jmc

We notice that 16 divides $78+82$ as well as $79+81$ and also 80. Therefore the sum is congruent to $$75+76+77(\mathrm{mod}16).$$Since these numbers are congruent to $-5$, $-4$, and $-3$ modulo 16, this can be computed as $$-5-4-3\equiv -12(\mathrm{mod}16).$$Finally, since $-12\equiv 4(\mathrm{mod}16)$ the remainder we seek is $\boxed{4}$.