2. Oleg and Sergey take turns writing down one digit from left to right until a nine-digit number is formed. At the same time, they cannot write down digits that have already been written. Oleg starts (and finishes). Oleg wins if the resulting number is divisible by 4; otherwise, Sergey wins. Who will win with correct play? (O. Dmitriev, R. Zhenodarov)
Answer. Sergey. First solution. In his first three moves, Sergey can ensure that the digits 2 and 6, and another even digit, for example, 0, are definitely used. With his last move, Sergey places any unused odd digit. If Oleg places an odd digit last, the number will be odd. If he places one of the remaining even digits - 4 or 8, the number will be even but not divisible by 4. In both cases, Oleg loses. Second solution. In his first three moves, Sergey can ensure that the digits: $4,8,0$ are definitely used. Then, before his last move, there are two possible scenarios. First: the digit 2 has not been used yet. Then he places the digit 2, and Oleg can only place the digit 6 at the end of the number (otherwise, the number is not divisible by 2), but such a number is not divisible by 4. Second: the digit 2 has been used. Then at most one even digit - 6 - remains. Sergey places it, and if it is not available, any remaining digit. Then Oleg is forced to place an odd digit, and the number does not divide by 2.