Finite State Machine for x/3
I think you already did pretty much the right thing. The table looks like this:

$$\begin{array}{r|cc}
&amp; \textrm{bits:}\\
 &amp; 0 &amp; \quad1\\
\hline
\textrm{states:}\quad0 &amp; 0:`0\textrm' &amp; \quad1:`0\textrm'\\
1 &amp; 2:`0\textrm' &amp; \quad 0:`1\textrm'\\
2 &amp; 1:`1\textrm' &amp; \quad 2:`1\textrm'
\end{array}$$

You enter (start) at state 0 and then index into the table as bits arrive. So the state diagram looks like:



I've added the "emit" parts, as this is what is needed for you to get the quotient. (Obviously, the remainder is the state itself.) You will need some method of accumulating bits that are emitted (shift register, for example.) 

Here are the results for all four-bit input values:

$$\begin{array}{r|cc}
\textrm{input} &amp; \textrm{emitted output} &amp; \textrm{remainder (state)}\\
\hline
0000 &amp; 0000 &amp; 0\\
0001 &amp; 0000 &amp; 1\\
0010 &amp; 0000 &amp; 2\\
0011 &amp; 0001 &amp; 0\\
0100 &amp; 0001 &amp; 1\\
0101 &amp; 0001 &amp; 2\\
0110 &amp; 0010 &amp; 0\\
0111 &amp; 0010 &amp; 1\\
1000 &amp; 0010 &amp; 2\\
1001 &amp; 0011 &amp; 0\\
1010 &amp; 0011 &amp; 1\\
1011 &amp; 0011 &amp; 2\\
1100 &amp; 0100 &amp; 0\\
1101 &amp; 0100 &amp; 1\\
1110 &amp; 0100 &amp; 2\\
1111 &amp; 0101 &amp; 0
\end{array}$$