Difference between revisions of "2017 AMC 8 Problems/Problem 5"
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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== | + | ==Solution 1== |
| − | We evaluate both the top and bottom: <math>\frac{40320}{36}</math>. This simplifies to <math>\boxed{\textbf{(B)}\ 1120}.</math> | + | |
| + | Directly calculating: | ||
| + | |||
| + | We evaluate both the top and bottom: <math>\frac{40320}{36}</math>. This simplifies to <math>\boxed{\textbf{(B)}\ 1120}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | ==Solution 3== | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | -[[User:elbertpark|elbertpark]] | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/cY4NYSAD0vQ | ||
==See Also== | ==See Also== | ||
| − | {{AMC8 box|year=2017|num-b= | + | {{AMC8 box|year=2017|num-b=4|num-a=6}} |
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:49, 23 November 2020
Problem 5
What is the value of the expression 
?
Solution 1
Directly calculating:
We evaluate both the top and bottom: 
. This simplifies to 
.
Solution 2
It is well known that the sum of all numbers from 
to 
is 
. Therefore, the denominator is equal to 
. Now we can cancel the factors of 
, 
, and 
from both the numerator and denominator, only leaving 
. This evaluates to .
Solution 3
First, we evaluate 
, to get 36. We notice that 36 is 6 squared, so we can factor the denominator like 
then cancel the 6s out, to get 
. Now that we have escaped fraction form, multiplying 
. Multiplying these, we get .
Video Solution
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
