Difference between revisions of "2016 AMC 8 Problems/Problem 8"
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− | + | ==Problem== | |
+ | |||
+ | 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== | + | ==Solutions== |
− | {{ | + | ===Solution 1=== |
+ | 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> | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2016|num-b=7|num-a=9}} | ||
+ | {{MAA Notice}} |
Latest revision as of 17:31, 21 April 2021
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
See Also
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.