PUMaC 2019 · 代数（B 组） · 第 3 题

PUMaC 2019 — Algebra (Division B) — Problem 3

Let x and y be positive real numbers that satisfy (log x ) + (log y ) = log( x ) + log( y ). 2 Compute the maximum possible value of (log xy ) . 2

解析

For all n > 0, we have f ( n ) = f ( n − 1) f ( b ) + 2 n − f ( b ) Find the sum of all possible values of f ( b + 100). Answer: 10201 Proposed by: Rahul Saha We’ll focus on condition 2. By AM-GM (or squaring and rearranging), √ √ 2 f ( a ) f ( b ) ≤ f ( a ) + f ( b ) ≤ 2 f ( a ) which implies f ( b ) ≤ 1. Since f ( b ) is as integer we must have f ( b ) = 0 , 1. Substituting in condition (3) gives us the possibilities f ( n ) = 2 n for n > 0 (for f ( b ) = 0) and 2 a recursion which easily amounts to f ( n ) = n + 1. For the first function, since f ( n ) = 2 n for n > 0 and f ( b ) = 0, we must necessarily have b = 0. So f ( b + 100) = f (100) = 200. 2 In the second case, similarly b = 0 and f ( b + 100) = 100 + 1 = 10001. Summing gives us the answer 10201 . Quick check: In both cases, if we have a = b = 0, condition 2 holds. Condition 1 works for both functions too. So our functions do satisfy the problem’s statement. 2 πi 2 πi 2017 2019