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

(Created page with "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 \...")
 
(16 intermediate revisions by 6 users not shown)
Line 1: Line 1:
 
Find the value of the expression
 
Find the value of the expression
<cmath>100-98+96-94+92-90+\cdots+8-6+4-2.</cmath>
+
<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>
<math>\textbf{(A) }20\qquad\textbf{(B) }40\qquad\textbf{(C) }50\qquad\textbf{(D) }80\qquad \textbf{(E) }100</math>
+
 
 +
==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 <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>
 +
 
 +
==Solution 2==
 +
Since our list does not end with one, we divide every number by 2 and we end up with
 +
<cmath>50-49+48-47+ \ldots +4-3+2-1</cmath>
 +
We can group each subtracting pair together:
 +
<cmath>(50-49)+(48-47)+(46-45)+ \ldots +(4-3)+(2-1).</cmath>
 +
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}}
 +
{{MAA Notice}}

Revision as of 08:30, 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 (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