Minimum states in DFA

Question

Number of final states in minimal DFA where $\sum = \{ a,b \}$

$L = \{ w| n_a(w)mod\ 3 \geq n_b(w)mod\ 2\}$

Answer 1

Let every state be represented as $(n_a(w)mod\;3,n_b(w)mod\;2)$, then the DFA will look like :

And clearly it contains $5$ final states. Drawing DFA is not necessary as we could form all $6$ possibilities of  $\left ( \bf{\color{blue}{n_a(w)mod\;3}},\bf{\color{red}{n_b(w)mod\;2}} \right )$ which are $\big \{\bf {\color{blue}{0}}{\color{red}{0}},{\color{blue}{0}}{\color{red}{1}},{\color{blue}{1}}{\color{red}{0}},{\color{blue}{1}}{\color{red}{1}},{\color{blue}{2}}{\color{red}{0}},{\color{blue}{2}}{\color{red}{1}} \big\}$  and each of these is a state in the DFA and look among these $6$ states that how many states satisfy constraint $n_a(w) mod\;3 \ge n_b(w) mod\;2$.

Comments

minimal..precise..quick !! nice answer !

Answer 2

If we draw DFA of  L = {n_a(w)mod 3  &&  n_b(w) mod 2 }.

minimum number of states = 6 = {00,01,10,11,20,21}; these are residue of a,3 and b,2 respectively 

So only one state where we will get n_a(w)mod 3  <  n_b(w) mod 2,

So Answer should be 6-1= 5.