Answer / mo saleem khan

In mod -N the no of flip flops gives by the following equation
2^(n-1)<=N<=2^n if the value of n satisfied both 2^n-1& 2^n then the value of N is correct ,otherwise not ,so in mod-10,mod-32
The value of n is 4 & 5 respectively bcz n=4 ,5 respectively satisfied both condition 2^n-1<=N<=2^n so N=4,5 respectively correct for mod-10,32

Answer / lakshmanan

6 flipflop needed

Answer / tousif

5

Answer / amar fulwade

5
Counter have 2^n output
Hence for modulo 19 we required 5 flip flop

Answer / kichus

AB=1

Answer / krishna

19 flipflops are used

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