HMMT 二月 2026 · 冲刺赛 · 第 11 题

HMMT February 2026 — Guts Round — Problem 11

[7] Compute the number of ordered pairs ( a, b ) of positive integers such that lcm( a, b ) + gcd( a, b ) = 2026 .

解析

[7] Compute the number of ordered pairs ( a, b ) of positive integers such that lcm( a, b )+gcd( a, b ) = 2026 . Proposed by: Marin Hristov Hristov Answer: 13 Solution: First, recall the identity gcd( a, b ) · lcm( a, b ) = ab . By factoring out gcd( a, b ) , we can write ′ ′ ′ ′ ′ ′ a = ka and b = kb , with gcd( a, b ) = k and gcd( a , b ) = 1 . In particular lcm( a, b ) = ka b . ′ ′ ′ ′ The equation now rewrites to finding triples of positive integers ( k, a , b ) with gcd( a , b ) = 1 , such that ′ ′ ka b + k = 2026 ′ ′ = ⇒ k ( a b + 1) = 2026 . In particular, k is a divisor of 2026 , so we do casework on k . ©2026 HMMT ′ ′ ′ ′ • If k = 1 , then a b = 2025 . Since a and b are coprime, every prime factor p dividing 2025 must ′ ′ 4 2 go to exactly one of a or b . This gives 2 choices for every prime. For 2025 = 3 · 5 , this gives 2 4 2 4 2 2 4 4 2 2 = 4 options: (1 , 3 · 5 ) , (3 , 5 ) , (5 , 3 ) , (3 · 5 , 1) . ′ ′ 2 • If k = 2 , then a b = 1012 = 2 · 11 · 23 . By the same argument presented above, we have a total 3 of 2 = 8 options. ′ ′ • If k = 1013 , then a b = 1 , and there is only one option. ′ ′ • If k = 2026 , then a b = 0 , which is impossible. In total, there are 4 + 8 + 1 + 0 = 13 pairs ( a, b ) .