HMMT 十一月 2021 · 冲刺赛 · 第 11 题

HMMT November 2021 — Guts Round — Problem 11

[8] Let n be a positive integer. Given that n has 861 positive divisors, find n .

解析

[8] Let n be a positive integer. Given that n has 861 positive divisors, find n . Proposed by: Sean Li Answer: 20 α α α 1 2 k Solution: If n = p p . . . p , we must have ( nα + 1)( nα + 1) . . . ( nα + 1) = 861 = 3 · 7 · 41. If 1 2 k 1 2 k 2 k = 1, we have n | 860, and the only prime powers dividing 860 are 2, 2 , 5, and 43, which are not solutions. Note that if nα + 1 = 3 or nα + 1 = 7 for some i , then n is either 1, 2, 3, or 6, which i i are not solutions. Therefore, we must have nα + 1 = 3 · 7 for some i . The only divisor of 20 that is i n/ 20 divisible by p for some prime p is 20, and it is indeed the solution. i i