jmc

Since $\varphi (49)=42$ (see Euler's totient function), Euler's Totient Theorem tells us that $a^{42}\equiv 1(\mathrm{mod}49)$ where $\text{gcd}(a,49)=1$. Thus $6^{83}+8^{83}\equiv 6^{2(42)-1}+8^{2(42)-1}$ $\equiv 6^{-1}+8^{-1}\equiv \frac{8+6}{48}$ $\equiv \frac{14}{-1}\equiv \boxed{35}(\mathrm{mod}49)$.