Find the remainder of the Euclidean division of $6^{83}+8^{83}$ by 49.
(Theorem of Euler and manipulations of inverses modulo an integer)

We denote $\phi$ as the Euler's totient function, and note that $\phi(49)=\phi\left(7^{2}\right)=7^{2}-7=42$. Thus, $6^{84} \equiv 1 \bmod 49$ and $8^{84} \equiv 1 \bmod 49$. Therefore, denoting $x^{-1}$ as the $^{84}$ inverse of $x$ modulo 49 when $x$ is coprime with 49, we have

$$
\begin{aligned}
6^{83}+8^{83} & \equiv\left(6^{84}\right) \cdot 6^{-1}+\left(8^{84}\right) \cdot 8^{-1} \quad \bmod 49 \\
& \equiv 6^{-1}+8^{-1} \bmod 49
\end{aligned}
$$

By factoring out $6^{-1} 8^{-1}$, we get:

$$
\begin{aligned}
6^{-1}+8^{-1} & \equiv(6+8) 6^{-1} 8^{-1} \quad \bmod 49 \\
& \equiv 14 \cdot 48^{-1} \quad \bmod 49 \\
& \equiv 14 \cdot(-1) \bmod 49 \\
& \equiv 35 \bmod 49
\end{aligned}
$$

The answer is therefore 35.