Find remainder of $\frac{17^{235}}{ 23}$ I need to find remainder of $\frac{17^{235}}{ 23}$.
This is supposed to be solved using the following method:


*

*$\varphi(23) = 22$

*${17}^{235} = (({17}^{22})^{10})\cdot {17}^{15}$

*${17}^{22}\equiv 1 \pmod{23}$

*Then, according to Euler theorem, $17^{235} \equiv 17^{15} \pmod {23}$


and then i get stuck.

$17^{15}=  17^1\times 17^2\times17^4\times 17^8$.
In the rhs every term is the square of the previous term. 
$17^2=289=230+69-10\equiv \mathbf{-10}\pmod{23}$.
$17^4\equiv(-10)^2=100\equiv\mathbf{8}\pmod{23} $
$17^8\equiv 8^2=64\equiv\mathbf{-5}\pmod{23} $
Now multiply $17.(-10).8.(-5)\equiv\mathbf{-8}\pmod{23}$
The answer is 15.
The answer is 15.
