Difference between revisions of "2017 AMC 8 Problems/Problem 5"
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==Solution 3== | ==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 | + | 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]] | ~[[User:elbertpark|elbertpark]] | ||
Revision as of 00:25, 6 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. 
