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   Pandigital Permutation and Divisibility Puzzle
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   Author  Topic: Pandigital Permutation and Divisibility Puzzle  (Read 1385 times)
K Sengupta
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Pandigital Permutation and Divisibility Puzzle  
« on: Mar 26th, 2010, 11:19am »
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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.  
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towr
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Re: Pandigital Permutation and Divisibility Puzzle  
« Reply #1 on: Mar 26th, 2010, 3:52pm »
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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

 
« Last Edit: Mar 26th, 2010, 3:54pm by towr » IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
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