32*. An integer decreases by 9 times when a certain digit is crossed out; at the same time, the resulting number is also divisible by 9.

a) Prove that to divide the resulting number by 9, it is also sufficient to cross out one digit in it.

b) Find all integers that satisfy the condition of the problem.
32. a) First of all, let us show that no number \( N \) can decrease by 9 times when a digit, other than the second one, is erased. Indeed, if this were not the case, denoting the digits of the number \( N \) by \( a_{0}, a_{1}, \ldots, a_{n} \), i.e.,

\[
a_{0} \cdot 10^{n} + a_{1} \cdot 10^{n-1} + \ldots + a_{n} = N
\]

we would conclude that the number \( \frac{N}{9} \) has \( n \) digits, the first two of which are \( a_{0} \) and \( a_{1} \), i.e.,

\[
a_{0} \cdot 10^{n-1} + a_{1} \cdot 10^{n-2} + \ldots = \frac{N}{9}
\]

Multiplying the last equation by 10 and subtracting the first, we get:

\[
\frac{N}{9} < 10^{n-1}
\]

which is impossible, since

\[
\frac{N}{10} = a_{0} \cdot 10^{n-1} + \ldots \geqslant 10^{n-1}
\]

On the other hand, from the divisibility rule by 9, it follows that if the number \( N \) is divisible by 9 simultaneously with the number obtained from \( N \) by erasing one digit, then this digit is 0 or 9. Thus, in our case, it is only possible that the first or second digit of the number \( N \) is 0 or 9, and erasing this digit is equivalent to dividing \( N \) by 9. But the first digit of \( N \) cannot be 0, and if it were 9, then the number \( \frac{N}{9} \) would have as many digits as \( N \) and could not be obtained from \( N \) by erasing one digit. Further, if the second digit of \( N \) is 9 and the number obtained from \( N \) by erasing this digit is \( \frac{N}{9} \), then we have:

\[
\begin{gathered}
N = a_{0} \cdot 10^{n} + 9 \cdot 10^{n-1} + a_{2} \cdot 10^{n-2} + \ldots + a_{n} \\
\frac{N}{9} = a_{0} \cdot 10^{n-1} + a_{2} \cdot 10^{n-2} + \ldots + a_{n}
\end{gathered}
\]

from which, multiplying the second equation by 10 and subtracting the first, we get:

\[
\frac{N}{9} < 10^{n-1}
\]

(since \( a_{2} \leqslant 9 \)). Thus, we are convinced that to reduce the number \( N \) by 9 times, one must erase the digit 0, standing in the second place.

Now we have:

\[
\begin{gathered}
N = a_{0} \cdot 10^{n} + a_{2} \cdot 10^{n-2} + \ldots + a_{n} \\
\frac{N}{9} = a_{0} \cdot 10^{n-1} + a_{2} \cdot 10^{n-2} + \ldots + a_{n}
\end{gathered}
\]

From this it follows that

\[
\frac{N}{9} = N - a_{0} \cdot 10^{n} + a_{0} \cdot 10^{n-1} = N - a_{0} \cdot 10^{n-1} \cdot 9
\]

and, finally,

\[
\frac{1}{9} \cdot \frac{N}{9} = \frac{N}{9} - a_{0} \cdot 10^{n-1}
\]

the latter means that to divide the number \( \frac{N}{9} \) by 9, it is sufficient to erase the first digit.

b) We have (see the solution to part a)):

\[
\frac{N}{2} = N - a_{0} \cdot 10^{n-1} \cdot 9
\]

from which it immediately follows that

\[
N = \frac{a_{1} \cdot 10^{n-1} \cdot 81}{8}
\]

Now, considering \( a_{0} \) to be 1, 2, 3, etc., we get that \( N \) can be one of the numbers 10125, 2025, 30375, 405, 50625, 6075, 70875, or differ from one of these numbers by some number of zeros at the end ( \( a_{0} \) cannot be 8 or 9, since in this case the second digit of \( N \) would already not be zero).