Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The remainder when the determinant
$\left|\begin{matrix}2014^{2014}&2015^{2015}&2016^{2016}\\ 2017^{2017}&2018^{2018}&2019^{2019}\\ 2020^{2020}&2021^{2021}&2022^{2022}\end{matrix}\right|$
is divided by $5$ is

KVPYKVPY 2015

Solution:

Let $D=\begin{vmatrix}2014^{2014}&2015^{2015}&2016^{2016}\\ 2017^{2017}&2018^{2018}&2019^{2019}\\ 2020^{2020}&2021^{2021}&2022^{2022}\end{vmatrix}$
$D=\begin{vmatrix}\left(2015-1\right)^{2014}&\left(2015\right)^{2015}&\left(2015+1\right)^{2016}\\ \left(2015+2\right)^{2017}&\left(2020-2\right)^{2018}&\left(2020-1\right)^{2019}\\ \left(2020\right)^{2020}&\left(2020+1\right)^{2021}&\left(2020+2\right)^{2022}\end{vmatrix}$
Remainder when divided by $5$, is
$D=\begin{vmatrix}1&0&1\\ 2^{2017}&2^{2018}&-1\\ 0&1&2^{2022}\end{vmatrix}$
$=1\left(2^{4040}+1\right)+2^{2017}$
$=\left(5-1\right)^{2020}+1+2\left(5-1\right)^{1008}$
$=1+1+2=4$