HMMT 二月 2017 · ALGNT 赛 · 第 8 题

HMMT February 2017 — ALGNT Round — Problem 8

Consider all ordered pairs of integers ( a, b ) such that 1 ≤ a ≤ b ≤ 100 and ( a + b )( a + b + 1) ab is an integer. Among these pairs, find the one with largest value of b . If multiple pairs have this maximal value of b , choose the one with largest a . For example choose (3 , 85) over (2 , 85) over (4 , 84). Note that your answer should be an ordered pair.

解析

Consider all ordered pairs of integers ( a, b ) such that 1 ≤ a ≤ b ≤ 100 and ( a + b )( a + b + 1) ab is an integer. Among these pairs, find the one with largest value of b . If multiple pairs have this maximal value of b , choose the one with largest a . For example choose (3 , 85) over (2 , 85) over (4 , 84). Note that your answer should be an ordered pair. Proposed by: Alexander Katz Answer: (35,90) 2 2 ( a + b )( a + b +1) a + b + a + b Firstly note that = 2+ . Let c be this fraction so that ( a + b )( a + b +1) = ab ( c +2) ab ab for some integers a, b, c . Suppose ( a, b ) with a ≥ b is a solution for some c . Consider the quadratic 2 2 x − ( bc − 1) x + b + b = 0 It has one root a , and the other root is therefore bc − a − 1. Furthermore the other root can also be 2 2 b + b b + b expressed as ≤ = b , so that 0 < bc − a − 1 ≤ b . In particular, ( b, bc − a − 1) is a solution as a b +1 well. 2 Thus all solutions ( a, b ) reduce to a solution where a = b , at which point c = 2 + . Since a, c are a positive integers we thus have a = 1 , 2, and so c = 3 , 4 . Through this jumping process, we iteratively find the solutions for c = 3: (2 , 2) → (2 , 3) → (3 , 6) → (6 , 14) → (14 , 35) → (35 , 90) and c = 4: (1 , 2) → (2 , 6) → (6 , 21) → (21 , 77) so that the desired pair is (35 , 90) .