Let's find a three-digit number such that the product of the digits in the tens and units place is equal to the two-digit number formed by the leftmost two digits.
Let the sought three-digit number be

$$
100 x+10 y+z
$$

then according to the condition

$$
y z=10 x+y
$$

or

$$
y(z-1)=10 x
$$

If the product $y(z-1)$ has one factor even, the other must be 5, because neither factor can be greater than 9, and they cannot both be even or equal to 5 at the same time.

If $y=5, z-1$ is even, that is, $z$ is odd and greater than 1, its only values are:

$$
3, \quad 5, \quad 7, \quad 9
$$

and the corresponding solutions are:

$$
153, \quad 255, \quad 357, \quad 459
$$

If $z-1=5$, that is, $z=6$, then $y$ is even and is one of the following numbers:

$$
2, \quad 4, \quad 6, \quad 8
$$

The other four solutions are therefore:

$$
126, \quad 246, \quad 366, \quad 486.
$$

The problem was solved by: Heymann Tivadar, real gymnasium VIII. grade, Györ, Imre János grammar school VIII. grade, Nyíregyháza.