Difference between revisions of "2018 AMC 8 Problems/Problem 5"
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We can see that the last numbers of each of the sets (even numbers and odd numbers) have a difference of two. So do the second last ones, and so on. Now, all we need to find is the number of integers in any of the sets (I chose even) to get <math>\boxed{\textbf{(E) }1010}</math> ~avamarora | We can see that the last numbers of each of the sets (even numbers and odd numbers) have a difference of two. So do the second last ones, and so on. Now, all we need to find is the number of integers in any of the sets (I chose even) to get <math>\boxed{\textbf{(E) }1010}</math> ~avamarora | ||
| + | |||
| + | ==Solution 3== | ||
| + | |||
| + | It is similar to the Solution 1: | ||
| + | Rearranging the terms, we get <math>1+(3-2)+(5-4)+(6-5)...(2017-2016)+(2019-2018)</math>, and our answer is <math>1+1009=\boxed{1010}, \textbf{(E)}</math> | ||
| + | ~LarryFlora | ||
==See Also== | ==See Also== | ||
Revision as of 13:23, 30 May 2021
Problem
What is the value of 
?
Solution 1
Rearranging the terms, we get 
, and our answer is 
Solution 2
We can see that the last numbers of each of the sets (even numbers and odd numbers) have a difference of two. So do the second last ones, and so on. Now, all we need to find is the number of integers in any of the sets (I chose even) to get 
~avamarora
Solution 3
It is similar to the Solution 1:
Rearranging the terms, we get 
, and our answer is 
~LarryFlora
See Also
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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. 
