Difference between revisions of "2016 AMC 8 Problems/Problem 8"

Revision as of 13:09, 5 July 2017

Find the value of the expression $100-98+96-94+92-90+\cdots+8-6+4-2.$ $\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$

Solution

We can group each subtracting pair together: $(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).$ After subtracting, we have: $2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).$ There are $50$ even numbers, therefore there are $\dfrac{50}{2}=25$ even pairs. Therefore the sum is $2 \cdot 25=\boxed{\textbf{(C) }50}$

Solution 2

Since our list does not end with one, we divide every number by 2 and we end up with $50-49+48-47+ \ldots +4-3+2-1$ We can group each subtracting pair together: $(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).$ 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 $\dfrac{50}{1}=\boxed{\textbf{(C) }50}$

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.

@@ Line 10: / Line 10: @@
 ==Solution 2==
-Since our list does not start at one, we divide every number by 2 and we end up with
+Since our list does not end with one, we divide every number by 2 and we end up with
 <cmath>50-49+48-47+ \ldots +4-3+2-1</cmath>
 We can group each subtracting pair together:

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Difference between revisions of "2016 AMC 8 Problems/Problem 8"

Revision as of 13:09, 5 July 2017

Solution

Solution 2