Flip-flops needed for a decade (MOD-10) counter How many flip-flops are required to construct a decade counter that counts 0 through 9 and then recycles?

Introduction / Context:
A decade counter advances through ten distinct states (0–9) before recycling to 0. Determining the minimum number of flip-flops needed is a classic design question and highlights the relationship between binary capacity (2^n states) and a desired modulus that is not a power of two. The design also illustrates how decoded resets create nonbinary moduli.

Given Data / Assumptions:

• Required modulus: 10 states (0 to 9 inclusive).
• Flip-flops are binary storage elements; an n-bit register can represent up to 2^n states.
• Counter may use synchronous or ripple architecture with decoding to reset after state 9.

Concept / Approach:

Find the smallest n such that 2^n ≥ 10. Since 2^3 = 8 < 10 and 2^4 = 16 ≥ 10, four flip-flops are the minimum to represent at least ten states. Additional combinational logic decodes state 10 (1010) to asynchronously/synchronously reset the counter to 0, thereby creating a MOD-10 sequence within a 16-state space.

Step-by-Step Solution:

Verification / Alternative check:

Simulation or truth-table enumeration confirms that states 0000 through 1001 appear in sequence, and when 1010 is reached the reset forces 0000, giving a 10-state loop.

Why Other Options Are Wrong:

10 and 8 flip-flops: far exceed the minimum; 8 yields astronomically more states than required.

5 flip-flops: also works but is not minimal; the question asks how many are required at minimum.

Common Pitfalls:

Forgetting the need for a reset decode when the modulus is not a power of two; miscounting valid states; or assuming 3 flip-flops suffice for 0–9 (they do not—only 8 states are available).

Final Answer:

4