PUMaC 2017 · 代数（A 组） · 第 4 题

PUMaC 2017 — Algebra (Division A) — Problem 4

Let a , a , . . . be a sequence of positive real numbers such that a = 11 a − n for all n > 1. 1 2 n n − 1 p The smallest possible value of a can be written as , where p and q are relatively prime 1 q positive integers. Find p + q ? 2

解析

Let b = a − a . Then, we have n n +1 n b = 10 a − ( n + 1) n n = 10(11 a − n ) − ( n + 1) n − 1 = 11(10 a − n ) − 1 n − 1 = 11 b − 1 . n − 1 1 Therefore, if b < , then the sequence b , b , . . . is decreasing, and in fact becomes arbitrarily 1 1 2 10 small, which means that the sequence a , a , . . . becomes arbitrarily small as well. Therefore, 1 2 1 1 21 b ≥ , so we have a − a ≥ , or equivalently a ≥ . Since the sequence a , a , . . . is 1 2 1 1 1 2 10 10 100 21 1 increasing if a = (because b = for all n ), our answer is 121 . 1 n 100 10 Problem written by Eric Neyman 3 3 2 2 2 2