1. How many numbers with at least four digits are divisible by 9 and can be formed from the digits $1,9,0,1,2,0,1,9$ (each digit can be used as many times as it is listed)?
1. Note that $1+9+0+1+2+0+1+9=23$. For the composed number to be divisible by 9, the sum of its digits must be divisible by 9. Therefore, the sum of the digits of the numbers we are considering must be 9 or 18, i.e., the sum of the unused digits must be 14 or 5. The unused digits, according to the problem, can be at most 4. Note that no 4 or fewer of these digits can sum to 14 (indeed, in that case, one of them would have to be the digit 9, and the sum of the remaining three digits would have to be 5, but the number 5 cannot be obtained as the sum of three of these digits). Further, the number 5 can be obtained as the sum of at most four of these digits in only one way: $2+1+1+1$. Therefore, the only way to form a number of at least four digits from these digits that is divisible by 9 is to discard the digits $2,1,1,1$ and form the number from the remaining digits: $9,9,0,0$. The first position from the left must be the digit 9, and for the next three positions, we have 3 possibilities (depending on which of them is the remaining digit 9). Thus, there are a total of 3 such numbers: 9900, 9090, and 9009.