Difference between revisions of "2013 AMC 8 Problems/Problem 3"
Dabthemath (talk | contribs) (→Problem) |
(→Solution) |
||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | |||
| − | |||
| − | + | . | |
| + | We see that if we group two numbers at a time, we see that it would make -1 on each. There are 500 pairs so the answer is -1 times 500 equals to \cdot 500= \boxed{\textbf{(E)}\ 2000}$. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=2|num-a=4}} | {{AMC8 box|year=2013|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:50, 2 November 2020
Problem
What is the value of 
?
Solution
.
We see that if we group two numbers at a time, we see that it would make -1 on each. There are 500 pairs so the answer is -1 times 500 equals to \cdot 500= \boxed{\textbf{(E)}\ 2000}$.
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. 
