A minimum state DFA for the language: L = \{ w \in \{0,1\}^* : \#0(w) \equiv 0 \mod 3,\; \#1(w) \equiv 0 \mod 5 \} has: (i) 15 states (ii) 11 states (

Question

A minimum state DFA for the language:

$$ L = \{ w \in \{0,1\}^* : \#0(w) \equiv 0 \mod 3,\; \#1(w) \equiv 0 \mod 5 \} $$

has:

(i) 15 states  
(ii) 11 states  
(iii) 10 states  
(iv) 9 states

Accepted Answer

## Answer
**Correct Option: (i) 15 states**

## Explanation
Language $L = \{ w \mid \#_0 \bmod 3 =0,\; \#_1 \bmod 5 =0\}$ is same as Q92 / Q112.

As before: DFA for $\#_0 \bmod 3$ needs 3 states, for $\#_1 \bmod 5$ needs 5 states. Product → $3 	imes 5 =15$ minimal states.

Transitions: $15 	imes 2 =30$.

Other options 11,10,9 would be for different moduli (e.g., 3+5=8) — incorrect; must multiply, not add.

### Final Answer
**15 states → Option (i)**.

Answer