You can produce a finite automaton to recognize binary strings divisible by $N$ for any $N>0$. Use $N+1$ states, $s$ (the start state) and $q_i$, where being in state $q_i$ corresponds to having seen input $B$ which is congruent to $i\bmod N$.
The transitions are then easy enough: in state $q_i$ and seeing a $0$ input the machine would pass to state $q_k$, where $k\equiv 2i\bmod N$, since appending a zero to a binary string doubles the number represented. Similarly $\delta(q_i, 1)=q_k$, where $k\equiv 2i+1\bmod N$.
The start state is there solely to avoid the question of what number the empty string represents, and obviously $q_0$ is the final state.