Eva had six cards with digits 1, 2, 3, 4, 5, 6 (each card had one digit). By rearranging all six cards, she formed groups of numbers and observed their properties. She found that she could arrange different groups of three numbers, each of which is divisible by six.

Determine all such groups of three numbers.

(E. Novotná)
A number is divisible by six if and only if it is even and the sum of its digits is divisible by three. The three created numbers must end in an even digit, and there are exactly three available:

$$
2, \quad 4, \quad 6 \text {. }
$$

The only freedom in creating the numbers is thus in the preceding positions.

Adding the digit $3 \mathrm{k}$ to any number does not affect its divisibility or non-divisibility by three. We can therefore prioritize adding the remaining digits 1 and 5 in such a way that divisibility by three is ensured. This can be done only as follows:

$$
12, \quad 54, \quad 6 .
$$

If adding 3 results in a three-digit number, we can swap the digits in the first two positions without compromising any of the controlled properties. In total, this gives us five valid options:

$$
\begin{array}{rrr}
312, & 54, & 6, \\
132, & 54, & 6, \\
12, & 354, & 6, \\
12, & 534, & 6, \\
12, & 54, & 36 .
\end{array}
$$

Evaluation. 2 points for preparatory observations on divisibility; 2 points for summarizing and listing the options; 2 points for the quality of the commentary.