PUMaC 2010 · 代数（B 组） · 第 7 题

PUMaC 2010 — Algebra (Division B) — Problem 7

Let f be a function such that f ( x ) + f ( x + 1) = 2 and f (0) = 2010. Find the last two digits of f (2010). √ ◦ ◦ ◦ ◦ 50

解析

Let f be a function such that f ( x ) + f ( x + 1) = 2 and f (0) = 2010. Find the last two digits of f (2010). Solution: 51. We have the sequence of equations x f ( x ) + f ( x + 1) = 2 x +1 x f ( x + 1) + f ( x + 2) = 2 = 2 · 2 · · · x + n − 1 n − 1 x f ( x + n − 1) + f ( x + n ) = 2 = 2 · 2 . Adding and subtracting alternate lines, we get a telescoping sum: ( ) n − 1 n − 1 n ∑ ∑ 1 − ( − 2) n +1 x k k x k x f ( x ) + ( − 1) f ( x + n ) = 2 2 ( − 1) = 2 ( − 2) = 2 . 3 k =0 k =0 Plug in x = 0 and n = 2010, so 2010 2 − 1 f (2010) = + 2010 . 3 2010 20 The last two digits of 2 are 24 (using Euler’s theorem with n = 25, we have 2 = 1 mod 2000 2010 10 2010 25, so 2 = 1 mod 25, so 2 = 2 mod 25, so 2 = 24 mod 100). Therefore the 2010 expression (2 − 1) / 3 has last digits 41, so overall the last two digits are 51. √ ◦ ◦ ◦ ◦ 50