Dividing by Dropping Digits

An integer decreases an integral number of times when its last D digits are dropped.  Obviously, there are infinite such integers, as any integer ending in D 0's decreases by a factor of 10^D when the training 0's are dropped.  Is the number of such integers finite, if the last D digits are no all 0's?  If it is finite, how many such integers exist for D=1,2,3,4?

Source: Sudipta Das.

Solutions were recieved from Nick McGrath, Joseph DeVincentis, Jeremy Galvagni, Alan O'Donnell, Adrian Atanasiu, Claudio Baiocchi, and Cleverley Graham.

Nick McGrath sent:

We need numbers of the form n=a*10^D+b such that a is a divisor of n .
n/a= 10^D + b/a   
 
Therefore a can take any value such that a is a divisor of b.
b can take any value from 1 to 10^D-1.
 
So, for example, when D=1.
We have the complete set of n's defined by:
b=1 a=1                   i.e  11
b=2; a=1,2                     12,22
b=3; a=1,3                     13,31
b=4; a=1,2,4                   14,24,44       
b=5; a=1,5                     15,55
b=6; a=1,2,3,6                 16,26,36,66
b=7; a=1,7                     17,77
b=8; a=1,2,4,8                 18,28,48,88
b=9; a=1,3,9                   19,39,99
 
for a total of 23.
 
So, for each D the number of integers that exist is simply the total number of divisors of all the numbers from 1 to 10^D-1.
 
Don't know how to do this except by writing a program but get
23, 473, 7053, 93643  for D=1,2,3,4 respectively.
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