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

m
Line 3: Line 3:
  
 
==Solution==
 
==Solution==
{{solution}}
+
We can group each subtracting pair together:
 +
<cmath>(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).</cmath>
 +
After subtracting, we have:
 +
<cmath>2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).</cmath>
 +
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 08:45, 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 $2 \cdot 25=\boxed{\textbf{(C) }50}$