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

m
Line 20: Line 20:
 
==Solution 4==
 
==Solution 4==
  
Note that the sum of consecutive odd numbers can be expressed as a square, namely <math>1+3+5+7+...+2017+2019 = 1010^2</math>. 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 <math>1010^2-1009^2-1010</math>. Using difference of squares, we obtain <math>(1010+1009)(1010-1009)-1009 = 2019-1009 = \boxed{2010}</math>
+
Note that the sum of consecutive odd numbers can be expressed as a square, namely <math>1+3+5+7+...+2017+2019 = 1010^2</math>. 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 <math>1010^2-1009^2-1009</math>. Using difference of squares, we obtain <math>(1010+1009)(1010-1009)-1009 = 2019-1009 = \boxed{2010}</math>
 
~SigmaPiE
 
~SigmaPiE
 
==See Also==
 
==See Also==

Revision as of 12:04, 23 October 2021

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 $1010^2-1009^2-1009$. Using difference of squares, we obtain $(1010+1009)(1010-1009)-1009 = 2019-1009 = \boxed{2010}$ ~SigmaPiE

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS