Question

Let $a>2$ be an integer. If there are just $18$ positive integers satisfying the inequality $(x-a)(x-2a)(x-a^2)<0,$ then the value of $a$ is

Answer 1

There are $18$ positive integers between $(0,a)$ and $(2a,a^2)$

$\Rightarrow a^2-2a-1+a-1=18\Rightarrow a^2-a-20=0$
$\Rightarrow a=5$ or $a=-4$
As $a>2,\therefore a=5$