Question

The remainder when

$$\begin{vmatrix} 2014^{2014} & 2015^{2015} & 2016^{2016} \\ 2017^{2017} & 2018^{2018} & 2019^{2019} \\ 2020^{2020} & 2021^{2021} & 2022^{2022} \end{vmatrix}$$

is divided by 5 is k, then $k$ is equal to ____.

Answer 1

4

Answer 2

We need to find the determinant modulo 5. First, we reduce each element modulo 5. $2014 \equiv 4 \pmod{5}$, $2015 \equiv 0 \pmod{5}$, $2016 \equiv 1 \pmod{5}$ $2017 \equiv 2 \pmod{5}$, $2018 \equiv 3 \pmod{5}$, $2019 \equiv 4 \pmod{5}$ $2020 \equiv 0 \pmod{5}$, $2021 \equiv 1 \pmod{5}$, $2022 \equiv 2 \pmod{5}$

For the exponents, we use Fermat's Little Theorem, which states that for a prime $p$, $a^{p-1} \equiv 1 \pmod{p}$ for any integer $a$ not divisible by $p$. Here $p=5$, so $a^4 \equiv 1 \pmod{5}$. We reduce the exponents modulo 4. $2014 \equiv 2 \pmod{4}$, $2015 \equiv 3 \pmod{4}$, $2016 \equiv 0 \pmod{4}$ $2017 \equiv 1 \pmod{4}$, $2018 \equiv 2 \pmod{4}$, $2019 \equiv 3 \pmod{4}$ $2020 \equiv 0 \pmod{4}$, $2021 \equiv 1 \pmod{4}$, $2022 \equiv 2 \pmod{4}$

Now we calculate each term modulo 5: $2014^{2014} \equiv 4^{2014} \equiv (-1)^{2014} \equiv 1 \pmod{5}$ $2015^{2015} \equiv 0^{2015} \equiv 0 \pmod{5}$ $2016^{2016} \equiv 1^{2016} \equiv 1 \pmod{5}$ $2017^{2017} \equiv 2^{2017} \equiv 2^{2016} \cdot 2^1 \equiv (2^4)^{504} \cdot 2 \equiv 1^{504} \cdot 2 \equiv 2 \pmod{5}$ $2018^{2018} \equiv 3^{2018} \equiv 3^{2016} \cdot 3^2 \equiv (3^4)^{504} \cdot 9 \equiv 1^{504} \cdot 9 \equiv 4 \pmod{5}$ $2019^{2019} \equiv 4^{2019} \equiv (-1)^{2019} \equiv -1 \equiv 4 \pmod{5}$ $2020^{2020} \equiv 0^{2020} \equiv 0 \pmod{5}$ $2021^{2021} \equiv 1^{2021} \equiv 1 \pmod{5}$ $2022^{2022} \equiv 2^{2022} \equiv 2^{2020} \cdot 2^2 \equiv (2^4)^{505} \cdot 4 \equiv 1^{505} \cdot 4 \equiv 4 \pmod{5}$

The determinant modulo 5 is:

$$\begin{vmatrix} 1 & 0 & 1 \\ 2 & 4 & 4 \\ 0 & 1 & 4 \end{vmatrix} \equiv 1(4 \cdot 4 - 4 \cdot 1) - 0(\dots) + 1(2 \cdot 1 - 4 \cdot 0) \pmod{5}$$ $$\equiv 1(16 - 4) + 1(2) \pmod{5}$$ $$\equiv 12 + 2 \pmod{5}$$ $$\equiv 14 \pmod{5}$$ $$\equiv 4 \pmod{5}$$

Thus, $k=4$.