Problem:
If 32+3+4=N1990+1991+1992, then N=
Answer Choices:
A. 3
B. 6
C. 1990
D. 1991
E. 1992
Solution:
Any fraction of the form k(k−1)+k+(k+1) equals 3, since (k−1)+ k+(k+1)=3k and k3k=3. The denominator of the fraction must equal the middle term of the numerator. Thus N=1991.
OR
Note that
32+3+4=39=33×3=3×33=3×1=3.
Also,
N1990+1991+1992=N3×1991=3×N1991=3×1=3.
Thus N must equal 1991.
Answer: D.
The problems on this page are the property of the MAA's American Mathematics Competitions