For $\Sigma = \{a, b\}$, let us consider the regular language $L = \{x \mid x = a^{2+3k} \text{ or } x = b^{10+12k}, k \ge 0\}$. Which one of the following can be a pumping length (the constant guaranteed by the pumping lemma) for $L$?

According to the Pumping Lemma for regular languages, there exists a pumping length $p$ such that every string $w \in L$ with $|w| \ge p$ can be written as $w = xyz$ satisfying:

1. $|y| \ge 1$

2. $|xy| \le p$

3. For all $i \ge 0$, $xy^i z \in L$.

Consider the component $L_2 = \{b^{10+12k} \mid k \ge 0\}$:

• The shortest string in $L_2$ is $w = b^{10}$ with $|w| = 10$.

• If $p \le 10$ (such as $p = 3, 5, 9$): Since $|w| = 10 \ge p$, $w = b^{10}$ must be pumpable. The condition $|xy| \le p$ implies $1 \le |y| \le p < 10$. When $w$ is pumped to $xy^2 z$, its new length is $10 + |y|$. For this string to be in $L_2$, the length must be of the form $10 + 12k^{\prime}$, requiring $|y|$ to be a positive multiple of $12$. But $|y| \le p < 10$, so no such $y$ exists. Thus, $p$ cannot be $3, 5,$ or $9$.

• If $p = 24$: The string $b^{10}$ has length $10 < 24$, so the pumping lemma condition does not apply to it. For any string in $L_2$ with length $\ge 24$ (e.g., $b^{34}$), we can choose $|y| = 12$ within the prefix of length $\le 24$ (since $12 \le 24$), ensuring that pumping $y$ produces lengths of the form $10 + 12k$, which belong to $L$. Similarly, for strings in $L_1 = \{a^{2+3k}\}$, any string of length $\ge 24$ can be pumped by choosing $|y| = 3$ or $12$.

Therefore, 24 is a valid pumping length.