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.

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
