$32 \cdot 42$ Suppose the $n$ digits of a certain $n$-digit positive integer are a permutation of $\{1,2, \cdots, n\}$, and if the integer formed by its first $k$ digits is divisible by $k$, where $k=1,2, \cdots, n$, then this $n$-digit number is called a “good number”. For example, 321 is a three-digit “good number”. Because 1 divides 3, 2 divides 32, and 3 divides 321. The number of six-digit “good numbers” is
(A) 0.
(B) 1.
(C) 2.
(D) 3.
(E) 4.
(42nd American High School Mathematics Examination, 1991)
[Solution] Let the six-digit number be $\overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}}$, where $a_{i} \in\{1,2,3,4,5,6\}, i=$ $1,2,3,4,5,6$.
According to the problem, we have
$1\left|a_{1}, 2\right| \overline{a_{1} a_{2}}, 3\left|\overline{a_{1} a_{2} a_{3}}, 4\right| \overline{a_{1} a_{2} a_{3} a_{4}}$,
$51 \overline{a_{1} a_{2} a_{3} a_{4} a_{5}}, 61 \overline{a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}}$.
Clearly, $a_{2}, a_{4}, a_{6}$ are even numbers, and $a_{5}$ is 5, so $a_{1}, a_{3}$ can only be 1 or 3.
Also, from 3 । $\overline{a_{1} a_{2} a_{3}}$ we know 3 । $\left(a_{1}+a_{2}+a_{3}\right)=4+a_{2}$, hence $a_{2}=2$.
This number could be $\overline{123 a_{4} 5 a_{6}}$ or $\overline{321 a_{4} 5 a_{6}}$.
That is, $123456,123654,321456,321654$.
Upon verification, 123654 and 321654 meet the conditions.
Therefore, the answer is $(C)$.