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. $4$
• D. $3$

Answer 1

The correct option is C $\left(4\right)$

Given $\left|\begin{array}{ccc}{2014}^{2014}& {2015}^{2015}& {2016}^{2016}\\ {\left(2017\right)}^{2017}& {\left(2018\right)}^{2018}& {\left(2019\right)}^{2019}\\ {2020}^{2020}& {2021}^{2021}& {2022}^{2022}\end{array}\right|$

$=\left|\begin{array}{ccc}(2015-1{)}^{2014}& {2015}^{2015}& (2015+1{)}^{2016}\\ {(2015+2)}^{2017}& {(2020-2)}^{2018}& {(2020-1)}^{2019}\\ {2020}^{2020}& {(2020+1)}^{2021}& {(2020+2)}^{2022}\end{array}\right|$

Since, the expression of the form $(5n-a{)}^k$, leaves the remainder $(-a{)}^k$ when it divided by $5$.
Similar way, $(2015-1{)}^{2014}$ leaves the remainder $(-1{)}^{2014}$, when it divided by $5$.
${2015}^{2015}$ leaves the remainder $0$ .
${(2015+2)}^{2017}$ leaves the remainder $2^{2017}$.
${(2020-1)}^{2019}$ leaves the remainder $(-1{)}^{2019}$.
Thus, Reminder $=\left|\begin{array}{ccc}1& 0& 1\\ (2{)}^{2017}& (-2{)}^{2018}& -1\\ 0& 1& (2{)}^{2022}\end{array}\right|$
Expanding along the first row, we get $1\times \left[\right(-2{)}^{2018}\times 2^{2022}+1]+2^{2017}$

$=\left[\right(2{)}^{2018}\times 2^{2022}]+2^{2017}+1$
$=2^{2018+2022}+2\times 2^{2016}+1$
$=2^{4040}+2(2^{\frac{2}{2}}{)}^{1008\times 2}+1$
$=(4{)}^{2020}+2(4{)}^{1008}+1$
$={(5-1)}^{2020}+2{(5-1)}^{1008}+1$
Remainder $=(-1{)}^{2020}+2{(-1)}^{1008}+1$ when divided by $5$.
$=1+2+1$

$=4$
Hence, the remainder is $4$ when divided by $5$.