Cute Determinant Question I stumbled across the following problem and found it cute. 
Problem: We are given that $19$ divides $23028$, $31882$, $86469$, $6327$, and $61902$. Show that $19$ divides the following determinant: 
$$\left|
 \begin{matrix}
  2 & 3&0&2&8 \\
  3 & 1&8&8&2\\
8&6&4&6&9\\
0&6&3&2&7\\
6&1&9&0&2
 \end{matrix}\right|$$
 A: Multiply the first column by $10^4$, the second by $10^3$, third by $10^2$ and fourth by $10$ - this will scale the value of the determinant by $10^{4+3+2+1}=10^{10}$, which is coprime to $19$. Now add the last four columns to the first one - this will not change the value of the determinant. Finally notice the first column now reads $23028, 31882, 86469, 6327$, and $61902$: each is a multiple of $19$ so we can factor a nineteen cleanly out the determinant.
A: Integer proof
Perform the column operation $C_5\leftarrow 10^4C_1+10^3C_2+10^3C_3+10C_4+C_5$: the coefficient of $C_5$ is $1$ so this doesn't change the determinant.
All elements of $C_5$ ($23028$, $31882$, $86469$, $6327$, and $61902$) are now divisible by $19$, so we can factor out $19$: hence the determinant is divisible by $19$.

Modular proof
In $\mathbb Z/19\mathbb Z$, the columns $10^4C_1+10^3C_2+10^3C_3+10C_4+C_5$ sum to $0$: hence the matrix is not invertible and has determinant $0$. So in $\mathbb Z$, the determinant is a multiple of $19$.
