$28 \cdot 34$ integers $a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ satisfy
$$
\frac{5}{7}=\frac{a_{2}}{2!}+\frac{a_{3}}{3!}+\frac{a_{4}}{4!}+\frac{a_{5}}{5!}+\frac{a_{6}}{6!}+\frac{a_{7}}{7!}
$$

For $i=2,3, \cdots, 7$, here $0 \leqslant a_{i}<i$. Then the value of $a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}$ is
(A) 8 .
(B) 9 .
(C) 10 .
(D) 11 .
(E) 12 .
(50th American High School Mathematics Examination, 1999)
[Solution] Clearly, $a_{2}$ can only be 0 or 1. However, if $a_{2}=0$, the maximum value of $\frac{a_{3}}{3!}+\frac{a_{4}}{4!}+\cdots+\frac{a_{7}}{7!}$ is less than $\frac{5}{7}$, which is not appropriate. Therefore, $a_{2}=1$. $a_{3}$ can be $0, 1, 2$. Note that $\frac{2}{3!}>\frac{5}{7}-\frac{1}{2}$, and the maximum value of $\frac{a_{4}}{4!}+\frac{a_{5}}{5!}+\frac{a_{6}}{6!}+\frac{a_{7}}{7!}$ is less than $\frac{5}{7}-\frac{1}{2}$, so $a_{3}=1$. Similarly, we can deduce that $a_{4}=1$.
Thus,
$$
\frac{a_{5}}{5!}+\frac{a_{6}}{6!}+\frac{a_{7}}{7!}=\frac{5}{7}-\frac{1}{2!}-\frac{1}{3!}-\frac{1}{4!}=\frac{1}{168},
$$
and since $\frac{1}{5!}>\frac{1}{168}$, we know $a_{5}=0$.
From $\frac{a_{6}}{6!}+\frac{a_{7}}{7!}=\frac{1}{168}$, we get $7 a_{6}+a_{7}=30$, leading to $a_{6}=4, a_{7}=2$.
Therefore,
$$
a_{2}+a_{3}+a_{4}+a_{5}+a_{6}+a_{7}=9.
$$

Hence, the answer is $(B)$.