Is the language $\{a^n b^m \mid 2n + 3m \le 1000 \}$ regular? We have a language  $$ L = \{a^n b^m \mid 2n + 3m \le 1000 \} $$
Is this language regular?
I'm trying to disprove this using the Pumping Lemma, but it didn't work.
assume I say x = $x=a^{h}$ and $y=a^{t}$ and $z =a^{n-t-h}b^m$
if I say i = 0 everything is okay because $L =a^{n-t}b^m$ and 2(n-t) + 3m &lt;= 1000
if I say i = 2  $L =a^{n+t}b^m$ and 2(n+t) + 3m &lt;= 1000 because I'm not sure about t value.
I think it didn't work. Is this language regular? How can I prove that?
The language is finite and, as such, it is regular (all finite languages are regular).
To see that the language is finite, notice that the maximum length of each word in $L$ is upper bounded by $500$.
Indeed, if $w \in L$, and $n$ (resp. $m$) is the number of $a$s (resp. $b$s) in $w$:
$$
2|w| = 2(n + m) \le 2n+ 3m \le 1000.
$$