We want to design a synchronous counter that counts the sequence 0-1-0-2-0-3…

A synchronous counter is a finite-state machine: on every clock edge it moves deterministically from one internal state to exactly one next state. Two occurrences of the same output value can share a single internal state only if they are always followed by the same next value — the moment one output value is followed by two different values at different points in the cycle, those occurrences must be distinct internal states. The minimum flip-flop count is then the smallest n with 2ⁿ ≥ (number of distinct states).

1. Write out the full cycle: 0 → 1 → 0 → 2 → 0 → 3 → back to the first 0.

2. Check every occurrence of the repeated value 0: the first 0 is followed by 1, the second 0 is followed by 2, and the third 0 is followed by 3 — three different successors, so these three 0's cannot be the same state.

3. Count the distinct states needed: one state for each of the three 0-occurrences, plus one each for 1, 2, and 3 — 6 distinct states in total.

4. Find the minimum n with 2ⁿ ≥ 6: 2² = 4 is too few, 2³ = 8 is enough, so the minimum is n = 3 flip-flops.

With 3 flip-flops, assign each of the 6 states one of the 8 available 3-bit codes (2 codes go unused) and derive the J-K excitation equations from the resulting state table — the flip-flop type (J-K, D, or T) changes only the excitation equations, never the number of flip-flops, which depends solely on the state count. Trying only 2 flip-flops (4 codes) cannot hold all 6 required states, confirming that 3 is the minimum.