# nayuki/Project-Euler-solutions

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 (* * Solution to Project Euler problem 100 * Copyright (c) Project Nayuki. All rights reserved. * * https://www.nayuki.io/page/project-euler-solutions * https://github.com/nayuki/Project-Euler-solutions *) (* * Suppose the box has b blue discs and r red discs. * The probability of taking 2 blue discs is [b / (b + r)] * [(b - 1) / (b + r - 1)], * which we want to be equal to 1/2. Rearrange the equation: * [b(b - 1)] / [(b + r)(b + r - 1)] = 1 / 2. * 2b(b - 1) = (b + r)(b + r - 1). * 2b^2 - 2b = b^2 + br - b + br + r^2 - r. * b^2 - b = r^2 + 2br - r. * b^2 - (2r + 1)b + (r - r^2) = 0. * Apply the quadratic equation to solve for b: * b = [(2r + 1) +/- sqrt((2r + 1)^2 - 4(r - r^2))] / 2 * = r + [1 +/- sqrt(8r^2 + 1)]/2 * = r + [sqrt(8r^2 + 1) + 1]/2. (Discard the minus solution because it would make b < r) * * For b to be an integer, we need sqrt(8r^2 + 1) to be odd, and also 8r^2 + 1 be a perfect square. * Assume 8y^2 + 1 = x^2 for some integer x > 0. * We can see this is in fact a Pell's equation: x^2 - 8y^2 = 1. * * Suppose we have the solution (x0, y0) such that x0 > 0 and x0 is as small as possible. * This is called the fundamental solution, and all other solutions be derived from it (proven elsewhere). * Suppose (x0, y0) and (x1, y1) are solutions. Then we have: * x0^2 - 8*y0^2 = 1. * (x0 + y0*sqrt(8))(x0 - y0*sqrt(8)) = 1. * (x1 + y1*sqrt(8))(x1 - y1*sqrt(8)) = 1. (Similarly) * Multiply them together: * [(x0 + y0*sqrt(8))(x0 - y0*sqrt(8))][(x1 + y1*sqrt(8))(x1 - y1*sqrt(8))] = 1 * 1. * [(x0 + y0*sqrt(8))(x1 + y1*sqrt(8))][(x0 - y0*sqrt(8))(x1 - y1*sqrt(8))] = 1. * [x0*x1 + x0*y1*sqrt(8) + x1*y0*sqrt(8) + 8y0*y1][x0*x1 - x0*y1*sqrt(8) - x1*y0*sqrt(8) + 8y0*y1] = 1. * [(x0*x1 + 8y0*y1) + (x0*y1 + x1*y0)*sqrt(8)][(x0*x1 + 8y0*y1) - (x0*y1 + x1*y0)*sqrt(8)] = 1. * (x0*x1 + 8y0*y1)^2 - 8*(x0*y1 + x1*y0)^2 = 1. * Therefore (x0*x1 + 8y0*y1, x0*y1 + x1*y0) is also a solution. * By inspection, the fundamental solution is (3, 1). *) (* Fundamental solution *) x0 = 3; y0 = 1; (* Current solution *) x = x0; y = y0; (* An alias for the number of red discs *) While[True, (* Check if this solution is acceptable *) sqrt = Sqrt[y^2 * 8 + 1]; blue = (sqrt + 1) / 2 + y; If[OddQ[sqrt] && blue + y > 10^12, ans = blue; Break[]]; nextx = x * x0 + y * y0 * 8; nexty = x * y0 + y * x0; x = nextx; y = nexty;] ans