# 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 $(a)\quad1\quad (b)\quad2\quad (c)\quad3\quad (d)\quad 4$

• @heather No calculators. Sep 18, 2016 at 12:04
• I doubt that you can make use of the multiple-choice nature of the question, since almost all possible answers are present in the options. Sep 18, 2016 at 12:05
• Your best best is to use Fermat's little theorem: reduce the exponents mod $4$ and the numbers being exponentiated mod $5$. Then compute the determinant as usual, and then reduce the result mod $5$. Sep 18, 2016 at 12:07
• @BarryCipra Do you see any pattern to compute the determinant easily while starting off? Sep 18, 2016 at 12:10
• @StubbornAtom, I don't. It's possible there is one, but if I were faced with this on an exam, I wouldn't spend a lot of time trying to find it. Sep 18, 2016 at 12:17

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 $$\begin{vmatrix} 1 & 0 & 1 \\ 2 & (-2)^2 & -1 \\ 0 & 1 & 2^2 \end{vmatrix}=\begin{vmatrix} 1 & 0 & 1 \\ 2 & -1 & -1 \\ 0 & 1 & -1 \end{vmatrix}=\begin{vmatrix} 1 & 0 & 1 \\ 0 & -1 & 2 \\ 0 & 1 & -1 \end{vmatrix}=(-1)^2-2=-1\equiv 4.$$
• I don't see what you have in mind. Could you explain more precisely? I just did two things: reduce each element in the determinant modulo $5$, then one row operation to simplify expansion along the first column Sep 18, 2016 at 12:39
• @HGSur Note that $4$ and $-1$ are congruent modulo $5:$ $4=-1+5.$