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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