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AMC 8 1991 Problem 12

  • Paper

If 2+3+43=1990+1991+1992N\frac{2+3+4}{3}=\frac{1990+1991+1992}{N}, then N=N= (A) 3(B) 6(C) 1990(D) 1991(E) 1992\text{(A)}\ 3 \text{(B)}\ 6 \text{(C)}\ 1990 \text{(D)}\ 1991 \text{(E)}\ 1992

Dataset text NuminaMath-1.5. Quoted with source.

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1991

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