PUMaC 2014 · 代数（A 组） · 第 6 题

PUMaC 2014 — Algebra (Division A) — Problem 6

[ 6 ] Given that x = , x = 25 , x = 11, it follows that = for some positive n +2 0 1 n 14 x 2 q n n =0 integers p, q with GCD ( p, q ) = 1. Find p + q . 3 3 3

解析

[ 6 ] Given that x = , x = 25 , x = 11, it follows that = for some positive n +2 0 1 n 14 x 2 q n n =0 integers p, q with GCD ( p, q ) = 1. Find p + q . Solution: 2 2 20 x 20 100 n +2 We have that x = = = . So we have that x = x and so: n +3 n +6 n 2 14 x 14 x 49 x n +1 n n ∞ ∞ ∞ ∑ ∑ ∑ x x x 3 n 6 n 6 n +3 = + n n n 2 4 2 · 4 n =0 n =0 n =0 ∞ ∞ ∑ ∑ 1 100 1 = x + 0 n n 4 98 x 4 0 n =0 n =0 4 x 200 0 = + 3 147 x 0 100 8 1636 = + = 3 147 49 So the answer is 1636 + 49 = 1685 . 3 3 3