Difference between revisions of "1987 AJHSME Problems/Problem 3"
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<math>2(81+83+85+87+89+91+93+95+97+99)</math> | <math>2(81+83+85+87+89+91+93+95+97+99)</math> | ||
| − | + | The answer is E | |
<math>\begin{align*} | <math>\begin{align*} | ||
& = 2\Big( (81+99) + (83+97) + (85+95) + (87+93) + (89+91) \Big) \\ | & = 2\Big( (81+99) + (83+97) + (85+95) + (87+93) + (89+91) \Big) \\ | ||
Revision as of 19:07, 26 October 2015
Problem
Solution
The answer is E
$\begin{align*}
& = 2\Big( (81+99) + (83+97) + (85+95) + (87+93) + (89+91) \Big) \\
& = 2(180+180+180+180+180) \\
& = 2\cdot 5\cdot 180 \\
& = 1800\rightarrow \boxed{\text{E}}
\end{align*}$ (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. 
