DFA that rejects $a^{23}$ but accepts $\{a^i|i\geq 24\}$
This is not possible. A DFA on a one symbol alphabet is a directed graph with out-degree 1. The walk one follows from the start state consists of a (possibly zero length) path followed by a cycle. As all sufficiently long strings are accepted, all states in the cycle must accept. If the DFA has at most 11 states, a walk of length 23 must be in the cycle, so it must accept.

More generally, if a DFA on one symbol rejects $a^n$ but accepts all $a^m$, $m&gt;n$, then it must have at least $n+1$ symbols and start with a path of length $\ge n$, the $n$th state rejecting, followed by a cycle of accepting states of length $\ge 1$.