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

(Problem 5)
Line 1: Line 1:
 
==Problem 5==
 
==Problem 5==
 
What is the value of <math>1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018</math>?
 
What is the value of <math>1+3+5+\cdots+2017+2019-2-4-6-\cdots-2016-2018</math>?
 +
 +
 +
==Solution==
 +
Rearranging the terms, we get <math>(1-2)+(3-4)+(5-6)+...(2017-2018)+2019</math>, and our answer is <math>-1009+2019=\boxed{1010}, \textbf{(E)}</math>
 +
 +
  
 
<math>\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010</math>
 
<math>\textbf{(A) }-1010\qquad\textbf{(B) }-1009\qquad\textbf{(C) }1008\qquad\textbf{(D) }1009\qquad \textbf{(E) }1010</math>
 
{{AMC8 box|year=2018|num-b=4|num-a=6}}
 
{{AMC8 box|year=2018|num-b=4|num-a=6}}

Revision as of 13:50, 21 November 2018

Problem 5

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


Solution

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


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

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