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