Digital Electronics - Counters - Discussion

2. 

Using four cascaded counters with a total of 16 bits, how many states must be deleted to achieve a modulus of 50,000?

[A]. 50,000
[B]. 65,536
[C]. 25,536
[D]. 15,536

Answer: Option D

Explanation:

No answer description available for this question.

Suri_493 said: (Jan 24, 2011)  
(2)^16=65,536
hence 65,536-50,000=15,536 it is the correct answer

Hari said: (Mar 23, 2011)  
For 4 FFs we have 2^4 = 16 different states. Here used 4 cascaded counters each has 4 FFs. So the number of different states are 2^16 = 65536

To achieve modulus of 50000 , the number of states deleted to be
65536 - 50000 = 15536

5G Iiit Basar said: (Mar 15, 2013)  
2^4=16 FFs.
For 4 bit counter 4 FFs used.
For 4 cascaded 4 bit counter 4*4=16 FFs used.

So,
2^16=65536.
But we have 50000.

So,
65536-50000=15536.

Rajeswari431 said: (Sep 19, 2013)  
Given 16 bit counter.

FOUR cascaded 4 bit counter 4*4 = 16.

2^n = 2^16 = 65536.

But the given modulus = 50000.

So we require modulus = 65536-50000.

= 15536.

Vijayalaxmi A said: (May 19, 2014)  
Each cascaded counter having 4 Flip Flops.

For four flip flops means 4-bit counter, for 4-bit counter possible states are 16.
2^4 = 16.

For four cascaded counter no.of bits are 4*4=16.

So, the possible states are 2^16 = 65536.

For achieving 50000 states, 15536 must be deleted.

65536-50000 = 15536.

Sriharsha said: (Nov 29, 2017)  
(2^16)-50,000 = 15,536.

Nico said: (Apr 10, 2018)  
16 bits means the total number of states is 2^16 = 65536,
65536 - 50000 = 15536,
15536 states have to deleted to reach a modulus of 50000.

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