Answer

Correct Option: (i) 15 states (Other options typically include 8, etc., correct is 15)

Explanation

Language $L = \{ w \in \{0,1\}^* \mid \#_0(w) \bmod 3 =0 \;\land\; \#_1(w) \bmod 5 =0 \}$.

This is intersection of two regular languages:

• $L_0 = \{ w \mid \#_0(w) \equiv 0 \pmod 3 \}$ → DFA with 3 states ($r_0=0,1,2$).
• $L_1 = \{ w \mid \#_1(w) \equiv 0 \pmod 5 \}$ → DFA with 5 states ($s_0=0..4$).

Intersection via product construction: states = $Q_0 \times Q_1$ → $3 \times 5 =15$ states.

Transitions: $\delta((r,s),0) = ((r+1)\bmod 3, s)$, $\delta((r,s),1) = (r, (s+1)\bmod5)$.

Initial $(0,0)$, accepting $(0,0)$ only. Minimal because all pairs distinguishable (Myhill-Nerode: $(i,j)
ot\equiv (i',j')$if$i
eq i'$or$j
eq j'$).

Final Answer

15 states (product DFA, minimal).