Libor is thinking of a three-digit natural number, all of whose digits are odd. If he adds 421 to it, he gets a three-digit number with no odd digits. Find all the numbers Libor could be thinking of.

(L. Šimünek)

Hint. Write the numbers one under the other and think as if you were doing written addition.
Let's denote the slices of the imagined number as $L_{1}, L_{2}, L_{3}$. The addition mentioned in the problem follows this scheme:

$$
\begin{gathered}
L_{1} L_{2} L_{3} \\
\begin{array}{ccc}
4 & 2 & 1 \\
\hline S_{1} & S_{2} & S_{3}
\end{array}
\end{gathered}
$$

For each odd $L_{3}$, the sum $L_{3}+1$ is even. However, for each odd $L_{2}$, the sum $L_{2}+2$ is odd. To still get an even digit in the tens place in the result, there must be a carry-over from the units place. Therefore, $L_{3}$ can only be 9.

For each odd $L_{1}$, the sum $L_{1}+4$ is odd. For the result to have an even digit in the hundreds place, there must be a carry-over from the tens place. Therefore, $L_{2}$ can only be 9 or 7.

There must not be a carry-over from the hundreds place, as the result is supposed to be a three-digit number. Therefore, $L_{1}$ can only be 1 or 3.

By grouping the permissible digits, we get four numbers that Libor could have thought of:

179, 199, 379, and 399.