Find a simple DFA (i.e. deterministic finite automaton) that accepts all natural numbers n for which n mod 3 = 0.
Hint: A natural number is divisible by 3 if its checksum (or sum of digits) is divisible by 3.
The Attempt at a Solution
I'm not sure how the hint can help for this question, even though I know what it means. For instance, 137 = 1+3+7 = 11 mod 3 ≠ 0 etc.
Any other useful hints or suggestions would be great. Thanks. :)