Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

$\text{determinant}\left|\begin{array}{ccc}{(2015-1)}^{2014}& {2015}^{2015}& {(2015+1)}^{2016}\\ {(2015+2)}^{2017}& {(2020-2)}^{2018}& {(2020-1)}^{2019}\\ {\left(2020\right)}^{2020}& {(2020+1)}^{2021}& {(2020+2)}^{2022}\end{array}\right|$
$\text{remainder of determinant}\left|\begin{array}{ccc}{(-1)}^{2014}& 0& 1^{2016}\\ {\left(2\right)}^{2017}& {(-2)}^{2018}& {(-1)}^{2019}\\ 0& {\left(1\right)}^{2021}& {\left(2\right)}^{2022}\end{array}\right|$
$\Rightarrow 1\{2^{4040}+1\}+1\left\{2^{2017}\right\}\Rightarrow \{4^{2020}+1\}+{2.4}^{1008}$
$\Rightarrow \left\{\right(5-1{)}^{2020}+1\}+2.(5-1{)}^{1008}$
remainder, $(-1{)}^{2020}+1+2.(-1{)}^{1008}$
$\therefore$ remainder $=1+1+2=4$