PUMaC 2016 · 代数（B 组） · 第 7 题
PUMaC 2016 — Algebra (Division B) — Problem 7
题目详情
- Define a sequence a as follows: a = 181 and for i ≥ 2, a = a − 1 if a is odd and i 1 i i − 1 i − 1 a = a / 2 if a is even. Find the least i such that a = 0. i i − 1 i − 1 i 2
解析
- By a computation, a = 2 − 1. If a = 2 − 1, then a = (2 − 1)2 , so a = 2 − 1. 5 i i +1 i + k +2 Eventually we get a = 0. Thus the answer is 105 . 105 Problem written by Zhuo Qun Song. ∏ 2