323. Find all two-digit numbers that are divisible by the product of their digits.
Let's denote a two-digit number as $\overline{a b}$. Then

$$
\overline{a b}: a b, \quad(10 a+b): a b
$$

Since the sum $10 a+b$ is divisible by $a$ and the first term $10 a$ is divisible by $a$, the second term $b$ is also divisible by $a: b=k a$, where $k$ is an integer non-negative number less than 10.

Similarly, the sum $10 a+b$ is divisible by $b$, and the second term is divisible by $b$, so the first term is also divisible by $b$:

$$
10 a: b, \quad 10 a: k a
$$

from which $10: k$. Therefore, $k$ can only take the values 1, 2, or 5. Let's consider all three cases.

1) Let $k=1$.

In this case, $b=a$. The divisibility $(10 a+b): a b$ turns into the divisibility $11 a: a^{2}$, i.e., $11: a$. Thus, $a=1, b=1$. We get the number $\overline{a b}=11$.

2) Let $k=2$.

Then $b=2 a$. Therefore,

$$
(10 a+2 a): 2 a \cdot a, \quad 12 a: 2 a^{2}, \quad 6: a
$$

From this, $a \in\{1,2,3,6\}$. The value $a=6$ should be immediately discarded, as then $b=12$. If $a=1,2$ or 3, then $b=2,4$ or 6, respectively. We get the numbers 12, 24, and 36.

3) Let $k=5$.

Then $b=5 a$. Thus,

$$
(10 a+5 a): 5 a^{2}, \quad 3: a
$$

from which $a=1, b=5$ (the case $a=3$ is impossible). Therefore, $\overline{a b}=15$.

Answer: $11,12,24,36,15$.