2016 AMC 8 Problems/Problem 8: Difference between revisions
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==Solution== | ==Solution== | ||
{{ | We can group each subtracting pair together: | ||
<cmath>(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).</cmath> | |||
After subtracting, we have: | |||
<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> | |||
Revision as of 08:45, 23 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 50 even numbers, therefore there are 
even pairs. Therefore the sum is 
