$10 \cdot 62$ Find all two-digit numbers that are divisible by the product of their digits.
(Kyiv Mathematical Olympiad, 1957)
［Solution］Let the required two-digit number be
$$
\overline{a b}=10 a+b,
$$

where $a, b \in\{0,1,2, \cdots, 9\}, a \neq 0$.
From the problem, we have
$$
10 a+b=k a b \text { ( } k \text { is a natural number). }
$$

From this equation, we get
$$
b=a(k b-10) .
$$

Thus,
$$
a \mid b \text {. }
$$

If $a=b$, then we have
$$
11 a=k a^{2}, \quad k a=11 .
$$

Thus,
$$
a=1, \quad k=11 .
$$

The only two-digit number obtained is 11.
If $a \neq b$, and $a \mid b$, then $a \leqslant 4$.
If $a=4$, then $b=8$ is the only possibility. However,
$$
48 \neq k \cdot 4 \cdot 8 \text {. }
$$

So there is no solution when $a=4$.
If $a=3$, then
$$
30=(3 k-1) b,
$$

By $a \mid b$ and $b \in\{0,1, \cdots, 9\}$, we know ${ }^{*} b=6$.
Clearly,
$$
36=2 \cdot 3 \cdot 6 \text {. }
$$

If $a=2$, then
$$
20=(2 k-1) b,
$$

By $a \mid b$ and $b \in\{0,1, \cdots, 9\}$, we know $b=4$.
Clearly,
$$
24=3 \cdot 2 \cdot 4 \text {. }
$$

If $a=1$, from
$$
10=(k-1) b
$$

we can find
$$
b=2, \quad b=5 .
$$

Clearly,
$$
12=6 \cdot 1 \cdot 2, \quad 15=3 \cdot 1 \cdot 5 .
$$

Therefore, the required two-digit numbers are 5 in total:
$$
11,12,15,24,36 .
$$