Difference between revisions of "2016 AMC 8 Problems/Problem 8"
| Line 7: | Line 7: | ||
After subtracting, we have: | After subtracting, we have: | ||
<cmath>2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).</cmath> | <cmath>2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).</cmath> | ||
| − | There are 50 even numbers, therefore there are <math>50/2=25</math> even pairs. Therefore the sum is <math>2 \cdot 25=\boxed{\textbf{(C) }50}</math> | + | There are <math>50</math> even numbers, therefore there are <math>50/2=25</math> even pairs. Therefore the sum is <math>2 \cdot 25=\boxed{\textbf{(C) }50}</math> |
{{AMC8 box|year=2016|num-b=7|num-a=9}} | {{AMC8 box|year=2016|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:21, 27 November 2016
Find the value of the expression
![\[100-98+96-94+92-90+\cdots+8-6+4-2.\]](http://latex.artofproblemsolving.com/0/2/7/027ea184e9e3fde749ba37eb702069fa8e201387.png)
Solution
We can group each subtracting pair together:
![\[(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).\]](http://latex.artofproblemsolving.com/c/f/f/cff0f3dd48ced41da7d2575921b5c0bf5f526e04.png)
After subtracting, we have:
![\[2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).\]](http://latex.artofproblemsolving.com/1/f/9/1f98cf231e1203e5b694c0eeb37a06c9a99763f5.png)
There are 
even numbers, therefore there are 
even pairs. Therefore the sum is 
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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. 
