Which of the following regular expressions represent(s) the set of all binary numbers that are divisible by three? Assume that the string $\epsilon$ is divisible by three.

To determine which regular expressions represent binary numbers divisible by $3$, construct the standard DFA with states $q_0, q_1, q_2$ representing the remainder modulo $3$ of the binary string read so far, where $q_0$ is both the initial state and the only accepting state.

The DFA transitions are:

• $\delta(q_0, 0) = q_0$, $\delta(q_0, 1) = q_1$

• $\delta(q_1, 0) = q_2$, $\delta(q_1, 1) = q_0$

• $\delta(q_2, 0) = q_1$, $\delta(q_2, 1) = q_2$

Option A: Eliminating state $q_2$ first:

• Path looping at $q_1$ through $q_2$: $01^*0$.

• Path from $q_0$ to $q_0$ through $q_1$: $1(01^*0)^*1$.

• Combined with the self-loop on $0$ at $q_0$, the regular expression is:

  $$(0 + 1(01^*0)^*1)^*$$

Hence, Option A is correct.

Option C: Using the regular expression algebraic identity $(r^*s^*)^* = (r+s)^*$, substituting $r = 0$ and $s = 1(01^*0)^*1$ gives:

This is completely equivalent to Option A. Hence, Option C is correct.

Option B: Eliminating state $q_1$ by identifying all simple paths from $q_0$ back to $q_0$:

• Direct loop on $0$: $0$

• Path $q_0 \xrightarrow{1} q_1 \xrightarrow{1} q_0$: $11$

• Path entering $q_2$ via $q_0 \xrightarrow{1} q_1 \xrightarrow{0} q_2$ (string $10$), looping in $q_2$ directly on $1$ or via $q_2 \xrightarrow{0} q_1 \xrightarrow{0} q_2$ (string $00$), giving $(1 + 00)^*$, and finally returning to $q_0$ via $q_2 \xrightarrow{0} q_1 \xrightarrow{1} q_0$ (string $01$): $10(1 + 00)^*01$.

• The resulting regular expression is:

  $$(0 + 11 + 10(1 + 00)^*01)^*$$

Hence, Option B is correct.

Option D: Option D can generate strings such as $11100_2 = 28_{10}$. Since $28 \bmod 3 = 1 \neq 0$, the expression generates numbers not divisible by $3$. Thus, Option D is incorrect.

Therefore, options A, B, and C are correct.