Difference between revisions of "1992 AJHSME Problems/Problem 1"
(→Solution) |
|||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | <math>\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2 | + | <math>\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2-3-4-5-6-7-8-9}=</math> |
<math>\text{(A)}\ -1 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10</math> | <math>\text{(A)}\ -1 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 10</math> | ||
| Line 8: | Line 8: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| − | \dfrac{10-9+8-7+6-5+4 | + | \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> | ||
