1991 AMC 8 Problem 12

Attempt Problem 12 of the 1991 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1991 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

12.

If 2+3+43=1990+1991+1992N, \begin{aligned} &\frac{2+3+4}{3} \\ &= \frac{1990+1991+1992}{N}, \end{aligned} then N=N =
33
66
19901990
19911991
19921992
Answer: D
Concepts:arithmetic sequencemean
Small hint:
Simplify the left side first
Big hint:
The sum of three consecutive integers divided by the middle one equals 33
Solution:
The left side is 93=3.\dfrac{9}{3} = 3. The right side is 5973N,\dfrac{5973}{N}, and setting it equal to 33 gives N=1991.N = 1991.
Equivalently, (k−1)+k+(k+1)=3k,(k-1) + k + (k+1) = 3k, so dividing by 33 leaves the middle term. Here the middle term is 1991.1991.
Thus, the correct answer is D .
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