Problem 5. On each of the six sides of a cube, a natural number is written. At each vertex, we write the product of the three numbers written on the sides that have the corresponding vertex in common. If the sum of the numbers written at the vertices is 2006, calculate the sum of the numbers written on the sides.
Solution. Let's denote the cube as $A B C D E F G H$. Let the number $a_{1}$ be written on the side $A B C D$, the number $a_{2}$ on the side $E F G H$, the number $b_{1}$ on $A B F E$, the number $b_{2}$ on $D C G H$, the number $c_{1}$ on $B C G F$, and the number $c_{2}$ on $A D H E$. Then, at vertex $A$ is written the number $a_{1} b_{1} c_{2}$, at $B$ the number $a_{1} b_{1} c_{1}$, at $C$ the number $a_{1} b_{2} c_{1}$, at $D$ the number $a_{1} b_{2} c_{2}$, at $E$ the number $a_{2} b_{1} c_{2}$, at $F$ the number $a_{2} b_{1} c_{1}$, at $G$ the number $a_{2} b_{2} c_{1}$, and at $H$ the number $a_{2} b_{2} c_{2}$. Then, from the condition in

![](https://cdn.mathpix.com/cropped/2024_06_05_d4fd26fa5c16caf575adg-9.jpg?height=348&width=355&top_left_y=845&top_left_x=1106)
the problem, we have

$$
a_{1} b_{1} c_{2} + a_{1} b_{1} c_{1} + a_{1} b_{2} c_{1} + a_{1} b_{2} c_{2} + a_{2} b_{1} c_{2} + a_{2} b_{1} c_{1} + a_{2} b_{2} c_{1} + a_{2} b_{2} c_{2} = 2006
$$

The expression on the left side of the last equation can be transformed into

$$
\begin{aligned}
a_{1} b_{1} c_{2} + a_{1} b_{1} c_{1} & + a_{1} b_{2} c_{1} + a_{1} b_{2} c_{2} + a_{2} b_{1} c_{2} + a_{2} b_{1} c_{1} + a_{2} b_{2} c_{1} + a_{2} b_{2} c_{2} = \\
& = a_{1} b_{1} (c_{2} + c_{1}) + a_{1} b_{2} (c_{1} + c_{2}) + a_{2} b_{1} (c_{2} + c_{1}) + a_{2} b_{2} (c_{1} + c_{2}) \\
& = (c_{1} + c_{2}) (a_{1} b_{1} + a_{1} b_{2} + a_{2} b_{1} + a_{2} b_{2}) \\
& = (c_{1} + c_{2}) (a_{1} (b_{1} + b_{2}) + a_{2} (b_{1} + b_{2})) \\
& = (c_{1} + c_{2}) (b_{1} + b_{2}) (a_{1} + a_{2})
\end{aligned}
$$

Thus,

$$
(a_{1} + a_{2}) (b_{1} + b_{2}) (c_{1} + c_{2}) = 2006 = 1 \cdot 2 \cdot 17 \cdot 59
$$

Since $a_{1}, a_{2}, b_{1}, b_{2}, c_{1}, c_{2}$ are natural numbers, it follows that $a_{1} + a_{2} > 1$, $b_{1} + b_{2} > 1$, and $c_{1} + c_{2} > 1$, and therefore, without loss of generality, we can take

$$
a_{1} + a_{2} = 2, \quad b_{1} + b_{2} = 17, \quad c_{1} + c_{2} = 59
$$

Then,

$$
a_{1} + a_{2} + b_{1} + b_{2} + c_{1} + c_{2} = 2 + 17 + 59 = 78
$$