KVPY 2015 7

Suppose $a$ , $b$ , $c$ $\ldots$ $h$ and $i$ are integers and let $a_{1}$ , $b_{1}$ , $c_{1}$ $\ldots$ $h_{1}$ and $i_{1}$ be the remainders when divided $p$ .

Then, $\begin{aligned} \left| \begin{bmatrix} a &b &c \\ d &e &f \\ g &h &i \end{bmatrix} \right| \mod p &\equiv a(ei-fh) - b(di-fg) + c(dh-eg) \mod p \\ &\equiv a_{1}(e_{1}i_{1}-f_{1}h_{1}) - b_{1}(d_{1}i_{1}-f_{1}g_{1}) + c_{1}(d_{1}h_{1}-e_{1}g_{1}) \\ &\equiv \left| \begin{bmatrix} a_{1} &b_{1} &c_{1} \\ d_{1} &e_{1} &f_{1} \\ g_{1} &h_{1} &i_{1} \end{bmatrix} \right| \mod p \end{aligned}$

Now by modular arithmetic, $2014^{2014} \equiv 1$ ; $2015^{2015} \equiv 0$ ; $2016^{2016} \equiv 1$ ; $2017^{2017} \equiv 2$ ; $2018^{2018} \equiv -1$ ; $2019^{2019} \equiv -1$ ; $2020^{2020} \equiv 0$ ; $2021^{2021} \equiv 1$ ; $2022^{2022} \equiv -1$ $\mod 5$ .

So, $\begin{aligned} \left| \begin{bmatrix} 2014^{2014} &2015^{2015} &2016^{2016} \\ 2017^{2017} &2018^{2018} &2019^{2019} \\2020^{2020} &2021^{2021} &2022^{2022} \end{bmatrix} \right| \mod 5 &\equiv \left| \begin{bmatrix} 1 &0 &1 \\ 2 &-1 &-1 \\ 0 &1 &-1 \end{bmatrix} \right| \mod 5 \\ &\equiv 4 \mod 5 \end{aligned}$

We can simply solve this by taking end number of term and solving its determinant and then dividing by 5 2014 power - 4 6 6 6 so we take 6 2015 power - 5 5 5 5 so we take 5 2016 power - 6 6 6 6 so we take 6 2017 power - 7 9 3 1 7 so we take 9 ( Since 2017 = 5n + 2 and here 2 for us will be 9 ) Similarly 2018 power - 8 4 2 6 8 so we take 6 2019 power - so we take 1 2020 power - so we take 0 2021 power - so we take 1 2022 power - so we take 4

and then simplifying determinant we get D= 264 264/5 = 5m+ 4 So Ans = 4