2025 HMMT November General 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.

Answers are checked by value, so any equivalent form is accepted: 1/2, \frac{1}{2} and 0.5 all count as the same answer.

Full credit goes to HMMT for authoring these problems. This problem is from the November 2025 contest; the official solution is available on the HMMT archive. HMMT is not affiliated with or endorsing TopsOJ in any way.