Difference between revisions of "2017 AMC 8 Problems/Problem 5"
(→Solution) |
(→Solution 2) |
||
| Line 9: | Line 9: | ||
==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 | + | 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>. |
==See Also== | ==See Also== | ||
Revision as of 15:54, 22 November 2017
Contents
Problem 5
What is the value of the expression 
?
Solution 1
We evaluate both the top and bottom: 
. This simplifies to 
.
Solution 2
It is well known that 
. 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 .
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. 
