Author |
Topic: Pandigital Permutation and Divisibility Puzzle (Read 1385 times) |
|
K Sengupta
Senior Riddler
   

Gender: 
Posts: 371
|
 |
Pandigital Permutation and Divisibility Puzzle
« on: Mar 26th, 2010, 11:19am » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
towr
wu::riddles Moderator Uberpuzzler
    
 Some people are average, some are just mean.
Gender: 
Posts: 13730
|
 |
Re: Pandigital Permutation and Divisibility Puzzle
« Reply #1 on: Mar 26th, 2010, 3:52pm » |
Quote Modify
|
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
|
|
|
|