Is there a quick way to find the remainder when this determinant is divided by $5$? 
Find the remainder when the determinant $\begin{vmatrix} { 2014 }^{ 2014 } & { 2015 }^{ 2015 } & { 2016 }^{ 2016 } \\ { 2017 }^{ 2017 } & { 2018 }^{ 2018 } & { 2019 }^{ 2019 } \\ { 2020 }^{ 2020 } & { 2021 }^{ 2021 } & { 2022 }^{ 2022 } \end{vmatrix}$ is divided by $5$.

I'm aware that this problem has a number-theoretic solution involving congruence relations. But considering that this was asked as a multiple-choice question in a test, what should be the best way to approach problems like this?
The options were $(a)\quad1\quad (b)\quad2\quad (c)\quad3\quad (d)\quad 4$

Note that, modulo $5$,
 \begin{align}
2014 &\equiv-1, & 2015 & \equiv 0, & 2016 &\equiv 1,\\ 
2017 & \equiv2, & 2018 &\equiv-2, & 2019 &\equiv-1, \\
 2020 &\equiv 0, & 2021 &\equiv 1, & 2022 & \equiv2.
\end{align}
Furthermore $2$ and $-2$ have order $4$ modulo $5$, so the determinant is congruent to
$$\begin{vmatrix}
1 & 0 & 1 \\ 2 & (-2)^2 & -1 \\ 0 & 1 & 2^2
\end{vmatrix}=\begin{vmatrix}
1 & 0 & 1 \\ 2 & -1 & -1 \\ 0 & 1 & -1
\end{vmatrix}=\begin{vmatrix}
1 & 0 & 1 \\ 0 & -1 & 2 \\ 0 & 1 & -1
\end{vmatrix}=(-1)^2-2=-1\equiv 4.$$
