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

Revision as of 12:59, 18 January 2021

Problem

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

Directly calculating:

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

Solution 2

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

Solution 3

First, we evaluate $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8$ , to get 36. We notice that 36 is 6 squared, so we can factor the denominator like $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{6 \cdot 6}$ then cancel the 6s out, to get $\frac{4 \cdot 5 \cdot 7 \cdot 8}{1}$ . Now that we have escaped fraction form, multiplying $4 \cdot 5 \cdot 7 \cdot 8 \cdot$ . Multiplying these, we get $\boxed{\textbf{(B)}\ 1120}$ .

-elbertpark

Video Solution

https://youtu.be/cY4NYSAD0vQ

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AJHSME/AMC 8 Problems and Solutions

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

Revision as of 17:49, 23 November 2020 (view source) Sugar rush (talk \| contribs) ← Older edit		Revision as of 12:59, 18 January 2021 (view source) Hashtagmath (talk \| contribs) Newer edit →
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−	==Problem 5==	+	==Problem==
	What is the value of the expression <math>\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}</math>?		What is the value of the expression <math>\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}</math>?

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