1991 AJHSME Problems/Problem 12

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 (ProblemsAnswer KeyResources)
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
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS