Difference between revisions of "2018 AMC 8 Problems/Problem 5"
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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> | ||
| − | == | + | ==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
Contents
Problem 5
What is the value of 
?
Solution
Rearranging the terms, we get 
, and our answer is 
Solution 2
We can rewrite the given expression as 
. The number of 
s is the same as the number of terms in 
. Thus the answer is 
See Also
| 2018 AMC 8 (Problems • Answer Key • Resources) | ||
| 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. 
