On each of the three cards, a digit different from zero is written (the digits on different cards are not necessarily different). We know that any three-digit number formed from these cards is divisible by six. Additionally, a three-digit number divisible by eleven can be formed from these cards.

What digits can be on the cards? Determine all possibilities.

(V. Hucíková)

Hint. Which digits cannot be on the cards?
A number is divisible by six if and only if it is divisible by two and simultaneously by three, i.e., if and only if it is even and the sum of its digits is divisible by three. Therefore, on all the cards, there must be even digits, and their sum must be divisible by three. The triplets of digits that meet these two requirements are (disregarding the order):

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

Now, it is necessary to test each triplet to see if it can form a three-digit number divisible by 11 when the digits are arranged in some way. This does not happen except for the third triplet, and for this triplet, two of the six possible arrangements give a number divisible by 11:

$$
462: 11=42, \quad 264: 11=24
$$

The digits on the cards can only be 2, 4, and 6.

Note. When verifying divisibility by 11, it is advantageous to use the following criterion: a number is divisible by 11 if and only if the difference between the sum of the digits in even positions and the sum of the digits in odd positions is divisible by 11. (Here $4-6+2=0$ and $2-6+4=0$ are divisible by 11, but for example, $2-4+6=4$ is not.)