Step 1: Recall the automaton for binary numbers divisible by 3.
Binary numbers divisible by $3$ are recognized by a DFA with three states corresponding to remainders $0$, $1$, and $2$ modulo $3$. The start and accepting state corresponds to remainder $0$. Any correct regular expression must generate exactly the strings accepted by this DFA, including the empty string $\epsilon$.
Step 2: Analyse option (A).
The expression $(0 + 1(01^*0)^*1)^*$ correctly captures cycles that return the automaton to the remainder-$0$ state. The outer Kleene star allows repetition of such cycles, including $\epsilon$. Hence, it generates exactly all binary strings divisible by $3$.
Step 3: Analyse option (B).
The expression $(0 + 11 + 10(1 + 00)^*01)^*$ also corresponds to concatenations of substrings that map the DFA from remainder $0$ back to remainder $0$. This structure correctly represents all valid transitions of the modulo-$3$ automaton. Thus, option (B) is correct.
Step 4: Analyse option (C).
The expression $(0^*(1(01^*0)^*1)^*)^*$ allows any number of leading zeros and valid remainder-$0$ cycles. Since leading zeros do not change divisibility by $3$, and the inner structure ensures modulo-$3$ correctness, this expression also represents the desired language. Hence, option (C) is correct.
Step 5: Analyse option (D).
The expression $(0 + 11 + 11(1 + 00)^*00)^*$ does not correctly correspond to all remainder-$0$ transitions in the modulo-$3$ DFA and fails to generate some valid binary multiples of $3$. Therefore, it is incorrect.
Step 6: Conclusion.
The regular expressions that correctly represent the set of all binary numbers divisible by $3$ are given in options (A), (B), and (C).
Match LIST-I with LIST-II \[\begin{array}{|c|c|c|}\hline \text{ } & \text{LIST-I} & \text{LIST-II} \\ \hline \text{A.} & \text{A Language L can be accepted by a Finite Automata, if and only if, the set of equivalence classes of $L$ is finite.} & \text{III. Myhill-Nerode Theorem} \\ \hline \text{B.} & \text{For every finite automaton M = $(Q, \Sigma, q_0, A, \delta)$, the language L(M) is regular.} & \text{II. Regular Expression Equivalence} \\ \hline \text{C.} & \text{Let, X and Y be two regular expressions over $\Sigma$. If X does not contain null, then the equation $R = Y + RX$ in R, has a unique solution (i.e. one and only one solution) given by $R = YX^*$.} & \text{I. Arden's Theorem} \\ \hline \text{D.} & \text{The regular expressions X and Y are equivalent if the corresponding finite automata are equivalent.} & \text{IV. Kleen's Theorem} \\ \hline \end{array}\]
\[\text{Matching List-I with List-II}\]
Choose the correct answer from the options given below:
if, then, else, a, b, c are the terminals.