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

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==Solution 2==
 
==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>
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W can see that the last numbers of each of the sets (even numbers and odd numbers) have a difference of one. So do the second last ones, and so on. Now, all we need to find is the number of intigers in any of the sets (I chose even) to get <math>\boxed{\textbf{(E) }1010}</math> ~avamarora
  
 
==See Also==
 
==See Also==

Revision as of 13:41, 1 November 2020

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 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

W can see that the last numbers of each of the sets (even numbers and odd numbers) have a difference of one. So do the second last ones, and so on. Now, all we need to find is the number of intigers in any of the sets (I chose even) to get $\boxed{\textbf{(E) }1010}$ ~avamarora

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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