2. On each face of the cube is written a natural number, and at each vertex is written the product of the numbers on the 3 faces that meet at that vertex. The sum of the numbers at the vertices of the cube is 70. What is the sum of the numbers on the faces of the cube?
Let's denote the numbers on the faces of the cube as $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$, and $a_{6}$. In the vertices of the cube, the numbers are $a_{1} a_{2} a_{5}, a_{2} a_{3} a_{5}, a_{3} a_{4} a_{5}, a_{4} a_{1} a_{5}, a_{1} a_{2} a_{6}, a_{2} a_{3} a_{6}, a_{3} a_{4} a_{6}$, and $a_{4} a_{1} a_{6}$, so 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 $70=2 \cdot 5 \cdot 7$ and all three factors are greater than 1, one of the factors is equal to 2, one is equal to 5, and one is equal to 7. Their sum is $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}=2+5+7=14$.