Minimum flip-flop count for a modulus-5 (divide-by-5) counter: What is the least number of flip-flops required to implement a mod-5 counter?

Introduction / Context:
Determining the minimum number of flip-flops helps size counters efficiently. Each flip-flop stores one bit, and the number of distinct states available is 2^n for n flip-flops. A mod-N counter must provide at least N distinct states, possibly discarding extras with gating when 2^n is not exactly N.

Given Data / Assumptions:

• Target modulus N = 5.
• Flip-flops are binary storage elements; state capacity is 2^n.
• It is acceptable to reset or skip unused states using combinational feedback.

Concept / Approach:
Find the smallest n such that 2^n ≥ N. For N = 5: 2^2 = 4 (insufficient), 2^3 = 8 (sufficient). Therefore, at least three flip-flops are needed. The counter can be designed to count 0–4 and then reset to 0, effectively creating a mod-5 sequence within the 8-state space of three flip-flops.

Step-by-Step Solution:

Verification / Alternative check:
Simulate a 3-bit synchronous counter with decode logic on state 101 to assert CLEAR. The output cycles through five states per period, confirming mod-5 behavior while using only three flip-flops.

Why Other Options Are Wrong:

• 5, 8, 10 flip-flops: Far exceed the minimum; 2^5, 2^8, and 2^10 states are unnecessary for mod-5.
• 2 flip-flops: Only 4 states available, which cannot realize a mod-5 sequence.

Common Pitfalls:
Forgetting to add gating to truncate the 8-state space; confusing Johnson/ring counters with binary counters; neglecting propagation delays when forming synchronous reset logic (glitch avoidance may require synchronous clear or additional gating).

Final Answer:
3