I know I am quite late, but I just wanted to add a few things to the already correct answer provided by @Grijesh. I'd like to just point out that the answer provided by @Grijesh does not produce the minimal DFA. While the answer surely is the right way to get a DFA, if you need the minimal DFA you will have to look into your divisor.
Like for example in binary numbers, if the divisor is a power of 2 (i.e. 2^n) then the minimum number of states required will be n+1. How would you design such an automaton? Just see the properties of binary numbers. For a number, say 8 (which is 2^3), all its multiples will have the last 3 bits as 0. For example, 40 in binary is 101000. Therefore for a language to accept any number divisible by 8 we just need an automaton which sees if the last 3 bits are 0, which we can do in just 4 states instead of 8 states. That's half the complexity of the machine.
In fact, this can be extended to any base. For a ternary base number system, if for example we need to design an automaton for divisibility with 9, we just need to see if the last 2 numbers of the input are 0. Which can again be done in just 3 states.
Although if the divisor isn't so special, then we need to go through with @Grijesh's answer only. Like for example, in a binary system if we take the divisors of 3 or 7 or maybe 21, we will need to have that many number of states only. So for any odd number n in a binary system, we need n states to define the language which accepts all multiples of n. On the other hand, if the number is even but not a power of 2 (only in case of binary numbers) then we need to divide the number by 2 till we get an odd number and then we can find the minimum number of states by adding the odd number produced and the number of times we divided by 2.
For example, if we need to find the minimum number of states of a DFA which accepts all binary numbers divisible by 20, we do :
20/2 = 10
10/2 = 5
Hence our answer is 5 + 1 + 1 = 7
. (The 1 + 1 because we divided the number 20 twice).