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 $n$. Thus,

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

is an integer if $n^2 \mid n!$, in other words if $n \mid (n-1)!$. This fails exactly for $n=4$ or $n$ prime (Wilson's Theorem). There are 15 primes between 1 and 50, inclusive, so the answer is $50-1-15=\boxed{\mathbf{(D)}\ 34}$.