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.

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.
