jmc

Note that $$\begin{array}{rl}1342& =1300+39+3\\ & =13(100+3)+3,\end{array}$$so $r=3$.

We are seeking the smallest multiple of $1342$ that is congruent to $0$, $1$, or $2$ modulo $13$.

We have $1342n\equiv 3n(\mathrm{mod}13)$, so the remainders of the first four multiples of $1342$ are $3,6,9,12$. The next number in this sequence is $15$, but $15$ reduces to $2$ modulo $13$. That is: $$5\cdot 1342\equiv 5\cdot 3\equiv 2(\mathrm{mod}13).$$Therefore, the number we are looking for is $5\cdot 1342=\boxed{6710}$.