Boolean circuit - 4 bits divisible by 3
General approach for \$m\$ bits
The general solution for a test for division by 3 is to sum up the even-numbered bits and separately sum up the odd-numbered bits, take the difference between these sums, and then see if the difference itself is divisible by 3. (There are a variety of approaches for this operation, but the one encountered first is usually via carry-save adders.)
For a binary value with \$m\$ bits, where \$m\$ is even, the difference will require at most \$\lceil\operatorname{ln}_2\frac{m}{2}\rceil\$ bits. For a binary value with \$m\$ bits, where \$m\$ is odd, the difference will require at most \$\lceil\operatorname{ln}_2\frac{m+1}{2}\rceil\$ bits. This difference result could itself then be submitted to a much smaller tier for, once again, computing the difference between the sums of even and odd numbered bits. (And repeat.)

Specific case where \$m=4\$
At this point it is pretty easy to see that the even and odd sums can be computed using a simple half-adder, each. The resulting table is:
$$
\begin{smallmatrix}\begin{array}{r|cccc}
&amp;\overline{C_\text{odd}}\:\overline{S_\text{odd}}&amp;\overline{C_\text{odd}}\:S_\text{odd}&amp;C_\text{odd}\:\overline{S_\text{odd}}\\
\hline
\overline{C_\text{even}}\:\overline{S_\text{even}}&amp;Y&amp;N&amp;N\\
\overline{C_\text{even}}\:S_\text{even}&amp;N&amp;Y&amp;N\\
C_\text{even}\:\overline{S_\text{even}}&amp;N&amp;N&amp;Y
\end{array}\end{smallmatrix}
$$
In this case, there is no need to worry about &quot;divisibility by 3&quot; of the difference. Instead, it's sufficient to compare the two sums for &quot;equal,&quot; as shown in the above table.
This should be very easy to implement:



simulate this circuit &ndash; Schematic created using CircuitLab
The half-adders are easily recognized above. In addition, their associated outputs are directly compared using a pair of XORs. The results of these two comparisons are then considered using a NOR for the final output.