How many minimum flip-flops are required for designing synchronous counter…

Concept: When a synchronous counter is designed for an explicitly listed counting sequence of decimal state values (rather than an abstract state diagram left free to be encoded any way), the flip-flop outputs must literally reproduce those decimal numbers in binary. So the minimum number of flip-flops n is the smallest n for which 2n exceeds the largest decimal value that appears anywhere in the sequence — not merely enough to tell the listed states apart.

1. List the given sequence: 0, 1, 6, 10, 4 (the counter then repeats back to 0).

2. Write each value in binary using the same register width: 0000, 0001, 0110, 1010, 0100.

3. Identify the largest value in the sequence: 10.

4. Check a 3-bit register: it spans 23 = 8 patterns, i.e. decimal 0 through 7. Since 10 > 7, 3 flip-flops cannot hold the value 10.

5. Check a 4-bit register: it spans 24 = 16 patterns, i.e. decimal 0 through 15. This comfortably holds every value in the sequence, including 10 (binary 1010).

6. So the minimum register width — and hence the minimum number of flip-flops — is 4.

Cross-check: if only the count of distinct states mattered (a free encoding with a separate output-decoding stage translating internal states to the displayed numbers), ⌈log2(5)⌉ = 3 flip-flops would be enough to distinguish the 5 states 0, 1, 6, 10, 4. But designing "a synchronous counter with counting sequence 0 ⇒ 1 ⇒ 6 ⇒ 10 ⇒ 4" is conventionally read as the register itself producing those decimal outputs directly, with no separate decoder — this is also how this exact question is solved in official exam solutions. Under that convention the register must be wide enough for the value 10, giving 4, not 3.

Result: the minimum number of flip-flops required is 4. Since 4 does not match any of 3, 6, or 5, the correct choice is "None of these".