The remainder when the determinant
`|(2014^(2014),2015^(2015),2016^(2016)),(2017^(2017),2018^(2018),2019^(2019)),(2020^(2020),2021^(2021),2022^(2022))|`
is divided by 5 is-

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To find the remainder when the determinant

\[ D = \begin{vmatrix} 2014^{2014} & 2015^{2015} & 2016^{2016} \\ 2017^{2017} & 2018^{2018} & 2019^{2019} \\ 2020^{2020} & 2021^{2021} & 2022^{2022} \end{vmatrix} \]

is divided by 5, we can simplify the calculations by reducing each base modulo 5.

### Step 1: Calculate each base modulo 5

- \(2014 \mod 5 = 4\)
- \(2015 \mod 5 = 0\)
- \(2016 \mod 5 = 1\)
- \(2017 \mod 5 = 2\)
- \(2018 \mod 5 = 3\)
- \(2019 \mod 5 = 4\)
- \(2020 \mod 5 = 0\)
- \(2021 \mod 5 = 1\)
- \(2022 \mod 5 = 2\)

### Step 2: Rewrite the determinant with reduced bases

Now we can rewrite the determinant \(D\) as:

\[ D \equiv \begin{vmatrix} 4^{2014} & 0^{2015} & 1^{2016} \\ 2^{2017} & 3^{2018} & 4^{2019} \\ 0^{2020} & 1^{2021} & 2^{2022} \end{vmatrix} \mod 5 \]

### Step 3: Simplify the powers

- \(0^{2015} = 0\) and \(0^{2020} = 0\), so the first column and the third row will contribute zeros to the determinant.

Thus, the determinant simplifies to:

\[ D \equiv \begin{vmatrix} 4^{2014} & 1 \\ 2^{2017} & 3^{2018} \end{vmatrix} \mod 5 \]

### Step 4: Calculate the determinant of the 2x2 matrix

The determinant of a 2x2 matrix is given by:

\[ \text{det} = ad - bc \]

For our matrix:

\[ D \equiv 4^{2014} \cdot 3^{2018} - 1 \cdot 2^{2017} \mod 5 \]

### Step 5: Calculate powers modulo 5

Now we need to calculate \(4^{2014} \mod 5\), \(3^{2018} \mod 5\), and \(2^{2017} \mod 5\).

Using Fermat's Little Theorem, we know that for any integer \(a\) not divisible by a prime \(p\):

\[ a^{p-1} \equiv 1 \mod p \]

Thus:

- \(4^{4} \equiv 1 \mod 5\) implies \(4^{2014} \equiv 4^{2} \equiv 16 \equiv 1 \mod 5\)
- \(3^{4} \equiv 1 \mod 5\) implies \(3^{2018} \equiv 3^{2} \equiv 9 \equiv 4 \mod 5\)
- \(2^{4} \equiv 1 \mod 5\) implies \(2^{2017} \equiv 2^{1} \equiv 2 \mod 5\)

### Step 6: Substitute back into the determinant

Now substituting these values back into the determinant:

\[ D \equiv (1 \cdot 4) - (1 \cdot 2) \mod 5 \]
\[ D \equiv 4 - 2 \equiv 2 \mod 5 \]

### Final Answer

The remainder when the determinant is divided by 5 is:

\[ \boxed{2} \]