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

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==Solution==
 
==Solution==
{{solution}}
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We can group each subtracting pair together:
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<cmath>(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).</cmath>
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After subtracting, we have:
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<cmath>2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).</cmath>
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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 09:45, 23 November 2016

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 $50/2=25$ even pairs. Therefore the sum is $2 \cdot 25=\boxed{\textbf{(C) }50}$

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