Designing a MOD-32 binary counter — number of flip-flops How many flip-flops are required to construct a MOD-32 binary counter (i.e., a counter that cycles through 32 distinct states)?

Introduction / Context:Binary counters implemented with flip-flops have a number of states equal to 2^n, where n is the number of flip-flops (bits). Determining the number of flip-flops needed for a target modulus is a foundational skill for digital timing and sequencing design.

Given Data / Assumptions:

• We need a counter with 32 distinct states (MOD-32).
• Each flip-flop contributes one binary digit (bit).
• States grow as powers of two: 2^n.

Concept / Approach:Find the smallest n such that 2^n ≥ 32. Because 2^5 = 32, five flip-flops are exactly sufficient and produce the required modulus without wasted states.

Step-by-Step Solution:

Verification / Alternative check:Simulate a 5-bit counter; it cycles through 00000 to 11111 (32 states) before repeating.

Why Other Options Are Wrong:

• 3 and 6 do not yield 32 states (2^3=8, 2^6=64).
• 45 is unrelated and not a number of flip-flops.

Common Pitfalls:Confusing decimal modulus with binary powers; forgetting that extra states appear if n is larger than necessary.

Final Answer:5