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

(Created page with "==Problem 5== 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>? <math>\textbf{(A) }1020\qqua...")
 
 
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==Problem 5==
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==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>?
  
 
<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>1120</math>, or <math>B</math>.
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Directly calculating:
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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 the sum of all numbers from <math>1</math> to <math>n</math> is <math>\frac{n(n+1)}{2}</math>. Therefore, the denominator is equal to <math>\frac{8 \cdot 9}{2} = 4 \cdot 9 = 2 \cdot 3 \cdot 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 \cdot 7 \cdot 5 \cdot 4 \cdot 1</math>. This evaluates to <math>\boxed{\textbf{(B)}\ 1120}</math>.
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==Solution 3==
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First, we evaluate <math>1 + 2 + 3 + 4 + 5 + 6 + 7 + 8</math>, to get 36. We notice that 36 is 6 squared, so we can factor the denominator like <math>\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{6 \cdot 6}</math> then cancel the 6s out, to get <math>\frac{4 \cdot 5 \cdot 7 \cdot 8}{1}</math>. Now that we have escaped fraction form, multiplying <math>4 \cdot 5 \cdot 7 \cdot 8 \cdot</math>. Multiplying these, we get <math>\boxed{\textbf{(B)}\ 1120}</math>.
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-[[User:elbertpark|elbertpark]]
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==Video Solution==
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https://youtu.be/cY4NYSAD0vQ
  
 
==See Also==
 
==See Also==
{{AMC8 box|year=2017|num-b=20|num-a=22}}
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{{AMC8 box|year=2017|num-b=4|num-a=6}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest 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

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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