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038.py
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038.py
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#!/usr/bin/python
# -*- coding: utf-8 -*-
# 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?
def find_largest_pandigital():
digits = "123456789"
i = 1
largest = 0
for i in xrange(1,100000):
r = []
l = 0
for j in xrange(1,10):
s = str(i*j)
r.append(s)
l += len(s)
if l >= 9:
break
if j > 1 and l == 9 and ''.join(sorted(''.join(r))) == digits:
n = int(''.join(r))
print i, j, r
if n > largest:
largest = n
return largest
if __name__ == '__main__':
largest = find_largest_pandigital()
print "Largest:", largest