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(D) To find the remainder when the determinant $D$ is divided by $5$,we evaluate the entries modulo $5$.$2014 \equiv -1 \pmod{5}$,$2015 \equiv 0 \pmod

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$4$

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(D) To find the remainder when the determinant $D$ is divided by $5$,we evaluate the entries modulo $5$.$2014 \equiv -1 \pmod{5}$,$2015 \equiv 0 \pmod{5}$,$2016 \equiv 1 \pmod{5}$,$2017 \equiv 2 \pmod{5}$,$2018 \equiv 3 \equiv -2 \pmod{5}$,$2019 \equiv 4 \equiv -1 \pmod{5}$,$2020 \equiv 0 \pmod{5}$,$2021 \equiv 1 \pmod{5}$,$2022 \equiv 2 \pmod{5}$.Substituting these values into the determinant modulo $5$:$D \equiv \left|\begin{array}{ccc} (-1)^{2014} & 0^{2015} & 1^{2016} \ 2^{2017} & (-2)^{2018} & (-1)^{2019} \ 0^{2020} & 1^{2021} & 2^{2022} \end{array}ight| \pmod{5}$$D \equiv \left|\begin{array}{ccc} 1 & 0 & 1 \ 2^{2017} & 2^{2018} & -1 \ 0 & 1 & 2^{2022} \end{array}ight| \pmod{5}$Expanding along the first row:$D \equiv 1(2^{2018} \cdot 2^{2022} - (-1)(1)) - 0 + 1(2^{2017} \cdot 1 - 0) \pmod{5}$$D \equiv 2^{4040} + 1 + 2^{2017} \pmod{5}$Using Fermat's Little Theorem,$a^4 \equiv 1 \pmod{5}$ for $a$ not divisible by $5$:$2^{4040} = (2^4)^{1010} \equiv 1^{1010} \equiv 1 \pmod{5}$$2^{2017} = (2^4)^{504} \cdot 2^1 \equiv 1^{504} \cdot 2 \equiv 2 \pmod{5}$Thus,$D \equiv 1 + 1 + 2 = 4 \pmod{5}$.The remainder is $4$.

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

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