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The minimum number of Flip-Flops required to construct a binary Modulo $n$ counter is ________

1. $n$
2. $n-1$
3. $2^n – 1$
4. $\lceil \log_2 n \rceil$

### 1 comment

plz any one explain???

n flip flops can be used to make modulo  2 counter.

So according to the question the answer must be  ceil(log2(n))

The "MOD" or "MODULUS" of a counter is the number of unique states. The MOD of the n flip flop ring counter is n.

But they did not mention its ring counter!!
Seeing options best on suited is n as we have already read the ring counter Ring counter was the best suited
confused with the question ?
no, the options were wrong. See now.
I think previously log n is not their right??
yes..

binary modulo n means the possible no.of states = n and mark them as 0,1,2,3,....n-1.

therefore you need to indicate n states with k flipflops

if n=2p then k=p

if n $\neq$ 2p then k= ⌈log2n⌉