HMMT 十一月 2025 · GEN 赛 · 第 7 题

HMMT November 2025 — GEN Round — Problem 7

A positive integer n is imbalanced if strictly more than 99 percent of the positive divisors of n are strictly less than 1 percent of n . Given that M is an imbalanced multiple of 2000, compute the minimum possible number of positive divisors of M .

解析

A positive integer n is imbalanced if strictly more than 99 percent of the positive divisors of n are strictly less than 1 percent of n . Given that M is an imbalanced multiple of 2000, compute the minimum possible number of positive divisors of M . Proposed by: Srinivas Arun Answer: 1305 M Solution: Note that M has at least 13 divisors which are at least , namely 100 M M M M M , , , , , 1 2 4 8 16 M M M M M , , , , , 5 10 20 40 80 M M M , , . 25 50 100 M Therefore, since more than 99 percent of M ’s divisors are less than , we know M must have at least 100 4 260 13 · 100 + 1 = 1301 divisors in total. Consider M = 2 · 5 . Note that M has (4 + 1)(260 + 1) = 1305 M divisors and the only divisors of M that are at least are the 13 divisors listed above. We now show 100 that the number of positive divisors of M can’t be between 1301 and 1304 inclusive. M Observe that 32 ∤ M . Indeed, if 32 | M , then would be a divisor of M , meaning there would be at 32 M least 14 divisors that are at least . This would mean M has at least 14 · 100 + 1 = 1401 divisors, 100 which is not minimal due to the construction above achieving 1305 divisors. As 2000 | M and 32 ∤ M , we get that ν ( M ) = 4. In particular, this means the number of divisors of M is divisible by 4 + 1 = 5, 2 so M can’t have between 1301 and 1304 divisors inclusive. This concludes the solution.