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Design a machine of the following language: 

Answer: 

As per the language given it is mod 2 and mod 3. 

So, we know in this case the number of states will be 2 * 3 = 6 (total remainder pair is 6)

We design a machine for ${\mathbf{n}}_{\mathbf{b}}\mathbf{\left(}\mathbf{\omega }\mathbf{\right)}\mathbf{ }\mathbf{mod}\mathbf{ }\mathbf{3}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0} :$

And we know machine of ${\mathbf{n}}_{\mathbf{a}}\mathbf{\left(}\mathbf{\omega }\mathbf{\right)}\mathbf{ }\mathbf{mod}\mathbf{ }\mathbf{2}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0} :$

Now machine for,

Here ${\mathbf{r}}_{\mathbf{a}}$ identified remainder of a’s and ${\mathbf{r}}_{\mathbf{b}}$ identified remainder of b’s. 

In the above machine, there are four grades,

So, the final state must the common state (Intersection) of ${\mathbf{n}}_{\mathbf{b}}\mathbf{\left(}\mathbf{\omega }\mathbf{\right)}\mathbf{ }\mathbf{mod}\mathbf{ }\mathbf{3}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}$states and ${\mathbf{n}}_{\mathbf{a}}\mathbf{\left(}\mathbf{\omega }\mathbf{\right)}\mathbf{ }\mathbf{mod}\mathbf{ }\mathbf{2}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}$.

Because language is,

 $\mathbf{\left(}{\mathbf{n}}_{\mathbf{a}}\mathbf{\right(}\mathbf{\omega }\mathbf{)}\mathbf{ }\mathbf{mod}\mathbf{ }\mathbf{2}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{)} \mathbf{AND} \mathbf{\left(}{\mathbf{n}}_{\mathbf{b}}\mathbf{\right(}\mathbf{\omega }\mathbf{)}\mathbf{ }\mathbf{mod}\mathbf{ }\mathbf{3}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0}\mathbf{)}$ here AND means intersection.