# Difference between revisions of "2019 AMC 10A Problems/Problem 25"

The following problem is from both the 2019 AMC 10A #25 and 2019 AMC 12A #24, so both problems redirect to this page.

## Problem

For how many integers $n$ between $1$ and $50$, inclusive, is $$\frac{(n^2-1)!}{(n!)^n}$$ an integer? (Recall that $0! = 1$.) $\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$

## Solution

The main insight is that $$\frac{(n^2)!}{(n!)^{n+1}}$$ is always an integer. This is true because it is precisely the number of ways to split up $n^2$ objects into $n$ unordered groups of size $n$, which has to be an integer. Thus, $$\frac{(n^2-1)!}{(n!)^n}$$ is not an integer if $n^2 \nmid n!$. This happens only for $n=4$ or $n$ prime. Thus, there are $16$ integers that do not work, which means the answer is $\boxed{\mathbf{(D)}\ 34}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 