I don't think so. Let the odd number be AnAn-1An-2...A3A2A1. This number can be odd, if AnAn-1...A3A2 is odd and A1 is even (0/2) OR AnAn-1...A3A2 is even and A1 is odd (1). So this clearly gives a simple DFA.
let Qs be the start state and Qe is the state DFA enters when an even number is seen so far and Qo be the state DFA enters when an odd number is seen so far. (Qs,0/2) --> Qe (Qs,1) --> Qo (Qe,0/2) --> Qe (Qe,1) --> Qo (Qo,0/2) --> Qo (Qo,1) --> Qe Here were are entering the digits from most significant first. So the set of odd numbers is regular in both notations. Any other ideas?
