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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>
  
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==Solution==
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==Solution 1==
 
The main insight is that  
 
The main insight is that  
  
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is potentially not an integer. It can be easily verified that the above expression is not an integer for <math>n=4</math> 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 <math>n=p</math>, 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 <math>50-16=\boxed{\mathbf{(D)}\ 34}</math>.
 
is potentially not an integer. It can be easily verified that the above expression is not an integer for <math>n=4</math> 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 <math>n=p</math>, 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 <math>50-16=\boxed{\mathbf{(D)}\ 34}</math>.
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==Solution 2==
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We can use the P-Adic Valuation of n to solve this problem (recall the P-Adic Valuation of 'n' is denoted by <math>v_p (n)</math> and is defined as the greatest power of some prime 'p' that divides n. For example, <math>v_2 (6)=1</math> or <math>v_7 (245)=2</math> .) Using Legendre's formula, we know that :
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<cmath> v_p (n!)= \sum_{i=1}^\infty  \lfloor \frac {n}{p^i} \rfloor </cmath>
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 +
Seeing factorials involved in the problem, this prompts us to use Legendre's formula where n is a power of a prime.
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We also know that , <math>v_p (m^n) = n \cdot v_p (m)</math> .
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Knowing that <math>a\mid b</math> iff <math>v_p (a) \le v_p (b)</math> , we have that :
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<cmath> n \cdot v_p (n!) \le v_p ((n^2 -1 )!) </cmath> and we must find all n for which this is true.
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If we plug in <math>n=p</math>, by Legendre's we get two equations:
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<cmath> v_p ((n^2 -1)!) = \sum_{i=1}^\infty  \lfloor \frac {n^2 -1}{p^i} \rfloor = (p-1)+0+...+0 = p-1 </cmath>
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And we also get :
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<cmath> 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 </cmath>
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But we are asked to prove that <math> n \cdot v_p (n!) \le v_p ((n^2 -1 )!) \Longrightarrow p \le p-1 </math> which is false for all 'n' where n is prime.
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 +
Now we try the same for n=p^2 , where p is a prime. By Legendre we arrive at:
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<cmath>v_p ((p^4 -1)!) = p^3 + p^2 + p -3</cmath> and <cmath>p^2 \cdot v_p (p^2 !) = p^3 + p^2 </cmath>
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Then we get:
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<cmath> p^2 \cdot v_p (p!) \le v_p ((n^4 -1)!) \Longrightarrow p^3 + p^2 \le p^3 + p^2 + p -3 </cmath> Which is true for all primes except for 2, so <math>2^2 = 4</math> doesn't work. It can easily be verified that for all <math>n=p^i</math> where i is an integer greater than 2, satisfies the inequality :<cmath> n \cdot v_p (n!) \le v_p ((n^2 -1 )!)</cmath>.
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Therefore, there are 16 values that don't work and <math> 50-16 = \boxed{\mathbf{(D)}\ 34}</math> values that work.
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~qwertysri987
  
 
==See Also==
 
==See Also==

Revision as of 11:51, 28 June 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 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 $n \mid (n-1)!$. 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 $50-16=\boxed{\mathbf{(D)}\ 34}$.

Solution 2

We can use the P-Adic Valuation 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 $a\mid b$ iff $v_p (a) \le v_p (b)$ , 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

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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