National mathematical Olympiad Final round

We shall prove that the solutions are the triples $(a, b, c)$ such that $a$ divides $b$ and $b$ divides $c$. We denote by $((x_0, y_0, z_0), p, q, r)$ the parallelepiped with a low right vertex $(x_0, y_0, z_0)$ and dimensions $p, q$ and $r$, at axes $Ox, Oy$ and $Oz$, respectively.

Let us assume that $b$ is not divisible by $a$, i.e. $b = ma + n$ for some $m, n \in \mathbb{N}$, $0 < n < a$. If $(p, q, r)$ is a permutation of $(a, b, c)$, then it follows from the condition for the parallelepipeds $((0, 0, 0), p, q, r)$ and $((0, 0, 1), p, q, r)$ that the parallelepipeds $((0, 0, 0), p, q, 1)$ and $((0, 0, r), p, q, 1)$ are filled with cubes of the same colors. This implies that the parallelepipeds $((0, 0, 0), c, a, 1)$ and $((0, 0, b), c, a, 1)$ are also filled with cubes of the same colors and the same is true for the parallelepipeds $((0, 0, 0), c, b, 1)$ and $((0, 0, ma), c, b, 1)$. Since $((0, 0, 0), c, b, 1)$ contains $((0, 0, 0), c, a, 1)$ and $((0, 0, ma), c, b, a)$ contains $((0, 0, ma), c, b, 1)$ and $((0, 0, b), c, a, 1)$, every color in $((0, 0, 0), c, a, 1)$ appears at least two times in $((0, 0, ma), c, b, a)$. This contradiction completes the proof that $a$ divides $b$. We analogously see that $b$ divides $c$.

Now let $a|b$ and $b|c$, as $b = p_1a, c = p_2b = p_1p_2a$, where $p_1, p_2 \in \mathbb{N}$. For every two positive integers $m$ and $n$ we denote by $R(m, n)$ the remainder of $m$ modulo $n$. The coordinates of a cube will be the coordinates of its low right vertex.

We determine the color the cube $(x, y, z)$ in remainders as follows:
$$(R(x, a); R(y, a); R(z, a); R(\lfloor \frac{x}{a} \rfloor + \lfloor \frac{y}{a} \rfloor, p_1); R(\lfloor \frac{y}{a} \rfloor + \lfloor \frac{z}{a} \rfloor, p_1); R(\lfloor \frac{x}{b} \rfloor + \lfloor \frac{y}{b} \rfloor + \lfloor \frac{z}{b} \rfloor, p_2)).$$
The counting of all possible remainders in the six coordinates shows that the number of the colors used is $a^3 p_1 p_1 p_2 = abc$.

Let us assume that two distinct cubes $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ lie in a parallelepiped of dimensions $a \times b \times c$. Then we have $|x_1 - x_2| \le \alpha$, $|y_1 - y_2| \le \beta$ and $|z_1 - z_2| \le \gamma$, where $(\alpha, \beta, \gamma)$ is a permutation of $(a, b, c)$. Since $|x_1 - x_2|$, $|y_1 - y_2|$ and $|z_1 - z_2|$ are divisible by $a$, one of these numbers equals $0$. Let us have $x_1 = x_2$. Then the fourth and fifth coordinates of that color show that $|y_1 - y_2|$ and $|z_1 - z_2|$ are divisible by $b$ and therefore one of these two numbers equals $0$. If, for example, $y_1 = y_2$, then the last coordinate shows that $|z_1 - z_2|$ is divisible by $c$, i.e. $z_1 = z_2$. We obtained $(x_1, y_1, z_1) \equiv (x_2, y_2, z_2)$, which is a contradiction.