Consider a DFA over $a,b$ accepting all strings having number of $a$ 's divible by 6 and number of $b$ 's divisble by 8. What is the minimum number of states in the resultant DFA ?
This problem can be solved by assuming 2 DFAs. One accepting the number of $a$ 's divible by 6 and the other accepting with number of $b$ 's divisble by 8. And then taking intersection of them. So what will be the number of states in the resultant DFA?
I tried drawing the DFA accepting all strings having number of $a$ 's divible by 3 and number of $b$ 's divisble by 3, and found it to have 3x3 (9) states. Can I asssume the reuslt to be 48 states for this case? Can it be inferred that the for string accepting number of $a$ 's divible by $m$ and number of $b$ 's divisble by $n$ , there will be $mn$ states? Or I need to draw it by hand? Or anything other? Any subtle hint will be very helpful.