# 1991 AJHSME Problems/Problem 12

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## Problem

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

$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 1990 \qquad \text{(D)}\ 1991 \qquad \text{(E)}\ 1992$

## Solution 1

Note that for all integers $n\neq 0$, $$\frac{(n-1)+n+(n+1)}{n}=3.$$ Thus, we must have $N=1991\rightarrow \boxed{\text{D}}$.

## Solution 2

As we know that $\frac{1990+1991+1992}{n}$ has to be some multiple of $\frac{2+3+4}{3}$, then we know that the first equation is $995$(1990/2) times bigger than the second one(in my solution), so the bottom must be $3\cdot995=\boxed{\text{(D)}1991}$-fn106068

## See Also

 1991 AJHSME (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

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