
Problem 1. The faces of an orthogonal parallelepiped whose dimensions are natural numbers are painted green. The parallelepiped is partitioned into unit cubes by planes parallel to its faces. Find the dimensions of the parallelepiped if the number of cubes having no green face is one third of the total number of cubes.

Sava Grozdev


Solution: Let $x \leq y \leq z$ be the dimensions of the parallelepiped. It follows from the conditions of the problem that $x \geq 3$ and $(x-$ 2) $(y-2)(z-2)=\frac{x y z}{3}$. Since $\frac{x-2}{x} \leq \frac{y-2}{y} \leq \frac{z-2}{z}$, when $x \geq 7$, we obtain that $\frac{(x-2)(y-2)(z-2)}{x y z} \geq\left(\frac{5}{7}\right)^{3}>\frac{1}{3}$. Therefore $x \leq 6$ and thus $x=3, x=4, x=5$ or $x=6$.

1. If $x=3$, then $(y-2)(z-2)=y z$, which is impossible.
2. If $x=4$, then $2(y-2)(z-2)=\frac{4 y z}{3}$, so $(y-6)(z-6)=24$. In this case the only solutions are $(4,7,30),(4,8,18),(4,9,14)$ and $(4,10,12)$.
3. If $x=5$, then $3(y-2)(z-2)=\frac{5 y z}{3}$, so $(2 y-9)(2 z-9)=45$. Therefore the solutions are $(5,5,27),(5,6,12)$ and $(5,7,9)$.
4. If $x=6$, then $4(y-2)(z-2)=2 y z$, so $(y-4)(z-4)=8$. Thus there is an unique solution $(6,6,8)$.

Answer: The problem has 8 solutions- $(4,7,30),(4,8,18),(4,9,14)$, $(4,10,12),(5,5,27),(5,6,12),(5,7,9)$ and $(6,6,8)$.
