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

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<math>\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35</math>
 
<math>\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35</math>
  
==Solution==
+
==Solutions==
 
===Solution 1===
 
===Solution 1===
 
The main insight is that  
 
The main insight is that  
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<cmath>\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}</cmath>
 
<cmath>\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}</cmath>
  
is an integer if <math>n^2 \mid n!</math>, or in other words, if <math>n \mid (n-1)!</math>. This condition is false precisely when <math>n=4</math> or <math>n</math> is prime, by [[Wilson's Theorem]]. There are <math>15</math> primes between <math>1</math> and <math>50</math>, inclusive, so there are <math>15 + 1 = 16</math> terms for which
+
is an integer if <math>n^2 \mid n!</math>, or in other words, if <math>\frac{(n-1)!}{n}</math>, is an integer. This condition is false precisely when <math>n=4</math> or <math>n</math> is prime, by [[Wilson's Theorem]]. There are <math>15</math> primes between <math>1</math> and <math>50</math>, inclusive, so there are <math>15 + 1 = 16</math> terms for which
  
 
<cmath>\frac{(n^2-1)!}{(n!)^{n}}</cmath>
 
<cmath>\frac{(n^2-1)!}{(n!)^{n}}</cmath>
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<cmath> v_p (n!)= \sum_{i=1}^\infty  \lfloor \frac {n}{p^i} \rfloor </cmath>
 
<cmath> v_p (n!)= \sum_{i=1}^\infty  \lfloor \frac {n}{p^i} \rfloor </cmath>
  
Seeing factorials involved in the problem, this prompts us to use Legendre's formula where n is a power of a prime.
+
Seeing factorials involved in the problem, this prompts us to use [[Legendre's Formula]] where n is a power of a prime.
  
 
We also know that , <math>v_p (m^n) = n \cdot v_p (m)</math> .
 
We also know that , <math>v_p (m^n) = n \cdot v_p (m)</math> .
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~qwertysri987
 
~qwertysri987
 
==Solution 3 (Guessing)==
 
 
First, we see that <math>n=1, 6, 8, 9, 10, 12, 14</math> work. This leads us to the conclusion of <math> 50-16 = \boxed{\textbf{(D)}\ 34}</math>
 
  
 
==See Also==
 
==See Also==

Revision as of 15:44, 23 October 2021

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 $0! = 1$.)

$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$

Solutions

Solution 1

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 $n^2 \mid n!$, or in other words, if $\frac{(n-1)!}{n}$, is an integer. This condition is false precisely when $n=4$ or $n$ is prime, by Wilson's Theorem. There are $15$ primes between $1$ and $50$, inclusive, so there are $15 + 1 = 16$ terms for which

\[\frac{(n^2-1)!}{(n!)^{n}}\]

is potentially not an integer. It can be easily verified that the above expression is not an integer for $n=4$ as there are more factors of $2$ in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime $n=p$, as there are more factors of p in the denominator than the numerator. Thus all $16$ values of n make the expression not an integer and the answer is $50-16=\boxed{\textbf{(D)}\ 34}$.

Solution 2

We can use the P-Adic Valuation (more info could be found here: Mathematicial notation) of n to solve this problem (recall the P-Adic Valuation of 'n' is denoted by $v_p (n)$ and is defined as the greatest power of some prime 'p' that divides n. For example, $v_2 (6)=1$ or $v_7 (245)=2$ .) Using Legendre's formula, we know that :

\[v_p (n!)= \sum_{i=1}^\infty   \lfloor \frac {n}{p^i} \rfloor\]

Seeing factorials involved in the problem, this prompts us to use Legendre's Formula where n is a power of a prime.

We also know that , $v_p (m^n) = n \cdot v_p (m)$ . Knowing that $a\mid b$ if $v_p (a) \le v_p (b)$ , we have that :

\[n \cdot v_p (n!) \le v_p ((n^2 -1 )!)\] and we must find all n for which this is true.

If we plug in $n=p$, by Legendre's we get two equations:

\[v_p ((n^2 -1)!) = \sum_{i=1}^\infty  \lfloor \frac {n^2 -1}{p^i} \rfloor = (p-1)+0+...+0 = p-1\]

And we also get :

\[v_p ((n!)^n) = n \cdot v_p (n!)= n \cdot \sum_{i=1}^\infty   \lfloor \frac {n}{p^i} \rfloor = p \cdot ( 1+0+...0) = p\]

But we are asked to prove that $n \cdot v_p (n!) \le v_p ((n^2 -1 )!) \Longrightarrow p \le p-1$ which is false for all 'n' where n is prime.

Now we try the same for $n=p^2$ , where p is a prime. By Legendre we arrive at:

\[v_p ((p^4 -1)!) = p^3 + p^2 + p -3\] and \[p^2 \cdot v_p (p^2 !) = p^3 + p^2\]

Then we get:

\[p^2 \cdot v_p (p!) \le v_p ((n^4 -1)!) \Longrightarrow p^3 + p^2 \le p^3 + p^2 + p -3\] Which is true for all primes except for 2, so $2^2 = 4$ doesn't work. It can easily be verified that for all $n=p^i$ where $i$ is an integer greater than 2, satisfies the inequality :\[n \cdot v_p (n!) \le v_p ((n^2 -1 )!).\]

Therefore, there are 16 values that don't work and $50-16 = \boxed{\mathbf{(D)}\ 34}$ values that work.

~qwertysri987

See Also

Video Solution by Richard Rusczyk: https://www.youtube.com/watch?v=9klaWnZojq0

2019 AMC 10A (ProblemsAnswer KeyResources)
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Problem 23
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Problem 25
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All AMC 12 Problems and Solutions

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