$29 \cdot 33$ How many integers $n$ between 1 and 100 allow $x^{2}+x-n$ to be factored into the product of two linear factors with integer coefficients
(A) 0 .
(B) 1 .
(C) 2 .
(D) 9 .
(E) 10 .
(40th American High School Mathematics Examination, 1989)
[Solution] Let $x^{2}+x+n=(x-a)(x-b)$,
then $a+b=-1$, and $ab=-n$.
That is, the absolute values of $a$ and $b$ differ by 1, and the absolute value of the negative integer is larger. Also, $ab=-n$, with $n$ ranging from 1 to 100. Thus, we have the table:
\begin{tabular}{r|rrrrrrrrr}
\hline$a$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline$b$ & -2 & -3 & -4 & -5 & -6 & -7 & -8 & -9 & -10 \\
\hline$n$ & 2 & 6 & 12 & 20 & 30 & 42 & 56 & 72 & 90 \\
\hline
\end{tabular}

Thus, there are 9 values of $n$ that satisfy the conditions.
Therefore, the answer is $(D)$.