Answer / k r md shafik

three FF.

Answer / umesh

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

Answer / anu

three ( 3 )

Answer / shirish

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

Answer / subhadip sikdar

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

Answer / saurabh kharakwal

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...

Answer / kausani

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

Answer / amit02

3 FF

Answer / shobha rani

for mod-6(mod-N) counter 3(n) ff are required...
i.e
2^n-1<N<=2^n
4<6<=8
that's why 3 FF are used.

Answer / vivek

here should be 2^n > N,
where n is no. of FF to be used. n should be taken the
least number.
here N=6,and 2^3>6,so no. of FF to be used is 3.

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