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