If the determinant is $0$ it is obvious that $19|0$.
Now suppose that the determinant is not $0$.
$$\begin{align*}
2\cdot10^4+3\cdot10^3+0\cdot10^2+2\cdot10+8\cdot1&=23028\\
3\cdot10^4+1\cdot10^3+8\cdot10^2+8\cdot10+2\cdot1&=31882\\
8\cdot10^4+6\cdot10^3+4\cdot10^2+6\cdot10+9\cdot1&=86469\\
0\cdot10^4+6\cdot10^3+3\cdot10^2+2\cdot10+7\cdot1&=06327\\
6\cdot10^4+1\cdot10^3+9\cdot10^2+0\cdot10+2\cdot1&=61902
\end{align*}$$
By Cramer's rule
$$1=\frac{\left|\begin{matrix}
2 & 3 & 0 & 2 & 23028 \\
3 & 1 & 8 & 8 & 31882 \\
8 & 6 & 4 & 6 & 86469 \\
0 & 6 & 3 & 2 & 06327 \\
6 & 1 & 9 & 0 & 61902
\end{matrix}\right|}{\left|\begin{matrix} 2 & 3 & 0 & 2 & 8 \\
3 & 1 & 8 & 8 & 2 \\
8 & 6 & 4 & 6 & 9 \\
0 & 6 & 3 & 2 & 7 \\
6 & 1 & 9 & 0 & 2\end{matrix}\right|}$$
Then
$$\left|\begin{matrix} 2 & 3 & 0 & 2 & 8 \\
3 & 1 & 8 & 8 & 2 \\
8 & 6 & 4 & 6 & 9 \\
0 & 6 & 3 & 2 & 7 \\
6 & 1 & 9 & 0 & 2\end{matrix}\right|=\left|\begin{matrix}
2 & 3 & 0 & 2 & 23028 \\
3 & 1 & 8 & 8 & 31882 \\
8 & 6 & 4 & 6 & 86469 \\
0 & 6 & 3 & 2 & 06327 \\
6 & 1 & 9 & 0 & 61902
\end{matrix}\right|$$
But last determinant is obviously divisible by $19$.