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AHSME 1999 Problem 1

    1−2+3−4+⋯−98+99=1 - 2 + 3 -4 + \cdots - 98 + 99 = (A) −50(B) −49(C) 0(D) 49(E) 50\mathrm{\textbf{(A)} \ -50 } \mathrm{\textbf{(B)} \ -49 } \mathrm{\textbf{(C)} \ 0 } \mathrm{\textbf{(D)} \ 49 } \mathrm{\textbf{(E)} \ 50 }

    Dataset text NuminaMath-1.5. Quoted with source.

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    50

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