# Digital Electronics - Counters - Discussion

### Discussion :: Counters - General Questions (Q.No.2)

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

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.