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 <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== | |
+ | Since our list does not start at 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> | ||
+ | As we can see, the list now starts at 1 and ends at 50, thus there are <math>50</math> numbers in total. | ||
{{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 17:01, 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
numbers in total.
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.