is the repeated digit, there are N(4, 1) = 9 such primes, + and the sum of these primes is S(4, 1) = 22275. It turns out that for d = + 0, it is only possible to have M(4, 0) = 2 repeated digits, but there are + N(4, 0) = 13 such cases. + + In the same way we obtain the following results for 4-digit primes. + + ┌──────────┬─────────┬─────────┬─────────┐ + │ Digit, d │ M(4, d) │ N(4, d) │ S(4, d) │ + ├──────────┼─────────┼─────────┼─────────┤ + │ 0 │ 2 │ 13 │ 67061 │ + ├──────────┼─────────┼─────────┼─────────┤ + │ 1 │ 3 │ 9 │ 22275 │ + ├──────────┼─────────┼─────────┼─────────┤ + │ 2 │ 3 │ 1 │ 2221 │ + ├──────────┼─────────┼─────────┼─────────┤ + │ 3 │ 3 │ 12 │ 46214 │ + ├──────────┼─────────┼─────────┼─────────┤ + │ 4 │ 3 │ 2 │ 8888 │ + ├──────────┼─────────┼─────────┼─────────┤ + │ 5 │ 3 │ 1 │ 5557 │ + ├──────────┼─────────┼─────────┼─────────┤ + │ 6 │ 3 │ 1 │ 6661 │ + ├──────────┼─────────┼─────────┼─────────┤ + │ 7 │ 3 │ 9 │ 57863 │ + ├──────────┼─────────┼─────────┼─────────┤ + │ 8 │ 3 │ 1 │ 8887 │ + ├──────────┼─────────┼─────────┼─────────┤ + │ 9 │ 3 │ 7 │ 48073 │ + └──────────┴─────────┴─────────┴─────────┘ + + For d = 0 to 9, the sum of all S(4, d) is 273700. + + Find the sum of all S(10, d). + + + Answer: cdf4d134a3b0caa10a69e2771ac4fd36 + + +Problem 112 +=========== + + + Working from left-to-right if no digit is exceeded by the digit to its + left it is called an increasing number; for example, 134468. + + Similarly if no digit is exceeded by the digit to its right it is called a + decreasing number; for example, 66420. + + We shall call a positive integer that is neither increasing nor decreasing + a "bouncy" number; for example, 155349. + + Clearly there cannot be any bouncy numbers below one-hundred, but just + over half of the numbers below one-thousand (525) are bouncy. In fact, the + least number for which the proportion of bouncy numbers first reaches 50% + is 538. + + Surprisingly, bouncy numbers become more and more common and by the time + we reach 21780 the proportion of bouncy numbers is equal to 90%. + + Find the least number for which the proportion of bouncy numbers is + exactly 99%. + + + Answer: e08c982713a1c2bd3637dd489199722e + + +Problem 113 +=========== + + + Working from left-to-right if no digit is exceeded by the digit to its + left it is called an increasing number; for example, 134468. + + Similarly if no digit is exceeded by the digit to its right it is called a + decreasing number; for example, 66420. + + We shall call a positive integer that is neither increasing nor decreasing + a "bouncy" number; for example, 155349. + + As n increases, the proportion of bouncy numbers below n increases such + that there are only 12951 numbers below one-million that are not bouncy + and only 277032 non-bouncy numbers below 10^10. + + How many numbers below a googol (10^100) are not bouncy? + + + Answer: a9e504ee704c87f9bddad6d3ffe39532 + + +Problem 114 +=========== + + + A row measuring seven units in length has red blocks with a minimum length + of three units placed on it, such that any two red blocks (which are + allowed to be different lengths) are separated by at least one black + square. There are exactly seventeen ways of doing this. + + ┌┬┬┬┬┬┬┐ ┌──┬┬┬┬┐ ┌┬──┬┬┬┐ + └┴┴┴┴┴┴┘ └──┴┴┴┴┘ └┴──┴┴┴┘ + ┌┬┬──┬┬┐ ┌┬┬┬──┬┐ ┌┬┬┬┬──┐ + └┴┴──┴┴┘ └┴┴┴──┴┘ └┴┴┴┴──┘ + ┌──┬┬──┐ ┌───┬┬┬┐ ┌┬───┬┬┐ + └──┴┴──┘ └───┴┴┴┘ └┴───┴┴┘ + ┌┬┬───┬┐ ┌┬┬┬───┐ ┌────┬┬┐ + └┴┴───┴┘ └┴┴┴───┘ └────┴┴┘ + ┌┬────┬┐ ┌┬┬────┐ ┌─────┬┐ + └┴────┴┘ └┴┴────┘ └─────┴┘ + ┌┬─────┐ ┌──────┐   + └┴─────┘ └──────┘ + + How many ways can a row measuring fifty units in length be filled? + + NOTE: Although the example above does not lend itself to the possibility, + in general it is permitted to mix block sizes. For example, on a row + measuring eight units in length you could use red (3), black (1), and red + (4). + + + Answer: de48ca72bf252a8be7e0aad762eadcf8 + + +Problem 115 +=========== + + + NOTE: This is a more difficult version of [1]Problem 114. + + A row measuring n units in length has red blocks with a minimum length of + m units placed on it, such that any two red blocks (which are allowed to + be different lengths) are separated by at least one black square. + + Let the fill-count function, F(m, n), represent the number of ways that a + row can be filled. + + For example, F(3, 29) = 673135 and F(3, 30) = 1089155. + + That is, for m = 3, it can be seen that n = 30 is the smallest value for + which the fill-count function first exceeds one million. + + In the same way, for m = 10, it can be verified that F(10, 56) = 880711 + and F(10, 57) = 1148904, so n = 57 is the least value for which the + fill-count function first exceeds one million. + + For m = 50, find the least value of n for which the fill-count function + first exceeds one million. + + + Visible links + 1. problem=114 + Answer: 006f52e9102a8d3be2fe5614f42ba989 + + +Problem 116 +=========== + + + A row of five black square tiles is to have a number of its tiles replaced + with coloured oblong tiles chosen from red (length two), green (length + three), or blue (length four). + + If red tiles are chosen there are exactly seven ways this can be done. + + ┌─╥╥╥┐ ┌╥─╥╥┐ ┌╥╥─╥┐ ┌╥╥╥─┐ + └─╨╨╨┘ └╨─╨╨┘ └╨╨─╨┘ └╨╨╨─┘ + + ┌─╥─╥┐ ┌─╥╥─┐ ┌╥─╥─┐   + └─╨─╨┘ └─╨╨─┘ └╨─╨─┘ + + If green tiles are chosen there are three ways. + + ┌──╥╥┐ ┌╥──╥┐ ┌╥╥──┐   + └──╨╨┘ └╨──╨┘ └╨╨──┘ + + And if blue tiles are chosen there are two ways. + + ┌╥───┐ ┌───╥┐ + └╨───┘ └───╨┘ + + Assuming that colours cannot be mixed there are 7 + 3 + 2 = 12 ways of + replacing the black tiles in a row measuring five units in length. + + How many different ways can the black tiles in a row measuring fifty units + in length be replaced if colours cannot be mixed and at least one coloured + tile must be used? + + NOTE: This is related to [1]Problem 117. + + + Visible links + 1. problem=117 + Answer: c21ca0ec54e6d1646a953a480f68feb4 + + +Problem 117 +=========== + + + Using a combination of black square tiles and oblong tiles chosen from: + red tiles measuring two units, green tiles measuring three units, and blue + tiles measuring four units, it is possible to tile a row measuring five + units in length in exactly fifteen different ways. + + ┌╥╥╥╥┐ ┌─╥╥╥┐ ┌╥─╥╥┐ ┌╥╥─╥┐ + └╨╨╨╨┘ └─╨╨╨┘ └╨─╨╨┘ └╨╨─╨┘ + + ┌╥╥╥─┐ ┌─╥─╥┐ ┌─╥╥─┐ ┌╥─╥─┐ + └╨╨╨─┘ └─╨─╨┘ └─╨╨─┘ └╨─╨─┘ + + ┌──╥╥┐ ┌╥──╥┐ ┌╥╥──┐ ┌─╥──┐ + └──╨╨┘ └╨──╨┘ └╨╨──┘ └─╨──┘ + + ┌──╥─┐ ┌───╥┐ ┌╥───┐   + └──╨─┘ └───╨┘ └╨───┘ + + How many ways can a row measuring fifty units in length be tiled? + + NOTE: This is related to [1]Problem 116. + + + Visible links + 1. problem=116 + Answer: 542612809b3dd08cf518b85450fce8d6 + + +Problem 118 +=========== + + + Using all of the digits 1 through 9 and concatenating them freely to form + decimal integers, different sets can be formed. Interestingly with the set + {2,5,47,89,631}, all of the elements belonging to it are prime. + + How many distinct sets containing each of the digits one through nine + exactly once contain only prime elements? + + + Answer: 080cc5a4ec71a747e260e274bdb13b64 + + +Problem 119 +=========== + + + The number 512 is interesting because it is equal to the sum of its digits + raised to some power: 5 + 1 + 2 = 8, and 8^3 = 512. Another example of a + number with this property is 614656 = 28^4. + + We shall define a[n] to be the nth term of this sequence and insist that a + number must contain at least two digits to have a sum. + + You are given that a[2] = 512 and a[10] = 614656. + + Find a[30]. + + + Answer: 72fddfa6c52a120892ade628f3819da4 + + +Problem 120 +=========== + + + Let r be the remainder when (a−1)^n + (a+1)^n is divided by a^2. + + For example, if a = 7 and n = 3, then r = 42: 6^3 + 8^3 = 728 ≡ 42 mod 49. + And as n varies, so too will r, but for a = 7 it turns out that r[max] = + 42. + + For 3 ≤ a ≤ 1000, find ∑ r[max]. + + + Answer: 0dd05ec40fe11279c2203b72e92a450a + + +Problem 121 +=========== + + + A bag contains one red disc and one blue disc. In a game of chance a + player takes a disc at random and its colour is noted. After each turn the + disc is returned to the bag, an extra red disc is added, and another disc + is taken at random. + + The player pays £1 to play and wins if they have taken more blue discs + than red discs at the end of the game. + + If the game is played for four turns, the probability of a player winning + is exactly 11/120, and so the maximum prize fund the banker should + allocate for winning in this game would be £10 before they would expect to + incur a loss. Note that any payout will be a whole number of pounds and + also includes the original £1 paid to play the game, so in the example + given the player actually wins £9. + + Find the maximum prize fund that should be allocated to a single game in + which fifteen turns are played. + + + Answer: 51de85ddd068f0bc787691d356176df9 + + +Problem 122 +=========== + + + The most naive way of computing n^15 requires fourteen multiplications: + + n × n × ... × n = n^15 + + But using a "binary" method you can compute it in six multiplications: + + n × n = n^2 + n^2 × n^2 = n^4 + n^4 × n^4 = n^8 + n^8 × n^4 = n^12 + n^12 × n^2 = n^14 + n^14 × n = n^15 + + However it is yet possible to compute it in only five multiplications: + + n × n = n^2 + n^2 × n = n^3 + n^3 × n^3 = n^6 + n^6 × n^6 = n^12 + n^12 × n^3 = n^15 + + We shall define m(k) to be the minimum number of multiplications to + compute n^k; for example m(15) = 5. + + For 1 ≤ k ≤ 200, find ∑ m(k). + + + Answer: b710915795b9e9c02cf10d6d2bdb688c + + +Problem 123 +=========== + + + Let p[n] be the nth prime: 2, 3, 5, 7, 11, ..., and let r be the remainder + when (p[n]−1)^n + (p[n]+1)^n is divided by p[n]^2. + + For example, when n = 3, p[3] = 5, and 4^3 + 6^3 = 280 ≡ 5 mod 25. + + The least value of n for which the remainder first exceeds 10^9 is 7037. + + Find the least value of n for which the remainder first exceeds 10^10. + + + Answer: 71497f728b86b55d965edbf1849cca8d + + +Problem 124 +=========== + + + The radical of n, rad(n), is the product of the distinct prime factors of + n. For example, 504 = 2^3 × 3^2 × 7, so rad(504) = 2 × 3 × 7 = 42. + + If we calculate rad(n) for 1 ≤ n ≤ 10, then sort them on rad(n), and + sorting on n if the radical values are equal, we get: + + Unsorted   Sorted + n rad(n) n rad(n) k + 1 1   1 1 1 + 2 2   2 2 2 + 3 3   4 2 3 + 4 2   8 2 4 + 5 5   3 3 5 + 6 6   9 3 6 + 7 7   5 5 7 + 8 2   6 6 8 + 9 3   7 7 9 + 10 10   10 10 10 + + Let E(k) be the kth element in the sorted n column; for example, E(4) = 8 + and E(6) = 9. + + If rad(n) is sorted for 1 ≤ n ≤ 100000, find E(10000). + + + Answer: f228d2e6f9099153388e9470180c8302 + + +Problem 125 +=========== + + + The palindromic number 595 is interesting because it can be written as the + sum of consecutive squares: 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2. + + There are exactly eleven palindromes below one-thousand that can be + written as consecutive square sums, and the sum of these palindromes is + 4164. Note that 1 = 0^2 + 1^2 has not been included as this problem is + concerned with the squares of positive integers. + + Find the sum of all the numbers less than 10^8 that are both palindromic + and can be written as the sum of consecutive squares. + + + Answer: 1b5635e8ab723e01570ca783129493dd + + +Problem 126 +=========== + + + The minimum number of cubes to cover every visible face on a cuboid + measuring 3 x 2 x 1 is twenty-two. + + If we then add a second layer to this solid it would require forty-six + cubes to cover every visible face, the third layer would require + seventy-eight cubes, and the fourth layer would require one-hundred and + eighteen cubes to cover every visible face. + + However, the first layer on a cuboid measuring 5 x 1 x 1 also requires + twenty-two cubes; similarly the first layer on cuboids measuring + 5 x 3 x 1, 7 x 2 x 1, and 11 x 1 x 1 all contain forty-six cubes. + + We shall define C(n) to represent the number of cuboids that contain n + cubes in one of its layers. So C(22) = 2, C(46) = 4, C(78) = 5, and C(118) + = 8. + + It turns out that 154 is the least value of n for which C(n) = 10. + + Find the least value of n for which C(n) = 1000. + + + p_126.gif + Answer: 387d6ae83cbc6fa0b9192b56bf095c49 + + +Problem 127 +=========== + + + The radical of n, rad(n), is the product of distinct prime factors of n. + For example, 504 = 2^3 × 3^2 × 7, so rad(504) = 2 × 3 × 7 = 42. + + We shall define the triplet of positive integers (a, b, c) to be an + abc-hit if: + +  1. GCD(a, b) = GCD(a, c) = GCD(b, c) = 1 +  2. a < b +  3. a + b = c +  4. rad(abc) < c + + For example, (5, 27, 32) is an abc-hit, because: + +  1. GCD(5, 27) = GCD(5, 32) = GCD(27, 32) = 1 +  2. 5 < 27 +  3. 5 + 27 = 32 +  4. rad(4320) = 30 < 32 + + It turns out that abc-hits are quite rare and there are only thirty-one + abc-hits for c < 1000, with ∑c = 12523. + + Find ∑c for c < 120000. + + + Answer: c6b1ae935b33c90a2c320b5f6ef3e4ba + + +Problem 128 +=========== + + + A hexagonal tile with number 1 is surrounded by a ring of six hexagonal + tiles, starting at "12 o'clock" and numbering the tiles 2 to 7 in an + anti-clockwise direction. + + New rings are added in the same fashion, with the next rings being + numbered 8 to 19, 20 to 37, 38 to 61, and so on. The diagram below shows + the first three rings. + + By finding the difference between tile n and each its six neighbours we + shall define PD(n) to be the number of those differences which are prime. + + For example, working clockwise around tile 8 the differences are 12, 29, + 11, 6, 1, and 13. So PD(8) = 3. + + In the same way, the differences around tile 17 are 1, 17, 16, 1, 11, and + 10, hence PD(17) = 2. + + It can be shown that the maximum value of PD(n) is 3. + + If all of the tiles for which PD(n) = 3 are listed in ascending order to + form a sequence, the 10th tile would be 271. + + Find the 2000th tile in this sequence. + + + p_128.gif + Answer: 93a1925da4792b4fa5d2dbb6ebb7c4a2 + + +Problem 129 +=========== + + + A number consisting entirely of ones is called a repunit. We shall define + R(k) to be a repunit of length k; for example, R(6) = 111111. + + Given that n is a positive integer and GCD(n, 10) = 1, it can be shown + that there always exists a value, k, for which R(k) is divisible by n, and + let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = + 5. + + The least value of n for which A(n) first exceeds ten is 17. + + Find the least value of n for which A(n) first exceeds one-million. + + + Answer: 82cd979a2b79600137aea54fa0bd944b + + +Problem 130 +=========== + + + A number consisting entirely of ones is called a repunit. We shall define + R(k) to be a repunit of length k; for example, R(6) = 111111. + + Given that n is a positive integer and GCD(n, 10) = 1, it can be shown + that there always exists a value, k, for which R(k) is divisible by n, and + let A(n) be the least such value of k; for example, A(7) = 6 and A(41) = + 5. + + You are given that for all primes, p > 5, that p − 1 is divisible by A(p). + For example, when p = 41, A(41) = 5, and 40 is divisible by 5. + + However, there are rare composite values for which this is also true; the + first five examples being 91, 259, 451, 481, and 703. + + Find the sum of the first twenty-five composite values of n for which + GCD(n, 10) = 1 and n − 1 is divisible by A(n). + + + Answer: 20594ea0ef7a2f4cf40d19a9b82a0beb + + +Problem 131 +=========== + + + There are some prime values, p, for which there exists a positive integer, + n, such that the expression n^3 + n^2p is a perfect cube. + + For example, when p = 19, 8^3 + 8^2×19 = 12^3. + + What is perhaps most surprising is that for each prime with this property + the value of n is unique, and there are only four such primes below + one-hundred. + + How many primes below one million have this remarkable property? + + + Answer: f7e6c85504ce6e82442c770f7c8606f0 + + +Problem 132 +=========== + + + A number consisting entirely of ones is called a repunit. We shall define + R(k) to be a repunit of length k. + + For example, R(10) = 1111111111 = 11×41×271×9091, and the sum of these + prime factors is 9414. + + Find the sum of the first forty prime factors of R(10^9). + + + Answer: 5df3a36faa173a393a04a022b2d5d49d + + +Problem 133 +=========== + + + A number consisting entirely of ones is called a repunit. We shall define + R(k) to be a repunit of length k; for example, R(6) = 111111. + + Let us consider repunits of the form R(10^n). + + Although R(10), R(100), or R(1000) are not divisible by 17, R(10000) is + divisible by 17. Yet there is no value of n for which R(10^n) will divide + by 19. In fact, it is remarkable that 11, 17, 41, and 73 are the only four + primes below one-hundred that can be a factor of R(10^n). + + Find the sum of all the primes below one-hundred thousand that will never + be a factor of R(10^n). + + + Answer: c1d33d79d08cde65eaa78e4583ea0594 + + +Problem 134 +=========== + + + Consider the consecutive primes p[1] = 19 and p[2] = 23. It can be + verified that 1219 is the smallest number such that the last digits are + formed by p[1] whilst also being divisible by p[2]. + + In fact, with the exception of p[1] = 3 and p[2] = 5, for every pair of + consecutive primes, p[2] > p[1], there exist values of n for which the + last digits are formed by p[1] and n is divisible by p[2]. Let S be the + smallest of these values of n. + + Find ∑ S for every pair of consecutive primes with 5 ≤ p[1] ≤ 1000000. + + + Answer: f12b07460d2586ea47b4d305ae0b0539 + + +Problem 135 +=========== + + + Given the positive integers, x, y, and z, are consecutive terms of an + arithmetic progression, the least value of the positive integer, n, for + which the equation, x^2 − y^2 − z^2 = n, has exactly two solutions is n = + 27: + + 34^2 − 27^2 − 20^2 = 12^2 − 9^2 − 6^2 = 27 + + It turns out that n = 1155 is the least value which has exactly ten + solutions. + + How many values of n less than one million have exactly ten distinct + solutions? + + + Answer: c457d7ae48d08a6b84bc0b1b9bd7d474 + + +Problem 136 +=========== + + + The positive integers, x, y, and z, are consecutive terms of an arithmetic + progression. Given that n is a positive integer, the equation, x^2 − y^2 − + z^2 = n, has exactly one solution when n = 20: + + 13^2 − 10^2 − 7^2 = 20 + + In fact there are twenty-five values of n below one hundred for which the + equation has a unique solution. + + How many values of n less than fifty million have exactly one solution? + + + Answer: 91db9e8e6cb2dbf9c07a6e0429697336 + + +Problem 137 +=========== + + + Consider the infinite polynomial series A[F](x) = xF[1] + x^2F[2] + + x^3F[3] + ..., where F[k] is the kth term in the Fibonacci sequence: 1, 1, + 2, 3, 5, 8, ... ; that is, F[k] = F[k−1] + F[k−2], F[1] = 1 and F[2] = 1. + + For this problem we shall be interested in values of x for which A[F](x) + is a positive integer. + + Surprisingly A[F](1/2)  =  (1/2).1 + (1/2)^2.1 + (1/2)^3.2 + (1/2)^4.3 + + (1/2)^5.5 + ... +    =  1/2 + 1/4 + 2/8 + 3/16 + 5/32 + ... +    =  2 + + The corresponding values of x for the first five natural numbers are shown + below. + + ┌─────────┬───────┐ + │x │A[F](x)│ + ├─────────┼───────┤ + │√2−1 │1 │ + ├─────────┼───────┤ + │1/2 │2 │ + ├─────────┼───────┤ + │(√13−2)/3│3 │ + ├─────────┼───────┤ + │(√89−5)/8│4 │ + ├─────────┼───────┤ + │(√34−3)/5│5 │ + └─────────┴───────┘ + + We shall call A[F](x) a golden nugget if x is rational, because they + become increasingly rarer; for example, the 10th golden nugget is + 74049690. + + Find the 15th golden nugget. + + + Answer: 44845aa0f47ec925a3b43e6460a55e27 + + +Problem 138 +=========== + + + Consider the isosceles triangle with base length, b = 16, and legs, L = + 17. + + By using the Pythagorean theorem it can be seen that the height of the + triangle, h = √(17^2 − 8^2) = 15, which is one less than the base length. + + With b = 272 and L = 305, we get h = 273, which is one more than the base + length, and this is the second smallest isosceles triangle with the + property that h = b ± 1. + + Find ∑ L for the twelve smallest isosceles triangles for which h = b ± 1 + and b, L are positive integers. + + + p_138.gif + Answer: f7524f4d0d6d042c0f92a0d6469aff85 + + +Problem 139 +=========== + + + Let (a, b, c) represent the three sides of a right angle triangle with + integral length sides. It is possible to place four such triangles + together to form a square with length c. + + For example, (3, 4, 5) triangles can be placed together to form a 5 by 5 + square with a 1 by 1 hole in the middle and it can be seen that the 5 by 5 + square can be tiled with twenty-five 1 by 1 squares. + + However, if (5, 12, 13) triangles were used then the hole would measure 7 + by 7 and these could not be used to tile the 13 by 13 square. + + Given that the perimeter of the right triangle is less than one-hundred + million, how many Pythagorean triangles would allow such a tiling to take + place? + + + p_139.gif + Answer: 1c343ba00e6d17d7239bf45869ffed0c + + +Problem 140 +=========== + + + Consider the infinite polynomial series A[G](x) = xG[1] + x^2G[2] + + x^3G[3] + ..., where G[k] is the kth term of the second order recurrence + relation G[k] = G[k−1] + G[k−2], G[1] = 1 and G[2] = 4; that is, 1, 4, 5, + 9, 14, 23, ... . + + For this problem we shall be concerned with values of x for which A[G](x) + is a positive integer. + + The corresponding values of x for the first five natural numbers are shown + below. + + ┌───────────┬───────┐ + │x │A[G](x)│ + ├───────────┼───────┤ + │(√5−1)/4 │1 │ + ├───────────┼───────┤ + │2/5 │2 │ + ├───────────┼───────┤ + │(√22−2)/6 │3 │ + ├───────────┼───────┤ + │(√137−5)/14│4 │ + ├───────────┼───────┤ + │1/2 │5 │ + └───────────┴───────┘ + + We shall call A[G](x) a golden nugget if x is rational, because they + become increasingly rarer; for example, the 20th golden nugget is + 211345365. + + Find the sum of the first thirty golden nuggets. + + + Answer: e5d75f96929ba250b2732aad52f3028c + + +Problem 141 +=========== + + + A positive integer, n, is divided by d and the quotient and remainder are + q and r respectively. In addition d, q, and r are consecutive positive + integer terms in a geometric sequence, but not necessarily in that order. + + For example, 58 divided by 6 has quotient 9 and remainder 4. It can also + be seen that 4, 6, 9 are consecutive terms in a geometric sequence (common + ratio 3/2). + We will call such numbers, n, progressive. + + Some progressive numbers, such as 9 and 10404 = 102^2, happen to also be + perfect squares. + The sum of all progressive perfect squares below one hundred thousand is + 124657. + + Find the sum of all progressive perfect squares below one trillion + (10^12). + + + Answer: 2aaefa1db80951be140183f9e8c0194e + + +Problem 142 +=========== + + + Find the smallest x + y + z with integers x > y > z > 0 such that x + y, x + − y, x + z, x − z, y + z, y − z are all perfect squares. + + + Answer: d3de282705508407532aa20ca8928e3b + + +Problem 143 +=========== + + + Let ABC be a triangle with all interior angles being less than 120 + degrees. Let X be any point inside the triangle and let XA = p, XC = q, + and XB = r. + + Fermat challenged Torricelli to find the position of X such that p + q + r + was minimised. + + Torricelli was able to prove that if equilateral triangles AOB, BNC and + AMC are constructed on each side of triangle ABC, the circumscribed + circles of AOB, BNC, and AMC will intersect at a single point, T, inside + the triangle. Moreover he proved that T, called the Torricelli/Fermat + point, minimises p + q + r. Even more remarkable, it can be shown that + when the sum is minimised, AN = BM = CO = p + q + r and that AN, BM and CO + also intersect at T. + + If the sum is minimised and a, b, c, p, q and r are all positive integers + we shall call triangle ABC a Torricelli triangle. For example, a = 399, b + = 455, c = 511 is an example of a Torricelli triangle, with p + q + r = + 784. + + Find the sum of all distinct values of p + q + r ≤ 120000 for Torricelli + triangles. + + + p_143_torricelli.gif + Answer: ec2d4c1a0c204d1f06ea5e2d189034f6 + + +Problem 144 +=========== + + + In laser physics, a "white cell" is a mirror system that acts as a delay + line for the laser beam. The beam enters the cell, bounces around on the + mirrors, and eventually works its way back out. + + The specific white cell we will be considering is an ellipse with the + equation 4x^2 + y^2 = 100 + + The section corresponding to −0.01 ≤ x ≤ +0.01 at the top is missing, + allowing the light to enter and exit through the hole. + + The light beam in this problem starts at the point (0.0,10.1) just outside + the white cell, and the beam first impacts the mirror at (1.4,-9.6). + + Each time the laser beam hits the surface of the ellipse, it follows the + usual law of reflection "angle of incidence equals angle of reflection." + That is, both the incident and reflected beams make the same angle with + the normal line at the point of incidence. + + In the figure on the left, the red line shows the first two points of + contact between the laser beam and the wall of the white cell; the blue + line shows the line tangent to the ellipse at the point of incidence of + the first bounce. + + The slope m of the tangent line at any point (x,y) of the given ellipse + is: m = −4x/y + + The normal line is perpendicular to this tangent line at the point of + incidence. + + The animation on the right shows the first 10 reflections of the beam. + + How many times does the beam hit the internal surface of the white cell + before exiting? + + + p_144_1.gif + p_144_2.gif + Answer: 8dd48d6a2e2cad213179a3992c0be53c + + +Problem 145 +=========== + + + Some positive integers n have the property that the sum [ n + reverse(n) ] + consists entirely of odd (decimal) digits. For instance, 36 + 63 = 99 and + 409 + 904 = 1313. We will call such numbers reversible; so 36, 63, 409, + and 904 are reversible. Leading zeroes are not allowed in either n or + reverse(n). + + There are 120 reversible numbers below one-thousand. + + How many reversible numbers are there below one-billion (10^9)? + + + Answer: 705e8444ad9c92e9a7589fb97515a9b6 + + +Problem 146 +=========== + + + The smallest positive integer n for which the numbers n^2+1, n^2+3, n^2+7, + n^2+9, n^2+13, and n^2+27 are consecutive primes is 10. The sum of all + such integers n below one-million is 1242490. + + What is the sum of all such integers n below 150 million? + + + Answer: 525bd2bf0e31b0f19b38a1d21f2f6a16 + + +Problem 147 +=========== + + + In a 3x2 cross-hatched grid, a total of 37 different rectangles could be + situated within that grid as indicated in the sketch. + + There are 5 grids smaller than 3x2, vertical and horizontal dimensions + being important, i.e. 1x1, 2x1, 3x1, 1x2 and 2x2. If each of them is + cross-hatched, the following number of different rectangles could be + situated within those smaller grids: + + 1x1: 1 + 2x1: 4 + 3x1: 8 + 1x2: 4 + 2x2: 18 + + Adding those to the 37 of the 3x2 grid, a total of 72 different rectangles + could be situated within 3x2 and smaller grids. + + How many different rectangles could be situated within 47x43 and smaller + grids? + + + p_147.gif + Answer: d0fca7d85d4a4df043a2ae5772ea472e + + +Problem 148 +=========== + + + We can easily verify that none of the entries in the first seven rows of + Pascal's triangle are divisible by 7: + +              1 +            1    1 +          1    2    1 +        1    3    3    1 +      1    4    6    4    1 +    1    5   10   10    5    1 + 1    6   15   20   15    6    1 + + However, if we check the first one hundred rows, we will find that only + 2361 of the 5050 entries are not divisible by 7. + + Find the number of entries which are not divisible by 7 in the first one + billion (10^9) rows of Pascal's triangle. + + + Answer: 8a631ab4e3d06baf88299bf4e501b837 + + +Problem 149 +=========== + + + Looking at the table below, it is easy to verify that the maximum possible + sum of adjacent numbers in any direction (horizontal, vertical, diagonal + or anti-diagonal) is 16 (= 8 + 7 + 1). + + ┌────┬────┬────┬─────┐ + │ −2 │ 5 │ 3 │ 2 │ + ├────┼────┼────┼─────┤ + │ 9 │ −6 │ 5 │ 1 │ + ├────┼────┼────┼─────┤ + │ 3 │ 2 │ 7 │ 3 │ + ├────┼────┼────┼─────┤ + │ −1 │ 8 │ −4 │   8 │ + └────┴────┴────┴─────┘ + + Now, let us repeat the search, but on a much larger scale: + + First, generate four million pseudo-random numbers using a specific form + of what is known as a "Lagged Fibonacci Generator": + + For 1 ≤ k ≤ 55, s[k] = [100003 − 200003k + 300007k^3] (modulo 1000000) − + 500000. + For 56 ≤ k ≤ 4000000, s[k] = [s[k−24] + s[k−55] + 1000000] (modulo + 1000000) − 500000. + + Thus, s[10] = −393027 and s[100] = 86613. + + The terms of s are then arranged in a 2000×2000 table, using the first + 2000 numbers to fill the first row (sequentially), the next 2000 numbers + to fill the second row, and so on. + + Finally, find the greatest sum of (any number of) adjacent entries in any + direction (horizontal, vertical, diagonal or anti-diagonal). + + + Answer: 96affc386f4b786c2521a32944424982 + + +Problem 150 +=========== + + + In a triangular array of positive and negative integers, we wish to find a + sub-triangle such that the sum of the numbers it contains is the smallest + possible. + + In the example below, it can be easily verified that the marked triangle + satisfies this condition having a sum of −42. + + We wish to make such a triangular array with one thousand rows, so we + generate 500500 pseudo-random numbers s[k] in the range ±2^19, using a + type of random number generator (known as a Linear Congruential Generator) + as follows: + + t := 0 + for k = 1 up to k = 500500: +     t := (615949*t + 797807) modulo 2^20 +     s[k] := t−2^19 + + Thus: s[1] = 273519, s[2] = −153582, s[3] = 450905 etc + + Our triangular array is then formed using the pseudo-random numbers thus: + + s[1] + s[2]  s[3] + s[4]  s[5]  s[6]  + s[7]  s[8]  s[9]  s[10] + ... + + Sub-triangles can start at any element of the array and extend down as far + as we like (taking-in the two elements directly below it from the next + row, the three elements directly below from the row after that, and so + on). + The "sum of a sub-triangle" is defined as the sum of all the elements it + contains. + Find the smallest possible sub-triangle sum. + + + p_150.gif + Answer: 1802939e514020769701c59b422c0498 + + +Problem 151 +=========== + + + A printing shop runs 16 batches (jobs) every week and each batch requires + a sheet of special colour-proofing paper of size A5. + + Every Monday morning, the foreman opens a new envelope, containing a large + sheet of the special paper with size A1. + + He proceeds to cut it in half, thus getting two sheets of size A2. Then he + cuts one of them in half to get two sheets of size A3 and so on until he + obtains the A5-size sheet needed for the first batch of the week. + + All the unused sheets are placed back in the envelope. + + At the beginning of each subsequent batch, he takes from the envelope one + sheet of paper at random. If it is of size A5, he uses it. If it is + larger, he repeats the 'cut-in-half' procedure until he has what he needs + and any remaining sheets are always placed back in the envelope. + + Excluding the first and last batch of the week, find the expected number + of times (during each week) that the foreman finds a single sheet of paper + in the envelope. + + Give your answer rounded to six decimal places using the format x.xxxxxx . + + + p_151.gif + Answer: fb84a530fa9a8199edfadd618727fb70 + + +Problem 152 +=========== + + + There are several ways to write the number 1/2 as a sum of inverse squares + using distinct integers. + + For instance, the numbers {2,3,4,5,7,12,15,20,28,35} can be used: + + In fact, only using integers between 2 and 45 inclusive, there are exactly + three ways to do it, the remaining two being: + {2,3,4,6,7,9,10,20,28,35,36,45} and {2,3,4,6,7,9,12,15,28,30,35,36,45}. + + How many ways are there to write the number 1/2 as a sum of inverse + squares using distinct integers between 2 and 80 inclusive? + + + p_152_sum.gif + Answer: 34ed066df378efacc9b924ec161e7639 + + +Problem 153 +=========== + + + As we all know the equation x^2=-1 has no solutions for real x. + If we however introduce the imaginary number i this equation has two + solutions: x=i and x=-i. + If we go a step further the equation (x-3)^2=-4 has two complex solutions: + x=3+2i and x=3-2i. + x=3+2i and x=3-2i are called each others' complex conjugate. + Numbers of the form a+bi are called complex numbers. + In general a+bi and a−bi are each other's complex conjugate. + + A Gaussian Integer is a complex number a+bi such that both a and b are + integers. + The regular integers are also Gaussian integers (with b=0). + To distinguish them from Gaussian integers with b ≠ 0 we call such + integers "rational integers." + A Gaussian integer is called a divisor of a rational integer n if the + result is also a Gaussian integer. + If for example we divide 5 by 1+2i we can simplify in the following + manner: + Multiply numerator and denominator by the complex conjugate of 1+2i: 1−2i. + The result is . + So 1+2i is a divisor of 5. + Note that 1+i is not a divisor of 5 because . + Note also that if the Gaussian Integer (a+bi) is a divisor of a rational + integer n, then its complex conjugate (a−bi) is also a divisor of n. + + In fact, 5 has six divisors such that the real part is positive: {1, 1 + + 2i, 1 − 2i, 2 + i, 2 − i, 5}. + The following is a table of all of the divisors for the first five + positive rational integers: + + ┌───┬──────────────────────────────┬───────────────┐ + │ n │ Gaussian integer divisors │ Sum s(n) of │ + │ │ with positive real part │ thesedivisors │ + ├───┼──────────────────────────────┼───────────────┤ + │ 1 │ 1 │ 1 │ + ├───┼──────────────────────────────┼───────────────┤ + │ 2 │ 1, 1+i, 1-i, 2 │ 5 │ + ├───┼──────────────────────────────┼───────────────┤ + │ 3 │ 1, 3 │ 4 │ + ├───┼──────────────────────────────┼───────────────┤ + │ 4 │ 1, 1+i, 1-i, 2, 2+2i, 2-2i,4 │ 13 │ + ├───┼──────────────────────────────┼───────────────┤ + │ 5 │ 1, 1+2i, 1-2i, 2+i, 2-i, 5 │ 12 │ + └───┴──────────────────────────────┴───────────────┘ + + For divisors with positive real parts, then, we have: . + + For 1 ≤ n ≤ 10^5, ∑ s(n)=17924657155. + + What is ∑ s(n) for 1 ≤ n ≤ 10^8? + + + p_153_formule1.gif + p_153_formule2.gif + p_153_formule5.gif + p_153_formule6.gif + Answer: 08ec9d6e6c2275d37e7a227fb2d1f06f + + +Problem 154 +=========== + + + A triangular pyramid is constructed using spherical balls so that each + ball rests on exactly three balls of the next lower level. + + Then, we calculate the number of paths leading from the apex to each + position: + + A path starts at the apex and progresses downwards to any of the three + spheres directly below the current position. + + Consequently, the number of paths to reach a certain position is the sum + of the numbers immediately above it (depending on the position, there are + up to three numbers above it). + + The result is Pascal's pyramid and the numbers at each level n are the + coefficients of the trinomial expansion (x + y + z)^n. + + How many coefficients in the expansion of (x + y + z)^200000 are multiples + of 10^12? + + + p_154_pyramid.gif + Answer: de866633fa075beb3897cbbc8abf2400 + + +Problem 155 +=========== + + + An electric circuit uses exclusively identical capacitors of the same + value C. + The capacitors can be connected in series or in parallel to form + sub-units, which can then be connected in series or in parallel with other + capacitors or other sub-units to form larger sub-units, and so on up to a + final circuit. + + Using this simple procedure and up to n identical capacitors, we can make + circuits having a range of different total capacitances. For example, + using up to n=3 capacitors of 60 F each, we can obtain the following 7 + distinct total capacitance values: + + If we denote by D(n) the number of distinct total capacitance values we + can obtain when using up to n equal-valued capacitors and the simple + procedure described above, we have: D(1)=1, D(2)=3, D(3)=7 ... + + Find D(18). + + Reminder : When connecting capacitors C[1], C[2] etc in parallel, the + total capacitance is C[T] = C[1] + C[2] +..., + whereas when connecting them in series, the overall capacitance is given + by: + + + p_155_capsmu.gif + p_155_capacitors1.gif + p_155_capsform.gif + Answer: da0a3fc900cc8ae42d514e280524ee39 + + +Problem 156 +=========== + + + Starting from zero the natural numbers are written down in base 10 like + this: + 0 1 2 3 4 5 6 7 8 9 10 11 12.... + + Consider the digit d=1. After we write down each number n, we will update + the number of ones that have occurred and call this number f(n,1). The + first values for f(n,1), then, are as follows: + + n f(n,1) + 0 0 + 1 1 + 2 1 + 3 1 + 4 1 + 5 1 + 6 1 + 7 1 + 8 1 + 9 1 + 10 2 + 11 4 + 12 5 + + Note that f(n,1) never equals 3. + So the first two solutions of the equation f(n,1)=n are n=0 and n=1. The + next solution is n=199981. + + In the same manner the function f(n,d) gives the total number of digits d + that have been written down after the number n has been written. + In fact, for every digit d ≠ 0, 0 is the first solution of the equation + f(n,d)=n. + + Let s(d) be the sum of all the solutions for which f(n,d)=n. + You are given that s(1)=22786974071. + + Find ∑ s(d) for 1 ≤ d ≤ 9. + + Note: if, for some n, f(n,d)=n for more than one value of d this value of + n is counted again for every value of d for which f(n,d)=n. + + + Answer: ac0c6b67ed28cebb02b802e7a204aaee + + +Problem 157 +=========== + + + Consider the diophantine equation ^1/[a]+^1/[b]= ^p/[10^n] with a, b, p, n + positive integers and a ≤ b. + For n=1 this equation has 20 solutions that are listed below: + +1/1+1/1=20/10 1/1+1/2=15/10 1/1+1/5=12/10 1/1+1/10=11/10 1/2+1/2=10/10 +1/2+1/5=7/10 1/2+1/10=6/10 1/3+1/6=5/10 1/3+1/15=4/10 1/4+1/4=5/10 +1/4+1/20=3/10 1/5+1/5=4/10 1/5+1/10=3/10 1/6+1/30=2/10 1/10+1/10=2/10 +1/11+1/110=1/10 1/12+1/60=1/10 1/14+1/35=1/10 1/15+1/30=1/10 1/20+1/20=1/10 + + How many solutions has this equation for 1 ≤ n ≤ 9? + + + Answer: c96fc71df4ef8f6420fda7958957538c + + +Problem 158 +=========== + + + Taking three different letters from the 26 letters of the alphabet, + character strings of length three can be formed. + Examples are 'abc', 'hat' and 'zyx'. + When we study these three examples we see that for 'abc' two characters + come lexicographically after its neighbour to the left. + For 'hat' there is exactly one character that comes lexicographically + after its neighbour to the left. For 'zyx' there are zero characters that + come lexicographically after its neighbour to the left. + In all there are 10400 strings of length 3 for which exactly one character + comes lexicographically after its neighbour to the left. + + We now consider strings of n ≤ 26 different characters from the alphabet. + For every n, p(n) is the number of strings of length n for which exactly + one character comes lexicographically after its neighbour to the left. + + What is the maximum value of p(n)? + + + Answer: 6070fa194890e52b2989af5b542aee90 + + +Problem 159 +=========== + + + A composite number can be factored many different ways. For instance, not + including multiplication by one, 24 can be factored in 7 distinct ways: + + 24 = 2x2x2x3 + 24 = 2x3x4 + 24 = 2x2x6 + 24 = 4x6 + 24 = 3x8 + 24 = 2x12 + 24 = 24 + + Recall that the digital root of a number, in base 10, is found by adding + together the digits of that number, and repeating that process until a + number is arrived at that is less than 10. Thus the digital root of 467 is + 8. + + We shall call a Digital Root Sum (DRS) the sum of the digital roots of the + individual factors of our number. + The chart below demonstrates all of the DRS values for 24. + + ┌─────────────┬────────────────┐ + │Factorisation│Digital Root Sum│ + ├─────────────┼────────────────┤ + │2x2x2x3 │ 9 │ + ├─────────────┼────────────────┤ + │2x3x4 │ 9 │ + ├─────────────┼────────────────┤ + │2x2x6 │ 10 │ + ├─────────────┼────────────────┤ + │4x6 │ 10 │ + ├─────────────┼────────────────┤ + │3x8 │ 11 │ + ├─────────────┼────────────────┤ + │2x12 │ 5 │ + ├─────────────┼────────────────┤ + │24 │ 6 │ + └─────────────┴────────────────┘ + + The maximum Digital Root Sum of 24 is 11. + The function mdrs(n) gives the maximum Digital Root Sum of n. So + mdrs(24)=11. + Find ∑mdrs(n) for 1 < n < 1,000,000. + + + Answer: 2ab79df40adc1028d1fa83a6333db907 + + +Problem 160 +=========== + + + For any N, let f(N) be the last five digits before the trailing zeroes in + N!. + For example, + + 9! = 362880 so f(9)=36288 + 10! = 3628800 so f(10)=36288 + 20! = 2432902008176640000 so f(20)=17664 + + Find f(1,000,000,000,000) + + + Answer: e51ada1e23f810eb1b51a18bb6825f85 + + +Problem 161 +=========== + + + A triomino is a shape consisting of three squares joined via the + edges.There are two basic forms: + + If all possible orientations are taken into account there are six: + + Any n by m grid for which nxm is divisible by 3 can be tiled with + triominoes. + If we consider tilings that can be obtained by reflection or rotation from + another tiling as different there are 41 ways a 2 by 9 grid can be tiled + with triominoes: + + In how many ways can a 9 by 12 grid be tiled in this way by triominoes? + + + p_161_trio1.gif + p_161_trio3.gif + p_161_k9.gif + Answer: 975ccc38bb5402c5b485f3de5928d919 + + +Problem 162 +=========== + + + In the hexadecimal number system numbers are represented using 16 + different digits: + + 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F + + The hexadecimal number AF when written in the decimal number system equals + 10x16+15=175. + + In the 3-digit hexadecimal numbers 10A, 1A0, A10, and A01 the digits 0,1 + and A are all present. + Like numbers written in base ten we write hexadecimal numbers without + leading zeroes. + + How many hexadecimal numbers containing at most sixteen hexadecimal digits + exist with all of the digits 0,1, and A present at least once? + Give your answer as a hexadecimal number. + + (A,B,C,D,E and F in upper case, without any leading or trailing code that + marks the number as hexadecimal and without leading zeroes , e.g. 1A3F and + not: 1a3f and not 0x1a3f and not $1A3F and not #1A3F and not 0000001A3F) + + + Answer: 049419b9fdad9af74d5888626fff56a3 + + +Problem 163 +=========== + + + Consider an equilateral triangle in which straight lines are drawn from + each vertex to the middle of the opposite side, such as in the size 1 + triangle in the sketch below. + + Sixteen triangles of either different shape or size or orientation or + location can now be observed in that triangle. Using size 1 triangles as + building blocks, larger triangles can be formed, such as the size 2 + triangle in the above sketch. One-hundred and four triangles of either + different shape or size or orientation or location can now be observed in + that size 2 triangle. + + It can be observed that the size 2 triangle contains 4 size 1 triangle + building blocks. A size 3 triangle would contain 9 size 1 triangle + building blocks and a size n triangle would thus contain n^2 size 1 + triangle building blocks. + + If we denote T(n) as the number of triangles present in a triangle of size + n, then + + T(1) = 16 + T(2) = 104 + + Find T(36). + + + p_163.gif + Answer: a4f66a42a5b5dc395d00463d77e0a0c6 + + +Problem 164 +=========== + + + How many 20 digit numbers n (without any leading zero) exist such that no + three consecutive digits of n have a sum greater than 9? + + + Answer: 6e96debf3bfe7cc132401bafe5a5d6d6 + + +Problem 165 +=========== + + + A segment is uniquely defined by its two endpoints. + By considering two line segments in plane geometry there are three + possibilities: + the segments have zero points, one point, or infinitely many points in + common. + + Moreover when two segments have exactly one point in common it might be + the case that that common point is an endpoint of either one of the + segments or of both. If a common point of two segments is not an endpoint + of either of the segments it is an interior point of both segments. + We will call a common point T of two segments L[1] and L[2] a true + intersection point of L[1] and L[2] if T is the only common point of L[1] + and L[2] and T is an interior point of both segments. + + Consider the three segments L[1], L[2], and L[3]: + + L[1]: (27, 44) to (12, 32) + L[2]: (46, 53) to (17, 62) + L[3]: (46, 70) to (22, 40) + + It can be verified that line segments L[2] and L[3] have a true + intersection point. We note that as the one of the end points of L[3]: + (22,40) lies on L[1] this is not considered to be a true point of + intersection. L[1] and L[2] have no common point. So among the three line + segments, we find one true intersection point. + + Now let us do the same for 5000 line segments. To this end, we generate + 20000 numbers using the so-called "Blum Blum Shub" pseudo-random number + generator. + + s[0] = 290797 + + s[n+1] = s[n]×s[n] (modulo 50515093) + + t[n] = s[n] (modulo 500) + + To create each line segment, we use four consecutive numbers t[n]. That + is, the first line segment is given by: + + (t[1], t[2]) to (t[3], t[4]) + + The first four numbers computed according to the above generator should + be: 27, 144, 12 and 232. The first segment would thus be (27,144) to + (12,232). + + How many distinct true intersection points are found among the 5000 line + segments? + + + Answer: b87b096af5a545b4f7a45cfed4e67c87 + + +Problem 166 +=========== + + + A 4x4 grid is filled with digits d, 0 ≤ d ≤ 9. + + It can be seen that in the grid + + 6 3 3 0 + 5 0 4 3 + 0 7 1 4 + 1 2 4 5 + + the sum of each row and each column has the value 12. Moreover the sum of + each diagonal is also 12. + + In how many ways can you fill a 4x4 grid with the digits d, 0 ≤ d ≤ 9 so + that each row, each column, and both diagonals have the same sum? + + + Answer: e4f39f61ee7f1bfe433a177c07f5512f + + +Problem 167 +=========== + + + For two positive integers a and b, the Ulam sequence U(a,b) is defined by + U(a,b)[1] = a, U(a,b)[2] = b and for k > 2,U(a,b)[k] is the smallest + integer greater than U(a,b)[(k-1)] which can be written in exactly one way + as the sum of two distinct previous members of U(a,b). + + For example, the sequence U(1,2) begins with + 1, 2, 3 = 1 + 2, 4 = 1 + 3, 6 = 2 + 4, 8 = 2 + 6, 11 = 3 + 8; + 5 does not belong to it because 5 = 1 + 4 = 2 + 3 has two representations + as the sum of two previous members, likewise 7 = 1 + 6 = 3 + 4. + + Find ∑U(2,2n+1)[k] for 2 ≤ n ≤10, where k = 10^11. + + + Answer: aa5b61f6f4d96cbaeb5944b8fcdf64a3 + + +Problem 168 +=========== + + + Consider the number 142857. We can right-rotate this number by moving the + last digit (7) to the front of it, giving us 714285. + It can be verified that 714285=5×142857. + This demonstrates an unusual property of 142857: it is a divisor of its + right-rotation. + + Find the last 5 digits of the sum of all integers n, 10 < n < 10^100, that + have this property. + + + Answer: 39e7aab76650b018578830bc6dba007a + + +Problem 169 +=========== + + + Define f(0)=1 and f(n) to be the number of different ways n can be + expressed as a sum of integer powers of 2 using each power no more than + twice. + + For example, f(10)=5 since there are five different ways to express 10: + + 1 + 1 + 8 + 1 + 1 + 4 + 4 + 1 + 1 + 2 + 2 + 4 + 2 + 4 + 4 + 2 + 8 + + What is f(10^25)? + + + Answer: d149d4836703a8908becea56ddd3ed42 + + +Problem 170 +=========== + + + Take the number 6 and multiply it by each of 1273 and 9854: + + 6 × 1273 = 7638 + 6 × 9854 = 59124 + + By concatenating these products we get the 1 to 9 pandigital 763859124. We + will call 763859124 the "concatenated product of 6 and (1273,9854)". + Notice too, that the concatenation of the input numbers, 612739854, is + also 1 to 9 pandigital. + + The same can be done for 0 to 9 pandigital numbers. + + What is the largest 0 to 9 pandigital 10-digit concatenated product of an + integer with two or more other integers, such that the concatenation of + the input numbers is also a 0 to 9 pandigital 10-digit number? + + + Answer: 6ffe65352f717c1731666a107ace96c1 + + +Problem 171 +=========== + + + For a positive integer n, let f(n) be the sum of the squares of the digits + (in base 10) of n, e.g. + + f(3) = 3^2 = 9, + f(25) = 2^2 + 5^2 = 4 + 25 = 29, + f(442) = 4^2 + 4^2 + 2^2 = 16 + 16 + 4 = 36 + + Find the last nine digits of the sum of all n, 0 < n < 10^20, such that + f(n) is a perfect square. + + + Answer: ff586db8c4a5699ec78c645fcb27db7b + + +Problem 172 +=========== + + + How many 18-digit numbers n (without leading zeros) are there such that no + digit occurs more than three times in n? + + + Answer: f5f260ee21ead7478403c2ccd18a1829 + + +Problem 173 +=========== + + + We shall define a square lamina to be a square outline with a square + "hole" so that the shape possesses vertical and horizontal symmetry. For + example, using exactly thirty-two square tiles we can form two different + square laminae: + + With one-hundred tiles, and not necessarily using all of the tiles at one + time, it is possible to form forty-one different square laminae. + + Using up to one million tiles how many different square laminae can be + formed? + + + p_173_square_laminas.gif + Answer: 177f825c89a68aefae37b8dec9bb8a9b + + +Problem 174 +=========== + + + We shall define a square lamina to be a square outline with a square + "hole" so that the shape possesses vertical and horizontal symmetry. + + Given eight tiles it is possible to form a lamina in only one way: 3x3 + square with a 1x1 hole in the middle. However, using thirty-two tiles it + is possible to form two distinct laminae. + + If t represents the number of tiles used, we shall say that t = 8 is type + L(1) and t = 32 is type L(2). + + Let N(n) be the number of t ≤ 1000000 such that t is type L(n); for + example, N(15) = 832. + + What is ∑ N(n) for 1 ≤ n ≤ 10? + + + p_173_square_laminas.gif + Answer: 73166006522ed7f51ed3e2ca66353b66 + + +Problem 175 +=========== + + Define f(0)=1 and f(n) to be the number of ways to write n as a sum of + powers of 2 where no power occurs more than twice. + + For example, f(10)=5 since there are five different ways to express 10: + 10 = 8+2 = 8+1+1 = 4+4+2 = 4+2+2+1+1 = 4+4+1+1 + + It can be shown that for every fraction p/q (p>0, q>0) there exists at + least one integer n such that + f(n)/f(n-1)=p/q. + + For instance, the smallest n for which f(n)/f(n-1)=13/17 is 241. + The binary expansion of 241 is 11110001. + Reading this binary number from the most significant bit to the least + significant bit there are 4 one's, 3 zeroes and 1 one. We shall call the + string 4,3,1 the Shortened Binary Expansion of 241. + + Find the Shortened Binary Expansion of the smallest n for which + f(n)/f(n-1)=123456789/987654321. + + Give your answer as comma separated integers, without any whitespaces. + + Answer: 796dddd004c3465229058072f5b4583e + + +Problem 176 +=========== + + + The four right-angled triangles with sides (9,12,15), (12,16,20), + (5,12,13) and (12,35,37) all have one of the shorter sides (catheti) equal + to 12. It can be shown that no other integer sided right-angled triangle + exists with one of the catheti equal to 12. + + Find the smallest integer that can be the length of a cathetus of exactly + 47547 different integer sided right-angled triangles. + + + Answer: c47c782ebaf8cdbb60eebfa86cd0003c + + +Problem 177 +=========== + + + Let ABCD be a convex quadrilateral, with diagonals AC and BD. At each + vertex the diagonal makes an angle with each of the two sides, creating + eight corner angles. + + For example, at vertex A, the two angles are CAD, CAB. + + We call such a quadrilateral for which all eight corner angles have + integer values when measured in degrees an "integer angled quadrilateral". + An example of an integer angled quadrilateral is a square, where all eight + corner angles are 45°. Another example is given by DAC = 20°, BAC = 60°, + ABD = 50°, CBD = 30°, BCA = 40°, DCA = 30°, CDB = 80°, ADB = 50°. + + What is the total number of non-similar integer angled quadrilaterals? + + Note: In your calculations you may assume that a calculated angle is + integral if it is within a tolerance of 10^-9 of an integer value. + + + p_177_quad.gif + Answer: d7a85236af930db0f7e84f2de8ee7ac2 + + +Problem 178 +=========== + + Consider the number 45656. + It can be seen that each pair of consecutive digits of 45656 has a + difference of one. + A number for which every pair of consecutive digits has a difference of + one is called a step number. + A pandigital number contains every decimal digit from 0 to 9 at least + once. + How many pandigital step numbers less than 10^40 are there? + + Answer: 2ffddfa898fa5df6321aebea84d4f33f + + +Problem 179 +=========== + + + Find the number of integers 1 < n < 10^7, for which n and n + 1 have the + same number of positive divisors. For example, 14 has the positive + divisors 1, 2, 7, 14 while 15 has 1, 3, 5, 15. + + + Answer: bafa0132bc7fc422a8d53bebb9d003c9 + + +Problem 180 +=========== + + + For any integer n, consider the three functions + + f[1,n](x,y,z) = x^n+1 + y^n+1 − z^n+1 + f[2,n](x,y,z) = (xy + yz + zx)*(x^n-1 + y^n-1 − z^n-1) + f[3,n](x,y,z) = xyz*(x^n-2 + y^n-2 − z^n-2) + + and their combination + + f[n](x,y,z) = f[1,n](x,y,z) + f[2,n](x,y,z) − f[3,n](x,y,z) + + We call (x,y,z) a golden triple of order k if x, y, and z are all rational + numbers of the form a / b with + 0 < a < b ≤ k and there is (at least) one integer n, so that f[n](x,y,z) = + 0. + + Let s(x,y,z) = x + y + z. + Let t = u / v be the sum of all distinct s(x,y,z) for all golden triples + (x,y,z) of order 35. + All the s(x,y,z) and t must be in reduced form. + + Find u + v. + + + Answer: 6459f69d151314c59df404868f45fa96 + + +Problem 181 +=========== + + + Having three black objects B and one white object W they can be grouped in + 7 ways like this: + + (BBBW) (B,BBW) (B,B,BW) (B,B,B,W) (B,BB,W) (BBB,W) (BB,BW) + + In how many ways can sixty black objects B and forty white objects W be + thus grouped? + + + Answer: 0e1233ecbc058dabf54a8602eac55d95 + + +Problem 182 +=========== + + + The RSA encryption is based on the following procedure: + + Generate two distinct primes p and q. + Compute n=pq and φ=(p-1)(q-1). + Find an integer e, 1 d + + For example, the best approximation to √13 for the denominator bound 20 is + 18/5 and the best approximation to √13 for the denominator bound 30 is + 101/28. + + Find the sum of all denominators of the best approximations to √n for the + denominator bound 10^12, where n is not a perfect square and 1 < n ≤ + 100000. + + + Answer: e5ec7d4b094709b1fcebbd73b10e6264 + + +Problem 193 +=========== + + + A positive integer n is called squarefree, if no square of a prime divides + n, thus 1, 2, 3, 5, 6, 7, 10, 11 are squarefree, but not 4, 8, 9, 12. + + How many squarefree numbers are there below 2^50? + + + Answer: ea29fcf755b560777b0b6d8714234d18 + + +Problem 194 +=========== + + + Consider graphs built with the units A: and B: , where the units are glued + alongthe vertical edges as in the graph . + + A configuration of type (a,b,c) is a graph thus built of a units A and b + units B, where the graph's vertices are coloured using up to c colours, so + that no two adjacent vertices have the same colour. + The compound graph above is an example of a configuration of type (2,2,6), + in fact of type (2,2,c) for all c ≥ 4. + + Let N(a,b,c) be the number of configurations of type (a,b,c). + For example, N(1,0,3) = 24, N(0,2,4) = 92928 and N(2,2,3) = 20736. + + Find the last 8 digits of N(25,75,1984). + + + p_194_GraphA.png + p_194_GraphB.png + p_194_Fig.png + Answer: e070561d568a80a0e45d7835e3817ba4 + + +Problem 195 +=========== + + + Let's call an integer sided triangle with exactly one angle of 60 degrees + a 60-degree triangle. + Let r be the radius of the inscribed circle of such a 60-degree triangle. + + There are 1234 60-degree triangles for which r ≤ 100. + Let T(n) be the number of 60-degree triangles for which r ≤ n, so + T(100) = 1234,  T(1000) = 22767, and  T(10000) = 359912. + + Find T(1053779). + + + Answer: 0fe232937a6d9f2a40825b86f568a38c + + +Problem 196 +=========== + + + Build a triangle from all positive integers in the following way: + +  1 +  2  3 +  4  5  6 +  7  8  9 10 + 11 12 13 14 15 + 16 17 18 19 20 21 + 22 23 24 25 26 27 28 + 29 30 31 32 33 34 35 36 + 37 38 39 40 41 42 43 44 45 + 46 47 48 49 50 51 52 53 54 55 + 56 57 58 59 60 61 62 63 64 65 66 + . . . + + Each positive integer has up to eight neighbours in the triangle. + + A set of three primes is called a prime triplet if one of the three primes + has the other two as neighbours in the triangle. + + For example, in the second row, the prime numbers 2 and 3 are elements of + some prime triplet. + + If row 8 is considered, it contains two primes which are elements of some + prime triplet, i.e. 29 and 31. + If row 9 is considered, it contains only one prime which is an element of + some prime triplet: 37. + + Define S(n) as the sum of the primes in row n which are elements of any + prime triplet. + Then S(8)=60 and S(9)=37. + + You are given that S(10000)=950007619. + + Find  S(5678027) + S(7208785). + + + Answer: fb6b6b0a4b7b31ba429152bc0b6bd037 + + +Problem 197 +=========== + + + Given is the function f(x) = ⌊2^30.403243784-x^2⌋ × 10^-9 ( ⌊ ⌋ is the + floor-function), + the sequence u[n] is defined by u[0] = -1 and u[n+1] = f(u[n]). + + Find u[n] + u[n+1] for n = 10^12. + Give your answer with 9 digits after the decimal point. + + + Answer: c98cbf87636906f2465d481be815e454 + + +Problem 198 +=========== + + + A best approximation to a real number x for the denominator bound d is a + rational number r/s (in reduced form) with s ≤ d, so that any rational + number p/q which is closer to x than r/s has q > d. + + Usually the best approximation to a real number is uniquely determined for + all denominator bounds. However, there are some exceptions, e.g. 9/40 has + the two best approximations 1/4 and 1/5 for the denominator bound 6.We + shall call a real number x ambiguous, if there is at least one denominator + bound for which x possesses two best approximations. Clearly, an ambiguous + number is necessarily rational. + + How many ambiguous numbers x = p/q,0 < x < 1/100, are there whose + denominator q does not exceed 10^8? + + + Answer: e59816f440fec9368c681314a127f3ee + + +Problem 199 +=========== + + + Three circles of equal radius are placed inside a larger circle such that + each pair of circles is tangent to one another and the inner circles do + not overlap. There are four uncovered "gaps" which are to be filled + iteratively with more tangent circles. + + At each iteration, a maximally sized circle is placed in each gap, which + creates more gaps for the next iteration. After 3 iterations (pictured), + there are 108 gaps and the fraction of the area which is not covered by + circles is 0.06790342, rounded to eight decimal places. + + What fraction of the area is not covered by circles after 10 iterations? + Give your answer rounded to eight decimal places using the format + x.xxxxxxxx . + + + p_199_circles_in_circles.gif + Answer: 0f8fd87159c28ae5fea6ac91a95d48dd + + +Problem 200 +=========== + + + We shall define a sqube to be a number of the form, p^2q^3, where p and q + are distinct primes. + For example, 200 = 5^22^3 or 120072949 = 23^261^3. + + The first five squbes are 72, 108, 200, 392, and 500. + + Interestingly, 200 is also the first number for which you cannot change + any single digit to make a prime; we shall call such numbers, prime-proof. + The next prime-proof sqube which contains the contiguous sub-string "200" + is 1992008. + + Find the 200th prime-proof sqube containing the contiguous sub-string + "200". + + + Answer: c911c8e346aa813da5f5ed4f8e9128d8 + + +Problem 201 +=========== + + + For any set A of numbers, let sum(A) be the sum of the elements of A. + Consider the set B = {1,3,6,8,10,11}. + There are 20 subsets of B containing three elements, and their sums are: + + sum({1,3,6}) = 10, + sum({1,3,8}) = 12, + sum({1,3,10}) = 14, + sum({1,3,11}) = 15, + sum({1,6,8}) = 15, + sum({1,6,10}) = 17, + sum({1,6,11}) = 18, + sum({1,8,10}) = 19, + sum({1,8,11}) = 20, + sum({1,10,11}) = 22, + sum({3,6,8}) = 17, + sum({3,6,10}) = 19, + sum({3,6,11}) = 20, + sum({3,8,10}) = 21, + sum({3,8,11}) = 22, + sum({3,10,11}) = 24, + sum({6,8,10}) = 24, + sum({6,8,11}) = 25, + sum({6,10,11}) = 27, + sum({8,10,11}) = 29. + + Some of these sums occur more than once, others are unique. + For a set A, let U(A,k) be the set of unique sums of k-element subsets of + A, in our example we find U(B,3) = {10,12,14,18,21,25,27,29} and + sum(U(B,3)) = 156. + + Now consider the 100-element set S = {1^2, 2^2, ... , 100^2}. + S has 100891344545564193334812497256 50-element subsets. + + Determine the sum of all integers which are the sum of exactly one of the + 50-element subsets of S, i.e. find sum(U(S,50)). + + + Answer: b7ad07c58c81a940b8ff067a13b2760d + + +Problem 202 +=========== + + + Three mirrors are arranged in the shape of an equilateral triangle, with + their reflective surfaces pointing inwards. There is an infinitesimal gap + at each vertex of the triangle through which a laser beam may pass. + + Label the vertices A, B and C. There are 2 ways in which a laser beam may + enter vertex C, bounce off 11 surfaces, then exit through the same vertex: + one way is shown below; the other is the reverse of that. + + There are 80840 ways in which a laser beam may enter vertex C, bounce off + 1000001 surfaces, then exit through the same vertex. + + In how many ways can a laser beam enter at vertex C, bounce off + 12017639147 surfaces, then exit through the same vertex? + + + p_201_laserbeam.gif + Answer: e9774949b5efad0d40d60ede379c5321 + + +Problem 203 +=========== + + + The binomial coefficients ^nC[k] can be arranged in triangular form, + Pascal's triangle, like this: + + 1 + 1 1 + 1 2 1 + 1 3 3 1 + 1 4 6 4 1 + 1 5 10 10 5 1 + 1 6 15 20 15 6 1 + 1 7 21 35 35 21 7 1 + + ......... + + It can be seen that the first eight rows of Pascal's triangle contain + twelve distinct numbers: 1, 2, 3, 4, 5, 6, 7, 10, 15, 20, 21 and 35. + + A positive integer n is called squarefree if no square of a prime divides + n.Of the twelve distinct numbers in the first eight rows of Pascal's + triangle, all except 4 and 20 are squarefree.The sum of the distinct + squarefree numbers in the first eight rows is 105. + + Find the sum of the distinct squarefree numbers in the first 51 rows of + Pascal's triangle. + + + Answer: d7ec16d216c923d3c927f46cfc914e92 + + +Problem 204 +=========== + + + A Hamming number is a positive number which has no prime factor larger + than 5. + So the first few Hamming numbers are 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15. + There are 1105 Hamming numbers not exceeding 10^8. + + We will call a positive number a generalised Hamming number of type n, if + it has no prime factor larger than n. + Hence the Hamming numbers are the generalised Hamming numbers of type 5. + + How many generalised Hamming numbers of type 100 are there which don't + exceed 10^9? + + + Answer: 4118ffb9edc56a033b5b27ca0bf34366 + + +Problem 205 +=========== + + + Peter has nine four-sided (pyramidal) dice, each with faces numbered 1, 2, + 3, 4. + Colin has six six-sided (cubic) dice, each with faces numbered 1, 2, 3, 4, + 5, 6. + + Peter and Colin roll their dice and compare totals: the highest total + wins. The result is a draw if the totals are equal. + + What is the probability that Pyramidal Pete beats Cubic Colin? Give your + answer rounded to seven decimal places in the form 0.abcdefg + + + Answer: ba6c6c3888227a0799eca38191b587be + + +Problem 206 +=========== + + + Find the unique positive integer whose square has the form + 1_2_3_4_5_6_7_8_9_0, + where each “_” is a single digit. + + + Answer: 09f9d87cb4b1ebb34e1f607e55a351d8 + + +Problem 207 +=========== + + + For some positive integers k, there exists an integer partition of the + form   4^t = 2^t + k, + where 4^t, 2^t, and k are all positive integers and t is a real number. + + The first two such partitions are 4^1 = 2^1 + 2 and 4^1.5849625... = + 2^1.5849625... + 6. + + Partitions where t is also an integer are called perfect. + For any m ≥ 1 let P(m) be the proportion of such partitions that are + perfect with k ≤ m. + Thus P(6) = 1/2. + + In the following table are listed some values of P(m) + +    P(5) = 1/1 +    P(10) = 1/2 +    P(15) = 2/3 +    P(20) = 1/2 +    P(25) = 1/2 +    P(30) = 2/5 +    ... +    P(180) = 1/4 +    P(185) = 3/13 + + Find the smallest m for which P(m) < 1/12345 + + + Answer: 3f17b264ed1717fe5fbde1e399bd501f + + +Problem 208 +=========== + + + A robot moves in a series of one-fifth circular arcs (72°), with a free + choice of a clockwise or an anticlockwise arc for each step, but no + turning on the spot. + + One of 70932 possible closed paths of 25 arcs starting northward is + + Given that the robot starts facing North, how many journeys of 70 arcs in + length can it take that return it, after the final arc, to its starting + position? + (Any arc may be traversed multiple times.) + + + p_208_robotwalk.gif + Answer: 3010e33173f30e0aac79e84835b48823 + + +Problem 209 +=========== + + + A k-input binary truth table is a map from k input bits(binary digits, 0 + [false] or 1 [true]) to 1 output bit. For example, the 2-input binary + truth tables for the logical AND and XOR functions are: + + ┌────┬────┬─────────┐ + │ x │ y │x AND y │ + ├────┼────┼─────────┤ + │ 0 │ 0 │ 0 │ + ├────┼────┼─────────┤ + │ 0 │ 1 │ 0 │ + ├────┼────┼─────────┤ + │ 1 │ 0 │ 0 │ + ├────┼────┼─────────┤ + │ 1 │ 1 │ 1 │ + └────┴────┴─────────┘ + + ┌────┬────┬─────────┐ + │ x │ y │x XOR y │ + ├────┼────┼─────────┤ + │ 0 │ 0 │ 0 │ + ├────┼────┼─────────┤ + │ 0 │ 1 │ 1 │ + ├────┼────┼─────────┤ + │ 1 │ 0 │ 1 │ + ├────┼────┼─────────┤ + │ 1 │ 1 │ 0 │ + └────┴────┴─────────┘ + + How many 6-input binary truth tables, τ, satisfy the formula + + τ(a, b, c, d, e, f) AND τ(b, c, d, e, f, a XOR (b AND c)) = 0 + + Answer: 954157aa4762df2ee29580ee5a351b13 + + +Problem 210 +=========== + + Consider the set S(r) of points (x,y) with integer coordinates satisfying + |x| + |y| ≤ r. + Let O be the point (0,0) and C the point (r/4,r/4). + Let N(r) be the number of points B in S(r), so that the triangle OBC has + an obtuse angle, i.e. the largest angle α satisfies 90°<α<180°. + So, for example, N(4)=24 and N(8)=100. + + What is N(1,000,000,000)? + + + Answer: 0c808b02789c4db462322ab2ac070bbb + + +Problem 211 +=========== + + + For a positive integer n, let σ[2](n) be the sum of the squares of its + divisors. For example, + + σ[2](10) = 1 + 4 + 25 + 100 = 130. + + Find the sum of all n, 0 < n < 64,000,000 such that σ[2](n) is a perfect + square. + + + Answer: 5fe0ed146690e7bca448687a94353a73 + + +Problem 212 +=========== + + + An axis-aligned cuboid, specified by parameters { (x[0],y[0],z[0]), + (dx,dy,dz) }, consists of all points (X,Y,Z) such that x[0] ≤ X ≤ x[0]+dx, + y[0] ≤ Y ≤ y[0]+dy and z[0] ≤ Z ≤ z[0]+dz. The volume of the cuboid is the + product, dx × dy × dz. The combined volume of a collection of cuboids is + the volume of their union and will be less than the sum of the individual + volumes if any cuboids overlap. + + Let C[1],...,C[50000] be a collection of 50000 axis-aligned cuboids such + that C[n] has parameters + + x[0] = S[6n-5] modulo 10000 + y[0] = S[6n-4] modulo 10000 + z[0] = S[6n-3] modulo 10000 + dx = 1 + (S[6n-2] modulo 399) + dy = 1 + (S[6n-1] modulo 399) + dz = 1 + (S[6n] modulo 399) + + where S[1],...,S[300000] come from the "Lagged Fibonacci Generator": + + For 1 ≤ k ≤ 55, S[k] = [100003 - 200003k + 300007k^3]   (modulo 1000000) + For 56 ≤ k, S[k] = [S[k-24] + S[k-55]]   (modulo 1000000) + + Thus, C[1] has parameters {(7,53,183),(94,369,56)}, C[2] has parameters + {(2383,3563,5079),(42,212,344)}, and so on. + + The combined volume of the first 100 cuboids, C[1],...,C[100], is + 723581599. + + What is the combined volume of all 50000 cuboids, C[1],...,C[50000] ? + + + Answer: 76650c9c077929e1ce5a80a1ac81fa96 + + +Problem 213 +=========== + + + A 30×30 grid of squares contains 900 fleas, initially one flea per square. + When a bell is rung, each flea jumps to an adjacent square at random + (usually 4 possibilities, except for fleas on the edge of the grid or at + the corners). + + What is the expected number of unoccupied squares after 50 rings of the + bell? Give your answer rounded to six decimal places. + + + Answer: f81ee7dd444a3d895a4a446f9d115bf8 + + +Problem 214 +=========== + + + Let φ be Euler's totient function, i.e. for a natural number n,φ(n) is the + number of k, 1 ≤ k ≤ n, for which gcd(k,n) = 1. + + By iterating φ, each positive integer generates a decreasing chain of + numbers ending in 1. + E.g. if we start with 5 the sequence 5,4,2,1 is generated. + Here is a listing of all chains with length 4: + + 5,4,2,1 + 7,6,2,1 + 8,4,2,1 + 9,6,2,1 + 10,4,2,1 + 12,4,2,1 + 14,6,2,1 + 18,6,2,1 + + Only two of these chains start with a prime, their sum is 12. + + What is the sum of all primes less than 40000000 which generate a chain of + length 25? + + + Answer: 1cefd865483c03552d5247ffb05685c7 + + +Problem 215 +=========== + + + Consider the problem of building a wall out of 2×1 and 3×1 bricks + (horizontal×vertical dimensions) such that, for extra strength, the gaps + between horizontally-adjacent bricks never line up in consecutive layers, + i.e. never form a "running crack". + + For example, the following 9×3 wall is not acceptable due to the running + crack shown in red: + + There are eight ways of forming a crack-free 9×3 wall, written W(9,3) = 8. + + Calculate W(32,10). + + + p_215_crackfree.gif + Answer: 60212c9ec4a6cd1d14277c32b6adf2d8 + + +Problem 216 +=========== + + + Consider numbers t(n) of the form t(n) = 2n^2-1 with n > 1. + The first such numbers are 7, 17, 31, 49, 71, 97, 127 and 161. + It turns out that only 49 = 7*7 and 161 = 7*23 are not prime. + For n ≤ 10000 there are 2202 numbers t(n) that are prime. + + How many numbers t(n) are prime for n ≤ 50,000,000 ? + + + Answer: e512153424a482deb9de401ac0465a72 + + +Problem 217 +=========== + + + A positive integer with k (decimal) digits is called balanced if its first + ⌈^k/[2]⌉ digits sum to the same value as its last ⌈^k/[2]⌉ digits, where + ⌈x⌉, pronounced ceiling of x, is the smallest integer ≥ x, thus ⌈π⌉ = 4 + and ⌈5⌉ = 5. + + So, for example, all palindromes are balanced, as is 13722. + + Let T(n) be the sum of all balanced numbers less than 10^n. + Thus: T(1) = 45, T(2) = 540 and T(5) = 334795890. + + Find T(47) mod 3^15 + + + Answer: 11bff97aac06892e1a07ebf7febfa8db + + +Problem 218 +=========== + + + Consider the right angled triangle with sides a=7, b=24 and c=25.The area + of this triangle is 84, which is divisible by the perfect numbers 6 and + 28. + Moreover it is a primitive right angled triangle as gcd(a,b)=1 and + gcd(b,c)=1. + Also c is a perfect square. + + We will call a right angled triangle perfect if + -it is a primitive right angled triangle + -its hypotenuse is a perfect square + + We will call a right angled triangle super-perfect if + -it is a perfect right angled triangle and + -its area is a multiple of the perfect numbers 6 and 28. + + How many perfect right-angled triangles with c≤10^16 exist that are not + super-perfect? + + + Answer: cfcd208495d565ef66e7dff9f98764da + + +Problem 219 +=========== + + + Let A and B be bit strings (sequences of 0's and 1's). + If A is equal to the leftmost length(A) bits of B, then A is said to be a + prefix of B. + For example, 00110 is a prefix of 001101001, but not of 00111 or 100110. + + A prefix-free code of size n is a collection of n distinct bit strings + such that no string is a prefix of any other. For example, this is a + prefix-free code of size 6: + + 0000, 0001, 001, 01, 10, 11 + + Now suppose that it costs one penny to transmit a '0' bit, but four pence + to transmit a '1'. + Then the total cost of the prefix-free code shown above is 35 pence, which + happens to be the cheapest possible for the skewed pricing scheme in + question. + In short, we write Cost(6) = 35. + + What is Cost(10^9) ? + + + Answer: 578c22ef288b88c60fbcf4541351aff5 + + +Problem 220 +=========== + + + Let D[0] be the two-letter string "Fa". For n≥1, derive D[n] from D[n-1] + by the string-rewriting rules: + + "a" → "aRbFR" + "b" → "LFaLb" + + Thus, D[0] = "Fa", D[1] = "FaRbFR", D[2] = "FaRbFRRLFaLbFR", and so on. + + These strings can be interpreted as instructions to a computer graphics + program, with "F" meaning "draw forward one unit", "L" meaning "turn left + 90 degrees", "R" meaning "turn right 90 degrees", and "a" and "b" being + ignored. The initial position of the computer cursor is (0,0), pointing up + towards (0,1). + + Then D[n] is an exotic drawing known as the Heighway Dragon of order n. + For example, D[10] is shown below; counting each "F" as one step, the + highlighted spot at (18,16) is the position reached after 500 steps. + + What is the position of the cursor after 10^12 steps in D[50] ? + Give your answer in the form x,y with no spaces. + + + p_220.gif + Answer: e2018d8efde8ea00319f1adc042f150b + + +Problem 221 +=========== + + + We shall call a positive integer A an "Alexandrian integer", if there + exist integers p, q, r such that: + + A = p · q · r    and   1 = 1 + 1 + 1 + A p q r + + For example, 630 is an Alexandrian integer (p = 5, q = −7, r = −18).In + fact, 630 is the 6^th Alexandrian integer, the first 6 Alexandrian + integers being: 6, 42, 120, 156, 420 and 630. + + Find the 150000^th Alexandrian integer. + + + Answer: cb000c24f653d9c8f78b74123e6515ab + + +Problem 222 +=========== + + + What is the length of the shortest pipe, of internal radius 50mm, that can + fully contain 21 balls of radii 30mm, 31mm, ..., 50mm? + + Give your answer in micrometres (10^-6 m) rounded to the nearest integer. + + + Answer: 6984ba429b968467619ec98a8ee51abf + + +Problem 223 +=========== + + + Let us call an integer sided triangle with sides a ≤ b ≤ c barely acute if + the sides satisfy + a^2 + b^2 = c^2 + 1. + + How many barely acute triangles are there with perimeter ≤ 25,000,000? + + + Answer: cb875e59736a1967c8da8fc8062a6bc5 + + +Problem 224 +=========== + + + Let us call an integer sided triangle with sides a ≤ b ≤ c barely obtuse + if the sides satisfy + a^2 + b^2 = c^2 - 1. + + How many barely obtuse triangles are there with perimeter ≤ 75,000,000? + + + Answer: c43cfb12750dee27b4b0d016261e831b + + +Problem 225 +=========== + + + The sequence 1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, 653, 1201 ... + is defined by T[1] = T[2] = T[3] = 1 and T[n] = T[n-1] + T[n-2] + T[n-3]. + + It can be shown that 27 does not divide any terms of this sequence. + In fact, 27 is the first odd number with this property. + + Find the 124^th odd number that does not divide any terms of the above + sequence. + + + Answer: f1981e4bd8a0d6d8462016d2fc6276b3 + + +Problem 226 +=========== + + + The blancmange curve is the set of points (x,y) such that 0 ≤ x ≤ 1 and , + where s(x) = the distance from x to the nearest integer. + + The area under the blancmange curve is equal to ½, shown in pink in the + diagram below. + + [1]blancmange curve + + Let C be the circle with centre (¼,½) and radius ¼, shown in black in the + diagram. + + What area under the blancmange curve is enclosed by C? + Give your answer rounded to eight decimal places in the form 0.abcdefgh + + + Visible links + p_226_formula.gif + p_226_scoop2.gif + Answer: ce6fd32d1d2fb58c4c0c1f7962c39f04 + + +Problem 227 +=========== + + + "The Chase" is a game played with two dice and an even number of players. + + The players sit around a table; the game begins with two opposite players + having one die each. On each turn, the two players with a die roll it. + If a player rolls a 1, he passes the die to his neighbour on the left; if + he rolls a 6, he passes the die to his neighbour on the right; otherwise, + he keeps the die for the next turn. + The game ends when one player has both dice after they have been rolled + and passed; that player has then lost. + + In a game with 100 players, what is the expected number of turns the game + lasts? + + Give your answer rounded to ten significant digits. + + + Answer: 7b87cd0a96f0f2f12f911cdc66608d95 + + +Problem 228 +=========== + + + Let S[n] be the regular n-sided polygon – or shape – whose vertices v[k] + (k = 1,2,…,n) have coordinates: + + x[k]   =   cos( ^2k-1/[n] ×180° ) + y[k]   =   sin( ^2k-1/[n] ×180° ) + + Each S[n] is to be interpreted as a filled shape consisting of all points + on the perimeter and in the interior. + + The Minkowski sum, S+T, of two shapes S and T is the result of adding + every point in S to every point in T, where point addition is performed + coordinate-wise: (u, v) + (x, y) = (u+x, v+y). + + For example, the sum of S[3] and S[4] is the six-sided shape shown in pink + below: + + [1]picture showing S_3 + S_4 + + How many sides does S[1864] + S[1865] + … + S[1909] have? + + + Visible links + p_228.png + Answer: 35d0195ddaf58e52e12400de1c9333d8 + + +Problem 229 +=========== + + + Consider the number 3600. It is very special, because + + 3600 = 48^2 +     36^2 + + 3600 = 20^2 + 2×40^2 + + 3600 = 30^2 + 3×30^2 + + 3600 = 45^2 + 7×15^2 + + Similarly, we find that 88201 = 99^2 + 280^2 = 287^2 + 2×54^2 = 283^2 + + 3×52^2 = 197^2 + 7×84^2. + + In 1747, Euler proved which numbers are representable as a sum of two + squares.We are interested in the numbers n which admit representations of + all of the following four types: + + n = a[1]^2 +   b[1]^2 + + n = a[2]^2 + 2 b[2]^2 + + n = a[3]^2 + 3 b[3]^2 + + n = a[7]^2 + 7 b[7]^2, + + where the a[k] and b[k] are positive integers. + + There are 75373 such numbers that do not exceed 10^7. + How many such numbers are there that do not exceed 2×10^9? + + + Answer: d68b5ec8df4a56991901f67afbdef24f + + +Problem 230 +=========== + + + For any two strings of digits, A and B, we define F[A,B] to be the + sequence (A,B,AB,BAB,ABBAB,...) in which each term is the concatenation of + the previous two. + + Further, we define D[A,B](n) to be the n^th digit in the first term of + F[A,B] that contains at least n digits. + + Example: + + Let A=1415926535, B=8979323846. We wish to find D[A,B](35), say. + + The first few terms of F[A,B] are: + 1415926535 + 8979323846 + 14159265358979323846 + 897932384614159265358979323846 + 14159265358979323846897932384614159265358979323846 + + Then D[A,B](35) is the 35^th digit in the fifth term, which is 9. + + Now we use for A the first 100 digits of π behind the decimal point: + + 14159265358979323846264338327950288419716939937510 + 58209749445923078164062862089986280348253421170679 + + and for B the next hundred digits: + + 82148086513282306647093844609550582231725359408128 + 48111745028410270193852110555964462294895493038196 . + + Find ∑[n = 0,1,...,17]   10^n× D[A,B]((127+19n)×7^n) . + + + Answer: 040735038021ff4704bbd3a0964369ef + + +Problem 231 +=========== + + + The binomial coefficient ^10C[3] = 120. + 120 = 2^3 × 3 × 5 = 2 × 2 × 2 × 3 × 5, and 2 + 2 + 2 + 3 + 5 = 14. + So the sum of the terms in the prime factorisation of ^10C[3] is 14. + + Find the sum of the terms in the prime factorisation of + ^20000000C[15000000]. + + + Answer: ef8bc4d9a843e71126bd10b5065132a5 + + +Problem 232 +=========== + + + Two players share an unbiased coin and take it in turns to play "The + Race". On Player 1's turn, he tosses the coin once: if it comes up Heads, + he scores one point; if it comes up Tails, he scores nothing. On Player + 2's turn, she chooses a positive integer T and tosses the coin T times: if + it comes up all Heads, she scores 2^T-1 points; otherwise, she scores + nothing. Player 1 goes first. The winner is the first to 100 or more + points. + + On each turn Player 2 selects the number, T, of coin tosses that maximises + the probability of her winning. + + What is the probability that Player 2 wins? + + Give your answer rounded to eight decimal places in the form 0.abcdefgh . + + + Answer: c8d5b243aa6e6b507725766f7c197a1d + + +Problem 233 +=========== + + + Let f(N) be the number of points with integer coordinates that are on a + circle passing through (0,0), (N,0),(0,N), and (N,N). + + It can be shown that f(10000) = 36. + + What is the sum of all positive integers N ≤ 10^11 such that f(N) = 420 ? + + + Answer: 7e80b27798170abb493e3b4671bd82ca + + +Problem 234 +=========== + + + For an integer n ≥ 4, we define the lower prime square root of n, denoted + by lps(n), as the largest prime ≤ √n and the upper prime square root of n, + ups(n), as the smallest prime ≥ √n. + + So, for example, lps(4) = 2 = ups(4), lps(1000) = 31, ups(1000) = 37. + Let us call an integer n ≥ 4 semidivisible, if one of lps(n) and ups(n) + divides n, but not both. + + The sum of the semidivisible numbers not exceeding 15 is 30, the numbers + are 8, 10 and 12. + 15 is not semidivisible because it is a multiple of both lps(15) = 3 and + ups(15) = 5. + As a further example, the sum of the 92 semidivisible numbers up to 1000 + is 34825. + + What is the sum of all semidivisible numbers not exceeding 999966663333 ? + + + Answer: c24a5d60f8ce5d04dec7466987c84d68 + + +Problem 235 +=========== + + + Given is the arithmetic-geometric sequence u(k) = (900-3k)r^k-1. + Let s(n) = Σ[k=1...n]u(k). + + Find the value of r for which s(5000) = -600,000,000,000. + + Give your answer rounded to 12 places behind the decimal point. + + + Answer: 41b13508789be1001308e065d4f83ea2 + + +Problem 236 +=========== + + + Suppliers 'A' and 'B' provided the following numbers of products for the + luxury hamper market: + + Product 'A' 'B' + Beluga Caviar 5248 640 + Christmas Cake 1312 1888 + Gammon Joint 2624 3776 + Vintage Port 5760 3776 + Champagne Truffles 3936 5664 + + Although the suppliers try very hard to ship their goods in perfect + condition, there is inevitably some spoilage - i.e. products gone bad. + + The suppliers compare their performance using two types of statistic: + + • The five per-product spoilage rates for each supplier are equal to the + number of products gone bad divided by the number of products + supplied, for each of the five products in turn. + • The overall spoilage rate for each supplier is equal to the total + number of products gone bad divided by the total number of products + provided by that supplier. + + To their surprise, the suppliers found that each of the five per-product + spoilage rates was worse (higher) for 'B' than for 'A' by the same factor + (ratio of spoilage rates), m>1; and yet, paradoxically, the overall + spoilage rate was worse for 'A' than for 'B', also by a factor of m. + + There are thirty-five m>1 for which this surprising result could have + occurred, the smallest of which is 1476/1475. + + What's the largest possible value of m? + Give your answer as a fraction reduced to its lowest terms, in the form + u/v. + + + Answer: 6e707fcffc510520d981ae16a29579bb + + +Problem 237 +=========== + + + Let T(n) be the number of tours over a 4 × n playing board such that: + + • The tour starts in the top left corner. + • The tour consists of moves that are up, down, left, or right one + square. + • The tour visits each square exactly once. + • The tour ends in the bottom left corner. + + The diagram shows one tour over a 4 × 10 board: + + T(10) is 2329. What is T(10^12) modulo 10^8? + + + p_237.gif + Answer: 0742988a3948491b15fb48e476c78a6a + + +Problem 238 +=========== + + + Create a sequence of numbers using the "Blum Blum Shub" pseudo-random + number generator: + + s[0] = 14025256 + s[n+1] = s[n]^2 mod 20300713 + + Concatenate these numbers  s[0]s[1]s[2]… to create a string w of infinite + length. + Then, w = 14025256741014958470038053646… + + For a positive integer k, if no substring of w exists with a sum of digits + equal to k, p(k) is defined to be zero. If at least one substring of w + exists with a sum of digits equal to k, we define p(k) = z, where z is the + starting position of the earliest such substring. + + For instance: + + The substrings 1, 14, 1402, … + with respective sums of digits equal to 1, 5, 7, … + start at position 1, hence p(1) = p(5) = p(7) = … = 1. + + The substrings 4, 402, 4025, … + with respective sums of digits equal to 4, 6, 11, … + start at position 2, hence p(4) = p(6) = p(11) = … = 2. + + The substrings 02, 0252, … + with respective sums of digits equal to 2, 9, … + start at position 3, hence p(2) = p(9) = … = 3. + + Note that substring 025 starting at position 3, has a sum of digits equal + to 7, but there was an earlier substring (starting at position 1) with a + sum of digits equal to 7, so p(7) = 1, not 3. + + We can verify that, for 0 < k ≤ 10^3, ∑ p(k) = 4742. + + Find ∑ p(k), for 0 < k ≤ 2·10^15. + + + Answer: 424ed6613a372ccb9a90dddb8961ca16 + + +Problem 239 +=========== + + + A set of disks numbered 1 through 100 are placed in a line in random + order. + + What is the probability that we have a partial derangement such that + exactly 22 prime number discs are found away from their natural positions? + (Any number of non-prime disks may also be found in or out of their + natural positions.) + + Give your answer rounded to 12 places behind the decimal point in the form + 0.abcdefghijkl. + + + Answer: 451fd2b8c19fbfec650a5c4699f6ef6e + + +Problem 240 +=========== + + + There are 1111 ways in which five 6-sided dice (sides numbered 1 to 6) can + be rolled so that the top three sum to 15. Some examples are: + + D[1],D[2],D[3],D[4],D[5] = 4,3,6,3,5 + D[1],D[2],D[3],D[4],D[5] = 4,3,3,5,6 + D[1],D[2],D[3],D[4],D[5] = 3,3,3,6,6 + D[1],D[2],D[3],D[4],D[5] = 6,6,3,3,3 + + In how many ways can twenty 12-sided dice (sides numbered 1 to 12) be + rolled so that the top ten sum to 70? + + + Answer: cb31a3106db3876e77cd160664cd683e + + +Problem 241 +=========== + + + For a positive integer n, let σ(n) be the sum of all divisors of n, so + e.g. σ(6) = 1 + 2 + 3 + 6 = 12. + + A perfect number, as you probably know, is a number with σ(n) = 2n. + + Let us define the perfection quotient of a positive integer p(n) =  σ(n) . + as n + + Find the sum of all positive integers n ≤ 10^18 for which p(n) has the + form k + ^1⁄[2], where k is an integer. + + + Answer: 556bfef2cacd1eff8af9126c5c13dcbc + + +Problem 242 +=========== + + + Given the set {1,2,...,n}, we define f(n,k) as the number of its k-element + subsets with an odd sum of elements. For example, f(5,3) = 4, since the + set {1,2,3,4,5} has four 3-element subsets having an odd sum of elements, + i.e.: {1,2,4}, {1,3,5}, {2,3,4} and {2,4,5}. + + When all three values n, k and f(n,k) are odd, we say that they make + an odd-triplet [n,k,f(n,k)]. + + There are exactly five odd-triplets with n ≤ 10, namely: + [1,1,f(1,1) = 1], [5,1,f(5,1) = 3], [5,5,f(5,5) = 1], [9,1,f(9,1) = 5] and + [9,9,f(9,9) = 1]. + + How many odd-triplets are there with n ≤ 10^12 ? + + + Answer: ba73cb75365ddca8f94a23e3fedfb6de + + +Problem 243 +=========== + + + A positive fraction whose numerator is less than its denominator is called + a proper fraction. + For any denominator, d, there will be d−1 proper fractions; for example, + with d = 12: + 1/12 , 2/12 , 3/12 , 4/12 , 5/12 , 6/12 , 7/12 , + 8/12 , 9/12 , 10/12 , 11/12 . + + We shall call a fraction that cannot be cancelled down a resilient + fraction. + Furthermore we shall define the resilience of a denominator, R(d), to be + the ratio of its proper fractions that are resilient; for example, R(12) = + 4/11 . + In fact, d = 12 is the smallest denominator having a resilience R(d) < + 4/10 . + + Find the smallest denominator d, having a resilience R(d) < 15499/94744 + . + + + Answer: 531721a10786c5c2a444b474fcf039f9 + + +Problem 244 +=========== + + + You probably know the game Fifteen Puzzle. Here, instead of numbered + tiles, we have seven red tiles and eight blue tiles. + + A move is denoted by the uppercase initial of the direction (Left, Right, + Up, Down) in which the tile is slid, e.g. starting from configuration (S), + by the sequence LULUR we reach the configuration (E): + + (S) , (E) + + For each path, its checksum is calculated by (pseudocode): + checksum = 0 + checksum = (checksum × 243 + m[1]) mod 100 000 007 + checksum = (checksum × 243 + m[2]) mod 100 000 007 +    … + checksum = (checksum × 243 + m[n]) mod 100 000 007 + where m[k] is the ASCII value of the k^th letter in the move sequence and + the ASCII values for the moves are: + + ┌──────┬─────┐ + │L │76 │ + ├──────┼─────┤ + │R │82 │ + ├──────┼─────┤ + │U │85 │ + ├──────┼─────┤ + │D │68 │ + └──────┴─────┘ + + For the sequence LULUR given above, the checksum would be 19761398. + + Now, starting from configuration (S),find all shortest ways to reach + configuration (T). + + (S) , (T) + + What is the sum of all checksums for the paths having the minimal length? + + + p_244_start.gif + p_244_example.gif + p_244_start.gif + p_244_target.gif + Answer: f8fd502ec1d0084a79d43d9dc5bd3a3d + + +Problem 245 +=========== + + + We shall call a fraction that cannot be cancelled down a resilient + fraction. + Furthermore we shall define the resilience of a denominator, R(d), to be + the ratio of its proper fractions that are resilient; for example, R(12) = + ^4⁄[11]. + + The resilience of a number d > 1 is φ(d) , where φ is Euler's totient + then d - 1 function. + + We further define the coresilience of a number n > 1 as C(n) =  n - φ(n) . + n - 1 + + The coresilience of a prime p is C(p) =  1 . + p - 1 + + Find the sum of all composite integers 1 < n ≤ 2×10^11, for which C(n) is + a unit fraction. + + + Answer: 0ebeb502fb0bd7157609835d27c266bc + + +Problem 246 +=========== + + + A definition for an ellipse is: + Given a circle c with centre M and radius r and a point G such that + d(G,M)0 and removes: + + • N stones from any single pile; or + • N stones from each of any two piles (2N total); or + • N stones from each of the three piles (3N total). + + The player taking the last stone(s) wins the game. + + A winning configuration is one where the first player can force a win. + For example, (0,0,13), (0,11,11) and (5,5,5) are winning configurations + because the first player can immediately remove all stones. + + A losing configuration is one where the second player can force a win, no + matter what the first player does. + For example, (0,1,2) and (1,3,3) are losing configurations: any legal move + leaves a winning configuration for the second player. + + Consider all losing configurations (x[i],y[i],z[i]) where x[i] ≤ y[i] ≤ + z[i] ≤ 100. + We can verify that Σ(x[i]+y[i]+z[i]) = 173895 for these. + + Find Σ(x[i]+y[i]+z[i]) where (x[i],y[i],z[i]) ranges over the losing + configurations + with x[i] ≤ y[i] ≤ z[i] ≤ 1000. + + + Answer: cab69719e6968409ba167707a09875cb + + +Problem 261 +=========== + + + Let us call a positive integer k a square-pivot, if there is a pair of + integers m > 0 and n ≥ k, such that the sum of the (m+1) consecutive + squares up to k equals the sum of the m consecutive squares from (n+1) on: + + (k-m)^2 + ... + k^2 = (n+1)^2 + ... + (n+m)^2. + + Some small square-pivots are + + • 4: 3^2 + 4^2 = 5^2 + • 21: 20^2 + 21^2 = 29^2 + • 24: 21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2 + • 110: 108^2 + 109^2 + 110^2 = 133^2 + 134^2 + + Find the sum of all distinct square-pivots ≤ 10^10. + + + Answer: d45ddf64010ed143228a6a6b84837de9 + + +Problem 262 +=========== + + + The following equation represents the continuous topography of a + mountainous region, giving the elevation h at any point (x,y): + + A mosquito intends to fly from A(200,200) to B(1400,1400), without leaving + the area given by 0 ≤ x, y ≤ 1600. + + Because of the intervening mountains, it first rises straight up to a + point A', having elevation f. Then, while remaining at the same elevation + f, it flies around any obstacles until it arrives at a point B' directly + above B. + + First, determine f[min] which is the minimum constant elevation allowing + such a trip from A to B, while remaining in the specified area. + Then, find the length of the shortest path between A' and B', while flying + at that constant elevation f[min]. + + Give that length as your answer, rounded to three decimal places. + + Note: For convenience, the elevation function shown above is repeated + below, in a form suitable for most programming languages: + h=( 5000-0.005*(x*x+y*y+x*y)+12.5*(x+y) ) * exp( + -abs(0.000001*(x*x+y*y)-0.0015*(x+y)+0.7) ) + + + p_262_formula1.gif + Answer: a5921e175a44d31e7f82f7f9a61a36af + + +Problem 263 +=========== + + + Consider the number 6. The divisors of 6 are: 1,2,3 and 6. + Every number from 1 up to and including 6 can be written as a sum of + distinct divisors of 6: + 1=1, 2=2, 3=1+2, 4=1+3, 5=2+3, 6=6. + A number n is called a practical number if every number from 1 up to and + including n can be expressed as a sum of distinct divisors of n. + + A pair of consecutive prime numbers with a difference of six is called a + sexy pair (since "sex" is the Latin word for "six"). The first sexy pair + is (23, 29). + + We may occasionally find a triple-pair, which means three consecutive sexy + prime pairs, such that the second member of each pair is the first member + of the next pair. + + We shall call a number n such that : + + • (n-9, n-3), (n-3,n+3), (n+3, n+9) form a triple-pair, and + • the numbers n-8, n-4, n, n+4 and n+8 are all practical, + + an engineers’ paradise. + + Find the sum of the first four engineers’ paradises. + + + Answer: 8fe3eb7196c69a080740e076cff9b4a1 + + +Problem 264 +=========== + + + Consider all the triangles having: + + • All their vertices on lattice points. + • Circumcentre at the origin O. + • Orthocentre at the point H(5, 0). + + There are nine such triangles having a perimeter ≤ 50. + Listed and shown in ascending order of their perimeter, they are: + + A(-4, 3), B(5, 0), C(4, -3) + A(4, 3), B(5, 0), C(-4, -3) + A(-3, 4), B(5, 0), C(3, -4) + + A(3, 4), B(5, 0), C(-3, -4) + A(0, 5), B(5, 0), C(0, -5) + A(1, 8), B(8, -1), C(-4, -7) + + A(8, 1), B(1, -8), C(-4, 7) + A(2, 9), B(9, -2), C(-6, -7) + A(9, 2), B(2, -9), C(-6, 7) + + The sum of their perimeters, rounded to four decimal places, is 291.0089. + + Find all such triangles with a perimeter ≤ 10^5. + Enter as your answer the sum of their perimeters rounded to four decimal + places. + + + p_264_TriangleCentres.gif + Answer: 287514a045a38be0a75a1786694c77ee + + +Problem 265 +=========== + + + 2^N binary digits can be placed in a circle so that all the N-digit + clockwise subsequences are distinct. + + For N=3, two such circular arrangements are possible, ignoring rotations: + + For the first arrangement, the 3-digit subsequences, in clockwise order, + are: + 000, 001, 010, 101, 011, 111, 110 and 100. + + Each circular arrangement can be encoded as a number by concatenating the + binary digits starting with the subsequence of all zeros as the most + significant bits and proceeding clockwise. The two arrangements for N=3 + are thus represented as 23 and 29: + 00010111 [2] = 23 + 00011101 [2] = 29 + + Calling S(N) the sum of the unique numeric representations, we can see + that S(3) = 23 + 29 = 52. + + Find S(5). + + + p_265_BinaryCircles.gif + Answer: c25cebbc8dce4bdcf96cb395a11afcc3 + + +Problem 266 +=========== + + + The divisors of 12 are: 1,2,3,4,6 and 12. + The largest divisor of 12 that does not exceed the square root of 12 is 3. + We shall call the largest divisor of an integer n that does not exceed the + square root of n the pseudo square root (PSR) of n. + It can be seen that PSR(3102)=47. + + Let p be the product of the primes below 190. + Find PSR(p) mod 10^16. + + + Answer: 32da1d501e539ab509f104e2db68d57a + + +Problem 267 +=========== + + + You are given a unique investment opportunity. + + Starting with £1 of capital, you can choose a fixed proportion, f, of your + capital to bet on a fair coin toss repeatedly for 1000 tosses. + + Your return is double your bet for heads and you lose your bet for tails. + + For example, if f = 1/4, for the first toss you bet £0.25, and if heads + comes up you win £0.5 and so then have £1.5. You then bet £0.375 and if + the second toss is tails, you have £1.125. + + Choosing f to maximize your chances of having at least £1,000,000,000 + after 1,000 flips, what is the chance that you become a billionaire? + + All computations are assumed to be exact (no rounding), but give your + answer rounded to 12 digits behind the decimal point in the form + 0.abcdefghijkl. + + + Answer: b8dd3306c2c64eacb0ac36b414892edb + + +Problem 268 +=========== + + + It can be verified that there are 23 positive integers less than 1000 that + are divisible by at least four distinct primes less than 100. + + Find how many positive integers less than 10^16 are divisible by at least + four distinct primes less than 100. + + + Answer: 6f84b20c10311cb24a824416a3c3e0a4 + + +Problem 269 +=========== + + + A root or zero of a polynomial P(x) is a solution to the equation P(x) = + 0. + Define P[n] as the polynomial whose coefficients are the digits of n. + For example, P[5703](x) = 5x^3 + 7x^2 + 3. + + We can see that: + + • P[n](0) is the last digit of n, + • P[n](1) is the sum of the digits of n, + • P[n](10) is n itself. + + Define Z(k) as the number of positive integers, n, not exceeding k for + which the polynomial P[n] has at least one integer root. + + It can be verified that Z(100 000) is 14696. + + What is Z(10^16)? + + + Answer: f7ba868cb52a9b9c7e58b1b92e230be8 + + +Problem 270 +=========== + + + A square piece of paper with integer dimensions N×N is placed with a + corner at the origin and two of its sides along the x- and y-axes. Then, + we cut it up respecting the following rules: + + • We only make straight cuts between two points lying on different sides + of the square, and having integer coordinates. + • Two cuts cannot cross, but several cuts can meet at the same border + point. + • Proceed until no more legal cuts can be made. + + Counting any reflections or rotations as distinct, we call C(N) the number + of ways to cut an N×N square. For example, C(1) = 2 and C(2) = 30 (shown + below). + + What is C(30) mod 10^8 ? + + + p_270_CutSquare.gif + Answer: 2a592dfd1e9e3e9e38578affa7c72605 + + +Problem 271 +=========== + + + For a positive number n, define S(n) as the sum of the integers x, for + which 1 1 coprime to 10 there is a positive divisibility + multiplier m < p which preserves divisibility by p for the following + function on any positive integer, n: + + f(n) = (all but the last digit of n) + (the last digit of n) * m + + That is, if m is the divisibility multiplier for p, then f(n) is divisible + by p if and only if n is divisible by p. + + (When n is much larger than p, f(n) will be less than n and repeated + application of f provides a multiplicative divisibility test for p.) + + For example, the divisibility multiplier for 113 is 34. + + f(76275) = 7627 + 5 * 34 = 7797 : 76275 and 7797 are both divisible by 113 + f(12345) = 1234 + 5 * 34 = 1404 : 12345 and 1404 are both not divisible by + 113 + + The sum of the divisibility multipliers for the primes that are coprime to + 10 and less than 1000 is 39517. What is the sum of the divisibility + multipliers for the primes that are coprime to 10 and less than 10^7? + + + Answer: ffd68ca67b9c3ea2653d375051e70288 + + +Problem 275 +=========== + + + Let us define a balanced sculpture of order n as follows: + + • A polyomino made up of n+1 tiles known as the blocks (n tiles) + and the plinth (remaining tile); + • the plinth has its centre at position (x = 0, y = 0); + • the blocks have y-coordinates greater than zero (so the plinth is the + unique lowest tile); + • the centre of mass of all the blocks, combined, has x-coordinate equal + to zero. + + When counting the sculptures, any arrangements which are simply + reflections about the y-axis, are not counted as distinct. For example, + the 18 balanced sculptures of order 6 are shown below; note that each pair + of mirror images (about the y-axis) is counted as one sculpture: + + There are 964 balanced sculptures of order 10 and 360505 of order 15. + How many balanced sculptures are there of order 18? + + + p_275_sculptures2.gif + Answer: a2a192f9790dcbfe4b450a82c4461d4a + + +Problem 276 +=========== + + + Consider the triangles with integer sides a, b and c with a ≤ b ≤ c. + An integer sided triangle (a,b,c) is called primitive if gcd(a,b,c)=1. + How many primitive integer sided triangles exist with a perimeter not + exceeding 10 000 000? + + + Answer: 29ae64e74ebfdf459dac56786e95c5d5 + + +Problem 277 +=========== + + + A modified Collatz sequence of integers is obtained from a starting value + a[1] in the following way: + + a[n+1] = a[n]/3 if a[n] is divisible by 3. We shall denote this as a large + downward step, "D". + + a[n+1] = (4a[n] + 2)/3 if a[n] divided by 3 gives a remainder of 1. We + shall denote this as an upward step, "U". + + a[n+1] = (2a[n] - 1)/3 if a[n] divided by 3 gives a remainder of 2. We + shall denote this as a small downward step, "d". + + The sequence terminates when some a[n] = 1. + + Given any integer, we can list out the sequence of steps. + For instance if a[1]=231, then the sequence + {a[n]}={231,77,51,17,11,7,10,14,9,3,1} corresponds to the steps + "DdDddUUdDD". + + Of course, there are other sequences that begin with that same sequence + "DdDddUUdDD....". + For instance, if a[1]=1004064, then the sequence is + DdDddUUdDDDdUDUUUdDdUUDDDUdDD. + In fact, 1004064 is the smallest possible a[1] > 10^6 that begins with the + sequence DdDddUUdDD. + + What is the smallest a[1] > 10^15 that begins with the sequence + "UDDDUdddDDUDDddDdDddDDUDDdUUDd"? + + + Answer: 9508afff135320c18d82c93a8b70cd11 + + +Problem 278 +=========== + + + Given the values of integers 1 < a[1] < a[2] <... < a[n], consider the + linear combination + q[1]a[1] + q[2]a[2] + ... + q[n]a[n] = b, using only integer values q[k] ≥ + 0. + + Note that for a given set of a[k], it may be that not all values of b are + possible. + For instance, if a[1] = 5 and a[2] = 7, there are no q[1] ≥ 0 and q[2] ≥ 0 + such that b could be + 1, 2, 3, 4, 6, 8, 9, 11, 13, 16, 18 or 23. + In fact, 23 is the largest impossible value of b for a[1] = 5 and a[2] = + 7. + We therefore call f(5, 7) = 23. + Similarly, it can be shown that f(6, 10, 15)=29 and f(14, 22, 77) = 195. + + Find ∑ f(p*q,p*r,q*r), where p, q and r are prime numbers and p < q < r < + 5000. + + + Answer: 7e680606b5e9890a19894dbdbbbd102a + + +Problem 279 +=========== + + + How many triangles are there with integral sides, at least one integral + angle (measured in degrees), and a perimeter that does not exceed 10^8? + + + Answer: 1f51455a8180fdeeb21285dfb6cba45f + + +Problem 280 +=========== + + + A laborious ant walks randomly on a 5x5 grid. The walk starts from the + central square. At each step, the ant moves to an adjacent square at + random, without leaving the grid; thus there are 2, 3 or 4 possible moves + at each step depending on the ant's position. + + At the start of the walk, a seed is placed on each square of the lower + row. When the ant isn't carrying a seed and reaches a square of the lower + row containing a seed, it will start to carry the seed. The ant will drop + the seed on the first empty square of the upper row it eventually reaches. + + What's the expected number of steps until all seeds have been dropped in + the top row? + Give your answer rounded to 6 decimal places. + + + Answer: 27f07f04d1908e5ce4fa6eac09881cc2 + + +Problem 281 +=========== + + + You are given a pizza (perfect circle) that has been cut into m·n equal + pieces and you want to have exactly one topping on each slice. + + Let f(m,n) denote the number of ways you can have toppings on the pizza + with m different toppings (m ≥ 2), using each topping on exactly n slices + (n ≥ 1). + Reflections are considered distinct, rotations are not. + + Thus, for instance, f(2,1) = 1, f(2,2) = f(3,1) = 2 and f(3,2) = 16. + f(3,2) is shown below: + + Find the sum of all f(m,n) such that f(m,n) ≤ 10^15. + + + p_281_pizza.gif + Answer: ceee6ced9d64aad844310c8ce2aae2b7 + + +Problem 282 +=========== + + + For non-negative integers m, n, the Ackermann function A(m, n) is defined + as follows: + + For example A(1, 0) = 2, A(2, 2) = 7 and A(3, 4) = 125. + + Find A(n, n) and give your answer mod 14^8. + + + p_282_formula.gif + Answer: a1cc665e127af4e907e13087ee777bd5 + + +Problem 283 +=========== + + + Consider the triangle with sides 6, 8 and 10. It can be seen that the + perimeter and the area are both equal to 24. So the area/perimeter ratio + is equal to 1. + Consider also the triangle with sides 13, 14 and 15. The perimeter equals + 42 while the area is equal to 84. So for this triangle the area/perimeter + ratio is equal to 2. + + Find the sum of the perimeters of all integer sided triangles for which + the area/perimeter ratios are equal to positive integers not exceeding + 1000. + + + Answer: 08afda4bc05c8f3ef71c9ffea1ddc0c8 + + +Problem 284 +=========== + + + The 3-digit number 376 in the decimal numbering system is an example of + numbers with the special property that its square ends with the same + digits: 376^2 = 141376. Let's call a number with this property a steady + square. + + Steady squares can also be observed in other numbering systems. In the + base 14 numbering system, the 3-digit number c37 is also a steady square: + c37^2 = aa0c37, and the sum of its digits is c+3+7=18 in the same + numbering system. The letters a, b, c and d are used for the 10, 11, 12 + and 13 digits respectively, in a manner similar to the hexadecimal + numbering system. + + For 1 ≤ n ≤ 9, the sum of the digits of all the n-digit steady squares in + the base 14 numbering system is 2d8 (582 decimal). Steady squares with + leading 0's are not allowed. + + Find the sum of the digits of all the n-digit steady squares in the base + 14 numbering system for + 1 ≤ n ≤ 10000 (decimal) and give your answer in the base 14 system using + lower case letters where necessary. + + + Answer: aff724582e583649876f518f9b340a69 + + +Problem 285 +=========== + + + Albert chooses a positive integer k, then two real numbers a, b are + randomly chosen in the interval [0,1] with uniform distribution. + The square root of the sum (k·a+1)^2 + (k·b+1)^2 is then computed and + rounded to the nearest integer. If the result is equal to k, he scores k + points; otherwise he scores nothing. + + For example, if k = 6, a = 0.2 and b = 0.85, then + (k·a+1)^2 + (k·b+1)^2 = 42.05. + The square root of 42.05 is 6.484... and when rounded to the nearest + integer, it becomes 6. + This is equal to k, so he scores 6 points. + + It can be shown that if he plays 10 turns with k = 1, k = 2, ..., k = 10, + the expected value of his total score, rounded to five decimal places, is + 10.20914. + + If he plays 10^5 turns with k = 1, k = 2, k = 3, ..., k = 10^5, what is + the expected value of his total score, rounded to five decimal places? + + + Answer: bbae95d0ce2999cae57782c3746aecb6 + + +Problem 286 +=========== + + + Barbara is a mathematician and a basketball player. She has found that the + probability of scoring a point when shooting from a distance x is exactly + (1 - ^x/[q]), where q is a real constant greater than 50. + + During each practice run, she takes shots from distances x = 1, x = 2, + ..., x = 50 and, according to her records, she has precisely a 2 % chance + to score a total of exactly 20 points. + + Find q and give your answer rounded to 10 decimal places. + + + Answer: cc5a1ef0deabf698733bcef4f1149498 + + +Problem 287 +=========== + + + The quadtree encoding allows us to describe a 2^N×2^N black and white + image as a sequence of bits (0 and 1). Those sequences are to be read from + left to right like this: + + • the first bit deals with the complete 2^N×2^N region; + • "0" denotes a split: + the current 2^n×2^n region is divided into 4 sub-regions of dimension + 2^n-1×2^n-1, + the next bits contains the description of the top left, top right, + bottom left and bottom right sub-regions - in that order; + • "10" indicates that the current region contains only black pixels; + • "11" indicates that the current region contains only white pixels. + + Consider the following 4×4 image (colored marks denote places where a + split can occur): + + This image can be described by several sequences, for example + :"001010101001011111011010101010", of length 30, or + "0100101111101110", of length 16, which is the minimal sequence for this + image. + + For a positive integer N, define D[N] as the 2^N×2^N image with the + following coloring scheme: + + • the pixel with coordinates x = 0, y = 0 corresponds to the bottom left + pixel, + • if (x - 2^N-1)^2 + (y - 2^N-1)^2 ≤ 2^2N-2 then the pixel is black, + • otherwise the pixel is white. + + What is the length of the minimal sequence describing D[24] ? + + + p_287_quadtree.gif + Answer: 6c2beec8a6c0bc788d5e45c317b0d7ca + + +Problem 288 +=========== + + + For any prime p the number N(p,q) is defined byN(p,q) = ∑[n=0 to q] + T[n]*p^n + with T[n] generated by the following random number generator: + + S[0] = 290797 + S[n+1] = S[n]^2 mod 50515093 + T[n] = S[n] mod p + + Let Nfac(p,q) be the factorial of N(p,q). + Let NF(p,q) be the number of factors p in Nfac(p,q). + + You are given that NF(3,10000) mod 3^20=624955285. + + Find NF(61,10^7) mod 61^10 + + + Answer: 192bf4aa33ea85e922d583f60fe99955 + + +Problem 289 +=========== + + + Let C(x,y) be a circle passing through the points (x, y), (x, y+1), + (x+1, y) and (x+1, y+1). + + For positive integers m and n, let E(m,n) be a configuration which + consists of the m·n circles: + { C(x,y): 0 ≤ x < m, 0 ≤ y < n, x and y are integers } + + An Eulerian cycle on E(m,n) is a closed path that passes through each arc + exactly once. + Many such paths are possible on E(m,n), but we are only interested in + those which are not self-crossing: A non-crossing path just touches itself + at lattice points, but it never crosses itself. + + The image below shows E(3,3) and an example of an Eulerian non-crossing + path. + + Let L(m,n) be the number of Eulerian non-crossing paths on E(m,n). + For example, L(1,2) = 2, L(2,2) = 37 and L(3,3) = 104290. + + Find L(6,10) mod 10^10. + + + p_289_euler.gif + Answer: 9fa32696df356b3d41faa7dd278c88a9 + + +Problem 290 +=========== + + + How many integers 0 ≤ n < 10^18 have the property that the sum of the + digits of n equals the sum of digits of 137n? + + + Answer: 8246684fec8ece9f0ee3c9898c8c9d6a + + +Problem 291 +=========== + + + A prime number p is called a Panaitopol prime if for some positive + integers + x and y. + + Find how many Panaitopol primes are less than 5×10^15. + + + p_291_formula.gif + Answer: 15d4b4d97452ca7d219e3fa72f6b7aef + + +Problem 292 +=========== + + + We shall define a pythagorean polygon to be a convex polygon with the + following properties: + + • there are at least three vertices, + • no three vertices are aligned, + • each vertex has integer coordinates, + • each edge has integer length. + + For a given integer n, define P(n) as the number of distinct pythagorean + polygons for which the perimeter is ≤ n. + Pythagorean polygons should be considered distinct as long as none is a + translation of another. + + You are given that P(4) = 1, P(30) = 3655 and P(60) = 891045. + Find P(120). + + + Answer: 27f50f02ef10f170379b144435e0144b + + +Problem 293 +=========== + + + An even positive integer N will be called admissible, if it is a power of + 2 or its distinct prime factors are consecutive primes. + The first twelve admissible numbers are 2,4,6,8,12,16,18,24,30,32,36,48. + + If N is admissible, the smallest integer M > 1 such that N+M is prime, + will be called the pseudo-Fortunate number for N. + + For example, N=630 is admissible since it is even and its distinct prime + factors are the consecutive primes 2,3,5 and 7. + The next prime number after 631 is 641; hence, the pseudo-Fortunate number + for 630 is M=11. + It can also be seen that the pseudo-Fortunate number for 16 is 3. + + Find the sum of all distinct pseudo-Fortunate numbers for admissible + numbers N less than 10^9. + + + Answer: db116b39f7a3ac5366079b1d9fe249a5 + + +Problem 294 +=========== + + + For a positive integer k, define d(k) as the sum of the digits of k in its + usual decimal representation.Thus d(42) = 4+2 = 6. + + For a positive integer n, define S(n) as the number of positive integers k + < 10^n with the following properties : + + • k is divisible by 23 and + • d(k) = 23. + + You are given that S(9) = 263626 and S(42) = 6377168878570056. + + Find S(11^12) and give your answer mod 10^9. + + + Answer: aefe049404a284c7d27fab3887c6c4a2 + + +Problem 295 +=========== + + + We call the convex area enclosed by two circles a lenticular hole if: + + • The centres of both circles are on lattice points. + • The two circles intersect at two distinct lattice points. + • The interior of the convex area enclosed by both circles does not + contain any lattice points. + + Consider the circles: + C[0]: x^2+y^2=25 + C[1]: (x+4)^2+(y-4)^2=1 + C[2]: (x-12)^2+(y-4)^2=65 + + The circles C[0], C[1] and C[2] are drawn in the picture below. + + C[0] and C[1] form a lenticular hole, as well as C[0] and C[2]. + + We call an ordered pair of positive real numbers (r[1], r[2]) a lenticular + pair if there exist two circles with radii r[1] and r[2] that form a + lenticular hole.We can verify that (1, 5) and (5, √65) are the lenticular + pairs of the example above. + + Let L(N) be the number of distinct lenticular pairs (r[1], r[2]) for which + 0 < r[1] ≤ r[2] ≤ N. + We can verify that L(10) = 30 and L(100) = 3442. + + Find L(100 000). + + + Answer: 5beaace6425205fe879116ee07dae961 + + +Problem 296 +=========== + + + Given is an integer sided triangle ABC with BC ≤ AC ≤ AB. + k is the angular bisector of angle ACB. + m is the tangent at C to the circumscribed circle of ABC. + n is a line parallel to m through B. + The intersection of n and k is called E. + + How many triangles ABC with a perimeter not exceeding 100 000 exist such + that BE has integral length? + + + Answer: 45986a4405b2dd6c163516319e0c4a1b + + +Problem 297 +=========== + + + Each new term in the Fibonacci sequence is generated by adding the + previous two terms. + Starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, + 34, 55, 89. + + Every positive integer can be uniquely written as a sum of nonconsecutive + terms of the Fibonacci sequence. For example, 100 = 3 + 8 + 89. + Such a sum is called the Zeckendorf representation of the number. + + For any integer n>0, let z(n) be the number of terms in the Zeckendorf + representation of n. + Thus, z(5) = 1, z(14) = 2, z(100) = 3 etc. + Also, for 0n. + + The sequence a(n) is defined by: + a(1)=next_prime(10^14) and a(n)=next_prime(a(n-1)) for n>1. + + The fibonacci sequence f(n) is defined by:f(0)=0, f(1)=1 and + f(n)=f(n-1)+f(n-2) for n>1. + + The sequence b(n) is defined as f(a(n)). + + Find ∑b(n) for 1≤n≤100 000. Give your answer mod 1234567891011. + + + Answer: 499427a3e4bf9ad34a6df3056604b4c1 + + +Problem 305 +=========== + + + Let's call S the (infinite) string that is made by concatenating the + consecutive positive integers (starting from 1) written down in base 10. + Thus, S = 1234567891011121314151617181920212223242... + + It's easy to see that any number will show up an infinite number of times + in S. + + Let's call f(n) the starting position of the n^th occurrence of n in S. + For example, f(1)=1, f(5)=81, f(12)=271 and f(7780)=111111365. + + Find ∑f(3^k) for 1≤k≤13. + + + Answer: 9def85298f598867d361e4afca8cdd96 + + +Problem 306 +=========== + + + The following game is a classic example of Combinatorial Game Theory: + + Two players start with a strip of n white squares and they take alternate + turns. + On each turn, a player picks two contiguous white squares and paints them + black. + The first player who cannot make a move loses. + + • If n = 1, there are no valid moves, so the first player loses + automatically. + • If n = 2, there is only one valid move, after which the second player + loses. + • If n = 3, there are two valid moves, but both leave a situation where + the second player loses. + • If n = 4, there are three valid moves for the first player; she can + win the game by painting the two middle squares. + • If n = 5, there are four valid moves for the first player (shown below + in red); but no matter what she does, the second player (blue) wins. + + So, for 1 ≤ n ≤ 5, there are 3 values of n for which the first player can + force a win. + Similarly, for 1 ≤ n ≤ 50, there are 40 values of n for which the first + player can force a win. + + For 1 ≤ n ≤ 1 000 000, how many values of n are there for which the first + player can force a win? + + + p_306_pstrip.gif + Answer: 394d602ba21e30693db90c9ecd4bd3a2 + + +Problem 307 +=========== + + + k defects are randomly distributed amongst n integrated-circuit chips + produced by a factory (any number of defects may be found on a chip and + each defect is independent of the other defects). + + Let p(k,n) represent the probability that there is a chip with at least 3 + defects. + For instance p(3,7) ≈ 0.0204081633. + + Find p(20 000, 1 000 000) and give your answer rounded to 10 decimal + places in the form 0.abcdefghij + + + Answer: 0c49094fa750365e13bb20ec4a158b6d + + +Problem 308 +=========== + + + A program written in the programming language Fractran consists of a list + of fractions. + + The internal state of the Fractran Virtual Machine is a positive integer, + which is initially set to a seed value. Each iteration of a Fractran + program multiplies the state integer by the first fraction in the list + which will leave it an integer. + + For example, one of the Fractran programs that John Horton Conway wrote + for prime-generation consists of the following 14 fractions: + + 17 , 78 , 19 , 23 , 29 , 77 , 95 , 77 , 1 , 11 , 13 , 15 , 1 , 55 . + 91 85 51 38 33 29 23 19 17 13 11 2 7 1 + + Starting with the seed integer 2, successive iterations of the program + produce the sequence: + 15, 825, 725, 1925, 2275, 425, ..., 68, 4, 30, ..., 136, 8, 60, ..., 544, + 32, 240, ... + + The powers of 2 that appear in this sequence are 2^2, 2^3, 2^5, ... + It can be shown that all the powers of 2 in this sequence have prime + exponents and that all the primes appear as exponents of powers of 2, in + proper order! + + If someone uses the above Fractran program to solve Project Euler Problem + 7 (find the 10001^st prime), how many iterations would be needed until the + program produces 2^10001st prime ? + + + Answer: 43e736dfc6478a52653814248a71771d + + +Problem 309 +=========== + + + In the classic "Crossing Ladders" problem, we are given the lengths x and + y of two ladders resting on the opposite walls of a narrow, level street. + We are also given the height h above the street where the two ladders + cross and we are asked to find the width of the street (w). + + Here, we are only concerned with instances where all four variables are + positive integers. + For example, if x = 70, y = 119 and h = 30, we can calculate that w = 56. + + In fact, for integer values x, y, h and 0 < x < y < 200, there are only + five triplets (x,y,h) producing integer solutions for w: + (70, 119, 30), (74, 182, 21), (87, 105, 35), (100, 116, 35) and (119, 175, + 40). + + For integer values x, y, h and 0 < x < y < 1 000 000, how many triplets + (x,y,h) produce integer solutions for w? + + + p_309_ladders.gif + Answer: 0875415a84bfe8bc237dcfc6b440d263 + + +Problem 310 +=========== + + + Alice and Bob play the game Nim Square. + Nim Square is just like ordinary three-heap normal play Nim, but the + players may only remove a square number of stones from a heap. + The number of stones in the three heaps is represented by the ordered + triple (a,b,c). + If 0≤a≤b≤c≤29 then the number of losing positions for the next player is + 1160. + + Find the number of losing positions for the next player if 0≤a≤b≤c≤100 + 000. + + + Answer: 6b94f848996393eef163add4d17360c7 + + +Problem 311 +=========== + + + ABCD is a convex, integer sided quadrilateral with 1 ≤ AB < BC < CD < AD. + BD has integer length. O is the midpoint of BD. AO has integer length. + We'll call ABCD a biclinic integral quadrilateral if AO = CO ≤ BO = DO. + + For example, the following quadrilateral is a biclinic integral + quadrilateral: + AB = 19, BC = 29, CD = 37, AD = 43, BD = 48 and AO = CO = 23. + + Let B(N) be the number of distinct biclinic integral quadrilaterals ABCD + that satisfy AB^2+BC^2+CD^2+AD^2 ≤ N. + We can verify that B(10 000) = 49 and B(1 000 000) = 38239. + + Find B(10 000 000 000). + + + p_311_biclinic.gif + Answer: 36115d4f7dc07eea106d78e8431868e6 + + +Problem 312 +=========== + + + - A Sierpiński graph of order-1 (S[1]) is an equilateral triangle. + - S[n+1] is obtained from S[n] by positioning three copies of S[n] so that + every pair of copies has one common corner. + + Let C(n) be the number of cycles that pass exactly once through all the + vertices of S[n]. + For example, C(3) = 8 because eight such cycles can be drawn on S[3], as + shown below: + + It can also be verified that : + C(1) = C(2) = 1 + C(5) = 71328803586048 + C(10 000) mod 10^8 = 37652224 + C(10 000) mod 13^8 = 617720485 + + Find C(C(C(10 000))) mod 13^8. + + + p_312_sierpinskyAt.gif + p_312_sierpinsky8t.gif + Answer: 535113d1a81f421fe814d48205dac570 + + +Problem 313 +=========== + + + In a sliding game a counter may slide horizontally or vertically into an + empty space. The objective of the game is to move the red counter from the + top left corner of a grid to the bottom right corner; the space always + starts in the bottom right corner. For example, the following sequence of + pictures show how the game can be completed in five moves on a 2 by 2 + grid. + + Let S(m,n) represent the minimum number of moves to complete the game on + an m by n grid. For example, it can be verified that S(5,4) = 25. + + There are exactly 5482 grids for which S(m,n) = p^2, where p < 100 is + prime. + + How many grids does S(m,n) = p^2, where p < 10^6 is prime? + + + p_313_sliding_game_1.gif + p_313_sliding_game_2.gif + Answer: 2468d42fa1c7f61547ce71c9826218ea + + +Problem 314 +=========== + + + The moon has been opened up, and land can be obtained for free, but there + is a catch. You have to build a wall around the land that you stake out, + and building a wall on the moon is expensive. Every country has been + allotted a 500 m by 500 m square area, but they will possess only that + area which they wall in. 251001 posts have been placed in a rectangular + grid with 1 meter spacing. The wall must be a closed series of straight + lines, each line running from post to post. + + The bigger countries of course have built a 2000 m wall enclosing the + entire 250 000 m^2 area. The [1]Duchy of Grand Fenwick, has a tighter + budget, and has asked you (their Royal Programmer) to compute what shape + would get best maximum enclosed-area/wall-length ratio. + + You have done some preliminary calculations on a sheet of paper.For a 2000 + meter wall enclosing the 250 000 m^2 area theenclosed-area/wall-length + ratio is 125. + Although not allowed , but to get an idea if this is anything better: if + you place a circle inside the square area touching the four sides the area + will be equal to π*250^2 m^2 and the perimeter will be π*500 m, so the + enclosed-area/wall-length ratio will also be 125. + + However, if you cut off from the square four triangles with sides 75 m, 75 + m and 75√2 m the total area becomes 238750 m^2 and the perimeter becomes + 1400+300√2 m. So this gives an enclosed-area/wall-length ratio of 130.87, + which is significantly better. + + Find the maximum enclosed-area/wall-length ratio. + Give your answer rounded to 8 places behind the decimal point in the form + abc.defghijk. + + + Visible links + 1. http://en.wikipedia.org/wiki/Grand_Fenwick + p_314_landgrab.gif + Answer: aa457cae6f67945d50683a85a9b70230 + + +Problem 315 +=========== + + + Sam and Max are asked to transform two digital clocks into two "digital + root" clocks. + A digital root clock is a digital clock that calculates digital roots step + by step. + + When a clock is fed a number, it will show it and then it will start the + calculation, showing all the intermediate values until it gets to the + result. + For example, if the clock is fed the number 137, it will show: "137" → + "11" → "2" and then it will go black, waiting for the next number. + + Every digital number consists of some light segments: three horizontal + (top, middle, bottom) and four vertical (top-left, top-right, bottom-left, + bottom-right). + Number "1" is made of vertical top-right and bottom-right, number "4" is + made by middle horizontal and vertical top-left, top-right and + bottom-right. Number "8" lights them all. + + The clocks consume energy only when segments are turned on/off. + To turn on a "2" will cost 5 transitions, while a "7" will cost only 4 + transitions. + + Sam and Max built two different clocks. + + Sam's clock is fed e.g. number 137: the clock shows "137", then the panel + is turned off, then the next number ("11") is turned on, then the panel is + turned off again and finally the last number ("2") is turned on and, after + some time, off. + For the example, with number 137, Sam's clock requires: + + "137" : (2 + 5 + 4) × 2 = 22 transitions ("137" on/off). + "11" : (2 + 2) × 2 = 8 transitions ("11" on/off). + "2" : (5) × 2 = 10 transitions ("2" on/off). + + For a grand total of 40 transitions. + + Max's clock works differently. Instead of turning off the whole panel, it + is smart enough to turn off only those segments that won't be needed for + the next number. + For number 137, Max's clock requires: + + 2 + 5 + 4 = 11 transitions ("137" on) + "137" : 7 transitions (to turn off the segments that are not needed for + number "11"). + 0 transitions (number "11" is already turned on correctly) + "11" : 3 transitions (to turn off the first "1" and the bottom part of + the second "1"; + the top part is common with number "2"). + 4 tansitions (to turn on the remaining segments in order to get a + "2" : "2") + 5 transitions (to turn off number "2"). + + For a grand total of 30 transitions. + + Of course, Max's clock consumes less power than Sam's one. + The two clocks are fed all the prime numbers between A = 10^7 and B = + 2×10^7. + Find the difference between the total number of transitions needed by + Sam's clock and that needed by Max's one. + + + p_315_clocks.gif + Answer: 79b587f9c25a72dbe95428e283628421 + + +Problem 316 +=========== + + + Let p = p[1] p[2] p[3] ... be an infinite sequence of random digits, + selected from {0,1,2,3,4,5,6,7,8,9} with equal probability. + It can be seen that p corresponds to the real number 0.p[1] p[2] p[3] .... + It can also be seen that choosing a random real number from the interval + [0,1) is equivalent to choosing an infinite sequence of random digits + selected from {0,1,2,3,4,5,6,7,8,9} with equal probability. + + For any positive integer n with d decimal digits, let k be the smallest + index such that + p[k, ]p[k+1], ...p[k+d-1] are the decimal digits of n, in the same order. + Also, let g(n) be the expected value of k; it can be proven that g(n) is + always finite and, interestingly, always an integer number. + + For example, if n = 535, then + for p = 31415926535897...., we get k = 9 + for p = 355287143650049560000490848764084685354..., we get k = 36 + etc and we find that g(535) = 1008. + + Given that , find + + Note: represents the floor function. + + p_316_decexp1.gif + p_316_decexp2.gif + p_316_decexp3.gif + Answer: 2495e8f6e9d4cdadbf0411144e7180b9 + + +Problem 317 +=========== + + + A firecracker explodes at a height of 100 m above level ground. It breaks + into a large number of very small fragments, which move in every + direction; all of them have the same initial velocity of 20 m/s. + + We assume that the fragments move without air resistance, in a uniform + gravitational field with g=9.81 m/s^2. + + Find the volume (in m^3) of the region through which the fragments move + before reaching the ground. Give your answer rounded to four decimal + places. + + + Answer: b0e2bec93bfe598ade5d3d1141f76bdd + + +Problem 318 +=========== + + + Consider the real number √2+√3. + When we calculate the even powers of √2+√3we get: + (√2+√3)^2 = 9.898979485566356... + (√2+√3)^4 = 97.98979485566356... + (√2+√3)^6 = 969.998969071069263... + (√2+√3)^8 = 9601.99989585502907... + (√2+√3)^10 = 95049.999989479221... + (√2+√3)^12 = 940897.9999989371855... + (√2+√3)^14 = 9313929.99999989263... + (√2+√3)^16 = 92198401.99999998915... + + It looks like that the number of consecutive nines at the beginning of the + fractional part of these powers is non-decreasing. + In fact it can be proven that the fractional part of (√2+√3)^2n approaches + 1 for large n. + + Consider all real numbers of the form √p+√q with p and q positive integers + and p 0. ( | is the bitwise-OR operator) + + It can be seen that eventually there will be an index N such that x[i] = + 2^32 -1 (a bit-pattern of all ones) for all i ≥ N. + + Find the expected value of N. + Give your answer rounded to 10 digits after the decimal point. + + + Answer: c8f8a7ab17a87f1b17a1f4a86c984ea7 + + +Problem 324 +=========== + + + Let f(n) represent the number of ways one can fill a 3×3×n tower with + blocks of 2×1×1. + You're allowed to rotate the blocks in any way you like; however, + rotations, reflections etc of the tower itself are counted as distinct. + + For example (with q = 100000007) : + f(2) = 229, + f(4) = 117805, + f(10) mod q = 96149360, + f(10^3) mod q = 24806056, + f(10^6) mod q = 30808124. + + Find f(10^10000) mod 100000007. + + + Answer: b8d91b06d43a2ef98a6fcb0be4a6d617 + + +Problem 325 +=========== + + + A game is played with two piles of stones and two players. At her turn, a + player removes a number of stones from the larger pile. The number of + stones she removes must be a positive multiple of the number of stones in + the smaller pile. + + E.g., let the ordered pair(6,14) describe a configuration with 6 stones in + the smaller pile and 14 stones in the larger pile, then the first player + can remove 6 or 12 stones from the larger pile. + + The player taking all the stones from a pile wins the game. + + A winning configuration is one where the first player can force a win. For + example, (1,5), (2,6) and (3,12) are winning configurations because the + first player can immediately remove all stones in the second pile. + + A losing configuration is one where the second player can force a win, no + matter what the first player does. For example, (2,3) and (3,4) are losing + configurations: any legal move leaves a winning configuration for the + second player. + + Define S(N) as the sum of (x[i]+y[i]) for all losing configurations + (x[i],y[i]), 0 < x[i] < y[i] ≤ N. We can verify that S(10) = 211 and + S(10^4) = 230312207313. + + Find S(10^16) mod 7^10. + + + Answer: 5b1ce9ac67e0ad6690c728ccba6f0070 + + +Problem 326 +=========== + + + Let a[n] be a sequence recursively defined by: . + + So the first 10 elements of a[n] are: 1,1,0,3,0,3,5,4,1,9. + + Let f(N,M) represent the number of pairs (p,q) such that: + + It can be seen that f(10,10)=4 with the pairs (3,3), (5,5), (7,9) and + (9,10). + + You are also given that f(10^4,10^3)=97158. + + Find f(10^12,10^6). + + + p_326_formula1.gif + p_326_formula2.gif + Answer: d95dff1a5ceee0064993d98defdd603e + + +Problem 327 +=========== + + + A series of three rooms are connected to each other by automatic doors. + + Each door is operated by a security card. Once you enter a room the door + automatically closes and that security card cannot be used again. A + machine at the start will dispense an unlimited number of cards, but each + room (including the starting room) contains scanners and if they detect + that you are holding more than three security cards or if they detect an + unattended security card on the floor, then all the doors will become + permanently locked. However, each room contains a box where you may safely + store any number of security cards for use at a later stage. + + If you simply tried to travel through the rooms one at a time then as you + entered room 3 you would have used all three cards and would be trapped in + that room forever! + + However, if you make use of the storage boxes, then escape is possible. + For example, you could enter room 1 using your first card, place one card + in the storage box, and use your third card to exit the room back to the + start. Then after collecting three more cards from the dispensing machine + you could use one to enter room 1 and collect the card you placed in the + box a moment ago. You now have three cards again and will be able to + travel through the remaining three doors. This method allows you to travel + through all three rooms using six security cards in total. + + It is possible to travel through six rooms using a total of 123 security + cards while carrying a maximum of 3 cards. + + Let C be the maximum number of cards which can be carried at any time. + + Let R be the number of rooms to travel through. + + Let M(C,R) be the minimum number of cards required from the dispensing + machine to travel through R rooms carrying up to a maximum of C cards at + any time. + + For example, M(3,6)=123 and M(4,6)=23. + And, ΣM(C,6)=146 for 3 ≤ C ≤ 4. + + You are given that ΣM(C,10)=10382 for 3 ≤ C ≤ 10. + + Find ΣM(C,30) for 3 ≤ C ≤ 40. + + + p_327_rooms_of_doom.gif + Answer: 2cd4c0ad8a00c5be99802188ee2628fb + + +Problem 328 +=========== + + + We are trying to find a hidden number selected from the set of integers + {1, 2, ..., n} by asking questions. Each number (question) we ask, has a + cost equal to the number asked and we get one of three possible answers: + + • "Your guess is lower than the hidden number", or + • "Yes, that's it!", or + • "Your guess is higher than the hidden number". + + Given the value of n, an optimal strategy minimizes the total cost (i.e. + the sum of all the questions asked) for the worst possible case. E.g. + + If n=3, the best we can do is obviously to ask the number "2". The answer + will immediately lead us to find the hidden number (at a total cost = 2). + + If n=8, we might decide to use a "binary search" type of strategy: Our + first question would be "4" and if the hidden number is higher than 4 we + will need one or two additional questions. + Let our second question be "6". If the hidden number is still higher than + 6, we will need a third question in order to discriminate between 7 and 8. + Thus, our third question will be "7" and the total cost for this + worst-case scenario will be 4+6+7=17. + + We can improve considerably the worst-case cost for n=8, by asking "5" as + our first question. + If we are told that the hidden number is higher than 5, our second + question will be "7", then we'll know for certain what the hidden number + is (for a total cost of 5+7=12). + If we are told that the hidden number is lower than 5, our second question + will be "3" and if the hidden number is lower than 3 our third question + will be "1", giving a total cost of 5+3+1=9. + Since 12>9, the worst-case cost for this strategy is 12. That's better + than what we achieved previously with the "binary search" strategy; it is + also better than or equal to any other strategy. + So, in fact, we have just described an optimal strategy for n=8. + + Let C(n) be the worst-case cost achieved by an optimal strategy for n, as + described above. + Thus C(1) = 0, C(2) = 1, C(3) = 2 and C(8) = 12. + Similarly, C(100) = 400 and C(n) = 17575. + + Find C(n). + + + p_328_sum1.gif + p_328_sum2.gif + Answer: 92a3220ad5b17a562c039e6e93d6df90 + + +Problem 329 +=========== + + + Susan has a prime frog. + Her frog is jumping around over 500 squares numbered 1 to 500.He can only + jump one square to the left or to the right, with equal probability, and + he cannot jump outside the range [1;500]. + (if it lands at either end, it automatically jumps to the only available + square on the next move.) + + When he is on a square with a prime number on it, he croaks 'P' (PRIME) + with probability 2/3 or 'N' (NOT PRIME) with probability 1/3 just before + jumping to the next square. + When he is on a square with a number on it that is not a prime he croaks + 'P' with probability 1/3 or 'N' with probability 2/3 just before jumping + to the next square. + + Given that the frog's starting position is random with the same + probability for every square, and given that she listens to his first 15 + croaks, what is the probability that she hears the sequence + PPPPNNPPPNPPNPN? + + Give your answer as a fraction p/q in reduced form. + + Answer: e392a8b1b053c83e68663e08456bb392 + + +Problem 330 +=========== + + An infinite sequence of real numbers a(n) is defined for all integers n as + follows: + + For example, + + a(0) = 1 + 1 + 1 + ... = e − 1 + 1! 2! 3! + + a(1) = e − 1 + 1 + 1 + ... = 2e − 3 + 1! 2! 3! + + a(2) = 2e − 3 + e − 1 + 1 + ... = 7 e − 6 + 1! 2! 3! 2 + + with e = 2.7182818... being Euler's constant. + + It can be shown that a(n) is of A(n) e + B(n) for integers A(n) and B(n). + the form n! + + For example a(10) = 328161643 e − 652694486 . + 10! + + Find A(10^9) + B(10^9) and give your answer mod 77 777 777. + + + p_330_formula.gif + Answer: d385d3fe0995b48a782a91477525b154 + + +Problem 331 +=========== + + + N×N disks are placed on a square game board. Each disk has a black side + and white side. + + At each turn, you may choose a disk and flip all the disks in the same row + and the same column as this disk: thus 2×N-1 disks are flipped. The game + ends when all disks show their white side. The following example shows a + game on a 5×5 board. + + It can be proven that 3 is the minimal number of turns to finish this + game. + + The bottom left disk on the N×N board has coordinates (0,0); + the bottom right disk has coordinates (N-1,0) and the top left disk has + coordinates (0,N-1). + + Let C[N] be the following configuration of a board with N×N disks: + A disk at (x,y) satisfying , shows its black side; otherwise, it shows its + white side. C[5] is shown above. + + Let T(N) be the minimal number of turns to finish a game starting from + configuration C[N] or 0 if configuration C[N] is unsolvable. + We have shown that T(5)=3. You are also given that T(10)=29 and T(1 + 000)=395253. + + Find . + + + p_331_crossflips3.gif + p_331_crossflips1.gif + p_331_crossflips2.gif + Answer: b609ccc578e71db9de0524fff94e1b70 + + +Problem 332 +=========== + + + A spherical triangle is a figure formed on the surface of a sphere by + three great circular arcs intersecting pairwise in three vertices. + + Let C(r) be the sphere with the centre (0,0,0) and radius r. + Let Z(r) be the set of points on the surface of C(r) with integer + coordinates. + Let T(r) be the set of spherical triangles with vertices in + Z(r).Degenerate spherical triangles, formed by three points on the same + great arc, are not included in T(r). + Let A(r) be the area of the smallest spherical triangle in T(r). + + For example A(14) is 3.294040 rounded to six decimal places. + + Find A(r). Give your answer rounded to six decimal places. + + + p_332_spherical.jpg + p_332_sum.gif + Answer: c2ae53ebfb15db373cfe5d71078ea1ca + + +Problem 333 +=========== + + + All positive integers can be partitioned in such a way that each and every + term of the partition can be expressed as 2^ix3^j, where i,j ≥ 0. + + Let's consider only those such partitions where none of the terms can + divide any of the other terms. + For example, the partition of 17 = 2 + 6 + 9 = (2^1x3^0 + 2^1x3^1 + + 2^0x3^2) would not be valid since 2 can divide 6. Neither would the + partition 17 = 16 + 1 = (2^4x3^0 + 2^0x3^0) since 1 can divide 16. The + only valid partition of 17 would be 8 + 9 = (2^3x3^0 + 2^0x3^2). + + Many integers have more than one valid partition, the first being 11 + having the following two partitions. + 11 = 2 + 9 = (2^1x3^0 + 2^0x3^2) + 11 = 8 + 3 = (2^3x3^0 + 2^0x3^1) + + Let's define P(n) as the number of valid partitions of n. For example, + P(11) = 2. + + Let's consider only the prime integers q which would have a single valid + partition such as P(17). + + The sum of the primes q <100 such that P(q)=1 equals 233. + + Find the sum of the primes q <1000000 such that P(q)=1. + + + Answer: 8408ff3a470a94dbfca1819249eb547d + + +Problem 334 +=========== + + + In Plato's heaven, there exist an infinite number of bowls in a straight + line. + Each bowl either contains some or none of a finite number of beans. + A child plays a game, which allows only one kind of move: removing two + beans from any bowl, and putting one in each of the two adjacent bowls. + The game ends when each bowl contains either one or no beans. + + For example, consider two adjacent bowls containing 2 and 3 beans + respectively, all other bowls being empty. The following eight moves will + finish the game: + + You are given the following sequences: + + t[0] = 123456. + + t[i-1] , if t[i-1] is even + t[i] = 2 + t[i-1] 926252, if t[i-1] is odd + 2 + where ⌊x⌋ is the floor function + and is the bitwise XOR operator. + + b[i] = ( t[i] mod 2^11) + 1. + + The first two terms of the last sequence are b[1] = 289 and b[2] = 145. + If we start with b[1] and b[2] beans in two adjacent bowls, 3419100 moves + would be required to finish the game. + + Consider now 1500 adjacent bowls containing b[1], b[2],..., b[1500] beans + respectively, all other bowls being empty. Find how many moves it takes + before the game ends. + + + p_334_beans.gif + p_334_cases.gif + p_334_lfloor.gif + p_334_rfloor.gif + p_334_oplus.gif + Answer: 71851da3058acf6b74e90251bdf4aa8f + + +Problem 335 +=========== + + + Whenever Peter feels bored, he places some bowls, containing one bean + each, in a circle. After this, he takes all the beans out of a certain + bowl and drops them one by one in the bowls going clockwise. He repeats + this, starting from the bowl he dropped the last bean in, until the + initial situation appears again. For example with 5 bowls he acts as + follows: + + So with 5 bowls it takes Peter 15 moves to return to the initial + situation. + + Let M(x) represent the number of moves required to return to the initial + situation, starting with x bowls. Thus, M(5) = 15. It can also be verified + that M(100) = 10920. + + Find M(2^k+1). Give your answer modulo 7^9. + + + p_335_mancala.gif + p_335_sum.gif + Answer: 9a519cfa0ebdd4d1dd318f14b5799eea + + +Problem 336 +=========== + + + A train is used to transport four carriages in the order: ABCD. However, + sometimes when the train arrives to collect the carriages they are not in + the correct order. + To rearrange the carriages they are all shunted on to a large rotating + turntable. After the carriages are uncoupled at a specific point the train + moves off the turntable pulling the carriages still attached with it. The + remaining carriages are rotated 180 degrees. All of the carriages are then + rejoined and this process is repeated as often as necessary in order to + obtain the least number of uses of the turntable. + Some arrangements, such as ADCB, can be solved easily: the carriages are + separated between A and D, and after DCB are rotated the correct order has + been achieved. + + However, Simple Simon, the train driver, is not known for his efficiency, + so he always solves the problem by initially getting carriage A in the + correct place, then carriage B, and so on. + + Using four carriages, the worst possible arrangements for Simon, which we + shall call maximix arrangements, are DACB and DBAC; each requiring him + five rotations (although, using the most efficient approach, they could be + solved using just three rotations). The process he uses for DACB is shown + below. + + It can be verified that there are 24 maximix arrangements for six + carriages, of which the tenth lexicographic maximix arrangement is DFAECB. + + Find the 2011^th lexicographic maximix arrangement for eleven carriages. + + + p_336_maximix.gif + Answer: 7968e48fc692ce25bf7f5494f4ab6814 + + +Problem 337 +=========== + + + Let {a[1], a[2],..., a[n]} be an integer sequence of length n such that: + + • a[1] = 6 + • for all 1 ≤ i < n : φ(a[i]) < φ(a[i+1]) < a[i] < a[i+1] ^1 + + Let S(N) be the number of such sequences with a[n] ≤ N. + For example, S(10) = 4: {6}, {6, 8}, {6, 8, 9} and {6, 10}. + We can verify that S(100) = 482073668 and S(10 000) mod 10^8 = 73808307. + + Find S(20 000 000) mod 10^8. + + ^1 φ denotes Euler's totient function. + + + Answer: a60bbbe1b90254043fb92820492a2f96 + + +Problem 338 +=========== + + + A rectangular sheet of grid paper with integer dimensions w × h is given. + Its grid spacing is 1. + When we cut the sheet along the grid lines into two pieces and rearrange + those pieces without overlap, we can make new rectangles with different + dimensions. + + For example, from a sheet with dimensions 9 × 4 , we can make rectangles + with dimensions 18 × 2, 12 × 3 and 6 × 6 by cutting and rearranging as + below: + + Similarly, from a sheet with dimensions 9 × 8 , we can make rectangles + with dimensions 18 × 4 and 12 × 6 . + + For a pair w and h, let F(w,h) be the number of distinct rectangles that + can be made from a sheet with dimensions w × h . + For example, F(2,1) = 0, F(2,2) = 1, F(9,4) = 3 and F(9,8) = 2. + Note that rectangles congruent to the initial one are not counted in + F(w,h). + Note also that rectangles with dimensions w × h and dimensions h × w are + not considered distinct. + + For an integer N, let G(N) be the sum of F(w,h) for all pairs w and h + which satisfy 0 < h ≤ w ≤ N. + We can verify that G(10) = 55, G(10^3) = 971745 and G(10^5) = 9992617687. + + Find G(10^12). Give your answer modulo 10^8. + + + p_338_gridpaper.gif + Answer: 99f4f702713f3422ced01dd7d3d79644 + + +Problem 339 +=========== + + + "And he came towards a valley, through which ran a river; and the borders + of the valley were wooded, and on each side of the river were level + meadows. And on one side of the river he saw a flock of white sheep, and + on the other a flock of black sheep. And whenever one of the white sheep + bleated, one of the black sheep would cross over and become white; and + when one of the black sheep bleated, one of the white sheep would cross + over and become black." + [1]en.wikisource.org + + Initially each flock consists of n sheep. Each sheep (regardless of + colour) is equally likely to be the next sheep to bleat. After a sheep has + bleated and a sheep from the other flock has crossed over, Peredur may + remove a number of white sheep in order to maximize the expected final + number of black sheep. Let E(n) be the expected final number of black + sheep if Peredur uses an optimal strategy. + + You are given that E(5) = 6.871346 rounded to 6 places behind the decimal + point. + Find E(10 000) and give your answer rounded to 6 places behind the decimal + point. + + + Visible links + 1. http://en.wikisource.org/wiki/The_Mabinogion/Peredur_the_Son_of_Evrawc + Answer: 0be02210b2d2212d37d026478093c457 + + +Problem 340 +=========== + + + For fixed integers a, b, c, define the crazy function F(n) as follows: + F(n) = n - c for all n > b + F(n) = F(a + F(a + F(a + F(a + n)))) for all n ≤ b. + + Also, define S(a, b, c) = . + + For example, if a = 50, b = 2000 and c = 40, then F(0) = 3240 and F(2000) + = 2040. + Also, S(50, 2000, 40) = 5204240. + + Find the last 9 digits of S(21^7, 7^21, 12^7). + + + p_340_formula.gif + Answer: fc838afe9ecde39bbe230923d7b50775 + + +Problem 341 +=========== + + + The Golomb's self-describing sequence {G(n)} is the only nondecreasing + sequence of natural numbers such that n appears exactly G(n) times in the + sequence. The values of G(n) for the first few n are + + n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 … + G(n) 1 2 2 3 3 4 4 4 5 5 5 6 6 6 6 … + + You are given that G(10^3) = 86, G(10^6) = 6137. + You are also given that ΣG(n^3) = 153506976 for 1 ≤ n < 10^3. + + Find ΣG(n^3) for 1 ≤ n < 10^6. + + + Answer: 7c163c3b4886943667b5c89db0a6cd02 + + +Problem 342 +=========== + + + Consider the number 50. + 50^2 = 2500 = 2^2 × 5^4, so φ(2500) = 2 × 4 × 5^3 = 8 × 5^3 = 2^3 × 5^3. + ^1 + So 2500 is a square and φ(2500) is a cube. + + Find the sum of all numbers n, 1 < n < 10^10 such that φ(n^2) is a cube. + + ^1 φ denotes Euler's totient function. + + + Answer: 0e9add0383d4116c7c5cb3dc73fc0536 + + +Problem 343 +=========== + + + For any positive integer k, a finite sequence a[i] of fractions x[i]/y[i] + is defined by: + a[1] = 1/k and + a[i] = (x[i-1]+1)/(y[i-1]-1) reduced to lowest terms for i>1. + When a[i] reaches some integer n, the sequence stops. (That is, when + y[i]=1.) + Define f(k) = n. + For example, for k = 20: + + 1/20 → 2/19 → 3/18 = 1/6 → 2/5 → 3/4 → 4/3 → 5/2 → 6/1 = 6 + + So f(20) = 6. + + Also f(1) = 1, f(2) = 2, f(3) = 1 and Σf(k^3) = 118937 for 1 ≤ k ≤ 100. + + Find Σf(k^3) for 1 ≤ k ≤ 2×10^6. + + + Answer: 0e10bd111425ad8e1343ac79dac7bb0e + + +Problem 344 +=========== + + + One variant of N.G. de Bruijn's silver dollar game can be described as + follows: + + On a strip of squares a number of coins are placed, at most one coin per + square. Only one coin, called the silver dollar, has any value. Two + players take turns making moves. At each turn a player must make either a + regular or a special move. + + A regular move consists of selecting one coin and moving it one or more + squares to the left. The coin cannot move out of the strip or jump on or + over another coin. + + Alternatively, the player can choose to make the special move of pocketing + the leftmost coin rather than making a regular move. If no regular moves + are possible, the player is forced to pocket the leftmost coin. + + The winner is the player who pockets the silver dollar. + + A winning configuration is an arrangement of coins on the strip where the + first player can force a win no matter what the second player does. + + Let W(n,c) be the number of winning configurations for a strip of n + squares, c worthless coins and one silver dollar. + + You are given that W(10,2) = 324 and W(100,10) = 1514704946113500. + + Find W(1 000 000, 100) modulo the semiprime 1000 036 000 099 (= 1 000 003 + · 1 000 033). + + + p_344_silverdollar.gif + Answer: 38e7b980b38fcac89b3e267e328cd292 + + +Problem 345 +=========== + + + We define the Matrix Sum of a matrix as the maximum sum of matrix elements + with each element being the only one in his row and column. For example, + the Matrix Sum of the matrix below equals 3315 ( = 863 + 383 + 343 + 959 + + 767): + +   7  53 183 439 863 + 497 383 563  79 973 + 287  63 343 169 583 + 627 343 773 959 943 + 767 473 103 699 303 + + Find the Matrix Sum of: + +   7  53 183 439 863 497 383 563  79 973 287  63 343 169 583 + 627 343 773 959 943 767 473 103 699 303 957 703 583 639 913 + 447 283 463  29  23 487 463 993 119 883 327 493 423 159 743 + 217 623   3 399 853 407 103 983  89 463 290 516 212 462 350 + 960 376 682 962 300 780 486 502 912 800 250 346 172 812 350 + 870 456 192 162 593 473 915  45 989 873 823 965 425 329 803 + 973 965 905 919 133 673 665 235 509 613 673 815 165 992 326 + 322 148 972 962 286 255 941 541 265 323 925 281 601  95 973 + 445 721  11 525 473  65 511 164 138 672  18 428 154 448 848 + 414 456 310 312 798 104 566 520 302 248 694 976 430 392 198 + 184 829 373 181 631 101 969 613 840 740 778 458 284 760 390 + 821 461 843 513  17 901 711 993 293 157 274  94 192 156 574 +  34 124   4 878 450 476 712 914 838 669 875 299 823 329 699 + 815 559 813 459 522 788 168 586 966 232 308 833 251 631 107 + 813 883 451 509 615  77 281 613 459 205 380 274 302  35 805 + + + Answer: cf3b784c8593890043b17e24088125d4 + + +Problem 346 +=========== + + + The number 7 is special, because 7 is 111 written in base 2, and 11 + written in base 6 + (i.e. 7[10] = 11[6] = 111[2]). In other words, 7 is a repunit in at least + two bases b > 1. + + We shall call a positive integer with this property a strong repunit. It + can be verified that there are 8 strong repunits below 50: + {1,7,13,15,21,31,40,43}. + Furthermore, the sum of all strong repunits below 1000 equals 15864. + + Find the sum of all strong repunits below 10^12. + + Answer: a17874b5a9ec9d7fc8c6489ab8ff29b9 + + +Problem 347 +=========== + + + The largest integer ≤ 100 that is only divisible by both the primes 2 and + 3 is 96, as 96=32*3=2^5*3.For two distinct primes p and q let M(p,q,N) be + the largest positive integer ≤N only divisibleby both p and q and + M(p,q,N)=0 if such a positive integer does not exist. + + E.g. M(2,3,100)=96. + M(3,5,100)=75 and not 90 because 90 is divisible by 2 ,3 and 5. + Also M(2,73,100)=0 because there does not exist a positive integer ≤ 100 + that is divisible by both 2 and 73. + + Let S(N) be the sum of all distinct M(p,q,N).S(100)=2262. + + Find S(10 000 000). + + + Answer: 96ce0eabcbe7a2b2eb1197a1bcc5d37b + + +Problem 348 +=========== + + + Many numbers can be expressed as the sum of a square and a cube. Some of + them in more than one way. + + Consider the palindromic numbers that can be expressed as the sum of a + square and a cube, both greater than 1, in exactly 4 different ways. + For example, 5229225 is a palindromic number and it can be expressed in + exactly 4 different ways: + + 2285^2 + 20^3 + 2223^2 + 66^3 + 1810^2 + 125^3 + 1197^2 + 156^3 + + Find the sum of the five smallest such palindromic numbers. + + + Answer: f286f9159fc20aeb97a8bf8396ba64de + + +Problem 349 +=========== + + + An ant moves on a regular grid of squares that are coloured either black + or white. + The ant is always oriented in one of the cardinal directions (left, right, + up or down) and moves from square to adjacent square according to the + following rules: + - if it is on a black square, it flips the color of the square to white, + rotates 90 degrees counterclockwise and moves forward one square. + - if it is on a white square, it flips the color of the square to black, + rotates 90 degrees clockwise and moves forward one square. + + Starting with a grid that is entirely white, how many squares are black + after 10^18 moves of the ant? + + + Answer: 412b0faec10b3adb415363d2df26530d + + +Problem 350 +=========== + + + A list of size n is a sequence of n natural numbers. + Examples are (2,4,6), (2,6,4), (10,6,15,6), and (11). + + The greatest common divisor, or gcd, of a list is the largest natural + number that divides all entries of the list. + Examples: gcd(2,6,4) = 2, gcd(10,6,15,6) = 1 and gcd(11) = 11. + + The least common multiple, or lcm, of a list is the smallest natural + number divisible by each entry of the list. + Examples: lcm(2,6,4) = 12, lcm(10,6,15,6) = 30 and lcm(11) = 11. + + Let f(G, L, N) be the number of lists of size N with gcd ≥ G and lcm ≤ L. + For example: + + f(10, 100, 1) = 91. + f(10, 100, 2) = 327. + f(10, 100, 3) = 1135. + f(10, 100, 1000) mod 101^4 = 3286053. + + Find f(10^6, 10^12, 10^18) mod 101^4. + + + Answer: cad3ce6a252568bbcb41ca627d7e58ae + + +Problem 351 +=========== + + + A hexagonal orchard of order n is a triangular lattice made up of points + within a regular hexagon with side n. The following is an example of a + hexagonal orchard of order 5: + + Highlighted in green are the points which are hidden from the center by a + point closer to it. It can be seen that for a hexagonal orchard of order + 5, 30 points are hidden from the center. + + Let H(n) be the number of points hidden from the center in a hexagonal + orchard of order n. + + H(5) = 30. H(10) = 138. H(1 000) = 1177848. + + Find H(100 000 000). + + + p_351_hexorchard.png + Answer: 338481092e945257756075a8d03978fd + + +Problem 352 +=========== + + + Each one of the 25 sheep in a flock must be tested for a rare virus, known + to affect 2% of the sheep population.An accurate and extremely sensitive + PCR test exists for blood samples, producing a clear positive / negative + result, but it is very time-consuming and expensive. + + Because of the high cost, the vet-in-charge suggests that instead of + performing 25 separate tests, the following procedure can be used instead: + + The sheep are split into 5 groups of 5 sheep in each group. For each + group, the 5 samples are mixed together and a single test is performed. + Then, + + • If the result is negative, all the sheep in that group are deemed to + be virus-free. + • If the result is positive, 5 additional tests will be performed (a + separate test for each animal) to determine the affected + individual(s). + + Since the probability of infection for any specific animal is only 0.02, + the first test (on the pooled samples) for each group will be: + + • Negative (and no more tests needed) with probability 0.98^5 = + 0.9039207968. + • Positive (5 additional tests needed) with probability 1 - 0.9039207968 + = 0.0960792032. + + Thus, the expected number of tests for each group is 1 + 0.0960792032 × 5 + = 1.480396016. + Consequently, all 5 groups can be screened using an average of only + 1.480396016 × 5 = 7.40198008 tests, which represents a huge saving of more + than 70% ! + + Although the scheme we have just described seems to be very efficient, it + can still be improved considerably (always assuming that the test is + sufficiently sensitive and that there are no adverse effects caused by + mixing different samples). E.g.: + + • We may start by running a test on a mixture of all the 25 samples. It + can be verified that in about 60.35% of the cases this test will be + negative, thus no more tests will be needed. Further testing will only + be required for the remaining 39.65% of the cases. + • If we know that at least one animal in a group of 5 is infected and + the first 4 individual tests come out negative, there is no need to + run a test on the fifth animal (we know that it must be infected). + • We can try a different number of groups / different number of animals + in each group, adjusting those numbers at each level so that the total + expected number of tests will be minimised. + + To simplify the very wide range of possibilities, there is one restriction + we place when devising the most cost-efficient testing scheme: whenever we + start with a mixed sample, all the sheep contributing to that sample must + be fully screened (i.e. a verdict of infected / virus-free must be reached + for all of them) before we start examining any other animals. + + For the current example, it turns out that the most cost-efficient testing + scheme (we'll call it the optimal strategy) requires an average of just + 4.155452 tests! + + Using the optimal strategy, let T(s,p) represent the average number of + tests needed to screen a flock of s sheep for a virus having probability p + to be present in any individual. + Thus, rounded to six decimal places, T(25, 0.02) = 4.155452 and T(25, + 0.10) = 12.702124. + + Find ΣT(10000, p) for p=0.01, 0.02, 0.03, ... 0.50. + Give your answer rounded to six decimal places. + + + Answer: 2e74b2fb574d6318cdbf2a41ad006de7 + + +Problem 353 +=========== + + + A moon could be described by the sphere C(r) with centre (0,0,0) and + radius r. + + There are stations on the moon at the points on the surface of C(r) with + integer coordinates. The station at (0,0,r) is called North Pole station, + the station at (0,0,-r) is called South Pole station. + + All stations are connected with each other via the shortest road on the + great arc through the stations. A journey between two stations is risky. + If d is the length of the road between two stations, (d/(π r))^2 is a + measure for the risk of the journey (let us call it the risk of the road). + If the journey includes more than two stations, the risk of the journey is + the sum of risks of the used roads. + + A direct journey from the North Pole station to the South Pole station has + the length πr and risk 1. The journey from the North Pole station to the + South Pole station via (0,r,0) has the same length, but a smaller risk: + (½πr/(πr))^2+(½πr/(πr))^2=0.5. + + The minimal risk of a journey from the North Pole station to the South + Pole station on C(r) is M(r). + + You are given that M(7)=0.1784943998 rounded to 10 digits behind the + decimal point. + + Find ∑M(2^n-1) for 1≤n≤15. + + Give your answer rounded to 10 digits behind the decimal point in the form + a.bcdefghijk. + + + Answer: 211b5626459be71baefc78478d18bdc3 + + +Problem 354 +=========== + + + Consider a honey bee's honeycomb where each cell is a perfect regular + hexagon with side length 1. + + One particular cell is occupied by the queen bee. + For a positive real number L, let B(L) count the cells with distance L + from the queen bee cell (all distances are measured from centre to + centre); you may assume that the honeycomb is large enough to accommodate + for any distance we wish to consider. + For example, B(√3) = 6, B(√21) = 12 and B(111 111 111) = 54. + + Find the number of L ≤ 5·10^11 such that B(L) = 450. + + + p_354_bee_honeycomb.png + Answer: e36240897614dc46e83405ae8cdf198c + + +Problem 355 +=========== + + + Define Co(n) to be the maximal possible sum of a set of mutually co-prime + elements from {1, 2, ..., n}. + For example Co(10) is 30 and hits that maximum on the subset + {1, 5, 7, 8, 9}. + + You are given that Co(30) = 193 and Co(100) = 1356. + + Find Co(200000). + + + Answer: 41cb97b6d02878d79f8b2e3b6c74920a + + +Problem 356 +=========== + + + Let a[n] be the largest real root of a polynomial g(x) = x^3 - 2^n·x^2 + + n. + For example, a[2] = 3.86619826... + + Find the last eight digits of. + + Note: represents the floor function. + + + p_356_cubicpoly1.gif + p_356_cubicpoly2.gif + Answer: ab2104e80fa7da630ce7fd835d8006ee + + +Problem 357 +=========== + + + Consider the divisors of 30: 1,2,3,5,6,10,15,30. + It can be seen that for every divisor d of 30, d+30/d is prime. + + Find the sum of all positive integers n not exceeding 100 000 000 + such thatfor every divisor d of n, d+n/d is prime. + + + Answer: ed25b13b18a21c1077fed00ef42f503b + + +Problem 358 +=========== + + + A cyclic number with n digits has a very interesting property: + When it is multiplied by 1, 2, 3, 4, ... n, all the products have exactly + the same digits, in the same order, but rotated in a circular fashion! + + The smallest cyclic number is the 6-digit number 142857 : + 142857 × 1 = 142857 + 142857 × 2 = 285714 + 142857 × 3 = 428571 + 142857 × 4 = 571428 + 142857 × 5 = 714285 + 142857 × 6 = 857142 + + The next cyclic number is 0588235294117647 with 16 digits : + 0588235294117647 × 1 = 0588235294117647 + 0588235294117647 × 2 = 1176470588235294 + 0588235294117647 × 3 = 1764705882352941 + ... + 0588235294117647 × 16 = 9411764705882352 + + Note that for cyclic numbers, leading zeros are important. + + There is only one cyclic number for which, the eleven leftmost digits are + 00000000137 and the five rightmost digits are 56789 (i.e., it has the form + 00000000137...56789 with an unknown number of digits in the middle). Find + the sum of all its digits. + + + Answer: 359e1ec8aeaa3932b54f2a5d20fa4f73 + + +Problem 359 +=========== + + + An infinite number of people (numbered 1, 2, 3, etc.) are lined up to get + a room at Hilbert's newest infinite hotel. The hotel contains an infinite + number of floors (numbered 1, 2, 3, etc.), and each floor contains an + infinite number of rooms (numbered 1, 2, 3, etc.). + + Initially the hotel is empty. Hilbert declares a rule on how the n^th + person is assigned a room: person n gets the first vacant room in the + lowest numbered floor satisfying either of the following: + + • the floor is empty + • the floor is not empty, and if the latest person taking a room in that + floor is person m, then m + n is a perfect square + + Person 1 gets room 1 in floor 1 since floor 1 is empty. + Person 2 does not get room 2 in floor 1 since 1 + 2 = 3 is not a perfect + square. + Person 2 instead gets room 1 in floor 2 since floor 2 is empty. + Person 3 gets room 2 in floor 1 since 1 + 3 = 4 is a perfect square. + + Eventually, every person in the line gets a room in the hotel. + + Define P(f, r) to be n if person n occupies room r in floor f, and 0 if no + person occupies the room. Here are a few examples: + P(1, 1) = 1 + P(1, 2) = 3 + P(2, 1) = 2 + P(10, 20) = 440 + P(25, 75) = 4863 + P(99, 100) = 19454 + + Find the sum of all P(f, r) for all positive f and r such that f × r = + 71328803586048 and give the last 8 digits as your answer. + + + Answer: 91525a22396940a99c496efcb75f2eee + + +Problem 360 +=========== + + + Given two points (x[1],y[1],z[1]) and (x[2],y[2],z[2]) in three + dimensional space, the Manhattan distance between those points is defined + as + |x[1]-x[2]|+|y[1]-y[2]|+|z[1]-z[2]|. + + Let C(r) be a sphere with radius r and center in the origin O(0,0,0). + Let I(r) be the set of all points with integer coordinates on the surface + of C(r). + Let S(r) be the sum of the Manhattan distances of all elements of I(r) to + the origin O. + + E.g. S(45)=34518. + + Find S(10^10). + + + Answer: 82ec91527315eafb7e3acc139eeeb8eb + + +Problem 361 +=========== + + + The Thue-Morse sequence {T[n]} is a binary sequence satisfying: + + • T[0] = 0 + • T[2n] = T[n] + • T[2n+1] = 1 - T[n] + + The first several terms of {T[n]} are given as follows: + 01101001100101101001011001101001.... + + We define {A[n]} as the sorted sequence of integers such that the binary + expression of each element appears as a subsequence in {T[n]}. + For example, the decimal number 18 is expressed as 10010 in binary. 10010 + appears in {T[n]} (T[8] to T[12]), so 18 is an element of {A[n]}. + The decimal number 14 is expressed as 1110 in binary. 1110 never appears + in {T[n]}, so 14 is not an element of {A[n]}. + + The first several terms of A[n] are given as follows: + + n 0 1 2 3 4 5 6 7 8 9 10 11 12 … + A[n] 0 1 2 3 4 5 6 9 10 11 12 13 18 … + + We can also verify that A[100] = 3251 and A[1000] = 80852364498. + + Find the last 9 digits of . + + + p_361_Thue-Morse1.gif + Answer: 6540278145900f1fa45b95cc2f9599f1 + + +Problem 362 +=========== + + + Consider the number 54. + 54 can be factored in 7 distinct ways into one or more factors larger than + 1: + 54, 2×27, 3×18, 6×9, 3×3×6, 2×3×9 and 2×3×3×3. + If we require that the factors are all squarefree only two ways remain: + 3×3×6 and 2×3×3×3. + + Let's call Fsf(n) the number of ways n can be factored into one or more + squarefree factors larger than 1, soFsf(54)=2. + + Let S(n) be ∑Fsf(k) for k=2 to n. + + S(100)=193. + + Find S(10 000 000 000). + + + Answer: b62f0d524bec8653ba7b8a2cab70260b + + +Problem 363 +=========== + + A cubic Bézier curve is defined by four points: P[0], P[1], P[2] and P[3]. + + The curve is constructed as follows: + On the segments P[0]P[1], P[1]P[2] and P[2]P[3] the points Q[0],Q[1] and + Q[2] are drawn such that + P[0]Q[0]/P[0]P[1]=P[1]Q[1]/P[1]P[2]=P[2]Q[2]/P[2]P[3]=t (t in [0,1]). + On the segments Q[0]Q[1] and Q[1]Q[2] the points R[0] and R[1] are drawn + such thatQ[0]R[0]/Q[0]Q[1]=Q[1]R[1]/Q[1]Q[2]=t for the same value of t. + On the segment R[0]R[1] the point B is drawn such that R[0]B/R[0]R[1]=t + for the same value of t.The Bézier curve defined by the points P[0], P[1], + P[2], P[3] is the locus of B as Q[0] takes all possible positions on the + segment P[0]P[1]. (Please note that for all points the value of t is the + same.) + + [1]Applet + + In the applet to the right you can drag the points P[0], P[1], P[2] and + P[3] to see what the Bézier curve (green curve) defined by those points + looks like. You can also drag the point Q[0] along the segment P[0]P[1]. + + From the construction it is clear that the Bézier curve will be tangent to + the segments P[0]P[1] in P[0] and P[2]P[3] in P[3]. + + A cubic Bézier curve with P[0]=(1,0), P[1]=(1,v), P[2]=(v,1) and + P[3]=(0,1) is used to approximate a quarter circle. + The value v>0 is chosen such that the area enclosed by the lines OP[0], + OP[3] and the curve is equal to ^π/[4] (the area of the quarter circle). + + By how many percent does the length of the curve differ from the length of + the quarter circle? + That is, if L is the length of the curve, calculate 100*^(L-π/2)/[(π/2)]. + Give your answer rounded to 10 digits behind the decimal point. + + + Visible links + 1. CabriJava.class + Answer: 2bc63386b7cccc64c67f90e719936143 + + +Problem 364 +=========== + + + There are N seats in a row. N people come after each other to fill the + seats according to the following rules: + +  1. If there is any seat whose adjacent seat(s) are not occupied take such + a seat. +  2. If there is no such seat and there is any seat for which only one + adjacent seat is occupied take such a seat. +  3. Otherwise take one of the remaining available seats. + + Let T(N) be the number of possibilities that N seats are occupied by N + people with the given rules. + The following figure shows T(4)=8. + + We can verify that T(10) = 61632 and T(1 000) mod 100 000 007 = 47255094. + + Find T(1 000 000) mod 100 000 007. + + + p_364_comf_dist.gif + Answer: d631977573d415a4766de9e6bd388cca + + +Problem 365 +=========== + + + The binomial coeffient C(10^18,10^9) is a number with more than 9 billion + (9×10^9) digits. + + Let M(n,k,m) denote the binomial coefficient C(n,k) modulo m. + + Calculate ∑M(10^18,10^9,p*q*r) for 1000 1/2 + + If the first player picks die B and the second player picks die C we get + P(second player wins) = 7/12 > 1/2 + + If the first player picks die C and the second player picks die A we get + P(second player wins) = 25/36 > 1/2 + + So whatever die the first player picks, the second player can pick another + die and have a larger than 50% chance of winning. + A set of dice having this property is called a nontransitive set of dice. + + We wish to investigate how many sets of nontransitive dice exist. We will + assume the following conditions: + + • There are three six-sided dice with each side having between 1 and N + pips, inclusive. + • Dice with the same set of pips are equal, regardless of which side on + the die the pips are located. + • The same pip value may appear on multiple dice; if both players roll + the same value neither player wins. + • The sets of dice {A,B,C}, {B,C,A} and {C,A,B} are the same set. + + For N = 7 we find there are 9780 such sets. + How many are there for N = 30 ? + + + Answer: c64df302990eb3738f8ec62ea6b66c0b + + +Problem 377 +=========== + + + There are 16 positive integers that do not have a zero in their digits and + that have a digital sum equal to 5, namely: + 5, 14, 23, 32, 41, 113, 122, 131, 212, 221, 311, 1112, 1121, 1211, 2111 + and 11111. + Their sum is 17891. + + Let f(n) be the sum of all positive integers that do not have a zero in + their digits and have a digital sum equal to n. + + Find . + Give the last 9 digits as your answer. + + + Answer: a915ccbae49de15208c88affba84d206 + + +Problem 378 +=========== + + + Let T(n) be the n^th triangle number, so T(n) = n (n+1) . + 2 + + Let dT(n) be the number of divisors of T(n). + E.g.:T(7) = 28 and dT(7) = 6. + + Let Tr(n) be the number of triples (i, j, k) such that 1 ≤ i < j < k ≤ n + and dT(i) > dT(j) > dT(k). + Tr(20) = 14, Tr(100) = 5772 and Tr(1000) = 11174776. + + Find Tr(60 000 000). + Give the last 18 digits of your answer. + + + Answer: 336745dc9d90928596237c4b471a8927 + + +Problem 379 +=========== + + + Let f(n) be the number of couples (x,y) with x and y positive integers, x + ≤ y and the least common multiple of x and y equal to n. + + Let g be the summatory function of f, i.e.: g(n) = ∑ f(i) for 1 ≤ i ≤ n. + + You are given that g(10^6) = 37429395. + + Find g(10^12). + + + Answer: de20f710cb6665c48795072197ad53e0 + + +Problem 380 +=========== + + + An m×n maze is an m×n rectangular grid with walls placed between grid + cells such that there is exactly one path from the top-left square to any + other square. + The following are examples of a 9×12 maze and a 15×20 maze: + + Let C(m,n) be the number of distinct m×n mazes. Mazes which can be formed + by rotation and reflection from another maze are considered distinct. + + It can be verified that C(1,1) = 1, C(2,2) = 4, C(3,4) = 2415, and C(9,12) + = 2.5720e46 (in scientific notation rounded to 5 significant digits). + Find C(100,500) and write your answer in scientific notation rounded to 5 + significant digits. + + When giving your answer, use a lowercase e to separate mantissa and + exponent.E.g. if the answer is 1234567891011 then the answer format would + be 1.2346e12. + + + Answer: c86d2f4c17c8134fbebed5d37a0f90d7 + + +Problem 381 +=========== + + + For a prime p let S(p) = (∑(p-k)!) mod(p) for 1 ≤ k ≤ 5. + + For example, if p=7, + (7-1)! + (7-2)! + (7-3)! + (7-4)! + (7-5)! = 6! + 5! + 4! + 3! + 2! = + 720+120+24+6+2 = 872. + As 872 mod(7) = 4, S(7) = 4. + + It can be verified that ∑S(p) = 480 for 5 ≤ p < 100. + + Find ∑S(p) for 5 ≤ p < 10^8. + + + Answer: 80c84973a9643e46d49d79d7284e7ff3 + + +Problem 382 +=========== + + + A polygon is a flat shape consisting of straight line segments that are + joined to form a closed chain or circuit. A polygon consists of at least + three sides and does not self-intersect. + + A set S of positive numbers is said to generate a polygon P if: + + • no two sides of P are the same length, + • the length of every side of P is in S, and + • S contains no other value. + + For example: + The set {3, 4, 5} generates a polygon with sides 3, 4, and 5 (a triangle). + The set {6, 9, 11, 24} generates a polygon with sides 6, 9, 11, and 24 (a + quadrilateral). + The sets {1, 2, 3} and {2, 3, 4, 9} do not generate any polygon at all. + + Consider the sequence s, defined as follows: + + • s[1] = 1, s[2] = 2, s[3] = 3 + • s[n] = s[n-1] + s[n-3] for n > 3. + + Let U[n] be the set {s[1], s[2], ..., s[n]}. For example, U[10] = {1, 2, + 3, 4, 6, 9, 13, 19, 28, 41}. + Let f(n) be the number of subsets of U[n] which generate at least one + polygon. + For example, f(5) = 7, f(10) = 501 and f(25) = 18635853. + + Find the last 9 digits of f(10^18). + + + Answer: 56a121bcf3bb674d0d3ce561b6b24ea5 + + +Problem 383 +=========== + + + Let f[5](n) be the largest integer x for which 5^x divides n. + For example, f[5](625000) = 7. + + Let T[5](n) be the number of integers i which satisfy f[5]((2·i-1)!) < + 2·f[5](i!) and 1 ≤ i ≤ n. + It can be verified that T[5](10^3) = 68 and T[5](10^9) = 2408210. + + Find T[5](10^18). + + + Answer: c1bc7c945344e1967bfaced9ade895a0 + + +Problem 384 +=========== + + + Define the sequence a(n) as the number of adjacent pairs of ones in the + binary expansion of n (possibly overlapping). + E.g.: a(5) = a(101[2]) = 0, a(6) = a(110[2]) = 1, a(7) = a(111[2]) = 2 + + Define the sequence b(n) = (-1)^a(n). + This sequence is called the Rudin-Shapiro sequence. + + Also consider the summatory sequence of b(n): . + + The first couple of values of these sequences are: + n        0     1     2     3     4     5     6     7 + a(n)     0     0     0     1     0     0     1     2 + b(n)     1     1     1    -1     1     1    -1     1 + s(n)     1     2     3     2     3     4     3     4 + + The sequence s(n) has the remarkable property that all elements are + positive and every positive integer k occurs exactly k times. + + Define g(t,c), with 1 ≤ c ≤ t, as the index in s(n) for which t occurs for + the c'th time in s(n). + E.g.: g(3,3) = 6, g(4,2) = 7 and g(54321,12345) = 1220847710. + + Let F(n) be the fibonacci sequence defined by: + F(0)=F(1)=1 and + F(n)=F(n-1)+F(n-2) for n>1. + + Define GF(t)=g(F(t),F(t-1)). + + Find ΣGF(t) for 2≤t≤45. + + + Answer: ea0bb1fff1a51b48971762b93aeed103 + + +Problem 385 +=========== + + + For any triangle T in the plane, it can be shown that there is a unique + ellipse with largest area that is completely inside T. + + For a given n, consider triangles T such that: + - the vertices of T have integer coordinates with absolute value ≤ n, and + - the foci^1 of the largest-area ellipse inside T are (√13,0) and + (-√13,0). + Let A(n) be the sum of the areas of all such triangles. + + For example, if n = 8, there are two such triangles. Their vertices are + (-4,-3),(-4,3),(8,0) and (4,3),(4,-3),(-8,0), and the area of each + triangle is 36. Thus A(8) = 36 + 36 = 72. + + It can be verified that A(10) = 252, A(100) = 34632 and A(1000) = 3529008. + + Find A(1 000 000 000). + + ^1The foci (plural of focus) of an ellipse are two points A and B such + that for every point P on the boundary of the ellipse, AP + PB is + constant. + + + Answer: a21c033d9e119c293e51966ea78c9950 + + +Problem 386 +=========== + + + Let n be an integer and S(n) be the set of factors of n. + + A subset A of S(n) is called an antichain of S(n) if A contains only one + element or if none of the elements of A divides any of the other elements + of A. + + For example: S(30) = {1, 2, 3, 5, 6, 10, 15, 30} + {2, 5, 6} is not an antichain of S(30). + {2, 3, 5} is an antichain of S(30). + + Let N(n) be the maximum length of an antichain of S(n). + + Find ΣN(n) for 1 ≤ n ≤ 10^8 + + + Answer: d1d893f7c50910aa10daec5e9352e86d + + +Problem 387 +=========== + + + A Harshad or Niven number is a number that is divisible by the sum of its + digits. + 201 is a Harshad number because it is divisible by 3 (the sum of its + digits.) + When we truncate the last digit from 201, we get 20, which is a Harshad + number. + When we truncate the last digit from 20, we get 2, which is also a Harshad + number. + Let's call a Harshad number that, while recursively truncating the last + digit, always results in a Harshad number a right truncatable Harshad + number. + + Also: + 201/3=67 which is prime. + Let's call a Harshad number that, when divided by the sum of its digits, + results in a prime a strong Harshad number. + + Now take the number 2011 which is prime. + When we truncate the last digit from it we get 201, a strong Harshad + number that is also right truncatable. + Let's call such primes strong, right truncatable Harshad primes. + + You are given that the sum of the strong, right truncatable Harshad primes + less than 10000 is 90619. + + Find the sum of the strong, right truncatable Harshad primes less than + 10^14. + + + Answer: a20cbd8639767decfa2c2c9955eb6be3 + + +Problem 388 +=========== + + + Consider all lattice points (a,b,c) with 0 ≤ a,b,c ≤ N. + + From the origin O(0,0,0) all lines are drawn to the other lattice points. + Let D(N) be the number of distinct such lines. + + You are given that D(1 000 000) = 831909254469114121. + + Find D(10^10). Give as your answer the first nine digits followed by the + last nine digits. + + + Answer: 2bab886c7d98d802d9249c9e12d72c25 + + +Problem 389 +=========== + + + An unbiased single 4-sided die is thrown and its value, T, is noted. + T unbiased 6-sided dice are thrown and their scores are added together. + The sum, C, is noted. + C unbiased 8-sided dice are thrown and their scores are added together. + The sum, O, is noted. + O unbiased 12-sided dice are thrown and their scores are added together. + The sum, D, is noted. + D unbiased 20-sided dice are thrown and their scores are added together. + The sum, I, is noted. + Find the variance of I, and give your answer rounded to 4 decimal places. + + + Answer: 79a080d38b837547b975c97b44764dfb + + +Problem 390 +=========== + + + Consider the triangle with sides √5, √65 and √68.It can be shown that this + triangle has area 9. + + S(n) is the sum of the areas of all triangles with sides √(1+b^2), + √(1+c^2) and √(b^2+c^2) (for positive integers b and c ) that have an + integral area not exceeding n. + + The example triangle has b=2 and c=8. + + S(10^6)=18018206. + + Find S(10^10). + + + Answer: ed7f2fbc05a2fd2033d80de671f35ea3 + + +Problem 391 +=========== + + + Let s[k] be the number of 1’s when writing the numbers from 0 to k in + binary. + For example, writing 0 to 5 in binary, we have 0, 1, 10, 11, 100, 101. + There are seven 1’s, so s[5] = 7. + The sequence S = {s[k] : k ≥ 0} starts {0, 1, 2, 4, 5, 7, 9, 12, ...}. + + A game is played by two players. Before the game starts, a number n is + chosen. A counter c starts at 0. At each turn, the player chooses a number + from 1 to n (inclusive) and increases c by that number. The resulting + value of c must be a member of S. If there are no more valid moves, the + player loses. + + For example: + Let n = 5. c starts at 0. + Player 1 chooses 4, so c becomes 0 + 4 = 4. + Player 2 chooses 5, so c becomes 4 + 5 = 9. + Player 1 chooses 3, so c becomes 9 + 3 = 12. + etc. + Note that c must always belong to S, and each player can increase c by at + most n. + + Let M(n) be the highest number the first player can choose at her first + turn to force a win, and M(n) = 0 if there is no such move. For example, + M(2) = 2, M(7) = 1 and M(20) = 4. + + Given Σ(M(n))^3 = 8150 for 1 ≤ n ≤ 20. + + Find Σ(M(n))^3 for 1 ≤ n ≤ 1000. + + + Answer: b2947548d4f5c4878c5f788f9849e750 + + +Problem 392 +=========== + + + A rectilinear grid is an orthogonal grid where the spacing between the + gridlines does not have to be equidistant. + An example of such grid is logarithmic graph paper. + + Consider rectilinear grids in the Cartesian coordinate system with the + following properties: + + • The gridlines are parallel to the axes of the Cartesian coordinate + system. + • There are N+2 vertical and N+2 horizontal gridlines. Hence there are + (N+1) x (N+1) rectangular cells. + • The equations of the two outer vertical gridlines are x = -1 and x = + 1. + • The equations of the two outer horizontal gridlines are y = -1 and y = + 1. + • The grid cells are colored red if they overlap with the unit circle, + black otherwise. + + For this problem we would like you to find the postions of the remaining N + inner horizontal and N inner vertical gridlines so that the area occupied + by the red cells is minimized. + + E.g. here is a picture of the solution for N = 10: + + The area occupied by the red cells for N = 10 rounded to 10 digits behind + the decimal point is 3.3469640797. + + Find the positions for N = 400. + Give as your answer the area occupied by the red cells rounded to 10 + digits behind the decimal point. + + + Answer: 3268b0bc489187db3d234c097040d909 + + +Problem 393 +=========== + + + An n×n grid of squares contains n^2 ants, one ant per square. + All ants decide to move simultaneously to an adjacent square (usually 4 + possibilities, except for ants on the edge of the grid or at the corners). + We define f(n) to be the number of ways this can happen without any ants + ending on the same square and without any two ants crossing the same edge + between two squares. + + You are given that f(4) = 88. + Find f(10). + + + Answer: 58e4990838fb3d1725872da30f9db748 + + +Problem 394 +=========== + + + Jeff eats a pie in an unusual way. + The pie is circular. He starts with slicing an initial cut in the pie + along a radius. + While there is at least a given fraction F of pie left, he performs the + following procedure: + - He makes two slices from the pie centre to any point of what is + remaining of the pie border, any point on the remaining pie border equally + likely. This will divide the remaining pie into three pieces. + - Going counterclockwise from the initial cut, he takes the first two pie + pieces and eats them. + When less than a fraction F of pie remains, he does not repeat this + procedure. Instead, he eats all of the remaining pie. + + For x ≥ 1, let E(x) be the expected number of times Jeff repeats the + procedure above with F = ^1/[x]. + It can be verified that E(1) = 1, E(2) ≈ 1.2676536759, and E(7.5) ≈ + 2.1215732071. + + Find E(40) rounded to 10 decimal places behind the decimal point. + + + Answer: f8ad575e1a03365a60b6429c3b7a64df + + +Problem 395 +=========== + + + The Pythagorean tree is a fractal generated by the following procedure: + + Start with a unit square. Then, calling one of the sides its base (in the + animation, the bottom side is the base): + +  1. Attach a right triangle to the side opposite the base, with the + hypotenuse coinciding with that side and with the sides in a 3-4-5 + ratio. Note that the smaller side of the triangle must be on the + 'right' side with respect to the base (see animation). +  2. Attach a square to each leg of the right triangle, with one of its + sides coinciding with that leg. +  3. Repeat this procedure for both squares, considering as their bases the + sides touching the triangle. + + The resulting figure, after an infinite number of iterations, is the + Pythagorean tree. + + It can be shown that there exists at least one rectangle, whose sides are + parallel to the largest square of the Pythagorean tree, which encloses the + Pythagorean tree completely. + + Find the smallest area possible for such a bounding rectangle, and give + your answer rounded to 10 decimal places. + + + p_395_pythagorean.gif + Answer: 505048b0c619161d05b9b3e492f3edc3 + + +Problem 396 +=========== + + + For any positive integer n, the nth weak Goodstein sequence {g[1], g[2], + g[3], ...} is defined as: + + • g[1] = n + • for k > 1, g[k] is obtained by writing g[k-1] in base k, interpreting + it as a base k + 1 number, and subtracting 1. + + The sequence terminates when g[k] becomes 0. + + For example, the 6th weak Goodstein sequence is {6, 11, 17, 25, ...}: + + • g[1] = 6. + • g[2] = 11 since 6 = 110[2], 110[3] = 12, and 12 - 1 = 11. + • g[3] = 17 since 11 = 102[3], 102[4] = 18, and 18 - 1 = 17. + • g[4] = 25 since 17 = 101[4], 101[5] = 26, and 26 - 1 = 25. + + and so on. + + It can be shown that every weak Goodstein sequence terminates. + + Let G(n) be the number of nonzero elements in the nth weak Goodstein + sequence. + It can be verified that G(2) = 3, G(4) = 21 and G(6) = 381. + It can also be verified that ΣG(n) = 2517 for 1 ≤ n < 8. + + Find the last 9 digits of ΣG(n) for 1 ≤ n < 16. + + + Answer: 4665c73fdca473ccc0643fc982f24e06 + + +Problem 397 +=========== + + + On the parabola y = x^2/k, three points A(a, a^2/k), B(b, b^2/k) and C(c, + c^2/k) are chosen. + + Let F(K, X) be the number of the integer quadruplets (k, a, b, c) such + that at least one angle of the triangle ABC is 45-degree, with 1 ≤ k ≤ K + and -X ≤ a < b < c ≤ X. + + For example, F(1, 10) = 41 and F(10, 100) = 12492. + Find F(10^6, 10^9). + + + Answer: 07f769df9543bc05e6318878c34d074d + + +Problem 398 +=========== + + + Inside a rope of length n, n-1 points are placed with distance 1 from each + other and from the endpoints. Among these points, we choose m-1 points at + random and cut the rope at these points to create m segments. + + Let E(n, m) be the expected length of the second-shortest segment.For + example, E(3, 2) = 2 and E(8, 3) = 16/7.Note that if multiple segments + have the same shortest length the length of the second-shortest segment is + defined as the same as the shortest length. + + Find E(10^7, 100).Give your answer rounded to 5 decimal places behind the + decimal point. + + + Answer: fa0a25d62fa225e05fd8736713a9bfc0 + + +Problem 399 +=========== + + + The first 15 fibonacci numbers are: + 1,1,2,3,5,8,13,21,34,55,89,144,233,377,610. + It can be seen that 8 and 144 are not squarefree: 8 is divisible by 4 and + 144 is divisible by 4 and by 9. + So the first 13 squarefree fibonacci numbers are: + 1,1,2,3,5,13,21,34,55,89,233,377 and 610. + + The 200th squarefree fibonacci number + is:971183874599339129547649988289594072811608739584170445. + The last sixteen digits of this number are: 1608739584170445 and in + scientific notation this number can be written as 9.7e53. + + Find the 100 000 000th squarefree fibonacci number. + Give as your answer its last sixteen digits followed by a comma followed + by the number in scientific notation (rounded to one digit after the + decimal point). + For the 200th squarefree number the answer would have been: + 1608739584170445,9.7e53 + + Note: + For this problem, assume that for every prime p, the first fibonacci + number divisible by p is not divisible by p^2 (this is part of Wall's + conjecture). This has been verified for primes ≤ 3·10^15, but has not been + proven in general. + If it happens that the conjecture is false, then the accepted answer to + this problem isn't guaranteed to be the 100 000 000th squarefree fibonacci + number, rather it represents only a lower bound for that number. + + + Answer: a0819cfe3be6a04645b8d4fe2345e184 + + +Problem 400 +=========== + + + A Fibonacci tree is a binary tree recursively defined as: + + • T(0) is the empty tree. + • T(1) is the binary tree with only one node. + • T(k) consists of a root node that has T(k-1) and T(k-2) as children. + + On such a tree two players play a take-away game. On each turn a player + selects a node and removes that node along with the subtree rooted at that + node. + The player who is forced to take the root node of the entire tree loses. + + Here are the winning moves of the first player on the first turn for T(k) + from k=1 to k=6. + + Let f(k) be the number of winning moves of the first player (i.e. the + moves for which the second player has no winning strategy) on the first + turn of the game when this game is played on T(k). + + For example, f(5) = 1 and f(10) = 17. + + Find f(10000). Give the last 18 digits of your answer. + + + Answer: 60aa790c07af1446c1e2deba72543a1f + + +Problem 401 +=========== + + + The divisors of 6 are 1,2,3 and 6. + The sum of the squares of these numbers is 1+4+9+36=50. + + Let sigma2(n) represent the sum of the squares of the divisors of n.Thus + sigma2(6)=50. + + Let SIGMA2 represent the summatory function of sigma2, that is + SIGMA2(n)=∑sigma2(i) for i=1 to n. + The first 6 values of SIGMA2 are: 1,6,16,37,63 and 113. + + Find SIGMA2(10^15) modulo 10^9. + + + Answer: 982a249d8b45ef10c98c32dabac00751 + + +Problem 402 +=========== + + + It can be shown that the polynomial n^4 + 4n^3 + 2n^2 + 5n is a multiple + of 6 for every integer n. It can also be shown that 6 is the largest + integer satisfying this property. + + Define M(a, b, c) as the maximum m such that n^4 + an^3 + bn^2 + cn is a + multiple of m for all integers n. For example, M(4, 2, 5) = 6. + + Also, define S(N) as the sum of M(a, b, c) for all 0 < a, b, c ≤ N. + + We can verify that S(10) = 1972 and S(10000) = 2024258331114. + + Let F[k] be the Fibonacci sequence: + F[0] = 0, F[1] = 1 and + F[k] = F[k-1] + F[k-2] for k ≥ 2. + + Find the last 9 digits of Σ S(F[k]) for 2 ≤ k ≤ 1234567890123. + + + Answer: fa7ae8e9243f01b0eac10ec5aaff1f42 + + +Problem 403 +=========== + + + For integers a and b, we define D(a, b) as the domain enclosed by the + parabola y = x^2 and the line y = a·x + b: + D(a, b) = { (x, y) | x^2 ≤ y ≤ a·x + b }. + + L(a, b) is defined as the number of lattice points contained in D(a, b). + For example, L(1, 2) = 8 and L(2, -1) = 1. + + We also define S(N) as the sum of L(a, b) for all the pairs (a, b) such + that the area of D(a, b) is a rational number and |a|,|b| ≤ N. + We can verify that S(5) = 344 and S(100) = 26709528. + + Find S(10^12). Give your answer mod 10^8. + + + Answer: 31c018e3781a3e170366f01e30f09602 + + +Problem 404 +=========== + + + E[a] is an ellipse with an equation of the form x^2 + 4y^2 = 4a^2. + E[a]' is the rotated image of E[a] by θ degrees counterclockwise around + the origin O(0, 0) for 0° < θ < 90°. + + b is the distance to the origin of the two intersection points closest to + the origin and c is the distance of the two other intersection points. + We call an ordered triplet (a, b, c) a canonical ellipsoidal triplet if a, + b and c are positive integers. + For example, (209, 247, 286) is a canonical ellipsoidal triplet. + + Let C(N) be the number of distinct canonical ellipsoidal triplets (a, b, + c) for a ≤ N. + It can be verified that C(10^3) = 7, C(10^4) = 106 and C(10^6) = 11845. + + Find C(10^17). + + + p_404_c_ellipse.gif + Answer: 2d1bc4b93bbc19d9e70c5b04338dea2e + + +Problem 405 +=========== + + + We wish to tile a rectangle whose length is twice its width. + Let T(0) be the tiling consisting of a single rectangle. + For n > 0, let T(n) be obtained from T(n-1) by replacing all tiles in the + following manner: + + The following animation demonstrates the tilings T(n) for n from 0 to 5: + + Let f(n) be the number of points where four tiles meet in T(n). + For example, f(1) = 0, f(4) = 82 and f(10^9) mod 17^7 = 126897180. + + Find f(10^k) for k = 10^18, give your answer modulo 17^7. + + + p_405_tile1.png + p_405_tile2.gif + Answer: 93b712426b768586f88d0bfe597842e6 + + +Problem 406 +=========== + + + We are trying to find a hidden number selected from the set of integers + {1, 2, ..., n} by asking questions. Each number (question) we ask, we get + one of three possible answers: + + • "Your guess is lower than the hidden number" (and you incur a cost of + a), or + • "Your guess is higher than the hidden number" (and you incur a cost of + b), or + • "Yes, that's it!" (and the game ends). + + Given the value of n, a, and b, an optimal strategy minimizes the total + cost for the worst possible case. + + For example, if n = 5, a = 2, and b = 3, then we may begin by asking "2" + as our first question. + + If we are told that 2 is higher than the hidden number (for a cost of + b=3), then we are sure that "1" is the hidden number (for a total cost of + 3). + If we are told that 2 is lower than the hidden number (for a cost of a=2), + then our next question will be "4". + If we are told that 4 is higher than the hidden number (for a cost of + b=3), then we are sure that "3" is the hidden number (for a total cost of + 2+3=5). + If we are told that 4 is lower than the hidden number (for a cost of a=2), + then we are sure that "5" is the hidden number (for a total cost of + 2+2=4). + Thus, the worst-case cost achieved by this strategy is 5. It can also be + shown that this is the lowest worst-case cost that can be achieved. So, in + fact, we have just described an optimal strategy for the given values of + n, a, and b. + + Let C(n, a, b) be the worst-case cost achieved by an optimal strategy for + the given values of n, a, and b. + + Here are a few examples: + C(5, 2, 3) = 5 + C(500, √2, √3) = 13.22073197... + C(20000, 5, 7) = 82 + C(2000000, √5, √7) = 49.63755955... + + Let F[k] be the Fibonacci numbers: F[k] = F[k-1] + F[k-2] with base cases + F[1] = F[2] = 1. + Find ∑[1≤k≤30] C(10^12, √k, √F[k]), and give your answer rounded to 8 + decimal places behind the decimal point. + + + Answer: 0766b1ee975f5674d30fd6c3c934c6e0 + + +Problem 407 +=========== + + + If we calculate a^2 mod 6 for 0 ≤ a ≤ 5 we get: 0,1,4,3,4,1. + + The largest value of a such that a^2 ≡ a mod 6 is 4. + Let's call M(n) the largest value of a < n such that a^2 ≡ a (mod n). + So M(6) = 4. + + Find ∑M(n) for 1 ≤ n ≤ 10^7. + + + Answer: f4da34a4b357123cb142739a52e010f2 + + +Problem 408 +=========== + + + Let's call a lattice point (x, y) inadmissible if x, y and x + y are all + positive perfect squares. + For example, (9, 16) is inadmissible, while (0, 4), (3, 1) and (9, 4) are + not. + + Consider a path from point (x[1], y[1]) to point (x[2], y[2]) using only + unit steps north or east. + Let's call such a path admissible if none of its intermediate points are + inadmissible. + + Let P(n) be the number of admissible paths from (0, 0) to (n, n). + It can be verified that P(5) = 252, P(16) = 596994440 and P(1000) mod + 1 000 000 007 = 341920854. + + Find P(10 000 000) mod 1 000 000 007. + + + Answer: 2c09e247c6144c16cae2358d316affd9 + + +Problem 409 +=========== + + + Let n be a positive integer. Consider nim positions where: + + • There are n non-empty piles. + • Each pile has size less than 2^n. + • No two piles have the same size. + + Let W(n) be the number of winning nim positions satisfying the + aboveconditions (a position is winning if the first player has a winning + strategy). For example, W(1) = 1, W(2) = 6, W(3) = 168, W(5) = 19764360 + and W(100) mod 1 000 000 007 = 384777056. + + Find W(10 000 000) mod 1 000 000 007. + + + Answer: 56c32e75a2656ec08ce177089bda2f53 + + +Problem 410 +=========== + + + Let C be the circle with radius r, x^2 + y^2 = r^2. We choose two points + P(a, b) and Q(-a, c) so that the line passing through P and Q is tangent + to C. + + For example, the quadruplet (r, a, b, c) = (2, 6, 2, -7) satisfies this + property. + + Let F(R, X) be the number of the integer quadruplets (r, a, b, c) with + this property, and with 0 < r ≤ R and 0 < a ≤ X. + + We can verify that F(1, 5) = 10, F(2, 10) = 52 and F(10, 100) = 3384. + Find F(10^8, 10^9) + F(10^9, 10^8). + + + Answer: 45826f3a23aa321f97acb1d2a8f2170b + + +Problem 411 +=========== + + + Let n be a positive integer. Suppose there are stations at the coordinates + (x, y) = (2^i mod n, 3^i mod n) for 0 ≤ i ≤ 2n. We will consider stations + with the same coordinates as the same station. + + We wish to form a path from (0, 0) to (n, n) such that the x and y + coordinates never decrease. + Let S(n) be the maximum number of stations such a path can pass through. + + For example, if n = 22, there are 11 distinct stations, and a valid path + can pass through at most 5 stations. Therefore, S(22) = 5.The case is + illustrated below, with an example of an optimal path: + + It can also be verified that S(123) = 14 and S(10000) = 48. + + Find ∑ S(k^5) for 1 ≤ k ≤ 30. + + + Answer: e351762bf2220ca1396e6a9ee3f6c84f + + +Problem 412 +=========== + + + For integers m, n (0 ≤ n < m), let L(m, n) be an m×m grid with the + top-right n×n grid removed. + + For example, L(5, 3) looks like this: + + We want to number each cell of L(m, n) with consecutive integers 1, 2, 3, + ... such that the number in every cell is smaller than the number below it + and to the left of it. + + For example, here are two valid numberings of L(5, 3): + + Let LC(m, n) be the number of valid numberings of L(m, n). + It can be verified that LC(3, 0) = 42, LC(5, 3) = 250250, LC(6, 3) = + 406029023400 and LC(10, 5) mod 76543217 = 61251715. + + Find LC(10000, 5000) mod 76543217. + + + Answer: 8919ccca34b7ccc293d33e06872c668d + + +Problem 413 +=========== + + + We say that a d-digit positive number (no leading zeros) is a one-child + number if exactly one of its sub-strings is divisible by d. + + For example, 5671 is a 4-digit one-child number. Among all its sub-strings + 5, 6, 7, 1, 56, 67, 71, 567, 671 and 5671, only 56 is divisible by 4. + Similarly, 104 is a 3-digit one-child number because only 0 is divisible + by 3. + 1132451 is a 7-digit one-child number because only 245 is divisible by 7. + + Let F(N) be the number of the one-child numbers less than N. + We can verify that F(10) = 9, F(10^3) = 389 and F(10^7) = 277674. + + Find F(10^19). + + + Answer: 569ad33af889215704df5a9e278aa004 + + +Problem 414 +=========== + + + 6174 is a remarkable number; if we sort its digits in increasing order and + subtract that number from the number you get when you sort the digits in + decreasing order, we get 7641-1467=6174. + Even more remarkable is that if we start from any 4 digit number and + repeat this process of sorting and subtracting, we'll eventually end up + with 6174 or immediately with 0 if all digits are equal. + This also works with numbers that have less than 4 digits if we pad the + number with leading zeroes until we have 4 digits. + E.g. let's start with the number 0837: + 8730-0378=8352 + 8532-2358=6174 + + 6174 is called the Kaprekar constant. The process of sorting and + subtracting and repeating this until either 0 or the Kaprekar constant is + reached is called the Kaprekar routine. + + We can consider the Kaprekar routine for other bases and number of digits. + Unfortunately, it is not guaranteed a Kaprekar constant exists in all + cases; either the routine can end up in a cycle for some input numbers or + the constant the routine arrives at can be different for different input + numbers. + However, it can be shown that for 5 digits and a base b = 6t+3≠9, a + Kaprekar constant exists. + E.g. base 15: (10,4,14,9,5)[15] + base 21: (14,6,20,13,7)[21] + + Define C[b] to be the Kaprekar constant in base b for 5 digits.Define the + function sb(i) to be + + • 0 if i = C[b] or if i written in base b consists of 5 identical digits + • the number of iterations it takes the Kaprekar routine in base b to + arrive at C[b], otherwise + + Note that we can define sb(i) for all integers i < b^5. If i written in + base b takes less than 5 digits, the number is padded with leading zero + digits until we have 5 digits before applying the Kaprekar routine. + + Define S(b) as the sum of sb(i) for 0 < i < b^5. + E.g. S(15) = 5274369 + S(111) = 400668930299 + + Find the sum of S(6k+3) for 2 ≤ k ≤ 300. + Give the last 18 digits as your answer. + + + Answer: 42f095bdfd71e1ae4ae0ceead1bb1802 + + +Problem 415 +=========== + + + A set of lattice points S is called a titanic set if there exists a line + passing through exactly two points in S. + + An example of a titanic set is S = {(0, 0), (0, 1), (0, 2), (1, 1), (2, + 0), (1, 0)}, where the line passing through (0, 1) and (2, 0) does not + pass through any other point in S. + + On the other hand, the set {(0, 0), (1, 1), (2, 2), (4, 4)} is not a + titanic set since the line passing through any two points in the set also + passes through the other two. + + For any positive integer N, let T(N) be the number of titanic sets S whose + every point (x, y) satisfies 0 ≤ x, y ≤ N.It can be verified that T(1) = + 11, T(2) = 494, T(4) = 33554178, T(111) mod 10^8 = 13500401 and + T(10^5) mod 10^8 = 63259062. + + Find T(10^11) mod 10^8. + + + Answer: 2357ad217832274f444cae2a6580b193 + + +Problem 416 +=========== + + + A row of n squares contains a frog in the leftmost square. By successive + jumps the frog goes to the rightmost square and then back to the leftmost + square. On the outward trip he jumps one, two or three squares to the + right, and on the homeward trip he jumps to the left in a similar manner. + He cannot jump outside the squares. He repeats the round-trip travel m + times. + + Let F(m, n) be the number of the ways the frog can travel so that at most + one square remains unvisited. + For example, F(1, 3) = 4, F(1, 4) = 15, F(1, 5) = 46, F(2, 3) = 16 and + F(2, 100) mod 10^9 = 429619151. + + Find the last 9 digits of F(10, 10^12). + + + Answer: 6f398386fdfec57ac166d4970c2bcad2 + + +Problem 417 +=========== + + + A unit fraction contains 1 in the numerator. The decimal representation of + the unit fractions with denominators 2 to 10 are given: + + 1/2 =  0.5 + 1/3 =  0.(3) + 1/4 =  0.25 + 1/5 =  0.2 + 1/6 =  0.1(6) + 1/7 =  0.(142857) + 1/8 =  0.125 + 1/9 =  0.(1) + 1/10 =  0.1 + + Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can + be seen that 1/7 has a 6-digit recurring cycle. + + Unit fractions whose denominator has no other prime factors than 2 and/or + 5 are not considered to have a recurring cycle. + We define the length of the recurring cycle of those unit fractions as 0. + + Let L(n) denote the length of the recurring cycle of 1/n.You are given + that ∑L(n) for 3 ≤ n ≤ 1 000 000 equals 55535191115. + + Find ∑L(n) for 3 ≤ n ≤ 100 000 000 + + + Answer: 93a7df08c972f1e7788516d056a7e016 + + +Problem 418 +=========== + + + Let n be a positive integer. An integer triple (a, b, c) is called a + factorisation triple of n if: + + • 1 ≤ a ≤ b ≤ c + • a·b·c = n. + + Define f(n) to be a + b + c for the factorisation triple (a, b, c) of n + which minimises c / a. One can show that this triple is unique. + + For example, f(165) = 19, f(100100) = 142 and f(20!) = 4034872. + + Find f(43!). + + + Answer: b032468ddb4847d8a2273789379753f5 + + +Problem 419 +=========== + + + The look and say sequence goes 1, 11, 21, 1211, 111221, 312211, 13112221, + 1113213211, ... + The sequence starts with 1 and all other members are obtained by + describing the previous member in terms of consecutive digits. + It helps to do this out loud: + 1 is 'one one' → 11 + 11 is 'two ones' → 21 + 21 is 'one two and one one' → 1211 + 1211 is 'one one, one two and two ones' → 111221 + 111221 is 'three ones, two twos and one one' → 312211 + ... + + Define A(n), B(n) and C(n) as the number of ones, twos and threes in the + n'th element of the sequence respectively. + One can verify that A(40) = 31254, B(40) = 20259 and C(40) = 11625. + + Find A(n), B(n) and C(n) for n = 10^12. + Give your answer modulo 2^30 and separate your values for A, B and C by a + comma. + E.g. for n = 40 the answer would be 31254,20259,11625 + + + Answer: b27db655498b3d64ad4338fcdc9d178f + + +Problem 420 +=========== + + + A positive integer matrix is a matrix whose elements are all positive + integers. + Some positive integer matrices can be expressed as a square of a positive + integer matrix in two different ways. Here is an example: + + We define F(N) as the number of the 2x2 positive integer matrices which + have a trace less than N and which can be expressed as a square of a + positive integer matrix in two different ways. + We can verify that F(50) = 7 and F(1000) = 1019. + + Find F(10^7). + + + p_420_matrix.gif + Answer: e265e34e34fc54e8ceecd5e4b94b1381 + + +Problem 421 +=========== + + + Numbers of the form n^15+1 are composite for every integer n > 1. + For positive integers n and m let s(n,m) be defined as the sum of the + distinct prime factors of n^15+1 not exceeding m. + + E.g. 2^15+1 = 3×3×11×331. + So s(2,10) = 3 and s(2,1000) = 3+11+331 = 345. + + Also 10^15+1 = 7×11×13×211×241×2161×9091. + So s(10,100) = 31 and s(10,1000) = 483. + + Find ∑ s(n,10^8) for 1 ≤ n ≤ 10^11. + + + Answer: 481fcc5ff16ccf1645fb136c123ed660 + + +Problem 422 +=========== + + + Let H be the hyperbola defined by the equation 12x^2 + 7xy - 12y^2 = 625. + + Next, define X as the point (7, 1). It can be seen that X is in H. + + Now we define a sequence of points in H, {P[i] : i ≥ 1}, as: + + • P[1] = (13, 61/4). + • P[2] = (-43/6, -4). + • For i > 2, P[i] is the unique point in H that is different from P[i-1] + and such that line P[i]P[i-1] is parallel to line P[i-2]X. It can be + shown that P[i] is well-defined, and that its coordinates are always + rational. + + You are given that P[3] = (-19/2, -229/24), P[4] = (1267/144, -37/12) and + P[7] = (17194218091/143327232, 274748766781/1719926784). + + Find P[n] for n = 11^14 in the following format: + If P[n] = (a/b, c/d) where the fractions are in lowest terms and the + denominators are positive, then the answer is (a + b + c + d) mod + 1 000 000 007. + + For n = 7, the answer would have been: 806236837. + + + Answer: 7034610688a8851f742f912143c1becf + + +Problem 423 +=========== + + + Let n be a positive integer. + A 6-sided die is thrown n times. Let c be the number of pairs of + consecutive throws that give the same value. + + For example, if n = 7 and the values of the die throws are + (1,1,5,6,6,6,3), then the following pairs of consecutive throws give the + same value: + (1,1,5,6,6,6,3) + (1,1,5,6,6,6,3) + (1,1,5,6,6,6,3) + Therefore, c = 3 for (1,1,5,6,6,6,3). + + Define C(n) as the number of outcomes of throwing a 6-sided die n times + such that c does not exceed π(n).^1 + For example, C(3) = 216, C(4) = 1290, C(11) = 361912500 and C(24) = + 4727547363281250000. + + Define S(L) as ∑ C(n) for 1 ≤ n ≤ L. + For example, S(50) mod 1 000 000 007 = 832833871. + + Find S(50 000 000) mod 1 000 000 007. + + ^1 π denotes the prime-counting function, i.e. π(n) is the number of + primes ≤ n. + + + Answer: e2add9d46ebd8ba59a07dca791cd629b + + +Problem 424 +=========== + + + The above is an example of a cryptic kakuro (also known as cross sums, or + even sums cross) puzzle, with its final solution on the right. (The common + rules of kakuro puzzles can be found easily on numerous internet sites. + Other related information can also be currently found at [1]krazydad.com + whose author has provided the puzzle data for this challenge.) + + The downloadable text file ([2]kakuro200.txt) contains the description of + 200 such puzzles, a mix of 5x5 and 6x6 types. The first puzzle in the file + is the above example which is coded as follows: + + 6,X,X,(vCC),(vI),X,X,X,(hH),B,O,(vCA),(vJE),X,(hFE,vD),O,O,O,O,(hA),O,I,(hJC,vB),O,O,(hJC),H,O,O,O,X,X,X,(hJE),O,O,X + + The first character is a numerical digit indicating the size of the + information grid. It would be either a 6 (for a 5x5 kakuro puzzle) or a 7 + (for a 6x6 puzzle) followed by a comma (,). The extra top line and left + column are needed to insert information. + + The content of each cell is then described and followed by a comma, going + left to right and starting with the top line. + X = Gray cell, not required to be filled by a digit. + O (upper case letter)= White empty cell to be filled by a digit. + A = Or any one of the upper case letters from A to J to be replaced by its + equivalent digit in the solved puzzle. + ( ) = Location of the encrypted sums. Horizontal sums are preceded by a + lower case "h" and vertical sums are preceded by a lower case "v". Those + are followed by one or two upper case letters depending if the sum is a + single digit or double digit one. For double digit sums, the first letter + would be for the "tens" and the second one for the "units". When the cell + must contain information for both a horizontal and a vertical sum, the + first one is always for the horizontal sum and the two are separated by a + comma within the same set of brackets, ex.: (hFE,vD). Each set of brackets + is also immediately followed by a comma. + + The description of the last cell is followed by a Carriage Return/Line + Feed (CRLF) instead of a comma. + + The required answer to each puzzle is based on the value of each letter + necessary to arrive at the solution and according to the alphabetical + order. As indicated under the example puzzle, its answer would be + 8426039571. At least 9 out of the 10 encrypting letters are always part of + the problem description. When only 9 are given, the missing one must be + assigned the remaining digit. + + You are given that the sum of the answers for the first 10 puzzles in the + file is 64414157580. + + Find the sum of the answers for the 200 puzzles. + + + Visible links + 1. http://krazydad.com/ + 2. kakuro200.txt + Answer: c412afe5b5d76dbfbb77443ed5836d89 + + +Problem 425 +=========== + + + Two positive numbers A and B are said to be connected (denoted by "A ↔ B") + if one of these conditions holds: + (1) A and B have the same length and differ in exactly one digit; for + example, 123 ↔ 173. + (2) Adding one digit to the left of A (or B) makes B (or A); for example, + 23 ↔ 223 and 123 ↔ 23. + + We call a prime P a 2's relative if there exists a chain of connected + primes between 2 and P and no prime in the chain exceeds P. + + For example, 127 is a 2's relative. One of the possible chains is shown + below: + 2 ↔ 3 ↔ 13 ↔ 113 ↔ 103 ↔ 107 ↔ 127 + However, 11 and 103 are not 2's relatives. + + Let F(N) be the sum of the primes ≤ N which are not 2's relatives. + We can verify that F(10^3) = 431 and F(10^4) = 78728. + + Find F(10^7). + + + Answer: 3d229894ba4c585138125e802af2d06e + + +Problem 426 +=========== + + + Consider an infinite row of boxes. Some of the boxes contain a ball. For + example, an initial configuration of 2 consecutive occupied boxes followed + by 2 empty boxes, 2 occupied boxes, 1 empty box, and 2 occupied boxes can + be denoted by the sequence (2, 2, 2, 1, 2), in which the number of + consecutive occupied and empty boxes appear alternately. + + A turn consists of moving each ball exactly once according to the + following rule: Transfer the leftmost ball which has not been moved to the + nearest empty box to its right. + + After one turn the sequence (2, 2, 2, 1, 2) becomes (2, 2, 1, 2, 3) as can + be seen below; note that we begin the new sequence starting at the first + occupied box. + + A system like this is called a Box-Ball System or BBS for short. + + It can be shown that after a sufficient number of turns, the system + evolves to a state where the consecutive numbers of occupied boxes is + invariant. In the example below, the consecutive numbers of occupied boxes + evolves to [1, 2, 3]; we shall call this the final state. + + We define the sequence {t[i]}: + + • s[0] = 290797 + • s[k+1] = s[k]^2 mod 50515093 + • t[k] = (s[k] mod 64) + 1 + + Starting from the initial configuration (t[0], t[1], …, t[10]), the final + state becomes [1, 3, 10, 24, 51, 75]. + Starting from the initial configuration (t[0], t[1], …, t[10 000 000]), + find the final state. + Give as your answer the sum of the squares of the elements of the final + state. For example, if the final state is [1, 2, 3] then 14 ( = 1^2 + 2^2 + + 3^2) is your answer. + + + p_426_baxball1.gif + p_426_baxball2.gif + Answer: b5d8157a351482da47da0512ca374007 + + +Problem 427 +=========== + + + A sequence of integers S = {s[i]} is called an n-sequence if it has n + elements and each element s[i] satisfies 1 ≤ s[i] ≤ n. Thus there are n^n + distinct n-sequences in total.For example, the sequence S = {1, 5, 5, 10, + 7, 7, 7, 2, 3, 7} is a 10-sequence. + + For any sequence S, let L(S) be the length of the longest contiguous + subsequence of S with the same value.For example, for the given sequence S + above, L(S) = 3, because of the three consecutive 7's. + + Let f(n) = ∑ L(S) for all n-sequences S. + + For example, f(3) = 45, f(7) = 1403689 and f(11) = 481496895121. + + Find f(7 500 000) mod 1 000 000 009. + + + Answer: ecb4da2c940b517c63d8d256814dd511 + + +Problem 428 +=========== + + + Let a, b and c be positive numbers. + Let W, X, Y, Z be four collinear points where |WX| = a, |XY| = b, |YZ| = c + and |WZ| = a + b + c. + Let C[in] be the circle having the diameter XY. + Let C[out] be the circle having the diameter WZ. + + The triplet (a, b, c) is called a necklace triplet if you can place k ≥ 3 + distinct circles C[1], C[2], ..., C[k] such that: + + • C[i] has no common interior points with any C[j] for 1 ≤ i, j ≤ k and + i ≠ j, + • C[i] is tangent to both C[in] and C[out] for 1 ≤ i ≤ k, + • C[i] is tangent to C[i+1] for 1 ≤ i < k, and + • C[k] is tangent to C[1]. + + For example, (5, 5, 5) and (4, 3, 21) are necklace triplets, while it can + be shown that (2, 2, 5) is not. + + Let T(n) be the number of necklace triplets (a, b, c) such that a, b and c + are positive integers, and b ≤ n.For example, T(1) = 9, T(20) = 732 and + T(3000) = 438106. + + Find T(1 000 000 000). + + + Answer: c6010c109b66b34bf3594e63eb58b446 + + +Problem 429 +=========== + + + A unitary divisor d of a number n is a divisor of n that has the property + gcd(d, n/d) = 1. + The unitary divisors of 4! = 24 are 1, 3, 8 and 24. + The sum of their squares is 1^2 + 3^2 + 8^2 + 24^2 = 650. + + Let S(n) represent the sum of the squares of the unitary divisors of n. + Thus S(4!)=650. + + Find S(100 000 000!) modulo 1 000 000 009. + + + Answer: ec4f87b0c01680e951326d9e85d2c03f + + +Problem 430 +=========== + + + N disks are placed in a row, indexed 1 to N from left to right. + Each disk has a black side and white side. Initially all disks show their + white side. + + At each turn, two, not necessarily distinct, integers A and B between 1 + and N (inclusive) are chosen uniformly at random. + All disks with an index from A to B (inclusive) are flipped. + + The following example shows the case N = 8. At the first turn A = 5 and B + = 2, and at the second turn A = 4 and B = 6. + + Let E(N, M) be the expected number of disks that show their white side + after M turns. + We can verify that E(3, 1) = 10/9, E(3, 2) = 5/3, E(10, 4) ≈ 5.157 and + E(100, 10) ≈ 51.893. + + Find E(10^10, 4000). + Give your answer rounded to 2 decimal places behind the decimal point. + + + p_430_flips.gif + Answer: 32b0825d7a110a1a220e80629c413411 + + +Problem 431 +=========== + + + Fred the farmer arranges to have a new storage silo installed on his farm + and having an obsession for all things square he is absolutely devastated + when he discovers that it is circular. Quentin, the representative from + the company that installed the silo, explains that they only manufacture + cylindrical silos, but he points out that it is resting on a square base. + Fred is not amused and insists that it is removed from his property. + + Quick thinking Quentin explains that when granular materials are delivered + from above a conical slope is formed and the natural angle made with the + horizontal is called the angle of repose. For example if the angle of + repose, $\alpha = 30$ degrees, and grain is delivered at the centre of the + silo then a perfect cone will form towards the top of the cylinder. In the + case of this silo, which has a diameter of 6m, the amount of space wasted + would be approximately 32.648388556 m^3. However, if grain is delivered at + a point on the top which has a horizontal distance of $x$ metres from the + centre then a cone with a strangely curved and sloping base is formed. He + shows Fred a picture. + + We shall let the amount of space wasted in cubic metres be given by + $V(x)$. If $x = 1.114785284$, which happens to have three squared decimal + places, then the amount of space wasted, $V(1.114785284) \approx 36$. + Given the range of possible solutions to this problem there is exactly one + other option: $V(2.511167869) \approx 49$. It would be like knowing that + the square is king of the silo, sitting in splendid glory on top of your + grain. + + Fred's eyes light up with delight at this elegant resolution, but on + closer inspection of Quentin's drawings and calculations his happiness + turns to despondency once more. Fred points out to Quentin that it's the + radius of the silo that is 6 metres, not the diameter, and the angle of + repose for his grain is 40 degrees. However, if Quentin can find a set of + solutions for this particular silo then he will be more than happy to keep + it. + + If Quick thinking Quentin is to satisfy frustratingly fussy Fred the + farmer's appetite for all things square then determine the values of $x$ + for all possible square space wastage options and calculate $\sum x$ + correct to 9 decimal places. + + + p_431_grain_silo.png + Answer: 5e5d81aa8bfaf92f68cdef0154c5c238 + + +Problem 432 +=========== + + + Let S(n,m) = ∑φ(n × i) for 1 ≤ i ≤ m. (φ is Euler's totient function) + You are given that S(510510,10^6 )= 45480596821125120. + + Find S(510510,10^11). + Give the last 9 digits of your answer. + + + Answer: e171c2872d650e47589842faa80f5707 + + +Problem 433 +=========== + + + Let E(x[0], y[0]) be the number of steps it takes to determine the + greatest common divisor of x[0] and y[0] with Euclid's algorithm. More + formally: + x[1] = y[0], y[1] = x[0] mod y[0] + x[n] = y[n-1], y[n] = x[n-1] mod y[n-1] + E(x[0], y[0]) is the smallest n such that y[n] = 0. + + We have E(1,1) = 1, E(10,6) = 3 and E(6,10) = 4. + + Define S(N) as the sum of E(x,y) for 1 ≤ x,y ≤ N. + We have S(1) = 1, S(10) = 221 and S(100) = 39826. + + Find S(5·10^6). + + + Answer: 0eeca9fa5cf25a2bfae01f1f04d6cd35 + + +Problem 434 +=========== + + + Recall that a graph is a collection of vertices and edges connecting the + vertices, and that two vertices connected by an edge are called adjacent. + Graphs can be embedded in Euclidean space by associating each vertex with + a point in the Euclidean space. + A flexible graph is an embedding of a graph where it is possible to move + one or more vertices continuously so that the distance between at least + two nonadjacent vertices is altered while the distances between each pair + of adjacent vertices is kept constant. + A rigid graph is an embedding of a graph which is not flexible. + Informally, a graph is rigid if by replacing the vertices with fully + rotating hinges and the edges with rods that are unbending and inelastic, + no parts of the graph can be moved independently from the rest of the + graph. + + The grid graphs embedded in the Euclidean plane are not rigid, as the + following animation demonstrates: + + However, one can make them rigid by adding diagonal edges to the cells. + For example, for the 2x3 grid graph, there are 19 ways to make the graph + rigid: + + Note that for the purposes of this problem, we do not consider changing + the orientation of a diagonal edge or adding both diagonal edges to a cell + as a different way of making a grid graph rigid. + + Let R(m,n) be the number of ways to make the m × n grid graph rigid. + E.g. R(2,3) = 19 and R(5,5) = 23679901 + + Define S(N) as ∑R(i,j) for 1 ≤ i, j ≤ N. + E.g. S(5) = 25021721. + Find S(100), give your answer modulo 1000000033 + + + Answer: f51d9fd41a8ce217682321a020be6fec + + +Problem 435 +=========== + + + The Fibonacci numbers {f[n], n ≥ 0} are defined recursively as f[n] = + f[n-1] + f[n-2] with base cases f[0] = 0 and f[1] = 1. + + Define the polynomials {F[n], n ≥ 0} as F[n](x) = ∑f[i]x^i for 0 ≤ i ≤ n. + + For example, F[7](x) = x + x^2 + 2x^3 + 3x^4 + 5x^5 + 8x^6 + 13x^7, and + F[7](11) = 268357683. + + Let n = 10^15. Find the sum [∑[0≤x≤100] F[n](x)] mod 1307674368000 (= + 15!). + + + Answer: 0f08231a97e872f565a085de75743a1c + + +Problem 436 +=========== + + + Julie proposes the following wager to her sister Louise. + She suggests they play a game of chance to determine who will wash the + dishes. + For this game, they shall use a generator of independent random numbers + uniformly distributed between 0 and 1. + The game starts with S = 0. + The first player, Louise, adds to S different random numbers from the + generator until S > 1 and records her last random number 'x'. + The second player, Julie, continues adding to S different random numbers + from the generator until S > 2 and records her last random number 'y'. + The player with the highest number wins and the loser washes the dishes, + i.e. if y > x the second player wins. + + For example, if the first player draws 0.62 and 0.44, the first player + turn ends since 0.62+0.44 > 1 and x = 0.44. + If the second players draws 0.1, 0.27 and 0.91, the second player turn + ends since 0.62+0.44+0.1+0.27+0.91 > 2 and y = 0.91.Since y > x, the + second player wins. + + Louise thinks about it for a second, and objects: "That's not fair". + What is the probability that the second player wins? + Give your answer rounded to 10 places behind the decimal point in the form + 0.abcdefghij + + + Answer: d797ed72189f045e8ea48aa960fec1f3 + + +Problem 437 +=========== + + + When we calculate 8^n modulo 11 for n=0 to 9 we get: 1, 8, 9, 6, 4, 10, 3, + 2, 5, 7. + As we see all possible values from 1 to 10 occur. So 8 is a primitive root + of 11. + But there is more: + If we take a closer look we see: + 1+8=9 + 8+9=17≡6 mod 11 + 9+6=15≡4 mod 11 + 6+4=10 + 4+10=14≡3 mod 11 + 10+3=13≡2 mod 11 + 3+2=5 + 2+5=7 + 5+7=12≡1 mod 11. + + So the powers of 8 mod 11 are cyclic with period 10, and 8^n + 8^n+1 ≡ + 8^n+2 (mod 11). + 8 is called a Fibonacci primitive root of 11. + Not every prime has a Fibonacci primitive root. + There are 323 primes less than 10000 with one or more Fibonacci primitive + roots and the sum of these primes is 1480491. + Find the sum of the primes less than 100,000,000 with at least one + Fibonacci primitive root. + + Answer: 98bb66462d635d8225416a644e4637b0 + + +Problem 438 +=========== + + + For an n-tuple of integers t = (a[1], ..., a[n]), let (x[1], ..., x[n]) be + the solutions of the polynomial equation x^n + a[1]x^n-1 + a[2]x^n-2 + ... + + a[n-1]x + a[n] = 0. + + Consider the following two conditions: + + • x[1], ..., x[n] are all real. + • If x[1], ..., x[n] are sorted, ⌊x[i]⌋ = i for 1 ≤ i ≤ n. (⌊·⌋: floor + function.) + + In the case of n = 4, there are 12 n-tuples of integers which satisfy both + conditions. + We define S(t) as the sum of the absolute values of the integers in t. + For n = 4 we can verify that ∑S(t) = 2087 for all n-tuples t which satisfy + both conditions. + + Find ∑S(t) for n = 7. + + + Answer: ? + + +Problem 439 +=========== + + + Let d(k) be the sum of all divisors of k. + We define the function S(N) = ∑[1≤i≤N] ∑[1≤j≤N] d(i·j). + For example, S(3) = d(1) + d(2) + d(3) + d(2) + d(4) + d(6) + d(3) + d(6) + + d(9) = 59. + + You are given that S(10^3) = 563576517282 and S(10^5) mod 10^9 = + 215766508. + Find S(10^11) mod 10^9. + + + Answer: ? + + +Problem 440 +=========== + + + We want to tile a board of length n and height 1 completely, with either 1 + × 2 blocks or 1 × 1 blocks with a single decimal digit on top: + + For example, here are some of the ways to tile a board of length n = 8: + + Let T(n) be the number of ways to tile a board of length n as described + above. + + For example, T(1) = 10 and T(2) = 101. + + Let S(L) be the triple sum ∑[a,b,c] gcd(T(c^a), T(c^b)) for 1 ≤ a, b, c ≤ + L. + For example: + S(2) = 10444 + S(3) = 1292115238446807016106539989 + S(4) mod 987 898 789 = 670616280. + + Find S(2000) mod 987 898 789. + + + Answer: ? + + +Problem 441 +=========== + + + For an integer M, we define R(M) as the sum of 1/(p·q) for all the integer + pairs p and q which satisfy all of these conditions: + + • 1 ≤ p < q ≤ M + • p + q ≥ M + • p and q are coprime. + + We also define S(N) as the sum of R(i) for 2 ≤ i ≤ N. + We can verify that S(2) = R(2) = 1/2, S(10) ≈ 6.9147 and S(100) ≈ 58.2962. + + Find S(10^7). Give your answer rounded to four decimal places. + + + Answer: 152cc265f5461c5055db95a122280416 + + +Problem 442 +=========== + + + An integer is called eleven-free if its decimal expansion does not contain + any substring representing a power of 11 except 1. + + For example, 2404 and 13431 are eleven-free, while 911 and 4121331 are + not. + + Let E(n) be the nth positive eleven-free integer. For example, E(3) = 3, + E(200) = 213 and E(500 000) = 531563. + + Find E(10^18). + + + Answer: c31bb13db787bce9a169dce600aec863 + + +Problem 443 +=========== + + + Let g(n) be a sequence defined as follows: + g(4) = 13, + g(n) = g(n-1) + gcd(n, g(n-1)) for n > 4. + + The first few values are: + + n 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ... + g(n) 13 14 16 17 18 27 28 29 30 31 32 33 34 51 54 55 60 ... + + You are given that g(1 000) = 2524 and g(1 000 000) = 2624152. + + Find g(10^15). + + + Answer: 28f9d9a9bf8fb3d606e0b711b59f42aa + + +Problem 444 +=========== + + + A group of p people decide to sit down at a round table and play a + lottery-ticket trading game. Each person starts off with a + randomly-assigned, unscratched lottery ticket. Each ticket, when + scratched, reveals a whole-pound prize ranging anywhere from £1 to £p, + with no two tickets alike. The goal of the game is for each person to + maximize his ticket winnings upon leaving the game. + + An arbitrary person is chosen to be the first player. Going around the + table, each player has only one of two options: + + 1. The player can scratch his ticket and reveal its worth to everyone at + the table. + 2. The player can trade his unscratched ticket for a previous player's + scratched ticket, and then leave the game with that ticket. The previous + player then scratches his newly-acquired ticket and reveals its worth to + everyone at the table. + + The game ends once all tickets have been scratched. All players still + remaining at the table must leave with their currently-held tickets. + + Assume that each player uses the optimal strategy for maximizing the + expected value of his ticket winnings. + + Let E(p) represent the expected number of players left at the table when + the game ends in a game consisting of p players (e.g. E(111) = 5.2912 when + rounded to 5 significant digits). + + Let S[1](N) = E(p) + Let S[k](N) = S[k-1](p) for k > 1 + + Find S[20](10^14) and write the answer in scientific notation rounded to + 10 significant digits. Use a lowercase e to separate mantissa and exponent + (e.g. S[3](100) = 5.983679014e5). + + + Answer: e6745c386ba3c0de1bf56897e453c7c8 + + +Problem 445 +=========== + + + For every integer n>1, the family of functions f[n,a,b] is defined by + f[n,a,b](x)≡ax+b mod n for a,b,x integer and 01, the family of functions f[n,a,b] is defined by + f[n,a,b](x)≡ax+b mod n for a,b,x integer and 01, the family of functions f[n,a,b] is defined by + f[n,a,b](x)≡ax+b mod n for a,b,x integer and 0 + 1). + + Here are the possible seating arrangements for N = 15: + + We see that if the first person chooses correctly, the 15 seats can seat + up to 7 people. + We can also see that the first person has 9 choices to maximize the number + of people that may be seated. + + Let f(N) be the number of choices the first person has to maximize the + number of occupants for N seats in a row. Thus, f(1) = 1, f(15) = 9, + f(20) = 6, and f(500) = 16. + + Also, ∑f(N) = 83 for 1 ≤ N ≤ 20 and ∑f(N) = 13343 for 1 ≤ N ≤ 500. + + Find ∑f(N) for 1 ≤ N ≤ 10^12. Give the last 8 digits of your answer. + + + Answer: ? + + +Problem 473 +=========== + + + Let $\varphi$ be the golden ratio: $\varphi=\frac{1+\sqrt{5}}{2}.$ + Remarkably it is possible to write every positive integer as a sum of + powers of $\varphi$ even if we require that every power of $\varphi$ is + used at most once in this sum. + Even then this representation is not unique. + We can make it unique by requiring that no powers with consecutive + exponents are used and that the representation is finite. + E.g: $2=\varphi+\varphi^{-2}$ and $3=\varphi^{2}+\varphi^{-2}$ + + To represent this sum of powers of $\varphi$ we use a string of 0's and + 1's with a point to indicate where the negative exponents start. + We call this the representation in the phigital numberbase. + So $1=1_{\varphi}$, $2=10.01_{\varphi}$, $3=100.01_{\varphi}$ and + $14=100100.001001_{\varphi}$. + The strings representing 1, 2 and 14 in the phigital number base are + palindromic, while the string representating 3 is not. + (the phigital point is not the middle character). + + The sum of the positive integers not exceeding 1000 whose phigital + representation is palindromic is 4345. + + Find the sum of the positive integers not exceeding $10^{10}$ whose + phigital representation is palindromic. + + + Answer: a4ea7a2040b6385b6d12863fd693e434 + + +Problem 474 +=========== + + + For a positive integer n and digits d, we define F(n, d) as the number of + the divisors of n whose last digits equal d. + For example, F(84, 4) = 3. Among the divisors of 84 (1, 2, 3, 4, 6, 7, 12, + 14, 21, 28, 42, 84), three of them (4, 14, 84) have the last digit 4. + + We can also verify that F(12!, 12) = 11 and F(50!, 123) = 17888. + + Find F(10^6!, 65432) modulo (10^16 + 61). + + + Answer: ? + + +Problem 475 +=========== + + + 12n musicians participate at a music festival. On the first day, they form + 3n quartets and practice all day. + + It is a disaster. At the end of the day, all musicians decide they will + never again agree to play with any member of their quartet. + + On the second day, they form 4n trios, each musician avoiding his previous + quartet partners. + + Let f(12n) be the number of ways to organize the trios amongst the 12n + musicians. + + You are given f(12) = 576 and f(24) mod 1 000 000 007 = 509089824. + + Find f(600) mod 1 000 000 007. + + + Answer: 6be2411783d9ca8e7ad174b269a85be5 + + +Problem 476 +=========== + + + Let R(a, b, c) be the maximum area covered by three non-overlapping + circles inside a triangle with edge lengths a, b and c. + + Let S(n) be the average value of R(a, b, c) over all integer triplets (a, + b, c) such that 1 ≤ a ≤ b ≤ c < a + b ≤ n + + You are given S(2) = R(1, 1, 1) ≈ 0.31998, S(5) ≈ 1.25899. + + Find S(1803) rounded to 5 decimal places behind the decimal point. + + + Answer: 4d6a99b2a0f22af561aeeb69c0126fef