2016 AMC 8 Problems/Problem 8
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 
Solution 2
Since our list does not start at one, we divide every number by 2 and we end up with
![\[50-49+48-47+ \ldots +4-3+2-1\]](http://latex.artofproblemsolving.com/d/4/2/d42ae896c5dc414cbf054c2400e60d91abdeffd0.png)
We can group each subtracting pair together:
![\[(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).\]](http://latex.artofproblemsolving.com/7/a/d/7ad418f42d4e4cb58369b973f60676ac04efabcc.png)
As we can see, the list now starts at 1 and ends at 50, thus there are 50 numbers in total. Since all the subtracting pairs are equal to one, the solution equals $\dfrac{50}{1}=
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
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Followed by Problem 9 | |
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