The height of a straight pyramid is $m$; this pyramid is to be divided into three equal volume parts by two planes parallel to the base. What are the heights of these parts?
Let the heights of the individual parts be $x$, $y$, and $z$. Since the volumes of similar pyramids are in the ratio of the cubes of their heights, we have

$$
\frac{x^{3}}{m^{3}}=\frac{1}{3}
$$

from which

$$
x=\frac{m}{\sqrt[3]{9}}=\frac{m}{3} \sqrt[3]{9}
$$

Furthermore,

$$
\frac{(m-z)^{3}}{m^{3}}=\frac{2}{3}
$$

from which

$$
z=m\left(1-\sqrt[3]{\frac{2}{3}}\right)=\frac{m}{3}(3-\sqrt[3]{18})
$$

Finally,

$$
y=m-(x+z)=\frac{m}{3}(\sqrt[3]{18}-\sqrt[3]{9})
$$

Number of solutions: 46.