1987 AJHSME Problems/Problem 3
Problem
Solution
Pair the least with the greatest, second least with the second greatest, etc, until you have five pairs, each adding up to 
= 
= 
= 
= 
= 
. Since we have 
pairs, we multiply by to get 
. But since we have to multiply by 2 (remember the 2 at the beginning of the parentheses!), we get 
, which is $\text{(E)$ (Error compiling LaTeX. Unknown error_msg).
See Also
| 1987 AJHSME (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. 
