Problem of the Week (POTW)

Over the summer, a problem will be posted each week for students to work on and submit their solutions. These problems are curated by the HMMT Problems staff — the same people who write the HMMT November and February tests — and are intended to increase in difficulty every week, resetting with the start of a new month. After the submission deadline closes, the problem writers will review submissions and post one of their favorites alongside the next POTW. Discussion of the POTW through any form of media is closed until the submission deadline passes.

This Week's Problem (November 29 to December 03)

This week's problem will be released on Monday, November 29 at 12 PM EDT (UTC-0400).

Submission Form

The link to the submission form will be available once the problem has been released.


The current POTW and past week’s solution will be posted on Monday, November 29 at 12 PM EDT (UTC-0400).
The deadline for submission is Friday, December 03 at 06 PM EDT (UTC-0400).

Last Week's Problem

In the spirit of this weekend’s theme round, here is another interesting problem stated in terms of its own answer!
Let $n$ be the answer to this question. Find the number of positive divisors of the integer \[\left(\frac{n-25}{75}\right)^{ \frac{n-25}{75}}\]

The answer to this week’s problem was $\boxed{1825}$. The solutions we received all used essentially the same method, which we will summarize below:
Let $a = \frac{n-25}{75}$. Then, we know that $a^a$ has $75a + 25$ divisors. Furthermore, let the prime factorization of $a$ be $\prod_{i = 1}^n p_i^{e_i}$. Then, we know that \[\prod_{i-1}^n (ae_i + 1) = 75a + 25.\]
The first observation to make here is that, after taking both sides modulo $a$, we have that $1 \equiv 25 \pmod a$, which implies that $24 \equiv 0 \pmod a$ and therefore that $a$ is a divisor of $24$. There are a few ways to proceed from here. One way is to take $p_1 = 2$, $p_2 = 3$, and $n = 2$ (allowing exponents to be $0$), so that our condition becomes \[(ae_1+1)(ae_2+1) = 75a+25 \iff (ae_1e_2+e_1+e_2-75)a = 24.\] This means that $ae_1e_2 + e_1 + e_2 > 75$. Since $e_1 \leq 3$ and $e_2 \leq 1$, this implies that $3a+4 > 75$. This rules out everything except $a = 24$, which ends up working since $ae_1e_2+e_1+e_2-75=72+3+1-75 = 1$. Therefore the answer is $75 \cdot 24 + 25 = 1825$.


Past POTWs and their solutions can be found in the archive.