jmc

Since $247,39,$ and $143$ are all divisible by $13$, the residues for $247+5\cdot 39+7\cdot 143$ is just $0$.

Therefore, $247+5\cdot 39+7\cdot 143+4\cdot 15\equiv 4\cdot 15\equiv 60\equiv \boxed{8}(\mathrm{mod}13)$.