If 2+3+43=1990+1991+1992N, then N=

Note that for all integers n 0, (n-1)+n+(n+1)n=3. Thus, we must have N=1991 1991.

jmc

algebra junior

Problem

\frac{2 + 3 + 4}{3} = \frac{1990 + 1991 + 1992}{N}

, then

N =

(A)

3

(B)

6

(C)

1990

(D)

1991

Solution

Note that for all integers

n \neq = 0

\frac{( n - 1 ) + n + ( n + 1 )}{n} = 3.

Thus, we must have

N = 1991 \to 1991

Final answer