$13 \cdot 68$ Let $x_{1}, x_{2}, \cdots, x_{n}$ be a sequence of numbers satisfying the following conditions:
(i) $-1 \leqslant x_{i} \leqslant 2$, for $i=1,2,3, \cdots, n$.
(ii) $x_{1}+x_{2}+\cdots+x_{n}=19$ and
(iii) $x_{1}^{2}+x_{2}^{2}+\cdots+x_{m}^{2}=99$.

Let $m$ and $M$ be the minimum and maximum values of $x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}$, respectively. Then the value of $M / m$ is
(A) 3.
(B) 4.
(C) 5.
(D) 6.
(E) 7.
(50th American High School Mathematics Examination, 1999)
[Solution] Given $-1 \leqslant x_{i} \leqslant 2$, let's assume among these $n$ $x_{i}$, there are $p$ $-1$s, $g$ $0$s, $r$ $1$s, and $s$ $2$s. At this point, we have
$$
\left\{\begin{array}{l}
-p+r+2 s=19 \\
p+r+4 s=99 .
\end{array}\right.
$$

From the above, we can deduce that $r+3 s=59, r+s=40$.
Since $p, q, r, s$ are all non-negative integers, then $s_{\text {min }}=0, s_{\text {max }}=19$.
And $x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=-p+r+8 s$,
Also, $\because \quad-p+r+8 s=19+6 s$ (from equation (1))
$\therefore \quad \frac{M}{m}=\frac{19+6 \cdot 19}{19+6 \cdot 0}=7$.
Therefore, the answer is $(E)$.