4
$\begingroup$

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

$\endgroup$
7
  • $\begingroup$ @heather No calculators. $\endgroup$ Sep 18, 2016 at 12:04
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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 $5$. $\endgroup$ Sep 18, 2016 at 12:07
  • $\begingroup$ @BarryCipra Do you see any pattern to compute the determinant easily while starting off? $\endgroup$ Sep 18, 2016 at 12:10
  • 1
    $\begingroup$ @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. $\endgroup$ Sep 18, 2016 at 12:17

1 Answer 1

5
$\begingroup$

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

$\endgroup$
11
  • $\begingroup$ Thanks for this, but do you think this can be done alternatively? $\endgroup$ 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? $\endgroup$ Sep 18, 2016 at 12:45
  • 1
    $\begingroup$ @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. $\endgroup$ Sep 18, 2016 at 12:59
  • 1
    $\begingroup$ @HGSur Note that $4$ and $-1$ are congruent modulo $5:$ $4=-1+5.$ $\endgroup$
    – mfl
    Oct 13, 2016 at 16:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.