Difference between revisions of "2016 AMC 8 Problems/Problem 8"
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+ | ==Problem== | ||
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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><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 1== | + | ==Solutions== |
+ | ===Solution 1=== | ||
We can group each subtracting pair together: | We can group each subtracting pair together: | ||
<cmath>(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).</cmath> | <cmath>(100-98)+(96-94)+(92-90)+ \ldots +(8-6)+(4-2).</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> | 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== | + | ===Solution 2=== |
Since our list does not end with one, we divide every number by 2 and we end up with | 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> | <cmath>50-49+48-47+ \ldots +4-3+2-1</cmath> |
Revision as of 02:53, 16 January 2021
Contents
[hide]Problem
Find the value of the expression
Solutions
Solution 1
We can group each subtracting pair together: After subtracting, we have: There are even numbers, therefore there are even pairs. Therefore the sum is
Solution 2
Since our list does not end with one, we divide every number by 2 and we end up with We can group each subtracting pair together: There are now pairs of numbers, and the value of each pair is . This sum is . However, we divided by originally so we will multiply to get the final answer of
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.