HMMT 二月 2006 · TEAM1 赛 · 第 14 题

HMMT February 2006 — TEAM1 Round — Problem 14

[40] A number n is called bummed out if there is exactly one ordered pair of positive integers ( x, y ) such that 2 2 b x /y c + b y /x c = n. Find all bummed out numbers.

解析

[40] A number n is called bummed out if there is exactly one ordered pair of positive integers ( x, y ) such that 2 2 b x /y c + b y /x c = n. Find all bummed out numbers. Answer: 2, 6, 8, 10 Solution: Suppose n is bummed out. If ( a, b ) is one solution for ( x, y ) to the given 2 2 equation b x /y c + b y /x c = n, then ( b, a ) is another, so the unique solution ( a, b ) better have the property that a = b and n = 2 a ≥ 2. In particular, n is an even positive integer. Now, if n = 2 a ≥ 12, then setting x = a − 1 ≥ 5, y = a + 1 ≥ 7, we have ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ ⌊ ⌋ 2 2 x y 4 4

= a − 3 + + a + 3 + = 2 a = n, y x a + 1 a − 1 so n cannot be bummed out. 2 2 Moreover, b 1 / 2 c + b 2 / 1 c = 4, so 4 is not bummed out. The only possibilities left are 2, 6, 8, and 10. To check these, note that ⌊ ⌋ ⌊ ⌋ 2 2 2 2 2 x y x x y 3 x √ n = + > − 2 + + + ≥ − 2 + 3 y x 2 y 2 y x 4 so √ 3 4 x < ( n + 2) < . 53( n + 2) , 3 and similarly for y . So we only have to check x, y ≤ b . 53(10 + 2) c = 6: x \ y 1 2 3 4 5 6 1 2 4 9 16 25 36 2 4 4 5 9 12 18 3 9 5 6 7 9 13 4 16 9 7 8 9 11 5 25 12 9 9 10 11 6 36 18 13 11 11 12