Difference between revisions of "1992 AJHSME Problems/Problem 1"
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9} &= \dfrac{(10-9)+(8-7)+(6-5)+(4-3)+(2-1)}{1+(-2+3)+(-4+5)+(-6+7)+(-8+9)} \\ | \dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9} &= \dfrac{(10-9)+(8-7)+(6-5)+(4-3)+(2-1)}{1+(-2+3)+(-4+5)+(-6+7)+(-8+9)} \\ | ||
| − | &= \dfrac{1+1+1+1+1}{1 | + | &= \dfrac{1+1+1+1+1}{1+1+1+1+1} \\ |
&= 1 \rightarrow \boxed{\text{B}}. | &= 1 \rightarrow \boxed{\text{B}}. | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
Latest revision as of 15:00, 9 November 2019
Problem
Solution
See Also
| 1992 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by First Problem |
Followed by Problem 2 | |
| 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. 
