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

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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>?
  
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==Solution 1==
 
==Solution 1==
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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>.
 
We evaluate both the top and bottom: <math>\frac{40320}{36}</math>. This simplifies to <math>\boxed{\textbf{(B)}\ 1120}</math>.
  
 
==Solution 2==
 
==Solution 2==
  
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 \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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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 <math>36</math> is <math>6</math> 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, we multiply <math>4 \cdot 5 \cdot 7 \cdot 8</math>. Multiplying these, we get <math>\boxed{\textbf{(B)}\ 1120}</math>.
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~megaboy6679 for minor edits
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==Video Solution (CREATIVE THINKING!!!)==
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https://youtu.be/O5thBbTXpeY
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/cY4NYSAD0vQ
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https://youtu.be/tGa0YTskTsw
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 +
~savannahsolver
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==Video Solution by OmegaLearn==
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https://youtu.be/TkZvMa30Juo?t=3529
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~pi_is_3.14
  
 
==See Also==
 
==See Also==

Latest revision as of 13:40, 2 November 2024

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, we multiply $4 \cdot 5 \cdot 7 \cdot 8$. Multiplying these, we get $\boxed{\textbf{(B)}\ 1120}$.

~megaboy6679 for minor edits

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/O5thBbTXpeY

~Education, the Study of Everything

Video Solution

https://youtu.be/cY4NYSAD0vQ

https://youtu.be/tGa0YTskTsw

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/TkZvMa30Juo?t=3529

~pi_is_3.14

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AJHSME/AMC 8 Problems and Solutions

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