A number $A$ has every digit, starting from the second one, 1 greater than the previous digit. If we multiply $A$ by 9 and add the number of its digits, the resulting number will have its last digit as 0, and all other digits will be 1. What could the number $A$ be?
Let $A$ be an $n (\geq 1)$-digit number, whose digits, according to the condition, are

$$
x, x+1, \ldots, x+n
$$

where $x+n-1 \leq 9$, hence $n \leq 9$.

We know that $9A + n$ ends in $0$, i.e., $10 \mid 9A + n = 10A + n - A$; this is divisible by 10 if $n - A$ is also divisible, meaning $A$ ends in the same digit as $n$.

Since $n$ is a single-digit number, we have $n = x + n - 1$, from which $x = 1$ follows, so the possible values of $A$ are:

$$
1, 12, 123, 1234, 12345, 123456, 1234567, 12345678, 123456789
$$

A simple calculation can verify that these nine numbers all satisfy the conditions (in the solution process, we only used the first condition!). The fact that this is not a coincidence is shown by the following sequence of transformations:

$$
\begin{gathered}
9A + n = 9 \cdot \overline{12 \ldots n} + n = 9\left(10^{n-1} + 2 \cdot 10^{n-2} + \ldots + (n-1)10 + n\right) + n = \\
= 9\left(\left(10^{n-1} + 10^{n-2} + \ldots + 1\right) + \left(10^{n-2} + 10^{n-3} + \ldots + 1\right) + (10 + 1) + 1\right) + n = \\
=\left(10^{n} - 1\right) + \left(10^{n-1} - 1\right) + \ldots + \left(10^{2} - 1\right) + (10 - 1) + n = \\
=10^{n} + 10^{n-1} + \ldots + 10 = \overline{11 \ldots 10}.
\end{gathered}
$$