Below I have the answer to the question solve: $23^{119} \ \mathrm{mod} \ 5$.
We have $119 = 29\cdot4 + 3$.
$23^{119} \ \mathrm{mod} \ 5 \\= (23^4)^{29}\cdot 23^3 \ \mathrm{mod} \ 5 \\ = 1 \cdot 23^3 \ \mathrm{mod} \ 5 \\ = 23^3 \ \mathrm{mod} \ 5 \\ = 3^3 \ \mathrm{mod} \ 5 \\ = 27 \ \mathrm{mod} \ 5 \\ = 2 \ \mathrm{mod} \ 5.$
I can follow the problem up to the point when we say that $23^3\ \mathrm{mod} \ 5 = 3^3 \ \mathrm{mod} \ 5$. I'm able to solve the problem by manually cubing $27$ but I don't understand how to make the logical jump from $23^3 \ \mathrm{mod} \ 5 = 3^3 \ \mathrm{mod} \ 5$.
$23 = 3 + 5(4)$, so \begin{align*} 23 &\cong 3 \pmod{5} \\ 23^2 \cong 23 \cdot 23 &\cong 3 \cdot 3 \cong 3^2 \pmod{5} \\ &\vdots \\ 23^{k} \cong 23^{k-1} \cdot 23 &\cong 3^{k-1} \cdot 3 \cong 3^{k} \pmod{5}, \end{align*} for any integer $k \geq 1$. (Actually true for any integer $k$.)