HMMT 二月 1998 · 代数 · 第 6 题
HMMT February 1998 — Algebra — Problem 6
题目详情
英文原题
- How many pairs of positive integers ( , a b ) with a ≤ b satisfy + = ?
a b 6
4 2
解析
英文解析
- Problem: How many pairs of positive integers ( a, b ) with ≤ b satisfy + = ?
a b 6
1 1 1 a + b 1
Solution: + = ⇒ = ⇒ ab = 6 a +6 b ⇒ ab − 6 a − 6 b = 0. Factoring yields ( a − b )( b − 6) − 36 = 0.
a b 6 ab 6
Then ( a − 6)( b − 6) = 36. Because a and b are positive integers, only the factor pairs of 36 are possible valuesof a − 6 and b − 6. The possible pairs are:
a − 6 = 1 , b − 6 = 36
a − 6 = 2 , b − 6 = 18
a − 6 = 3 , b − 6 = 12
a − 6 = 4 , b − 6 = 9
a − 6 = 6 , b − 6 = 6
Because a ≤ b , the symmetric cases, such as a − 6 = 12 , b − 6 = 3 are not applicable. Then there are 5
possible pairs.
4 2