HMMT 二月 2006 · TEAM1 赛 · 第 14 题
HMMT February 2006 — TEAM1 Round — Problem 14
题目详情
英文原题
- [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
英文解析
- [40] A number n is called bummed out if there is exactly one ordered pair of positiveintegers ( 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 ) betterhave the property that a = b and n = 2 a ≥ 2. In particular, n is an even positiveinteger.
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 +
y x 2 y 2 y x 3
so 4
√
x < ( n + 2) < . 53( n + 2) ,43
and similarly for y . So we only have to check x, y ≤ b . 53(10 + 2) c = 6:3
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