3. Given the set $A=\{x \mid 5 x-a \leqslant 0\}, B=$ $\{x \mid 6 x-b>0\}, a, b \in \mathbf{N}$, and $A \cap B \cap \mathbf{N}=$ $\{2,3,4\}$. Then the number of integer pairs $(a, b)$ is ( ).
(A) 20
(B) 25
(C) 30
(D) 42
3.C.

From $5 x-a \leqslant 0$, we get $x \leqslant \frac{a}{5}$;
From $6 x-b>0$, we get $x>\frac{b}{6}$.
To make $A \cap B \cap \mathbf{N}=\{2,3,4\}$, it should be
$$
\left\{\begin{array}{l}
1 \leqslant \frac{b}{6} < 2, \\
4 \leqslant \frac{a}{5} < 5,
\end{array} \text { i.e., } \left\{\begin{array}{l}
6 \leqslant b<12, \\
20 \leqslant a<25 .
\end{array}\right.\right.
$$

Therefore, the number of pairs $(a, b)$ is $\mathrm{C}_{6}^{1} \mathrm{C}_{5}^{1}=30$.