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

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(Solution)
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<math>\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360</math>  
 
<math>\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360</math>  
  
==Solution==
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==Solution 1==
We evaluate both the top and bottom: <math>\frac{40320}{36}</math>. This simplifies to <math>\boxed{\textbf{(B)}\ 1120}.</math>
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We evaluate both the top and bottom: <math>\frac{40320}{36}</math>. This simplifies to <math>\boxed{\textbf{(B)}\ 1120}</math>.
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==Solution 2==
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It is well known that <math>1 + 2 + \cdots + n = \frac{n(n+1)}{2}</math>. Therefore, the denominator is equal to <math>\frac{8*9}{2} = 4*9 = 2*3*6</math>. Now we can cancel the factors of <math>2</math>, <math>3</math>, and <math>6</math> from both the numerator and denominator, only leaving <math>8*7*5*4*1</math>. This evaluates to <math>\boxed{\textbf{(B)}\ 1120}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 15:52, 22 November 2017

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

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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