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wu :: forums    riddles    easy (Moderators: Eigenray, ThudnBlunder, Icarus, towr, SMQ, william wu, Grimbal)    Pandigital Permutation and Divisibility Puzzle « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    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 Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------=> easy   - medium   - hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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