# 2020 AMC 8 Problems/Problem 12

For a positive integer $n$, the factorial notation $n!$ represents the product of the integers from $n$ to $1$. What value of $N$ satisfies the following equation? $$5!\cdot 9!=12\cdot N!$$

$\textbf{(A) }10\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }14\qquad$

## Solution 1

Notice that $5!$ = $2*3*4*5,$ and we can combine the numbers to create a larger factorial. To turn $9!$ into $10!,$ we need to multiply $9!$ by $2*5,$ which equals to $10!.$

Therefore, we have

$$10!*12=12*N!.$$ We can cancel the $12$'s, since we are multiplying them on both sides of the equation.

We have

$$10!=N!.$$ From here, it is obvious that $N=\boxed{10\textbf{(A)}}.$

-iiRishabii

## Solution 2

$5!\cdot 9!=12\cdot N!$
$120\cdot 9!=12\cdot N!$
$12\cdot 10\cdot 9!=12\cdot N!$
$12 \cdot 10!=12\cdot N!$
$N=10 \implies\boxed{\textbf{(A) }10}$.
~junaidmansuri

## Solution 3 (Non-rigorous)

We can see that the answers B through E have the factor 11, but there is no 11 in $5!\cdot9!$. Therefore, the answer must be the only answer without a $11$ factor, $A$.

~Windigo

## Solution 4

Notice that $5!\cdot 9!=12\cdot 10\cdot 9!=12\cdot 10!$. We are also told that $12\cdot 10!=12*N!$ from where it is obvious that $N=\textbf{(A)}10$.

-franzliszt

## Solution 5

We see that $5!\cdot9! = 5\cdot4\cdot3\cdot2\cdot1\cdot9! = 12\cdot{N!}$. Notice that $12 = 3\cdot4$, so: $$5\cdot2\cdot1\cdot9! = N!$$ We see that $5\cdot2\cdot1\cdot9! = 10\cdot9! = 10! = N!$. So $N = \boxed{10} = \textbf{(A)}10$.

~savannahsolver