3. (48th Slovenian Mathematical Olympiad) We write a positive integer on each face of a cube, and then, at each vertex, we write a number that is the product of the integers on the three faces meeting at that vertex. It is known that the sum of the numbers at the vertices of the cube is 70. Find the sum of the numbers on the faces of the cube.
3. Let the numbers on each face of the cube be $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$, and the numbers written at the vertices be $a_{1} a_{2} a_{5}, a_{2} a_{3} a_{5}, a_{3} a_{4} a_{5}, a_{4} a_{1} a_{5}, a_{1} a_{2} a_{6}, a_{2} a_{3} a_{6}, a_{3} a_{4} a_{6}, a_{4} a_{1} a_{6}$.
Thus, we have
$$
\begin{aligned}
70 & =a_{1} a_{2} a_{5}+a_{2} a_{3} a_{5}+a_{3} a_{4} a_{5}+a_{4} a_{1} a_{5}+a_{1} a_{2} a_{6}+a_{2} a_{3} a_{6}+a_{3} a_{4} a_{6}+a_{4} a_{1} a_{6} \\
& =\left(a_{5}+a_{6}\right)\left(a_{1} a_{2}+a_{2} a_{3}+a_{3} a_{4}+a_{4} a_{1}\right) \\
& =\left(a_{1}+a_{3}\right)\left(a_{2}+a_{4}\right)\left(a_{5}+a_{6}\right) .
\end{aligned}
$$

Since $70=2 \times 5 \times 7$, it is clear that in the product
$$
\left(a_{1}+a_{3}\right)\left(a_{2}+a_{4}\right)\left(a_{5}+a_{6}\right)
$$

one factor is 2, one factor is 5, and another factor is 7.
Therefore, the required sum is
$$
a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=2+5+7=14 .
$$