Consider the product $A . B$.

On an infinite tape, numbers are written in a row. The first is one, and each subsequent number is obtained by adding to the previous number its smallest non-zero digit in its decimal representation. How many digits are in the decimal representation of the number that stands in the $9 \cdot 1000^{1000}-$th position in this sequence?
Since each number in the sequence, starting from the second, is 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 the $n$-th number in the sequence be denoted by $a_{n}$, and let $k$ be the smallest index for which the number $a_{k}$ has 3002 digits.

Consider the numbers from 0 to $10^{3001}-1$ that do not contain the digit 1 in their decimal representation. By padding each number 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 $a_{1}, \ldots, a_{k-1}$ (all of which do not exceed $10^{3001}-1$), there are no more than $9^{3001}$ numbers that do not contain the digit 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 it follows that

$k \geq 10^{3001} - 8 \cdot 9^{3001} > 9 \cdot 10^{3000}$. Indeed, to prove this, it is sufficient to show that $8 \cdot 9^{3001} > 72$. According to Bernoulli's inequality (see problem 30899), $\left(\frac{10}{9}\right)^{3000} > 1 + \frac{3000}{9} > 334$.

Thus, the number with 3002 digits appears in the sequence beyond the position specified in the problem.

## Answer

3001 digits.