PUMaC 2014 · 代数（A 组） · 第 3 题

PUMaC 2014 — Algebra (Division A) — Problem 3

[ 4 ]A function f has its domain equal to the set of integers 0, 1, ..., 11, and f ( n ) ≥ 0 for all such n , and f satisfies f (0) = 0 f (6) = 1 f ( x )+ f ( y ) If x ≥ 0 , y ≥ 0 , andx + y ≤ 11, then f ( x + y ) = 1 − f ( x ) f ( y ) 2 2 Find f (2) + f (10) . (⌊ ⌋) (⌈ ⌉) n n

解析

[ 4 ]A function f has its domain equal to the set of integers 0, 1, ..., 11, and f ( n ) ≥ 0 for all such n , and f satisfies f (0) = 0 1 f (6) = 1 f ( x )+ f ( y ) If x ≥ 0 , y ≥ 0 , andx + y ≤ 11, then f ( x + y ) = 1 − f ( x ) f ( y ) 2 2 Find f (2) + f (10) . Solution: Represent f (4) as a function of f (2) using the identity. Represent f (6) as a function of f (4) and f (2), and substitute in the first result to get f (6) as a function of f (2). This gives us that √ f (2) = 2 − 3. Then, using f (2) and f (6) to get f (8), and then using f (2) and f (8) to get √ √ √ 2 2 2 2 f (10) = 2 + 3, we get that f (2) + f (10) = (2 − 3) + (2 + 3) = 14 . (⌊ ⌋) (⌈ ⌉) n n