Jóska - observing the license plate numbers - saw an interesting three-digit number. The number could be split into two by placing a dash between the digits, such that three times the product of the resulting two numbers gave back the original number. Let's find all three-digit numbers with this property.
Let's denote the three-digit number in the usual way as $100a + 10b + c$, where $a \neq 0$ and $0 \leq b, c \leq 9$. The division can be done in two ways: either the beginning is a two-digit number and the end is a one-digit number, or vice versa.

In the first case, according to the problem statement, the following equality holds:

$$
3(10a + b)c = 100a + 10b + c
$$

Rearranging, we get:

$$
(10a + b)(3c - 10) = c
$$

$3c - 10 \neq 0$, because $c$ is an integer and $10a + b$ is definitely a two-digit number ($a \neq 0$), so $c$ cannot be 0. However, multiplying a two-digit number by a non-zero integer cannot result in a one-digit number. Therefore, there is no solution in this case.

If, however, we divide the number such that the beginning is a one-digit number, we get the equation:

$$
3(10b + c)a = 100a + 10b + c
$$

Factoring, we get:

$$
(3a - 1)(10b + c) = 100a
$$

This will only be an integer if $3a - 1$ is a divisor of 100, since $a$ and $(3a - 1)$ have no common divisor greater than 1. It is easy to show that this only holds for $a = 1$, $a = 2$, and $a = 7$.

If $a = 1$, then $10b + c = \frac{100}{2} = 50$, so $b = 5$ and $c = 0$. The three-digit number is 150, and indeed $50 \cdot 3 \cdot 1 = 150$.

If $a = 2$, then $10b + c = \frac{200}{5} = 40$, so $b = 4$ and $c = 0$. The three-digit number is 240, and $40 \cdot 3 \cdot 2 = 240$.

Finally, if $a = 7$, then $10b + c = \frac{700}{20} = 35$, so $b = 3$ and $c = 5$, and the three-digit number is 735.

(Indeed, $35 \cdot 3 \cdot 7 = 735$.)

There are no other solutions.

Újhelyi Tamara (Zalaegerszeg, Zrínyi M. Gimn., IV.o.t.)

Note. Two submissions only sent a computer program. According to the September competition rules, a computer program alone is not accepted as a solution. (L. 1996/6. issue, p. 360.)