DFA for $L=\{w: w\in \{a,b\}^*, |w|_a\ \equiv |w|_b \equiv 0$ $(mod$ $5) \}$
The minimal DFA for $L$ contains 25 states, as can easily be shown using Myhill–Nerode theory.

Indeed, consider the 25 words $\{a^ib^j : 0 \leq i,j &lt; 5\}$. I will show that when a DFA reads any two different words from this list, it cannot end up in the same state. To show this, for any two different words $x,y$ from the list, I will give a word $z$ such that $xz \in L$ but $yz \notin L$.

Let $x = a^i b^j$ and $y = a^k b^\ell$ be two different words from the list. Since $x \neq y$, either $i \neq k$ or $j \neq \ell$ (or both). Assume without loss of generality that $i \neq k$. Take $z = a^{5-i} b^{5-j}$. Then $|xz|_a = |xz|_b = 5$, and so $xz \in L$. In contrast, $|yz|_a = 5-i+k \not\equiv 0 \pmod{5}$, since $i \not\equiv k \pmod{5}$ by assumption, showing that $yz \notin L$.