Difference between revisions of "1992 AJHSME Problems/Problem 1"

Revision as of 20:24, 3 October 2009

Problem

$\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9}=$

$\text{(A)}\ -1 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10$

Solution

$\begin{align*} \dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9} &= \dfrac{(10-9)+(8-7)+(6-5)+(4-3)+(2-1)}{1+(-2+3)+(-4+5)+(-6+7)+(-8+9)} \\ &= \dfrac{1+1+1+1+1}{1+1+1+1+1} \\ &= \boxed{1}. \end{align*}$

1992 AJHSME (Problems • Answer Key • Resources)
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Difference between revisions of "1992 AJHSME Problems/Problem 1"

Revision as of 20:24, 3 October 2009

Problem

Solution

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