A positive integer decreases an integral number of times when its last digit is deleted. Find all such numbers 
A positive integer decreases an integral number of times when its last digit is deleted. Find all such numbers.    

My Working
Let original number $n$ be $\ a_1+10a_2+10^2a_3+...+10^{k-1}a_k$.
The new number $n'$ is given by $\ a_2+10a_3+10^2a_4+...+10^{k-2}a_k$.   
$n'$|$n$   
$\frac{n-a_1}{10} |\ n$
$n=m \frac{n-a_1}{10} $
$a_1+10a_2+10^2a_3+...+10^{k-1}a_k=m \frac{(a_1+10a_2+10^2a_3+...+10^{k-1}a_k)-a_1}{10} $
$a_1+10a_2+10^2a_3+...+10^{k-1}a_k= \frac{m(10a_2+10^2a_3+...+10^{k-1}a_k)}{10} $
$a_1= \frac{m(10a_2+10^2a_3+...+10^{k-1}a_k)-10(10a_2+10^2a_3+...+10^{k-1}a_k)}{10} $
$a_1= \frac{(m-10)(10a_2+10^2a_3+...+10^{k-1}a_k)}{10} $
$a_1= (m-10)(a_2+10a_3+...+10^{k-2}a_k) $   
For 
$a_1\ |\ n'$, or $n'=ka_1$, the above equation is true*.
Original number is given by
$n=10n'+a_1$
$n=10ka_1+a_1$
$n=(10k+1)a_1$    
Problem
If we put values into this equation we don't get numbers that satisfy the condition. For example if $a_1=1$ and $k=1$, The number comes out to be 201 which does not satisfy the condition. I am not able to figure out where I went wrong. 
*My equation might not give all solutions but the ones it gives should satisfy the condition.

Assume that $n$ has at least two digits, then $n=10a+b$ where $a\geq 1$ and $b\in \{0,1,2,3,4,5,6,7,8,9\}$. 
By assumption $a$ divides $n$, therefore $a$ divides also $b=n-10a$. 
Hence if $b\not=0$ then $a\leq b\leq 9$, which implies that $n$ has only two digits.
It follows that the number has this property iff it ends with a $0$ or it is in this finite list:
$$1,2,3,4,5,6,7,8,9,11, 12, 13, 14, 15, 16, 17, 18, 19,\\ 22, 24, 26, 28, 33, 36, 39,  44, 48, 55,  66,  77,  88,  99.$$
