Věrka used three given digits to form different three-digit numbers, using all three digits in each number. She formed all possible numbers and when she added them up, she got 1221.

What digits did Věrka use? Determine five possibilities.

(K. Pazourek)

Hint. Could any of the digits be 0?
For Věrka to be able to form a three-digit number from the given digits, at least one digit had to be non-zero. For the sum of all formed numbers to be a four-digit number, she had to form at least two three-digit numbers. From this, it follows that the given digits could not be the same, and at most one of them could be zero.

In the following, we will check all possibilities (different letters denote different non-zero digits):

- With one digit being zero and the other two being different non-zero digits, the numbers $\overline{a b 0}, \overline{b a 0}, \overline{a 0 b}, \overline{b 0 a}$ can be formed. Their sum is equal to

$$
(2 a+2 b) \cdot 100+(a+b) \cdot 10+(a+b)=211(a+b) .
$$

The number 1221, however, is not divisible by 211, so it is not possible to obtain the required sum in this way.

- With one digit being zero and the other two being the same, the numbers $\overline{a a 0}, \overline{a 0 a}$ can be formed. Their sum is equal to

$$
2 a \cdot 100+a \cdot 10+a=211 a .
$$

For the same reason as in the previous case, it is not possible to obtain the required sum in this way.

- With three mutually different non-zero digits, the numbers $\overline{a b c}, \overline{a c b}, \overline{b a c}, \overline{b c a}$, $\overline{c a b}, \overline{c b a}$ can be formed. Their sum is equal to

$$
(2 a+2 b+2 c) \cdot 100+(2 a+2 b+2 c) \cdot 10+(2 a+2 b+2 c)=222(a+b+c) .
$$

The number 1221, however, is not divisible by 222, so it is not possible to obtain the required sum in this way.

- With two identical non-zero digits and a third different non-zero digit, the numbers $\overline{a a b}, \overline{a b a}, \overline{b a a}$ can be formed. Their sum is equal to

$$
(2 a+b) \cdot 100+(2 a+b) \cdot 10+(2 a+b)=111(2 a+b) .
$$

Since $1221=111 \cdot 11$, the required sum can be obtained precisely when $2 a+b=11$ or $b=11-2 a$. Among the digits from 1 to 9, the pairs that satisfy this condition are:

$$
\begin{aligned}
& a=1,2,3,4,5 \\
& b=9,7,5,3,1
\end{aligned}
$$

Věrka could use any of the following triplets of digits:

$$
(1,1,9), \quad(2,2,7), \quad(3,3,5), \quad(4,4,3), \quad(5,5,1) .
$$