If \beginaligned &(2+3+4)/(3) \ &= (1990+1991+1992)/(N), \endaligned then N =

1991 AMC 8 Problem 12 — Solution

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12.

If $\begin{aligned} &\frac{2+3+4}{3} \\ &= \frac{1990+1991+1992}{N}, \end{aligned}$ then $N =$

$3$

$6$

$1990$

$1991$

$1992$

Difficulty rating: 820

The left side is $\dfrac{9}{3} = 3.$ The right side is $\dfrac{5973}{N},$ and setting it equal to $3$ gives $N = 1991.$

Equivalently, $(k-1) + k + (k+1) = 3k,$ so dividing by $3$ leaves the middle term. Here the middle term is $1991.$

Thus, the correct answer is D .