Yielding pandigitals

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 1^st digit, d₂ be the 2^nd 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.