PUMaC 2017 · 代数（B 组） · 第 6 题

PUMaC 2017 — Algebra (Division B) — Problem 6

Let the sequence a , a , . . . be defined recursively as follows: a = 11 a − n . If all terms of 1 2 n n − 1 m the sequence are positive, the smallest possible value of a can be written as , where m and 1 n n are relatively prime positive integers. What is m + n ? 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 goes to −∞ , 1 1 2 10 which means that the sequence a , a , . . . does the same, and in particular becomes negative. 1 2 1 1 21 Therefore, b ≥ , so we have a − a ≥ , or equivalently a ≥ . Since the sequence 1 2 1 1 10 10 100 21 1 a , a , . . . is increasing if a = (because b = for all n ), our answer is 121 . 1 2 1 n 100 10 Problem written by Eric Neyman 3 3 2 2 2 2