determinant divisible by 13 Question:
Given: $195,403$ and $247$ are divisible by 13. 
Prove (without actually calculating the determinant) that 
$$\det \begin{bmatrix} 1 & 9 & 5 \\ 2 & 4 & 7 \\ 4 & 0 & 3 \end{bmatrix}$$
is divisible by 13.
What I did:
Apart from calculating the determinant and seeing that it's true, I couldn't really think of anything else...
Thanks
 A: Hint:  If you multiply the first column by $100$, how does that change the determinant?  Then if you add the third column to the first, how does that change the determinant?
A: *

*We start by proving that $x$ is divisible by 13 if and only if $100x$ is also divisible by 13:

*$\Rightarrow $ If x is divisible by 13 then $x=13n \Rightarrow 100x=100(13n)=13(100n)$ So it's divisible as 
well.

*$\Leftarrow$ If 100x is divisible by 13 then $100x=13n$ but since 100 is not divisible by 13, $x$ must have 13 as a factor. Therefore $x$ is divisible by 13.  

*Now we prove what was asked:
$100 detA=100\cdot det \begin{bmatrix} 1 & 9 & 5 \\ 2 & 4 & 7 \\ 4 & 0 & 3 \end{bmatrix}$= 
$det \begin{bmatrix} 100 & 9 & 5 \\ 200 & 4 & 7 \\ 400 & 0 & 3 \end{bmatrix}$= 
(elementary column operations that don't change the determinant)
=$det \begin{bmatrix} 195 & 9 & 5 \\ 247 & 4 & 7 \\ 403 & 0 & 3 \end{bmatrix}$=
(extracting 13 from the 1st column)
$13 det\hat A$ ($det \hat A$ is the determinant after the extraction of 13 from the 1st column)

*Now since $100detA$ is divisible by 13, $detA$ is also divisible by 13. 
