8. (I. Bogdanov) On an infinite tape, numbers are written in a row. The first is one, and each subsequent number is obtained from the previous one by adding to it the smallest non-zero digit of its decimal representation. How many digits are in the decimal representation of the number standing in this sequence at the $9 \cdot 1000^{1000}$-th position?
Answer: 3001. Solution. Since each number in the sequence, starting from the second, is at least one greater than the previous one, the $9 \cdot 1000^{1000}$-th number in the sequence is greater than $9 \cdot 1000^{1000}$, meaning it has at least 3001 digits. Let's denote the $n$-th number in the sequence by $a_{n}$, and let $k$ be the smallest index such that the number $a_{k}$ has 3002 digits. If we prove that $k > 9 \cdot 1000^{1000}$, then we will have that the $9 \cdot 1000^{1000}$-th number in the sequence has no more than 3001 digits, meaning it has exactly 3001 digits.

Consider the numbers from 0 to $10^{3001}-1$ that do not contain the digit 1 in their decimal representation. By padding each of these numbers with leading zeros to 3001 digits, we obtain all sequences of length 3001 consisting of digits other than 1. There are $9^{3001}$ such sequences. Therefore, among the numbers $a_{1}, \ldots, a_{k-1}$, there are no more than $9^{3001}$ numbers that do not contain the digit 1 (since all of them are less than or equal to $10^{3001}-1$).

Now consider the process of obtaining the number $a_{k}$ from $a_{1}$. On each of the $k-1$ steps, a number from 1 to 9 is added, and the number of steps on which a number other than 1 is added does not exceed $9^{3001}$. Therefore,

$$
10^{3001}-1 \leq a_{k}-a_{1} \leq 9 \cdot 9^{3001} + 1 \cdot (k-1-9^{3001}) = k-1 + 8 \cdot 9^{3001}
$$

from which $k \geq 10^{3001} - 8 \cdot 9^{3001}$. It remains to show that $10^{3001} - 8 \cdot 9^{3001} > 9 \cdot 10^{3000}$. To do this, it is sufficient to prove that $9^{3002} < 10^{3000}$. Note that $9^{7} = 4782969 < 5 \cdot 10^{6}$, from which $9^{28} < 5^{4} \cdot 10^{24} < 10^{27}$ and $9^{56} < 10^{54}$. Therefore, $9^{3002} = 9^{56} \cdot 9^{2946} < 10^{54} \cdot 10^{2946} = 10^{3000}$.