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Difference between revisions of "2019 AMC 10A Problems/Problem 25"

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<cmath>\frac{(n^2)!}{(n!)^{n+1}}</cmath>  
 
<cmath>\frac{(n^2)!}{(n!)^{n+1}}</cmath>  
  
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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,  
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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>. Thus,
  
−
<cmath>\frac{(n^2)!}{(n!)^{n+1}}=\frac{n^2}{n!}\cdot\frac{(n^2-1)!}{(n!)^n}</cmath>  
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<cmath>\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}</cmath>
  
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is not an integer if <math>n^2 \nmid n!</math>. This happens only for <math>n=4</math> or <math>n</math> prime (by Wilson's Theorem). Thus, there are <math>16</math> integers that do not work, which means the answer is <math>\boxed{\mathbf{(D)}\ 34}</math>.
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is an integer if <math>n^2 \mid n!</math>, in other words if <math>n \mid (n-1)!</math>. This fails exactly for <math>n=4</math> or <math>n</math> prime (Wilson's Theorem). There are <math>15</math> primes between 1 and 50, inclusive, so the answer is <math>50-1-15=\boxed{\mathbf{(D)}\ 34}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 21:05, 24 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 $n$. 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 $50-1-15=\boxed{\mathbf{(D)}\ 34}$.

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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