Art of Problem Solving
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2018 AMC 8 Problems/Problem 5

Revision as of 12:41, 1 November 2020 by Avamarora (talk | contribs) (Solution 2)

Problem 5

What is the value of $1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018$?

$\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010$

Solution 1

Rearranging the terms, we get $(1-2)+(3-4)+(5-6)+...(2017-2018)+2019$, and our answer is $-1009+2019=\boxed{1010}, \textbf{(E)}$

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 $\boxed{\textbf{(E) }1010}$ ~avamarora

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
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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