10.1. Natural numbers starting from 1 are written in a row. This results in a sequence of digits: 1234567891011121314... What digit is in the 2021st position?
Answer. 1.

Solution. Note that the sum of the digits of all single-digit and two-digit numbers is $1892021$. Therefore, the digit in the 2021st position belongs to the recording of some three-digit number.

Let $x$ be some three-digit number, then the sum of the digits in the sequence from 1 to $x$ is $n=189+3(x-99)$. We have $189+3(x-99)<2021, 3x<2129, x<709 \frac{2}{3}$. The sequence of numbers from 1 to 709 contains $189+3 \cdot(709-99)=2019$ digits. Therefore, the digit in the 2021st position is the second digit of the number 710, which is 1.