Question

Let $N=\alpha \alpha \alpha \alpha \alpha \alpha$ be a $6$ digit number (all digits repeated) and $N$ is divisible by $924$ and let $\alpha ,\beta$ be the roots of the equation $x^2-11x+\lambda =0$. If the products of all possible values of $\lambda$ is $168k$, then value of $k$ is

Answer 1

$N=\alpha \alpha \alpha \alpha \alpha \alpha =\alpha \left(111111\right)$ is divisible by $3\times 7\times 11$

Hence for $N$ to be divisible by $924,\alpha =4\text{or}8$
and $\alpha$ and $\beta$ are roots of the equation $x^2-11x+\lambda =0\alpha +\beta =11(\alpha ,\beta )=(4,7)\text{or}(8,3)$

possible value of $\lambda =28,24$
The product of $\lambda =672$