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Difference between revisions of "2019 AMC 10A Problems/Problem 25"
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==Solution==
 
==Solution==
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The main insight is that <cmath>\frac{(n^2)!}{(n!)^{n+1}}</cmath> is always an integer. This is true because it is precisely the number of ways to split up <math>n^2</math> objects into <math>n</math> unordered groups of size <math>n</math>, which has to be an integer. Thus, <cmath>\frac{(n^2-1)!}{(n!)^n}</cmath> is not an integer iff <math>n^2 \nmid n!</math>. This happens only for <math>n=4</math> or <math>n</math> prime. Thus, there are <math>16</math> integers that do not work, which means the answer is <math>\boxed{\mathbf{(D)}\ 34}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 18:26, 9 February 2019

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 iff $n^2 \nmid n!$. 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 (Problems • Answer Key • Resources)
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 (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

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