How many numbers have the form $10n+d$ where $d$ is a non-zero digit? How many numbers that do not end in a series of zeros are such that if we erase the last digit, the resulting number will divide the original?
I was even thinking of excluding the possibility of being multiples of 5, and by the divisibility criterion of 4, excluding it too (if the number ends in 2 zeros). Maybe if it's on another number base it might help in some other way.
 A: If the last digit is $r$ and what you get when you remove the $r$ is $n$ then then first number is $N=10n + r$.
And you want $n|N=10n+r$.  As $n|10n$ and $n|10n+r$ that means $n|r$.  
Assuming $r\ne 0$ then were can have $n|r$ and $r=1,....,9$...
If $n=1$ we can have $r$ be anything.  $N= 11,12,.....,19$.
If $n=2$ we can have $r=2,4,6,8$.  $N = 22,24,26,28$.
If $n=3$ we can have $r=3,6,9$.  $N=33,36,39$.
If $n=4$ we can have $r=4,8$.  $N=44,48$.
If $n \ge 5$ we can only have $r=n$.  $N = 55,66,77,88,99$.
And that's it. Of course if $r = 0$ you have $n|10n+ 0$ always.
And if $n \ge 10$ you NEVER have $n|10n + r$ where $r = 1....9$ because $n\not\mid r$.
[BTW:
$0$ doesn't divide into anything excepts $0$ because $0*k \ne m$ for any $k$ so $0|m$.
And everything divides into $0$ because for any $m$ we have $m*0 = 0$ so $m|0$.
So if $n =1$ and $N=r$ we can not have any one digit solution except $N=0$.
And if $r=0$ you always have $n|r$ so $n|10n+r$.
]
A: HINT:  
$a \mid (10a+b) \implies a\mid b$
A: Let the number be $(a_{1}a_{2}.....a_{n})_{10}$$=k$
so $k=$$(a_{1}a_{2}.....a_{n})_{10}=$$10$$(a_{1}a_{2}.....a_{n-1})_{10}$$+a_{n}$
now $(a_{1}a_{2}.....a_{n-1})_{10}$ divides $(a_{1}a_{2}.....a_{n})_{10}$
which means $(a_{1}a_{2}.....a_{n-1})_{10}$ divides $a_{n}$ which is only possible for 2 digit numbers
so just a little bit observation tells us the numbers are $10,11,12,13,14,15,16,17,18,19,20,22,24,26,28,30,33,36,39,40,44,48,50,55,60,66,70,77,80,88,90,99$
and trivially any number $(a_{1}a_{2}.....a_{n})_{10}$ where $a_n$ is 0 and $a_{n-1}$ is non zero is applicable
A: You can write it formally as $n|10n+r$, where $r$ is the remainder from division by 10. We need to solve: $10n+r \equiv r \equiv 0 \pmod{n}$. If $n>9$, then the only solution is $r=0$. 
Conclusion: there is finite set of solutions, the largest is 99, which is divisible by 9.
