Difference between revisions of "1987 AJHSME Problems/Problem 3"
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
2(5 \cdot 180) &= 10 \cdot 180\\ | 2(5 \cdot 180) &= 10 \cdot 180\\ | ||
| − | &= 1800 & \text{ Thus \boxed{E} is the correct answer} | + | &= 1800 & \text{ Thus \boxed{\text{E}} is the correct answer} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Revision as of 13:31, 15 March 2018
Contents
Problem
Solution 1
Find that
![\[(81+83+85+87+89+91+93+95+97+99) = 5 \cdot 180\]](http://latex.artofproblemsolving.com/2/0/3/2037d7cda59bdba39c56561bb33dbe8ae863c61e.png)
Which gives us
Solution 2
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 
.
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. 
