A3. The faces of a cube contain the numbers $1,2,3,4,5,6$ such that the sum of the numbers on each pair of opposite faces is 7 . For each of the cube's eight corners, we multiply the three numbers on the faces incident to that corner, and write down its value. (In the diagram, the value of the indicated corner is $1 \times 2 \times 3=6$.) What is the sum of the eight values assigned to the cube's corners?
Solution: The answer is 343 .
Solution 1: The left picture shows the corners and the faces touching the side labeled 1 and the right picture shows the opposite side of the die, whose label is 6 , which is incident to the other four corners.

We compute the eight numbers individually and sum the eight numbers. The eight triplets of integers at the eight corners are
$$
(1,2,3),(1,2,4),(1,3,5),(1,4,5),(6,2,3),(6,2,4),(6,3,5),(6,4,5) .
$$

These eight values are
$$
\begin{array}{rlr}
1 \times 2 \times 3 & =6 \\
1 \times 2 \times 4 & = \\
1 \times 3 \times 5= & 15 \\
1 \times 4 \times 5= & 20 \\
6 \times 2 \times 3= & 36 \\
6 \times 2 \times 4=48 \\
6 \times 3 \times 5=90 \\
6 \times 4 \times 5=120
\end{array} .
$$

The sum of these eight positive integers is 343 .

Solution 2: Since no corner contains two numbers that sum to 7 , the sum in solution 1 can be computed as
$$
(1+6)(2+5)(3+4)=7^{3}=343 \text {. }
$$