2018 AMC 8 Problems/Problem 5: Difference between revisions
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==Solution 2== | ==Solution 2== | ||
W can see that the last numbers of each of the sets (even numbers and odd numbers) have a difference of one. So do the second last ones, and so on. Now, all we need to find is the number of intigers in any of the sets (I chose even) to get <math>\boxed{\textbf{(E) }1010}</math> ~avamarora | |||
==See Also== | ==See Also== | ||
Revision as of 12:41, 1 November 2020
Problem 5
What is the value of 
?
Solution 1
Rearranging the terms, we get 
, and our answer is 
Solution 2
W can see that the last numbers of each of the sets (even numbers and odd numbers) have a difference of one. So do the second last ones, and so on. Now, all we need to find is the number of intigers in any of the sets (I chose even) to get 
~avamarora
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
