A reduced state table has 18 rows. The minimum number of flip-flops needed to…

Each flip-flop can independently store a 0 or a 1, so a group of m flip-flops together can represent at most 2m distinct binary codes. To implement a sequential machine whose reduced state table has N states, the flip-flops used for state encoding must supply at least one distinct binary code per state, so the minimum number of flip-flops is the smallest integer m for which 2m is greater than or equal to N.

1. The reduced state table given has N = 18 rows, i.e. 18 states to encode.

2. Try m = 4 flip-flops: 24 = 16 distinct codes, which is fewer than the 18 states needed, so 4 flip-flops are not enough.

3. Try m = 5 flip-flops: 25 = 32 distinct codes, which is at least the 18 states needed, so 5 flip-flops are enough (with 32 − 18 = 14 codes left unused).

4. Since 4 flip-flops fall short and 5 flip-flops suffice, the minimum number of flip-flops required is 5.

This matches the logarithmic form of the same rule: the minimum flip-flop count is ⌈log2 18⌉ = ⌈4.17⌉ = 5, since log2 16 = 4 and log2 32 = 5 bracket 18 between them. Hence 5 flip-flops is the minimum needed, matching the option value 5.