PUMaC 2014 · 代数（B 组） · 第 2 题
PUMaC 2014 — Algebra (Division B) — Problem 2
题目详情
- [ 3 ] f is a function whose domain is the set of nonnegative integers and whose range is contained in the set of nonnegative integers. f satisfies the condition that f ( f ( n )) + f ( n ) = 2 n + 3 for all nonnegative integers n . Find f (2014).
解析
- [ 3 ] f is a function whose domain is the set of nonnegative integers and whose range is contained in the set of nonnegative integers. f satisfies the condition that f ( f ( n )) + f ( n ) = 2 n + 3 for all nonnegative integers n . Find f (2014). Solution: If f ( m ) = f ( n ), then left-hand-sides must be equal, so right-hand-sides must be equal, so m = n . So f is injective. Let f (0) = x . Then f ( x )+ x = 3. If x = 0, then 0+0 = 3, contradiction. If x = 2, then we can get f (1) = 6, and substituting n = 1 gets us f (6) = − 1 < 0, contradiction. So x = 1. This means by induction, f ( n ) = n + 1. So f (2014) = 2015 .