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

m (→‎Solution 2)
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We can group each subtracting pair together:
 
We can group each subtracting pair together:
 
<cmath>(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).</cmath>
 
<cmath>(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).</cmath>
−
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 <math>\dfrac{50}{1}=\boxed{\textbf{(C) }50}</math>
+
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>
 
{{AMC8 box|year=2016|num-b=7|num-a=9}}
 
{{AMC8 box|year=2016|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 07:30, 6 August 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 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 $\boxed{\textbf{(C) }50}$

2016 AMC 8 (Problems • Answer Key • Resources)
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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