HMMT 十一月 2025 · 冲刺赛 · 第 19 题

HMMT November 2025 — Guts Round — Problem 19

[11] Compute the number of ordered triples of positive integers ( a, b, c ) such that b is a divisor of 2025 a b a and + = . b c c

解析

[11] Compute the number of ordered triples of positive integers ( a, b, c ) such that b is a divisor of 2025 a b a and + = . b c c © 2025 HMMT Proposed by: Derek Liu, Pitchayut Saengrungkongka Answer: 105 2 2 Solution: Multiplying both sides by bc , we get ac + b = ab , so b = a ( b − c ). Note that it is necessary 2 2 for d = b − c to be a divisor of b strictly less than b . Conversely, for any b and d < b dividing b , there 2 exists exactly one such solution ( a, b, c ), namely, ( b /d, b, b − d ). It remains to sum up the number of 2 divisors of b less than b for all divisors b of 2025. 2 2 The divisors of b other than b can be paired up with each pair multiplying to b . Each pair contains 2 one divisor greater than b and one divisor less than b , so if b has s divisors, exactly ( s − 1) / 2 of them x y are strictly less than b . Any divisor b | 2025 is of the form 3 5 for nonnegative integers x ≤ 4 and 2 y ≤ 2, in which case b has (2 x + 1)(2 y + 1) divisors. Thus, the answer is P P 4 2 4 2 (2 x + 1)(2 y + 1) − 15 X X x =0 y =0 (2 x + 1)(2 y + 1) − 1 = 2 2 x =0 y =0 (1 + 3 + 5 + 7 + 9)(1 + 3 + 5) − 15 = 2 = 105 . 2 τ ( n ) − τ ( n ) Remark. If 2025 is replaced by n , the answer is , where τ ( n ) is the number of divisors of n . 2 This follows from the identity X 2 2 τ ( d ) = τ ( n ) , d | n which can be easily shown by noting that both sides are multiplicative, so it suffices to check this at prime powers, which is easy.