16. Find all real values of $m$ such that the inequality
$$
m x^{2}+8(m-1) x+7 m-16 \leqslant 0
$$

has at most 6 integer solutions, and one of them is 2.
(2002, Bulgarian Spring Mathematical Competition)
Solution: Let $f(x)=m x^{2}+8(m-1) x+7 m-16$. Since the inequality $f(x) \leqslant 0$ has a finite number of integer solutions, it must be that $m>0$.

Since 2 is a solution, then $f(2) \leqslant 0$, which means $m \leqslant \frac{32}{27}$. Also, because $f(-1)=-80, f(3)>\frac{48}{55}$. Therefore, $m \in\left(\frac{48}{55}, 1\right)$.
(2) $m \in\left(1, \frac{32}{27}\right], f(-3)<0$, and $f(-4)$ $=16-9 m>0$.
In summary, $m \in\left(\frac{48}{55}, 1\right) \cup\left(1, \frac{32}{27}\right]$.