$32 \cdot 38 \quad 1$ to 1990, the number of integers $N$ for which $\frac{N^{2}+7}{N+4}$ is not in lowest terms is
(A) 0.
(B) 86.
(C) 90.
(D) 104.
(E) 105.
(41st American High School Mathematics Examination, 1990)
[Solution] From $\frac{N^{2}+7}{N+4}=\frac{N^{2}-16+23}{N+4}=N-4+\frac{23}{N+4}$,
Since $23$ is a prime number, if $\frac{N^{2}+7}{N+4}$ is not a reduced fraction, then $23 \mid (N+4)$.
When $1 \leqslant N \leqslant 1990$ or $5 \leqslant N+4 \leqslant 1994$, the number of cases where $23 \mid (N+4)$ is
$$
\left[\frac{1994}{23}\right]-\left[\frac{5}{23}\right]=86 \text{ (cases). }
$$

Therefore, the answer is $(B)$.