# Difference between revisions of "2018 AMC 8 Problems/Problem 5"

## Problem

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

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

## Solution 3

It is similar to the Solution 1: Rearranging the terms, we get $1+(3-2)+(5-4)+(6-5)...(2017-2016)+(2019-2018)$, and our answer is $1+1009=\boxed{1010}, \textbf{(E)}$ ~LarryFlora

## Solution 4

Note that the sum of consecutive odd numbers can be expressed as a square, namely $1+3+5+7+...+2017+2019 = 1010^2$. We can modify the negative numbers in the same way by adding 1 to each negative term, factoring a negative sign, and accounting for the extra 1's by subtracting 1009. We then have $1010^2-1009^2-1010$. Using difference of squares, we obtain $(1010+1009)(1010-1009)-1009 = 2019-1009 = \boxed{2010}$ ~SigmaPiE