Use congruence's to find the reminder when $2^{50}$ and $41^{65}$ are divided by $7$ Use congruence's to find the reminder when $2^{50}$ and $41^{65}$ are divided by 7
$2^{50}$
$50=(7)^2+1$
$2^{50}=2^{7\cdot7+1}$
and I'm not sure where to go from here?

Note that, $$\begin{align} & 2^3\equiv 1 \pmod7 \\ \implies & (2^3)^{16}\equiv 1^{16} \pmod7 \\ \implies & 2^{48}\equiv 1 \pmod7 \\ \implies & 2^{48}\cdot 2^2\equiv 1\cdot 2^2 \pmod7 \\ \implies & \color{blue}{2^{50}\equiv 4 \pmod7}\end{align}$$
Also note that $$\begin{align} & 41\equiv -1\pmod7 \\ \implies & 41^{65}\equiv (-1)^{65}\equiv -1\pmod7 \\ \implies & \color{blue}{41^{65}\equiv 6\pmod7}\end{align}$$
