3. (2 points) Petya, Vasya, and Tёma are playing a game. Petya moves first, then Vasya, then Tёma, and then Petya again, and so on. Initially, the number $123456789 \ldots 123456789$ (the sequence 123456789 repeats 2015 times) is written on the board. On their turn, each player can erase one of the digits of the number written on the board and add it to the resulting number. The game ends when only one digit remains on the board. Petya wins if the remaining digit is 1, 4, or 7; Vasya wins if it is 2, 5, or 8; otherwise, Tёma wins. Who will win with correct play?
Answer: Tёma.

Solution: Note that the original number is divisible by three: the sum of its digits is $45 \cdot 2015$.

The allowed moves do not change divisibility by 3. Indeed, the number $a \cdot 10^{k+1}+b \cdot 10^{k}+c$ transforms into $a \cdot 10^{k}+b+c$. The difference between these numbers is $9 a \cdot 10^{k}+\left(10^{k}-1\right) b$, which is divisible by 3.

Thus, at the end, one of the digits divisible by 3 is obtained, meaning Tёma wins.