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        riddles >> easy >> Pandigital Permutation and Divisibility Puzzle
        
(Message started by: K Sengupta on Mar 26th, 2010, 11:19am)
Title: Pandigital Permutation and Divisibility Puzzle
Post by K Sengupta on Mar 26th, 2010, 11:19am
40320 different 8- digit base ten positive integers X1, X2, ....... X40320 are such that each Xi is formed by using each of the digits from 1 to 8 exactly once.

Determine the remainder when X1 + X2+ .......+ X40320 is divided by 13.
Title: Re: Pandigital Permutation and Divisibility Puzzle
Post by towr on Mar 26th, 2010, 3:52pm
[hide]Every digit occurs in every position 5040 times
So we have 11111111 * 5040 * (8+7+6+5+4+3+2+1)
So the remainder is 2
[/hide]


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