2. Honza has three cards, each with a different non-zero digit. The sum of all three-digit numbers that can be formed from these cards is a number 6 greater than three times one of them. What digits are on the cards?
2. Let $\overline{a b c}$ denote the three-digit number whose triple is mentioned in the text of the problem. Thus, the equation

$$
3 \overline{a b c}+6=\overline{a b c}+\overline{a c b}+\overline{b a c}+\overline{b c a}+\overline{c a b}+\overline{c b a}
$$

holds. Since each of the digits $a, b, c$ appears twice in the units, tens, and hundreds places on the right side, we can rewrite the equation as

$$
300 a+30 b+3 c+6=222 a+222 b+222 c \quad \text { or } \quad 78 a+6=192 b+219 c .
$$

Dividing by 3, we get the equation $26 a+2=64 b+73 c$, from which we see that $c$ is an even digit. Therefore, $c \geq 2$, which, together with the obvious inequality $b \geq 1$ (we recall that all three unknown digits are non-zero), leads to the estimate

$$
64 b+73 c \geq 64+146=210 .
$$

Thus, $26 a+2 \geq 210$, from which $a \geq (210-2): 26=8$, so the digit $a$ is either 8 or 9. For $a=8$, however, equality holds in the inequality from the previous sentence, so it must be $b=1$ and $c=2$ (and the equation given in the problem is then satisfied). For $a=9$, we get the equation

$$
64 b+73 c=26 \cdot 9+2=236
$$

from which it follows that $c$ is both divisible by four and less than 4, which cannot happen simultaneously.

Answer: The digits on the cards are 8, 2, and 1.

Note. The equation $26 a+2=64 b+73 c$ for the unknown (non-zero and distinct) digits $a, b, c$ can be solved by more complete and systematic methods, we have only presented one of them.

For a complete solution, award 6 points, of which up to 3 points for correctly setting up the equation, expanding the decimal representations of the numbers, and simplifying to a linear equation in $a, b, c$. The remaining up to 3 points should be awarded for the process of finding the solution to the derived equation, with 1 point for simply guessing the desired triplet.