1.1. Find the smallest 12-digit natural number that is divisible by 36 and contains each of the 10 digits at least once.
Answer: 100023457896.

Solution. The number must be divisible by 4 and by 9. Since the sum of ten different digits is 45, the sum of the two remaining digits must be 0, 9, or 18. We need the smallest number, so we add two digits 0 to the digits $0,1, \ldots, 9$ and place the digits 10002345 at the beginning of the desired number (this is the smallest possible "start" of the number). The remaining digits $6,7,8,9$ must ensure divisibility by 4. Therefore, the last two digits can be 76, 96, or 68. The smallest option: 7896.