$32 \cdot 48$ Among the integers $n$ between 1 and 1990, the number of $n$ that allows $x^{2}+x-3 n$ to be factored into the product of two linear factors with integer coefficients is
(A) 1990.
(B) 75.
(C) 50.
(D) 44.
(2nd "Five Sheep Cup" Junior High School Mathematics Competition, 1990)
[Solution] Let $x^{2}+x-3 n=(x+a)(x+b)$, where $a, b$ are integers.
Then
$$
a+b=1, \quad a b=-3 n .
$$

Thus, $a, b$ have opposite signs. Without loss of generality, assume $a>0$.
Also, $n=-\frac{a b}{3}=\frac{a(a-1)}{3}$, so in $1 \leqslant n \leqslant 1990$, $a$ can take the values
$$
3,4 ; 6,7 ; 9,10 ; \cdots ; 72,73 ; 75,76 \text {. }
$$

There are a total of 50 possible values.
Therefore, the answer is $(C)$.