Find all positive integers n for which the number obtained by erasing the last digit of n is a divisor of n? I know, through this, , that all numbers ending on 0 and 11, 12..19, 22, 24, 26, 28, 33, 36, 39, 44, 48, 55, 66, 77, 88, 99 are solutions. But how to prove that all 3- and more-digit numbers which do not end on 0 are not the solution?
 A: We are told that $a$ divides $n$, so write $n=ka$.  Then $ka=10a+b, (k-10)a=b.$   If $b \gt 0, k-10 \gt 0,$ so $a$ has to divide into $b$.  Since $b$ is a single digit it cannot be larger than $9$, so $a$ cannot be larger than $9$ and the original number has at most two digits.
A: $$a_na_{n-1}.....a_1a_0=10(a_na_{n-1}.....a_1)+a_0\\a_na_{n-1}.....a_1|10(a_na_{n-1}.....a_1)+a_0\Rightarrow 10(a_na_{n-1}.....a_1)+a_0=k(a_na_{n-1}.....a_1)$$ It follows $$\frac{a_0}{a_na_{n-1}.....a_1}\in\mathbb N$$
All digit cannot divide a two digit number; in other words
$$a_0=b(a_1)$$ has the solutions given by the OP but $$a_0=b(a_2a_1)$$ has no solution for $a_2\ne 0$
A: First case: 
$ 100 \leq n $. 
In this case we $\color{Green}{\text{claim}}$ that
$\color{Green}{\text{the last digit is equal to zero}}$ , 
conversly every integer with the last digit equal to zero 
has the above property.

Let 
$$ 100 \leq 
n=\overline{  a_m  a_{m-1}  ...  a_1  a_0  }= 
a_m10^m + a_{m-1}10^{m-1} + ... + a_110 + a_0  \  
; \ \ \   
\text{i.e.} \ \ 
2 \leq m $$ 
with $a_m \neq 0$, 
also on the otherhand let 
$$n ^ {\prime}=\overline{  a_m  a_{m-1}  ...  a_1  }= 
a_m10^{m-1} + a_{m-1}10^{m-2} + ... + a_210 + a_1  \  .$$ 

then one can see easily that: 
$$\color{Blue} {n=10n^{\prime}+a_0} ,$$ 
also notice that 
$$ a_0 < 10 
\ \ \ \ \ \ \ \ \ \ \ \ 
\text{and} 
\ \ \ \ \ \ \ \ \ \ \ \ 
10 \leq n^{\prime},$$ 
so we can conclude that: 
$$a_0  < n^{\prime} 
\ \ \ \ \ \ \ \ 
,\text{i.e.}   
\ \ \ \ \ \ \ \ \ \ \ \ 
0  \leq  \dfrac  {a_0}  {n^{\prime}}  < 1 .$$

Now notice that: 
$$       \dfrac{n}{n^{\prime}}     = 
\dfrac{    10    n^{\prime}    +    a_0}    {    n^{\prime}    } = 
\dfrac{    10    n^{\prime}            }    {    n^{\prime}    } + 
\dfrac{                             a_0}    {    n^{\prime}    } = 
10 + 
\dfrac{                             a_0}    {    n^{\prime}    } 
\ \ \ \ \ \ \ \ 
\Longrightarrow 
\\ 
\color{Red} 
{       10 \leq \dfrac{n}{n^{\prime}} < 10+1=11       } 
, $$ 
so we must have: $\color{Red} 
{       \dfrac{n}{n^{\prime}}=10       }$ , 
i.e. $\color{Blue} {n=10n^{\prime}+0}$ ; 
which implies that 
$\color{Green}{a_0=0}$ . 




Second case : $n < 100$, which can be done by a simple calculation! 
