PUMaC 2012 · 代数（A 组） · 第 7 题

PUMaC 2012 — Algebra (Division A) — Problem 7

[ 7 ] Let a be a sequence such that a = 1 and a = b a + a + c , where b x c denotes the n 1 n +1 n n 2 greatest integer less than or equal to x . What are the last four digits of a ? 2012 k k +1

解析

[ 7 ] Let a be a sequence such that a = 1 and a = b a + a + c , where b x c denotes the n 1 n +1 n n 2 greatest integer less than or equal to x . What are the last four digits of a ? 2012 Solution: Computing some particular cases suggests that the function f is defined by the fol- lowing: n n + 1 a = 1 + b cb c , n 2 2 for all natural numbers n . We will show this hypothesis is true by induction. We assume it’s true for n and prove it for n + 1. 2 If n = 2 m , then a = 1 + m . Thus, n √ √ 1 1 2 2 a = b a + a + c = b m + 1 + m + 1 + c n +1 n n 2 2 n + 1 n + 2 2 = m + 1 + m = 1 + m ( m + 1) = 1 + b cb c 2 2 4 If n = 2 m + 1, then a = 1 + m ( m + 1), thus n √ 1 a = b 1 + m ( m + 1) + 1 + m ( m + 1) + c = 1 + m ( m + 1) + m + 1 n +1 2 n + 1 n + 2 2 = 1 + ( m + 1) = 1 + b cb c 2 2 The induction is complete, so that is the solution to the equation. In conclusion, a = 2012 2012 2013 1 + b cb c = 1012037, so the answer is 2037. 2 2 Answer: 2037 k k +1