Gate2019 cs Q25
- Option : B
- Explanation : L = {a2 + 3k or b10 + 12k} for k ≥ 0 = a2 (a3)* or b10 (b12)* = {a2, a5, a8, ..., b10, b22, b234 .....} The pumping length is p, than for any string w ∈ L with ⏐w⏐≥ p must have a repetition i.e. such a string must be breakable into w = xyz such that ⏐y⏐≥ 0 and y can be pumped indefinitely, which is same as saying xyz ∈ L ⇒ xy*z ∈ L. The minimum pumping length in this language is clearly 11, since b10 is a string which has no repetition number, so upto 10 no number can serve as a pumping length. Minimum pumping length is 11. Any number at or above minimum pumping length can serve as a pumping length. The only number at or above 11, in the choice given is 24. So correct answer is option (B)
