Use congruence's to find the reminder when $2^{50}$ and $41^{65}$ are divided by $7$

Question

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?

Answer 1

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 $$\begin{align} & 41\equiv -1\pmod7 \\ \implies & 41^{65}\equiv (-1)^{65}\equiv -1\pmod7 \\ \implies & \color{blue}{41^{65}\equiv 6\pmod7}\end{align}$$

Answer 2

$\color\red{2^3}\equiv\color\red{1}\pmod7\implies$

$2^{50}\equiv2^{3\cdot16+2}\equiv(\color\red{2^3})^{16}\cdot2^2\equiv\color\red{1}^{16}\cdot2^2\equiv1\cdot4\equiv4\pmod7$

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$\color\red{41}\equiv\color\red{-1}\pmod7\implies$

$\color\red{41}^{65}\equiv(\color\red{-1})^{65}\equiv(-1)\equiv6\pmod7$