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

Line 8: Line 8:
 
<cmath>2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).</cmath>
 
<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>
 
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>
 +
 +
{{AMC8 box|year=2016|num-b=7|num-a=9}}
 +
{{MAA Notice}}

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 $2 \cdot 25=\boxed{\textbf{(C) }50}$

2016 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS