TITLE: A positive integer gets reduced by 9 times when one of its digits is deleted....
QUESTION [8 upvotes]: 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, $81|a$

REPLY [4 votes]: 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 $9$.