for mod-6(mod-N) counter 3(n) ff are required... i.e 2^n-1<N<=2^n

mod 6 counter : no. of FF used.

Question Posted / shobha rani

Answer Posted / 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.

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