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

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  • $\begingroup$ @heather No calculators. $\endgroup$
    – StubbornAtom
    Sep 18, 2016 at 12:04
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    $\begingroup$ 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. $\endgroup$
    – lisyarus
    Sep 18, 2016 at 12:05
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    $\begingroup$ 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 $5$. $\endgroup$
    – Barry Cipra
    Sep 18, 2016 at 12:07
  • $\begingroup$ @BarryCipra Do you see any pattern to compute the determinant easily while starting off? $\endgroup$
    – StubbornAtom
    Sep 18, 2016 at 12:10
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    $\begingroup$ @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. $\endgroup$
    – Barry Cipra
    Sep 18, 2016 at 12:17

1 Answer 1

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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 $$\begin{vmatrix} 1 & 0 & 1 \\ 2 & (-2)^2 & -1 \\ 0 & 1 & 2^2 \end{vmatrix}=\begin{vmatrix} 1 & 0 & 1 \\ 2 & -1 & -1 \\ 0 & 1 & -1 \end{vmatrix}=\begin{vmatrix} 1 & 0 & 1 \\ 0 & -1 & 2 \\ 0 & 1 & -1 \end{vmatrix}=(-1)^2-2=-1\equiv 4.$$

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  • $\begingroup$ Thanks for this, but do you think this can be done alternatively? $\endgroup$
    – StubbornAtom
    Sep 18, 2016 at 12:25
  • $\begingroup$ 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 $\endgroup$
    – Bernard
    Sep 18, 2016 at 12:39
  • $\begingroup$ No I mean is there a solution that does not use congruence? $\endgroup$
    – StubbornAtom
    Sep 18, 2016 at 12:45
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    $\begingroup$ @Stub, the question is asking you to evaluate a congruence. Anything you do to solve it is just going to be using congruence, either openly or secretly. $\endgroup$
    – Gerry Myerson
    Sep 18, 2016 at 12:59
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    $\begingroup$ @HGSur Note that $4$ and $-1$ are congruent modulo $5:$ $4=-1+5.$ $\endgroup$
    – mfl
    Oct 13, 2016 at 16:58

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