# Difference between revisions of "2020 AMC 8 Problems/Problem 12"

## Problem

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

We have $5! = 2 \cdot 3 \cdot 4 \cdot 5$, and $2 \cdot 5 \cdot 9! = 10 \cdot 9! = 10!$. Therefore the equation becomes $3 \cdot 4 \cdot 10! = 12 \cdot N!$, and so $12 \cdot 10! = 12 \cdot N!$. Cancelling the $12$s, it is clear that $N=\boxed{\textbf{(A) }10}$.

## Solution 2 (variant of Solution 1)

Since $5! = 120$, we obtain $120\cdot 9!=12\cdot N!$, which becomes $12\cdot 10\cdot 9!=12\cdot N!$ and thus $12 \cdot 10!=12\cdot N!$. We therefore deduce $N=\boxed{\textbf{(A) }10}$.

## Solution 3 (using answer choices and elimination)

We can see that the answers $\textbf{(B)}$ to $\textbf{(E)}$ contain a factor of $11$, but there is no such factor of $11$ in $5! \cdot 9!$. The factor 11 is in every answer choice after $\boxed{\textbf{(A) }10}$, so four of the possible answers are eliminated. Therefore, the answer must be $\boxed{\textbf{(A) }10}$. ~edited by HW73

## Solution 4

We notice that $5! \cdot 9! = (5!)^2 \cdot (9 \cdot 8 \cdot 7 \cdot 6).$

We know that $5! = 120,$ so we have $120(5! \cdot 9 \cdot 8 \cdot 7 \cdot 6) = 12 \cdot N!$

Isolating $N!$ we have $N! = 10 \cdot 5! \cdot 9 \cdot 8 \cdot 7 \cdot 6 \Rightarrow N! = 10! \Rightarrow N = \boxed{\textbf{(A) }10}.$

~mathboy282

## Video Solution by North America Math Contest Go Go Go

~North America Math Contest Go Go Go

~savannahsolver

~Interstigation