HMMT 二月 2026 · 冲刺赛 · 第 19 题

HMMT February 2026 — Guts Round — Problem 19

[11] Compute the smallest positive integer n for which n has exactly 10 positive integer divisors and n + 1 has exactly 9 positive integer divisors.

解析

[11] Compute the smallest positive integer n for which n has exactly 10 positive integer divisors and n + 1 has exactly 9 positive integer divisors. Proposed by: Derek Liu Answer: 6723 √ Solution: Since n + 1 has an odd number of divisors, it must be a perfect square. Let k = n + 1 , so 9 4 n = ( k − 1)( k + 1) . For n to have 10 divisors, it must be of the form p or p q for primes p and q . 2 First assume k − 1 and k + 1 share a prime factor. As they differ by 2 , this prime must be 2 , so 2 | n . 4 9 9 Thus n = 2 q for some prime q (or 2 , but 2 + 1 is not a square). Exactly one of k − 1 and k + 1 is 3 3 divisible by 4 , so k − 1 and k + 1 must be 2 and 2 q in some order (or 2 and 2 q , which clearly differ 2 by more than 2 ). Thus k = 3 or k = 5 . Neither of these work, as k = n + 1 must have 9 divisors. 9 Thus, k − 1 and k + 1 are relatively prime, which implies their product cannot be p , so they are 4 4 p and q in some order, and q = p ± 2 . The next-smallest possible p is p = 3 , with q = 79 or 83 , 2 2 and thus k = 80 or 82 . While 80 has more than 9 divisors, 82 has exactly 9 , making our answer 2 82 − 1 = 6723 .