3. $n$ is a natural number, $19 n+14$ and $10 n+3$ are both multiples of some natural number $d$ not equal to 1, then $d=$ $\qquad$ .
3. 83.

Since $\frac{19 n+14}{10 n+3}=1+\frac{9 n+11}{10 n+3}$,
and $\frac{10 n+3}{9 n+11}=1+\frac{n-8}{9 n+11}$,
also $\frac{9 n+11}{n-8}=9+\frac{83}{n-8}$,
since 83 is a prime number, if 83 and $n-8$ are coprime, then $n=91$.
When $n=91$,
$$
\begin{array}{l}
19 n+14=19 \times 91+14=83 \times 21, \\
10 n+3=10 \times 91+3=83 \times 11, \\
\therefore d=(19 n+14,10 n+3)=83 .
\end{array}
$$