nayuki/Project-Euler-solutions

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 (* * Solution to Project Euler problem 120 * Copyright (c) Project Nayuki. All rights reserved. * * https://www.nayuki.io/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions *) (* * For a given a, what is the n that maximizes the remainder, and what is the value of this remainder? * * Let's simplify one term, mod a^2: * (a+1)^n = 1^n + (n choose 1) a 1^(n-1) + (n choose 2) a^2 1^(n-2) + ... (by the binomial theorem) * = 1 + an + 0. (remaining addends are 0 because they have a to the power of 2 or more, mod a^2) * Similarly for the other term, mod a^2: * (a-1)^n = (-1)^n + (n choose 1) a (-1)^(n-1) + ... * = (-1)^(n-1) (-1 + an + 0) * = if n is even then (1 - an) else (an - 1). * Therefore, adding the two terms: * (a+1)^n + (a-1)^n * = if n is even then 2 else 2an. * We can always make 2an >= 2 by taking n=1, for example. So we can disregard the "n is even" case. * Maximizing 2an mod a^2 for n is the same as maximizing 2n mod a for n. * If a is even, then the maximum achievable value is a - 2 by setting n = a/2 - 1. * Else a is odd, then the maximum achievable value is a - 1 by setting n = (a - 1) / 2. * In conclusion, if a is even, the maximum remainder is a(a-2); otherwise a is odd, the maximum remainder is a(a-1). *) Sum[If[Mod[a, 2] == 0, a(a - 2), a(a - 1)], {a, 3, 1000}]