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

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

Revision as of 14:23, 30 May 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

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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