HMMT 十一月 2025 · 冲刺赛 · 第 10 题

HMMT November 2025 — Guts Round — Problem 10

[8] Compute the number of positive divisors of 10 that leave a remainder of 1 when divided by 9.

解析

[8] Compute the number of positive divisors of 10 that leave a remainder of 1 when divided by 9. Proposed by: Rohan Bodke Answer: 75 20 a b Solution: Note that all divisors of 10 can be written as 2 5 , where a and b are nonnegative integers with 0 ≤ a ≤ 20, 0 ≤ b ≤ 20. For this divisor to leave a remainder of 1 when divided by 9, we need a b 2 5 ≡ 1 (mod 9) a − 1 b = ⇒ 2 ≡ (5 ) (mod 9) a b = ⇒ 2 ≡ 2 (mod 9) a − b = ⇒ 2 ≡ 1 (mod 9) = ⇒ a ≡ b (mod 6) . © 2025 HMMT For a ≡ b ≡ 0 , 1 , 2 (mod 6), there are four possible values of a and four possible values of b . For a ≡ b ≡ 3 , 4 , 5 (mod 6), there are three possible values of a and three possible values of b . Thus, the answer is 3(4 · 4 + 3 · 3) = 75 .