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

(Solution 1)
m
(4 intermediate revisions by 3 users not shown)
Line 7: Line 7:
 
<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>
  
==Solutions==
+
==Solution 1==
===Solution 1===
 
 
The main insight is that  
 
The main insight is that  
  
Line 23: Line 22:
 
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 <math>2</math> 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 <math>16</math> values of n make the expression not an integer and the answer is <math>50-16=\boxed{\textbf{(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 <math>2</math> 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 <math>16</math> values of n make the expression not an integer and the answer is <math>50-16=\boxed{\textbf{(D)}\ 34}</math>.
  
SideNote: Another method to prove that <cmath>\frac{(n^2)!}{(n!)^{n+1}}</cmath> is always an integer is instead as follows using Number Theory. Notice that <math>n</math> will divide the numerator <math>n+1</math> times, since <math>n^2 = n \times n</math> contains not one but two factors of <math>n.</math> Also, for <math>a < n,</math> notice that <math>a</math> divides <math>(n^2)!</math> at least <cmath>\lfloor \frac{n^2}{a} \rfloor \ge \lfloor \frac{n^2}{n-1} \rfloor \ge \lfloor \frac{n^2 - 1}{n-1} \rfloor \ge n+1</cmath> times. Thus, each integer from <math>1</math> to <math>n</math> will divide <math>(n^2)!</math> at least <math>n</math> times. <math>\square{}</math>
+
SideNote: Another method to prove that <cmath>\frac{(n^2)!}{(n!)^{n+1}}</cmath> is always an integer is instead as follows using Number Theory. Notice that <math>n</math> will divide the numerator <math>n+1</math> times, since <math>n^2 = n \times n</math> contains not one but two factors of <math>n.</math> Also, for <math>a < n,</math> notice that <math>a</math> divides <math>(n^2)!</math> at least <cmath>\lfloor \frac{n^2}{a} \rfloor \ge \lfloor \frac{n^2}{n-1} \rfloor \ge \lfloor \frac{n^2 - 1}{n-1} \rfloor \ge n+1</cmath> times. Thus, each integer from <math>1</math> to <math>n</math> will divide <math>(n^2)!</math> at least <math>n+1</math> times, which proves such a lemma. <math>\square{}</math>
  
===Solution 2===
+
==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 <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 :
 
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 <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 :
  
Line 49: Line 48:
 
Now we try the same for <math>n=p^2</math> , where p is a prime. By Legendre we arrive at:
 
Now we try the same for <math>n=p^2</math> , where p is a prime. By Legendre we arrive at:
  
<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>
+
<cmath>v_p ((p^4 -1)!) = p^3 + p^2 + p -3</cmath>
 +
 
 +
(as <math>v_p(p^4!) = p^3 + p^2 + p + 1</math> and <math>p^4</math> contains 4 factors of <math>p</math>) and
 +
 
 +
<cmath>p^2 \cdot v_p (p^2 !) = p^3 + p^2 </cmath>
  
 
Then we get:
 
Then we get:

Revision as of 17:52, 26 March 2023

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 $\frac{(n-1)!}{n}$, 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 $50-16=\boxed{\textbf{(D)}\ 34}$.

SideNote: Another method to prove that \[\frac{(n^2)!}{(n!)^{n+1}}\] is always an integer is instead as follows using Number Theory. Notice that $n$ will divide the numerator $n+1$ times, since $n^2 = n \times n$ contains not one but two factors of $n.$ Also, for $a < n,$ notice that $a$ divides $(n^2)!$ at least \[\lfloor \frac{n^2}{a} \rfloor \ge \lfloor \frac{n^2}{n-1} \rfloor \ge \lfloor \frac{n^2 - 1}{n-1} \rfloor \ge n+1\] times. Thus, each integer from $1$ to $n$ will divide $(n^2)!$ at least $n+1$ times, which proves such a lemma. $\square{}$

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:

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

(as $v_p(p^4!) = p^3 + p^2 + p + 1$ and $p^4$ contains 4 factors of $p$) 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)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png