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,

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

is not an integer if $n^2 \nmid n!$ . This happens only for $n=4$ or $n$ prime. 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 by Problem 24	Followed by Last Problem
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All AMC 10 Problems and Solutions

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 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

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

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2019 AMC 10A Problems/Problem 25

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