2. Given that $n$ is a natural number, $11 n+10$ and $10 n+3$ are both multiples of some natural number $d$ greater than 1. Then $d=$ $\qquad$
2.67.

From $11 n+10$ and $10 n+3$ both being multiples of $d$, we know
$$
10(11 n+10)-11(10 n+3)=67
$$

is still a multiple of $d$. But 67 is a prime number, $d \neq 1$, so $d=67$.