# 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,

$$\frac{(n^2)!}{(n!)^{n+1}}=\frac{n^2}{n!}\cdot\frac{(n^2-1)!}{(n!)^n}$$

is not an integer if $n^2 \nmid n!$. This happens only for $n=4$ or $n$ prime (by Wilson's Theorem). Thus, there are $16$ integers that do not work, which means the answer is $\boxed{\mathbf{(D)}\ 34}$.

 2019 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast Problem 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
 2019 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 23 Followed byProblem 25 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions