Difference between revisions of "2016 AMC 8 Problems/Problem 8"
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<cmath>2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).</cmath> | <cmath>2+2+2+\ldots+2+2=2(1+1+1+\ldots+1+1).</cmath> | ||
There are 50 even numbers, therefore there are <math>50/2=25</math> even pairs. Therefore the sum is <math>2 \cdot 25=\boxed{\textbf{(C) }50}</math> | There are 50 even numbers, therefore there are <math>50/2=25</math> even pairs. Therefore the sum is <math>2 \cdot 25=\boxed{\textbf{(C) }50}</math> | ||
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+ | {{AMC8 box|year=2016|num-b=7|num-a=9}} | ||
+ | {{MAA Notice}} |
Revision as of 09:45, 23 November 2016
Find the value of the expression
Solution
We can group each subtracting pair together: After subtracting, we have: There are 50 even numbers, therefore there are even pairs. Therefore the sum is
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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All AJHSME/AMC 8 Problems and Solutions |
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