Deleting states for a target modulus You have four cascaded counters providing a total of 16 bits (overall natural count = 65,536 states). To realize an exact modulus of 50,000, how many states must be deleted (skipped)?

Difficulty: Medium

Correct Answer: 15,536

Explanation:


Introduction / Context:
When the native modulus of a counter chain exceeds the required count range, designers implement truncation by decoding a terminal count and forcing a synchronous/asynchronous clear. Determining how many states to delete is a straightforward arithmetic exercise but must be done precisely to avoid off-by-one errors in the division ratio.


Given Data / Assumptions:

  • Total available states from 16 bits: 2^16 = 65,536.
  • Desired modulus: 50,000.
  • We delete states above the desired terminal state so the sequence repeats correctly.


Concept / Approach:
Deleted states = native_states − desired_modulus. After reaching the desired terminal count (49,999 if starting at 0), the counter is forced back to 0, skipping the remaining high states so that exactly 50,000 pulses form a complete cycle, giving a divide-by-50,000 output.


Step-by-Step Solution:

Compute native_states = 2^16 = 65,536.Compute deletion count = 65,536 − 50,000.Perform subtraction: 65,536 − 50,000 = 15,536.Implement by decoding the first unwanted state and asserting reset.


Verification / Alternative check:
Simulate the counter: confirm that after 50,000 input clocks, the output sequence repeats, demonstrating the correct modulus and hence the correctness of the deletion count.


Why Other Options Are Wrong:

  • 50,000 / 65,536: These are the desired and native states themselves, not the difference.
  • 25,536: Arithmetic error (confuses difference with midpoint or mixes digits).


Common Pitfalls:
Off-by-one errors when the sequence starts at 1 instead of 0; neglecting synchronous clear timing which can add one extra count if not aligned with the clock edge.


Final Answer:
15,536

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