Question

If a,b,c are integers such that b is always twice the sum of a and c lies between 1 and 9 and 6a2, 5b2, 4c2 are three digit numbers than value of $\left|\begin{array}{ccc}6& 6a2& a\\ 5& 5b2& b\\ 4& 4c2& c\end{array}\right|$ equals.

• A. $a+b+c$
• B. $a-b+c$
• C. $2a-b+c$
• D. $0$

Answer 1

Answer: (D)

$0$

$a,b,c$ are integers

$2b=a+cϵ(1,9)$

$\left|\begin{array}{ccc}6& 6a2& a\\ 5& 5b2& b\\ 4& 4c2& c\end{array}\right|$

In a special case

$a=6,b=5,c=4$ also satisfies the given condition

In that case $C_1$ and $C_2$ are same

$\Rightarrow det=0$

No other option with $a,b,c$ values satisfies $det=0$

$\therefore$ Option D is correct