Question

If $x_i=a_i{b}_i{c}_i$, $i=1,2,3$ are three-digit positive integers such that each $x_i$ is a multiple of $19$, then for some integer $n$ prove that $\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$ is divisible by $19$

Answer 1

$x_i=a_i{b}_i{c}_i$

then $x_i=a_i\times 100+b_i\times 10+c_i$

$x_i$ is divisible by $19$

$\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|=P$

If $P$ is divisible by $19$ then $1000P$ should also be divisible by $19$

$1000P=10\times 100\times \left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$

$1000P=10\times 100\times \left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$

$=\left|\begin{array}{ccc}{100a}_1+10b_1+c_1& 100a_2+10b_2+c_2& 100a_3+10b_3+c_3\\ {10b}_1& {10b}_2& 10b_3\\ c_1& c_2& c_3\end{array}\right|(R_1=R_1+R_2+R_3)$

$=\left|\begin{array}{ccc}{x}_1& x_2& x_3\\ {10b}_1& 10b_2& 10b_3\\ c_1& c_2& c_3\end{array}\right|$

$x_1,x_2,x_3$ are all divisible by $19$

so $1000P$ is divisible by $19$

Hence $P=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$ is also divisible by $19$