Difference between revisions of "2018 AMC 8 Problems/Problem 5"
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==Solution 2== | ==Solution 2== | ||
− | We can see that the last numbers of each of the sets (even numbers and odd numbers) have a difference of | + | 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 |
==See Also== | ==See Also== |
Revision as of 12:26, 9 November 2020
Contents
Problem 5
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
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 |
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