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

Revision as of 21:48, 10 May 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 $0! = 1$ .)

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

Solution

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 $50-16=\boxed{\textbf{(D)}\ 34}$ .

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

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

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

@@ Line 56: / Line 56: @@
 ~qwertysri987
-==Solution 3==
-Complementary Counting. Looking at the problem, it says that it is from 1-50 inclusive. Worst case, we'll have to check through all of them. First, try plugging in some numbers and you find that 2,3,4,5(all the primes and 4) don't work. There are 15 primes less than 50, and add 1 because of the 4. There are 16 cases, and you subtract them from 50. <math>50-16=\boxed{\textbf{(D) } 34}</math>
 ==See Also==

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