3. If the set $A=\{x \mid x=6 n-1, n \in \mathbf{N}\}$,
$$
B=\{x \mid x=8 n+3, n \in \mathbf{N}\},
$$

then the number of elements in $A \cap B$ that are less than 2010 is $\qquad$
3. 84 .

According to the problem, if $x \in A$, then $x \equiv 5(\bmod 6)$; if $x \in B$, then $x \equiv 3(\bmod 8)$. Therefore, if $x \in A \cap B$, then $x \equiv 11(\bmod 24)$, i.e., $x=24 k+11(k \in \mathbf{N})$.
Thus, there are 84 elements that satisfy the condition.