Question 25

The language is $L = L_1 \cup L_2$ where:

• $L_1 = \{a^{2 + 3k} \mid k \ge 0\} = \{a^2, a^5, a^8, a^{11}, \dots\}$
• $L_2 = \{b^{10 + 12k} \mid k \ge 0\} = \{b^{10}, b^{22}, b^{34}, \dots\}$

According to the Pumping Lemma for regular languages, if $p$ is a valid pumping length, then for every string $s \in L$ with $|s| \ge p$, $s$ can be written as $s = uvw$ such that:

1. $|uv| \le p$
2. $|v| \ge 1$
3. $uv^i w \in L$ for all $i \ge 0$.

Consider the string $s = b^{10} \in L$. Its length is $|s| = 10$.

• If $p \le 10$, then $|s| \ge p$, so $s$ must be pumpable.
• By condition 1, $|uv| \le p \le 10$, so $v = b^m$ with $1 \le m \le p \le 10$.
• Pumping with $i = 2$ gives $s' = uv^2 w$, whose length is $|s'| = 10 + m$.
• Since $1 \le m \le p \le 10$, the length of $s'$ satisfies $11 \le |s'| \le 20$.
• But the valid lengths of strings of $b$ in $L$ are $10, 22, 34, \dots$. None of the integers between $11$ and $20$ is of the form $10 + 12k$ (the next valid length after 10 is 22).
• Furthermore, $s'$ consists purely of $b$'s, so it cannot belong to $L_1$.

Hence, $s' \notin L$, which contradicts the pumping lemma! Therefore, no value $p \le 10$ can be a valid pumping length. This immediately rules out 3, 5, and 9.
Among the given options, only $p = 24$ can be a valid pumping length (Option D).