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 <math>50</math> | + | 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 50/1 or <math>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 17:03, 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 50/1 or
.
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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