HMMT 二月 1998 · 代数 · 第 6 题

HMMT February 1998 — Algebra — Problem 6

How many pairs of positive integers ( , a b ) with a ≤ b satisfy + = ? a b 6 4 2

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 values of 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