3. If a positive integer is written on each face of a cube, and then a number is written at each vertex, which is equal to the product of the two integers on the faces passing through that vertex, then, when the sum of the numbers at the vertices of the cube is 290, the sum of the numbers on the faces of the cube is
3. 36 .

Let the numbers on each face of the cube be $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$, $x_{5}$, $x_{6}$, and the numbers written at the vertices be
$$
\begin{array}{l}
x_{1} x_{2} x_{5}, x_{2} x_{3} x_{5}, x_{3} x_{4} x_{5}, x_{4} x_{1} x_{5}, x_{1} x_{2} x_{6}, \\
x_{2} x_{3} x_{6}, x_{3} x_{4} x_{6}, x_{4} x_{1} x_{6} .
\end{array}
$$
$$
\begin{array}{l} 
\text { Then } 290=x_{1} x_{2} x_{5}+x_{2} x_{3} x_{5}+x_{3} x_{4} x_{5}+x_{4} x_{1} x_{5}+ \\
\quad x_{1} x_{2} x_{6}+x_{2} x_{3} x_{6}+x_{3} x_{4} x_{6}+x_{4} x_{1} x_{6} \\
=\left(x_{5}+x_{6}\right)\left(x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{4}+x_{4} x_{1}\right) \\
=\left(x_{1}+x_{3}\right)\left(x_{2}+x_{4}\right)\left(x_{5}+x_{6}\right) .
\end{array}
$$

Since $290=2 \times 5 \times 29$, it is clear that in the product $\left(x_{1}+x_{3}\right)\left(x_{2}+x_{4}\right)\left(x_{5}+x_{6}\right)$, one factor is 2, one factor is 5, and the other factor is 29.
Therefore, $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=2+5+29=36$.