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

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Find the value of the expression
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==Problem==
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Find the value of the expression
 
<cmath>100-98+96-94+92-90+\cdots+8-6+4-2.</cmath><math>\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100</math>
 
<cmath>100-98+96-94+92-90+\cdots+8-6+4-2.</cmath><math>\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100</math>
  
==Solution==
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==Solutions==
{{solution}}
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===Solution 1===
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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 <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===
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Since our list does not end with one, we divide every number by 2 and we end up with
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<cmath>50-49+48-47+ \ldots +4-3+2-1</cmath>
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We can group each subtracting pair together:
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<cmath>(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).</cmath>
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There are now <math>25</math> pairs of numbers, and the value of each pair is <math>1</math>.  This sum is <math>25</math>. However, we divided by <math>2</math> originally so we will multiply <math>2*25</math> to get the final answer of <math>\boxed{\textbf{(C) }50}</math>
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==See Also==
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{{AMC8 box|year=2016|num-b=7|num-a=9}}
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{{MAA Notice}}

Latest revision as of 17:31, 21 April 2021

Problem

Find the value of the expression \[100-98+96-94+92-90+\cdots+8-6+4-2.\]$\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100$

Solutions

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 $\boxed{\textbf{(C) }50}$

See Also

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

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