Lenka had a paper flower with eight petals. On each petal, there was exactly one digit, and no digit was repeated on any other petal. When Lenka played with the flower, she noticed several things:

- It was possible to tear off four petals such that the sum of the numbers written on them would be the same as the sum of the numbers on the remaining petals.
- It was also possible to tear off four petals such that the sum of the numbers written on them would be twice the sum of the numbers on the remaining petals.
- It was even possible to tear off four petals such that the sum of the numbers written on them would be four times the sum of the numbers on the remaining petals.

Determine which digits could have been written on the petals.

(E. Novotná)
From the first condition, we know that the tickets can be divided into two groups such that the sum of the numbers in one group is the same as the sum of the other. This implies that the sum of all the numbers written on the tickets must be divisible by two. A similar consideration from the second condition leads us to conclude that the sum of all the written numbers is divisible by three; from the last condition, it follows that this sum must also be divisible by five. Therefore, the sum of all the written numbers is divisible by 30.

The sum of all usable digits is $0+1+\ldots+8+9=45$, and thus the above-discussed sum is exactly 30. Therefore, the digits on the tickets could not have included any two digits whose sum was 15. These could have been the digits 7 and 8, or 6 and 9. In both cases, we need to verify whether it is possible to tear off the tickets with the other digits so that the conditions in the problem are met.

- Let's verify the possibility where the digits 7 and 8 were not on the flower:

After tearing according to the first condition, the remaining digits could be, for example, $0+1+5+9=15$,

after tearing according to the second condition, the remaining digits could be, for example, $0+1+3+6=10$, after tearing according to the third condition, the remaining digits could be $0+1+2+3=6$.

- Let's verify the possibility where the digits 6 and 9 were not on the flower:

After tearing according to the first condition, the remaining digits could be, for example, $0+2+5+8=15$, after tearing according to the second condition, the remaining digits could be, for example, $0+1+2+7=10$, after tearing according to the third condition, the remaining digits could be $0+1+2+3=6$.

Therefore, the digits on the petals could have been all the digits except 7 and 8, or all except 6 and 9.

Evaluation. 2 points for deriving that the total sum of the written numbers is divisible by 30; 2 points for determining both pairs of digits that could have been omitted; 2 points for checking whether it is possible to tear off the tickets according to the problem statement.

For a solution containing only one possibility without the principle of divisibility by 30, award no more than 3 points.