# 2016 AMC 8 Problems/Problem 8

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 1

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: $$(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).$$ There are now $25$ pairs of numbers, and the value of each pair is $1$. This sum is $25$. However, we divided by $2$ originally so we will multiply $2*25$ to get the final answer of $\boxed{\textbf{(C) }50}$

 2016 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 7 Followed byProblem 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

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