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, $81|a$
Best Answer
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.