HMMT 二月 2025 · ALGNT 赛 · 第 1 题

HMMT February 2025 — ALGNT Round — Problem 1

Compute the sum of the positive divisors (including 1) of 9! that have units digit 1. √

解析

Compute the sum of the positive divisors (including 1) of 9! that have units digit 1. Proposed by: Jackson Dryg Answer: 103 7 4 Solution: The prime factorization of 9! is 2 · 3 · 5 · 7. Every divisor of 9! has prime factorization a b c d 2 · 3 · 5 · 7 , where 0 ≤ a ≤ 7, 0 ≤ b ≤ 4, 0 ≤ c ≤ 1, and 0 ≤ d ≤ 1. If the divisor has units digit 1, it cannot be divisible by 2 or 5, so a = c = 0. Now take cases on the value of d : b • If d = 0, then the divisor is 3 for some 0 ≤ b ≤ 4. The possible divisors are 1, 3, 9, 27, and 81, of which 1 and 81 work. b • If d = 1, then the divisor is 3 · 7 for some 0 ≤ b ≤ 4. The possible divisors are then 7, 3 · 7, 9 · 7, 27 · 7, and 81 · 7. Of these, only 3 · 7 = 21 works. The answer is 1 + 21 + 81 = 103 . √