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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.
*)
lim = 10^6;
A[n_] := If[GCD[n, 10] != 1, 0,
Block[{k = 1, s = 1, p = 1},
While[Mod[s, n] != 0,
k++;
p = Mod[p * 10, n];
s = Mod[s + p, n]];
k]]
For[n = lim, A[n] <= lim, n++]
n