A positive integer gets reduced by 9 times when one of its digits is deleted.... 
A positive integer gets reduced by $9$ times when one of its digits is deleted and the resultant number is divisible by $9$. Prove that to divide the resultant number by $9$, it is again sufficient to delete one of it's digits. Find all such numbers.

I am completely clueless as to how this question can be solved. I require a hint to start solving this.
Note:
The only thing I can think of is that the number deleted the first time is either $0$ or $9$. According to the divisibility rule the sum of the digits should be a multiple of $9$ if the no. is divisible by $9$. If the sum of the digits of both the 1st number and the 2nd number are a multiple of $9$ then the deleted digit is surely $9$ or $0$.
let the 3 nos. be $a,b,c$.
$a=9b$
$b=9c$
Therefore, $81|a$
 A: Write the first number as $10^{n+1}a+10^nb+c$ where $b$ is the digit that will be deleted, $c$ has $n$ digits, and $a$ can have multiple digits.  We are told that $10^{n+1}a+10^nb+c=9(10^na+c)$ with $10^{n-1} \le c \lt 10^n$.  This gives $8c=10^n(a+b)$, which shows $a+b \le 7$.  The fact that deleting a digit does not spoil the divisibility by $9$ shows that $b=0$ as $b=9$ is prohibited.  If we take $a=1,b=0$ we find the number to be $10125$ with as many trailing zeros as desired.  Similarly we find the solutions $2025,30375,405,50626,6025,70875$, all of which can be multiplied by $10^k$.  You delete the second digit $0$ to do the first division by $9$ and the first digit for the second division by $9$.
A: Having established that the original number is divisible by 9, the deleted digit must be either 0 or 9 because only these two digits can leave the digit sum divisible by 9. Call the deleted digit $d$ and assume it is in the $10^k$ place:
$$aaa\dots adb\dots bbb=A\cdot10^{k+1}+d\cdot10^k+B$$
$$aaa\dots ab\dots bbb=A\cdot10^k+B$$
$$A\cdot10^{k+1}+d\cdot10^k+B=9(A\cdot10^k+B)$$
$$10A\cdot10^k+d\cdot10^k+B=9A\cdot10^k+9B$$
$$A\cdot10^k+d\cdot10^k=8B$$
$$8B=(A+d)\cdot10^k$$
$$B=(A+d)\cdot\frac{10^k}8$$
Yet by our construction above we must have $B<10^k$, so
$$(A+d)\cdot\frac{10^k}8<10^k$$
$$A+d<8$$
If $d=9$ then $A$ would be forced to be negative, which is impossible. Therefore $d=0$, $A<8$ and all numbers satisfying the conditions in the first part of the question are of the form
$$N=A\cdot10^{k+1}+A\cdot\frac{10^k}8,\ 0<A<8,\ k\ge3-\log_2\gcd(A,8)$$
The restriction on $k$ ensures that $A\cdot\frac{10^k}8$ is an integer. $A$ cannot be zero because $N$ would then start with a zero.
The numbers $N$ fall into seven classes depending on what $A$ is:
$$A=1: N=10125\cdot10^{k-3}$$
$$A=2: N=2025\cdot10^{k-2}$$
$$A=3: N=30375\cdot10^{k-3}$$
$$A=4: N=405\cdot10^{k-1}$$
$$A=5: N=50625\cdot10^{k-3}$$
$$A=6: N=6075\cdot10^{k-2}$$
$$A=7: N=70875\cdot10^{k-3}$$
Regardless of what $k$ is, division by 9 will not touch the trailing zeros, so we can ignore them. Dividing $N$ by 9 removes the zero that is second from left, producing the following prefixes, and dividing by 9 again can be accomplished by deleting the leftmost digit:
$$\require{cancel}A=1:\cancel1125\ldots\to125\dots$$
$$A=2:\cancel225\ldots\to25\dots$$
$$A=3:\cancel3375\ldots\to375\dots$$
$$A=4:\cancel45\ldots\to5\dots$$
$$A=5:\cancel5625\ldots\to625\dots$$
$$A=6:\cancel675\ldots\to75\dots$$
$$A=7:\cancel7875\ldots\to875\dots$$
Hence the proof asked for by the question, that $\frac N{81}$ can be reached by deleting a single digit from $\frac N9$, has been shown.
