1. A student sequentially writes down the natural numbers from 1 to 1000. Which digit stands at the 1992nd position?
Solution. By counting the single-digit and double-digit numbers, the student counted $9 \cdot 1 + 90 \cdot 2 = 189$ digits. Thus, there are still $1992 - 189 = 1803$ digits left, and $1803 \div 3 = 601$, meaning the student counted the first 601 three-digit numbers. Therefore, the 1992nd digit is the third digit of the 601st three-digit number, which is 700. Consequently, the required digit is 0.