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Nine, One, MinusOne solution in Speedy category for Numbers Factory by hrvoje
def factorize(n, t=[]):
for i in range(9,1,-1):
if n % i == 0:
n, t = n/i, t+[str(i)]
return factorize(n, t)
return (n, t)
def checkio(number):
residue, factors = factorize(number, [])
return int(''.join(sorted(factors))) if residue==1 else 0
July 9, 2015
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