Difference between revisions of "2016 AMC 8 Problems/Problem 8"
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<cmath>100-98+96-94+92-90+\cdots+8-6+4-2.</cmath><math>\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100</math> | <cmath>100-98+96-94+92-90+\cdots+8-6+4-2.</cmath><math>\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100</math> | ||
| − | ==Solution== | + | ==Solution 1== |
We can group each subtracting pair together: | We can group each subtracting pair together: | ||
<cmath>(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).</cmath> | <cmath>(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).</cmath> | ||
Revision as of 22:06, 1 November 2020
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 1
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 end with 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)
There are now 
pairs of numbers, and the value of each pair is 
. This sum is . However, we divided by 
originally so we will multiply 
to get the final answer of 
| 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. 
