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AMC 8 1996 Problem 4

  • Paper

2+4+6+⋯+343+6+9+⋯+51=\dfrac{2+4+6+\cdots + 34}{3+6+9+\cdots+51}= (A) 13(B) 23(C) 32(D) 173(E) 343\text{(A)}\ \dfrac{1}{3} \text{(B)}\ \dfrac{2}{3} \text{(C)}\ \dfrac{3}{2} \text{(D)}\ \dfrac{17}{3} \text{(E)}\ \dfrac{34}{3}

Dataset text NuminaMath-1.5. Quoted with source.

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\frac{2}{3}

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