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.


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


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,


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


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.


Solution 3 (Non-Rigorous)

Notice all $15$ primes don't work as there are $n$ factors of $n$ in the denominator and $n-1$ factors of $n$ in the numerator. Further experimentation finds that $n=4$ does not work as there are 11 factors of 2 in the numerator and 12 in the denominator. We also find that it seems that all other values of $n$ work. So we get $50-15-1=\boxed{34}$ and that happens to be right.


See Also

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

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