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

m (Solution)
(Alternate Solution)
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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>
 
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>
  
==Alternate Solution==
+
==Solution 2==
  
 
We can rewrite the given expression as <math>1+(3-2)+(5-4)+\cdots +(2017-2016)+(2019-2018)=1+1+1+\cdots+1</math>. The number of <math>1</math>s is the same as the number of terms in <math>1,3,5,7\dots ,2017,2019</math>. Thus the answer is <math>\boxed{\textbf{(E) }1010}</math>
 
We can rewrite the given expression as <math>1+(3-2)+(5-4)+\cdots +(2017-2016)+(2019-2018)=1+1+1+\cdots+1</math>. The number of <math>1</math>s is the same as the number of terms in <math>1,3,5,7\dots ,2017,2019</math>. Thus the answer is <math>\boxed{\textbf{(E) }1010}</math>

Revision as of 15:44, 23 November 2018

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

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 rewrite the given expression as $1+(3-2)+(5-4)+\cdots +(2017-2016)+(2019-2018)=1+1+1+\cdots+1$. The number of $1$s is the same as the number of terms in $1,3,5,7\dots ,2017,2019$. Thus the answer is $\boxed{\textbf{(E) }1010}$

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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