Tricky Problem: Designing a Mod-6 Counter and Handling Its Unused States

Connect the required-bit-width formula (ceil(log2(N))) to the existence of unused states above N-1, and recognize that a self-correcting counter needs explicit decode-and-reset logic rather than relying on 'natural' behavior.

Design a Mod-6 counter. Answer: (1) how many flip-flops are required, (2) which states are valid, (3) what happens to unused states, and (4) how the counter can return to a valid state. Key calculation given: ceil(log2(6)) = 3.

Why This Problem Is Tricky

Three flip-flops give 2^3 = 8 possible states, but a Mod-6 counter only needs 6 of them (0 through 5). The trap is assuming that because the counter is 'designed to count 0 to 5,' it will automatically and safely avoid states 6 and 7 (110 and 111) without any extra circuitry. A plain 3-bit binary counter with no added logic will happily count through all 8 states (0-7, i.e., mod 8), not mod 6 — reaching a true Mod-6 count sequence requires deliberately detecting a terminal count and forcing a reset, and separately reasoning about what happens if the counter is ever forced into an unused state (for example, by a power-up glitch).

Step-by-Step Worked Solution

Flip-flops required: a counter needs enough bits to represent every valid state from 0 up to N-1. For N=6, the number of bits is ceil(log2(6)). Since 2^2=4 is too few (only 4 states) and 2^3=8 is enough (8 >= 6), ceil(log2(6)) = 3. Three flip-flops are required.

Valid states: with 3 bits, the 6 valid Mod-6 states are 000, 001, 010, 011, 100, 101 (decimal 0 through 5).

Unused states: 3 bits provide 8 total combinations, but only 6 are used, leaving 110 and 111 (decimal 6 and 7) as unused/invalid states that should never appear during normal, correct operation.

Returning to a valid state: because a plain binary counter does not know to stop at 5, explicit decode logic must be added — typically a gate (e.g., an AND gate) that detects when the count reaches the terminal valid state (101, i.e., 5) and asserts a synchronous clear, forcing the count back to 000 on the very next clock edge instead of incrementing to 110. The same or similar decode logic should also detect the unused states 110 and 111 directly (in case the counter is ever forced into one, such as by a power-up condition) and force a clear from there too, guaranteeing the counter self-corrects back into the valid 0-5 cycle no matter what state it starts in.

Why the Obvious Wrong Answer Is Wrong

The common wrong assumption is that using exactly 3 flip-flops 'automatically' produces mod-6 counting, or that the unused states 110 and 111 simply 'don't happen' without any specific circuitry to prevent them. In reality, a bare 3-bit binary counter counts 0 through 7 and wraps via ordinary overflow (111 -> 000) — that is mod-8 behavior, not mod-6. To get a true Mod-6 sequence, the designer must add decode logic that recognizes the terminal valid count and forces an early, synchronous reset to 0, and must also make sure that logic (or an equivalent safety mechanism) recovers the counter if it somehow starts up in one of the two unused states.

Code Examples

Each of the four sub-questions from the problem is answered in turn: the bit-width calculation, the enumerated valid states, the enumerated unused states, and the specific decode-and-reset mechanism that guarantees the counter cannot get stuck outside its intended 0-5 cycle. Expected result: Mod-6 counter fully specified.

3 flip-flops; valid states 0-5; unused states 6-7; decode-and-reset recovers to 000.

Flip-flops needed: ceil(log2(6)) = 3   (2^2=4 too few, 2^3=8 is enough)

Valid states (decimal 0-5):
  000, 001, 010, 011, 100, 101

Unused states (decimal 6-7):
  110, 111

Recovery mechanism:
  Decode logic detects count = 101 (5) -> forces synchronous clear to 000
  Decode logic also detects 110 or 111 directly -> forces clear to 000
  (guarantees the counter self-corrects into the valid 0-5 cycle from any state)

Mod-6 Counter State Diagram

000 -> 001 -> 010 -> 011 -> 100 -> 101 --+
    ^                                        |
    +---------(synchronous clear)------------+

   110, 111 = unused states (outside the 0-5 cycle)
   decode logic detects them and forces a clear back to 000

Why Self-Correcting Counter Design Matters

Every mod-N counter used in real hardware — a seconds counter that resets every 60, an hours counter that resets every 12 or 24, a traffic-light-timer counter — relies on this exact decode-and-reset technique. Forgetting to handle the unused states leaves a design vulnerable to a rare but real failure mode: a power-on glitch or a transient upset (e.g., from noise or radiation) that lands the counter in an unused state can leave it permanently stuck or producing a wrong sequence if there is no explicit path back to the valid cycle.

Assuming a 3-flip-flop counter automatically counts mod-6 with no extra logic, and that states 110/111 simply never occur

Add explicit decode-and-reset logic that forces a synchronous clear to 000 when the count reaches 101 (5), and that also recovers from states 110/111 if the counter is ever forced into them

A plain 3-bit binary counter naturally counts mod-8 (0 through 7), not mod-6. Achieving a true Mod-6 sequence, and guaranteeing recovery from an invalid power-up state, both require deliberately added decode logic — they are not automatic consequences of using 3 flip-flops.

How many flip-flops does a Mod-6 counter require, and how many unused states does it have? A. 2 flip-flops, 0 unused states B. 3 flip-flops, 2 unused states C. 3 flip-flops, 0 unused states D. 4 flip-flops, 10 unused states

B. 3 flip-flops, 2 unused states — ceil(log2(6)) = 3, since 2^2=4 is too few but 2^3=8 is enough. Three flip-flops give 8 total states, of which only 6 (000-101) are used for counting 0-5, leaving 2 unused states (110 and 111) that must be handled by explicit decode-and-reset logic.

A Mod-10 (BCD) counter requires how many flip-flops, and how many unused states does it have? A. 3 flip-flops, 0 unused states B. 4 flip-flops, 6 unused states C. 4 flip-flops, 0 unused states D. 5 flip-flops, 22 unused states

B. 4 flip-flops, 6 unused states — ceil(log2(10)) = 4, since 2^3=8 is too few but 2^4=16 is enough. Four flip-flops give 16 total states, of which 10 are used (0-9) for BCD counting, leaving 16-10=6 unused states (1010 through 1111) that require decode-and-reset logic, exactly as in the Mod-6 case.

Design a Mod-5 counter: how many flip-flops are needed, which states are valid, which are unused, and how should the unused states be handled?

ceil(log2(5)) = 3 flip-flops (since 2^2=4 < 5 <= 2^3=8). Valid states: 000-100 (0-4). Unused states: 101, 110, 111 (5, 6, 7). Add decode logic that detects the terminal valid count (100, i.e., 4) and forces a synchronous clear to 000 on the next clock edge, and that also detects any of the three unused states directly to force recovery back to 000.

Compute ceil(log2(5)) the same way as for Mod-6.