Let M be a nondeterministic finite automaton (NFA) with 6 states over a finite alphabet. Which of the following options CANNOT be the number of states in the minimal deterministic finite automaton (DFA) that is equivalent to M?

The correct option is 65.

To understand why this is the correct answer, we can analyze the relationship between the number of states in a Nondeterministic Finite Automaton (NFA) and its equivalent Minimal Deterministic Finite Automaton (DFA) using the subset construction method.

Given an NFA with $n$ states, the standard power-set construction (or subset construction) algorithm creates an equivalent DFA. Since each state in the DFA corresponds to a subset of the states of the NFA, and a set with $n$ elements has $2^n$ possible subsets, the maximum number of states in the equivalent DFA is:
$$S_{\text{max}}=2^n$$
This upper bound represents the maximum possible states before minimization. Since a minimal DFA is obtained by merging equivalent states of the DFA, its number of states must be less than or equal to the number of states in the DFA obtained via subset construction.

For an NFA $M$ with $n=6$ states:
$$S_{\text{max}}=2^6=64$$
Therefore, the equivalent minimal DFA can have at most 64 states.

Comparing the given options:
- 32 is less than or equal to 64, so it is a possible number of states.
- 1 is less than or equal to 64, so it is a possible number of states.
- 65 is strictly greater than 64, making it mathematically impossible to be the number of states in the minimal DFA.
(Note: While 128 is also greater than 64, the option 65 is identified as the correct choice here.)