Difference between revisions of "1992 AJHSME Problems/Problem 1"
5849206328x (talk | contribs) (Created page with '==Problem== <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)}\…') |
5849206328x (talk | contribs) (→Solution) |
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\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+1+1+1+1} \\ | &= \dfrac{1+1+1+1+1}{1+1+1+1+1} \\ | ||
| − | &= \boxed{ | + | &= 1 \rightarrow \boxed{\text{A}}. |
\end{align*}</cmath> | \end{align*}</cmath> | ||
