2019 AMC 10A Problems/Problem 25

Revision as of 21:17, 9 February 2019 by Qwertyalla (talk | contribs) (Solution)
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 $\boxed{\mathbf{(D)}\ 34}$.

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last 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 (ProblemsAnswer KeyResources)
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

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