HMMT 十一月 2019 · GEN 赛 · 第 6 题
HMMT November 2019 — GEN Round — Problem 6
题目详情
- Find all ordered pairs ( a, b ) of positive integers such that 2 a + 1 divides 3 b − 1 and 2 b + 1 divides 3 a − 1.
解析
- Find all ordered pairs ( a, b ) of positive integers such that 2 a + 1 divides 3 b − 1 and 2 b + 1 divides 3 a − 1. Proposed by: Milan Haiman Answer: (2 , 2) , (12 , 17) , (17 , 12) This is equivalent to the existence of nonnegative integers c and d such that 3 b − 1 = c (2 a + 1) and 3 a − 1 = d (2 b + 1). Then (3 b − 1)(3 a − 1) 3 a − 1 3 b − 1 3 3 cd = = · < · = 2 . 25 . (2 a + 1)(2 b + 1) 2 a + 1 2 b + 1 2 2 1 1 Neither c nor d can equal 0 since that would give a = or b = , so cd ≤ 2 . 25 implies ( c, d ) ∈ 3 3 { (1 , 1) , (2 , 1) , (1 , 2) } . Substituting ( c, d ) back in gives three systems of equations and the three solutions: (2 , 2) , (12 , 17) , (17 , 12).