nayuki/Project-Euler-solutions

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 {- - Solution to Project Euler problem 129 - Copyright (c) Project Nayuki. All rights reserved. - - https://www.nayuki.io/page/project-euler-solutions - https://github.com/nayuki/Project-Euler-solutions -} {- - Let n >= 1 be arbitrary but assume that it is coprime with 10. - We want to find the smallest k such that R(k) = 0 mod n, and we'll show that 1 <= k <= n. - - Let "the sequence" of n values be (R(1) mod n, R(2) mod n, R(3) mod n, ..., R(n) mod n). - For the sake of contradiction, assume that none of the values in the sequence are 0. - - Each number in the sequence is an integer in the range [1, n). - The range has n - 1 elements, but there are n elements in the sequence. - Hence by the pigeonhole principle, there exist two distinct indexes - in the sequence where the elements have the same value. - - Suppose the two distinct indexes (1-based) are i and j. - So the two values in question are R(i) mod n and R(j) mod n. - Suppose WLOG that j > i. Then clearly R(j) - R(i) = 0 mod n, - and so R(j) - R(i) = 1...10...0 = R(j - i) * 10^i = 0 mod n. - - Since 10 is coprime with n, 10 (and its powers) are invertible modulo n. - Multiply everything in the equation by 10^-i, and we get R(j - i) = 1...1 = 0 mod n. - - We know 1 <= j - i <= n - 1. Then R(i - j) mod n, which is 0, is in the sequence. - This contradicts our assumption that none of (R(1), R(2), ... R(n)) is 0 mod n. - - Therefore if we want to find an n whose solution k is such that - k > 1000000, then we need to have n > 1000000. -} limit = (10^6) :: Int main = putStrLn (show ans) cond n = (leastDivisibleRepunit n) > limit ans = head (filter cond [limit..]) -- Computes the smallest k such that R(k) is divisible by n. leastDivisibleRepunit :: Int -> Int leastDivisibleRepunit n | (mod n 2) == 0 || (mod n 5) == 0 = 0 | otherwise = let func 0 _ k = k func s p k = func (mod (s + p * 10) n) (mod (p * 10) n) (k + 1) in func 1 1 1