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 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

1991 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AJHSME/AMC 8 Problems and Solutions

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1991 AJHSME Problems/Problem 12

Contents

Problem

Solution 1

Solution 2

See Also