Let a > 2 be a constant. If there are just 18 positive integers satisfying the inequality $(\mathrm{x}-\mathrm{a})(\mathrm{x}-2\mathrm{a})\left(\mathrm{x}-{\mathrm{a}}^2\right)<0$, then the value of a is ________.

As a > 2,

${\mathrm{a}}^2>2\mathrm{a}>\mathrm{a}>2$

Now, $(\mathrm{x}-\mathrm{a})(\mathrm{x}-2\mathrm{a})\left(\mathrm{x}-{\mathrm{a}}^2\right)<0$

Thus, the solution set is as shown. 

Between 0, a, there are (a - 1) positive integers and between 2a, a², there are ${\mathrm{a}}^2-2\mathrm{a}-1$ integers. Therefore, 

${\mathrm{a}}^2-2\mathrm{a}-1+\mathrm{a}-1=18$

or ${\mathrm{a}}^2-\mathrm{a}-20=0$

$\begin{array}{l}\mathrm{or} (\mathrm{a}-5)(\mathrm{a}+4)=0\\ \therefore \mathrm{a}=5\end{array}$