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AHSME 1979 Problem 6

    32+54+98+1716+3332+6564−7=\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7= (A) −164(B) −116(C) 0(D) 116(E) 164\textbf{(A) }-\frac{1}{64} \textbf{(B) }-\frac{1}{16} \textbf{(C) }0 \textbf{(D) }\frac{1}{16} \textbf{(E) }\frac{1}{64}

    Dataset text NuminaMath-1.5. Quoted with source.

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    -\frac{1}{64}

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