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 $\boxed{\textbf{(B)}\ 1120}$ .

2017 AMC 8 (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

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2017 AMC 8 Problems/Problem 5

Contents

Problem 5

Solution 1

Solution 2

See Also