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

(Solution 2)
(Solution)
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<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 <math>50</math> even numbers, therefore there are <math>\dfrac{50}{2}=25</math> even pairs. Therefore the sum is <math>2 \cdot 25=\boxed{\textbf{(C) }50}</math>
 
There are <math>50</math> even numbers, therefore there are <math>\dfrac{50}{2}=25</math> even pairs. Therefore the sum is <math>2 \cdot 25=\boxed{\textbf{(C) }50}</math>
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==Solution 2==
 
==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 start at one, we divide every number by 2 and we end up with

Revision as of 17:03, 23 June 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 start at 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 50/1 or $50$.

2016 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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All AJHSME/AMC 8 Problems and Solutions

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