three FF.

buddy this is a simple question
for mode we have a set of binary digits 2 4 8 16 and so on
u have to count upto 6 the it comes under 8 
8=2^3
so three flip flop u have to use plus a circuit that reset
after counting upto 6

no.of ff used are 4.

three..!

with one ff we can count maximum of 2(0 and 1)
with two ff we can count maximum of 4
with three ff we can count maximum of 8
so on.. in short with n ff we can count max upto 2^n

four flip flop

three  ( 3 )

three FF r used in MOD 6 Counter..
this can be found from the equation 2^n>=N>=2^n-1

3 flip flops are required........because 6<8 and 8 is 
obtained by multipling 2 three times...and moreover 8 is 
use and why not 16 because of the difference between 8to 6 
and 6 to 16...ok...

3 F/F used for mod 6 counter.
for n no.of F/F we get 2^n counter.
for 2 F/F we get the no of counter=2^2=4<6
for 3 F/F we get 2^3=8>6 counter

n no.of ff are use for n no.of bits so 6 ffs are use for
mod-6 counter.