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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

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}}.

See Also

1991 AJHSME (Problems • Resources)
Preceded by
Problem 11 		Followed by
Problem 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
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