Difference between revisions of "2016 AMC 8 Problems/Problem 8"
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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 | + | 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> |
{{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 16:06, 23 June 2017
Find the value of the expression
Solution
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 start at one, we divide every number by 2 and we end up with We can group each subtracting pair together: 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
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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