= Solution
Use three <D flip-flops> whose stored <bits> $A,B,C$ represent $4A+2B+C$. Initialize them to $000$. The <Moore machine>'s output is this stored <binary representation>, and its <state transition table> is:
|| Present $ABC$
|| Next $A^+B^+C^+$
| $000$
| $001$
| $001$
| $010$
| $010$
| $011$
| $011$
| $100$
| $100$
| $101$
| $101$
| $000$
Treat the two unused <state encodings> $110,111$ as <don't-care conditions> when minimizing the <sum of products>. With juxtaposition for AND, $+$ for OR and a bar for NOT, one minimal choice is
$$
\boxed{D_A=A\overline C+BC,\qquad D_B=B\overline C+\overline A\,\overline B C,\qquad D_C=\overline C.}
$$
For $D_A$, the required ones at $011$ and $100$ cannot share one valid <implicant>: their differing <bits> would force inclusion of required zeros. Pair $011$ with the unused $111$ to get $BC$, and $100$ with unused $110$ to get $A\overline C$. Neither term can lose a <Boolean literal> without covering a required zero.
For $D_B$, state $001$ is isolated from every other required one and from the unused states by required zeros, so it forces the three-<Boolean literal> term $\overline A\,\overline B C$. State $010$ pairs with unused $110$ to give $B\overline C$. Finally $D_C=\overline C$ is a single <Boolean literal>. Thus the three expressions are minimal <sum of products> expressions under the stated unused-state convention. In fact these choices also recover from $110\to111\to100$, so initialization into an unused state eventually joins the intended <modulo-six counter> cycle.
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