# Difference between revisions of "2017 AMC 8 Problems/Problem 5"

## Problem 5

What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$?

$\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$

## Solution 1

We evaluate both the top and bottom: $\frac{40320}{36}$. This simplifies to $\boxed{\textbf{(B)}\ 1120}$.

## Solution 2

It is well known that $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$. Therefore, the denominator is equal to $\frac{8*9}{2} = 4*9 = 2*3*6$. Now we can cancel the factors of $2$, $3$, and $6$ from both the numerator and denominator, only leaving $8*7*5*4*1$. This evaluates to $\boxed{\textbf{(B)}\ 1120}$.