2019 AMC 10A Problems/Problem 25

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

2019 AMC 10A (Problems • Answer Key • Resources)
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2019 AMC 10A Problems/Problem 25

Contents

Problem

Solution 1

Solution 2

See Also