# 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 # import eulerlib # 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). def compute(): # 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 = eulerlib.sqrt(y**2 * 8 + 1) if sqrt % 2 == 1: # Is odd blue = (sqrt + 1) // 2 + y if blue + y > 10**12: return str(blue) # Create the next bigger solution nextx = x * x0 + y * y0 * 8 nexty = x * y0 + y * x0 x = nextx y = nexty if __name__ == "__main__": print(compute())