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. This is a thing that I first encountered while solving Project Euler problems and it keeps on turning up – most recently (for me) in 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 d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following:
d2d3d4=406 is divisible by 2
d3d4d5=063 is divisible by 3
d4d5d6=635 is divisible by 5
d5d6d7=357 is divisible by 7
d6d7d8=572 is divisible by 11
d7d8d9=728 is divisible by 13
d8d9d10=289 is divisible by 17
Find the sum of all 0 to 9 pandigital numbers with this property.
My eventual approach was (mainly) to brute-force my way through all of the possible pandigitals and testing each against the divisible criteria listed. This, however, raised its own problem: How do I quickly generate a list of all nine digit pandigitals?
The answer turned out to be a rather elegant application of the itertools module.
#!/usr/bin/env python import itertools def yield_pandigitals(length): """ Yields all 0 to length pandigitals length is an integer between 1 and 9 """ for pandigital in list(itertools.permutations(range(length + 1))): yield int(''.join(map(str, pandigital)))
Well, I was rather pleased with it.