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 )