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AMC 8 1990 Problem 16

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1990−1980+1970−1960+⋯−20+10=1990-1980+1970-1960+\cdots -20+10 = (A) −990(B) −10(C) 990(D) 1000(E) 1990\text{(A)}\ -990 \text{(B)}\ -10 \text{(C)}\ 990 \text{(D)}\ 1000 \text{(E)}\ 1990

Dataset text NuminaMath-1.5. Quoted with source.

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1000

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