$14 \cdot 5$ If a number is randomly selected from a set of five-digit numbers, where the sum of its digits is 43, what is the probability that this number is divisible by 11?
(A) $\frac{2}{5}$.
(B) $\frac{1}{5}$.
(C) $\frac{1}{6}$.
(D) $\frac{1}{11}$.
(E) $\frac{1}{15}$.
(21st American High School Mathematics Examination, 1970)
[Solution] In decimal, the maximum digit is 9, so the sum of the five digits $d_{1}+d_{2}+d_{3}+d_{4}+d_{5}$ can be at most 45. If the sum of the digits is 43, there are the following cases:
(1) One of the digits is 7, and the rest are 9, giving 5 possibilities:
79999, 97999, 99799, 99979, 99997.
(2) Two of the digits are 8, and the rest are 9, giving 10 possibilities:
$$
\begin{array}{l}
\text { 88999, 89899, 89989, 89998, } 98899 \text {, } \\
\text { 98989, 98998, 99889, 99898, } 99988 . \\
\end{array}
$$

Among the above numbers, only $97999, 99979, 98989$ are divisible by 11.
In summary, the probability is
$$
p=\frac{3}{15}=\frac{1}{5} \text {. }
$$

Therefore, the answer is $(B)$.