Difference between revisions of "2013 AMC 8 Problems/Problem 3"
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==Solution== | ==Solution== | ||
Notice that we can pair up every two numbers to make a sum of 1: | Notice that we can pair up every two numbers to make a sum of 1: | ||
| − | <cmath> \begin{eqnarray*}(-1 + 2 - 3 + 4 - \cdots + 1000) &=& ((-1 + 2) + (-3 + 4) + \cdots + (-999 + 1000)) \\ &=& (1 + 1 + \cdots + 1) \\ &=& 500</cmath> | + | <cmath> \begin{eqnarray*}(-1 + 2 - 3 + 4 - \cdots + 1000) &=& ((-1 + 2) + (-3 + 4) + \cdots + (-999 + 1000)) \\ &=& (1 + 1 + \cdots + 1) \\ &=& 500\end{eqnarray*}</cmath> |
Therefore, the answer is <math>4 \cdot 500= \boxed{\textbf{(E)}\ 2000}</math>. | Therefore, the answer is <math>4 \cdot 500= \boxed{\textbf{(E)}\ 2000}</math>. | ||
Revision as of 18:01, 10 March 2015
Problem
What is the value of 
?
Solution
Notice that we can pair up every two numbers to make a sum of 1:
Therefore, the answer is 
.
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
