Problem 4. On each side of a cube, a natural number is written. At each vertex (corner) of the cube, the product of the three numbers written on the sides that form the vertex (corner) is written. The sum of the eight thus obtained products is 385. Determine the sum of the numbers written on the sides of the cube.
Solution. Let $a, b, c, d, e, f$ be the natural numbers written on the faces of the cube, such that $a$ and $b$ are numbers written on opposite faces, the numbers $c$ and $d$ are written on opposite faces, and the numbers $e$ and $f$ are written on opposite faces. At the vertices of the cube, the following numbers are written: $ace, acf, ade, adf, bce, bcf, bde, bdf$. From the condition of the problem, we get

$$
a c e + a c f + a d e + a d f + b c e + b c f + b d e + b d f = (a + b)(c + d)(e + f) = 385
$$

The only way to write the number 385 as a product of three factors greater than 1 is $385 = 5 \cdot 7 \cdot 11$, from which it follows that

$$
a + b + c + d + e + f = (a + b) + (c + d) + (e + f) = 5 + 7 + 11 = 23
$$

## VIII Department