DFA for words with $a$s not a multiple of 3 and at least one $c$ Consider this language: $$L = \{w \in \{a,c\}^* \mid 3\nmid\#a(w)\land\#c(w)&gt;0\}$$.

Here is an automaton for the first part of language, but I do not know how to devise and attach the second part of the condition $\#c(w) &gt; 0$. Because $C$ can come in any state.
The product construction is a way of taking two DFAs for languages $L_1,L_2$, and constructing a new DFA for the language $L_1 \cup L_2$ or $L_1 \cap L_2$. In your case, $L_1$ consists of all words in which the number of $a$s is not a multiple of 3, and $L_2$ consists of all words containing at least one $c$. You have already constructed a DFA for $L_1$. Construct one for $L_2$, and use the product construction to construct one for $L_1 \cap L_2$.