Question

If $C_i,i=1,2,....9$ are perfect odd squares, then $\left|\begin{array}{ccc}{C}_1& C_2& C_3\\ C_4& C_5& C_6\\ C_7& C_8& C_9\end{array}\right|$ is always a multiple of

• A. 14
• B. 7
• C. 16
• D. 5

Answer 1

The correct option is C 16

$\begin{array}{c}{C}_1=(2a_1+1{)}^2\\ C_2=(2a_2+1{)}^2\\ .\\ .\\ Soon\end{array}$
$\left|\begin{array}{ccc}(2a_1+1{)}^2& (2a_2+1{)}^2& (2a_3+1{)}^2\\ (2a_4+1{)}^2& (2a_5+1{)}^2& (2a_6+1{)}^2\\ (2a_7+1{)}^2& (2a_8+1{)}^2& (2a_9+1{)}^2\end{array}\right|$
$C_1\to C_1-C_2,C_2\to C_2-C_3$
$\left|\begin{array}{ccc}4(a_1-a_2)(a_1+a_2+1)& 4(a_2-a_3)(a_2+a_3+1)& (2a_3+1{)}^2\\ 4(a_4-a_5)(a_4+a_5+1)& 4(a_5-a_6)(a_5+a_6+1)& (2a_6+1{)}^2\\ 4(a_7-a_8)(a_7+a_8+1)& 4(a_8-a_9)(a_8+a_9+1)& (2a_9+1{)}^2\end{array}\right|$
=16(K)