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AMC 8 1992 Problem 1

  • Paper

10−9+8−7+6−5+4−3+2−11−2+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}= (A) −1(B) 1(C) 5(D) 9(E) 10\text{(A)}\ -1 \text{(B)}\ 1 \text{(C)}\ 5 \text{(D)}\ 9 \text{(E)}\ 10

Dataset text NuminaMath-1.5. Quoted with source.

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