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.
You can, by fast exponentiation, check that $17^{11}\equiv-1\mod 23$. Thus, as $23$ is prime, Little Fermat entails that:
$$17^{235}\equiv17^{235\bmod22}= 17^{15}\equiv -17^4=-(17^2)^2\equiv-13^2\equiv -8\equiv 15.$$