Problem 2. A die is a cube with an edge of 1, having the numbers from 1 to 6 inscribed on its faces, such that the sum of the numbers on any two opposite faces is 7.

Using 27 dice, we construct a cube with an edge of 3.

Determine what values the sum of all the numbers on the six faces of the cube with an edge of 3 can take.
Solution. We call dice of type I those which have only one face visible (those dice in the centers of the large cube's faces), type II those with two faces visible (those in the middles of the edges), and type III those with three faces visible (the dice in the corners). Any cube of edge 3 contains 6 dice of type I, 12 dice of type II, and 8 dice of type III.

The minimum sum on the six faces is obtained when the type I dice show 1, the type II dice show 1 and 2, and the type III dice show 1, 2, and 3. The minimum sum is $6 \cdot 1 + 12 \cdot 3 + 8 \cdot 6 = 90$.

The maximum is obtained when the type I dice show 6, the type II dice show 5 and 6, and the type III dice show 4, 5, and 6. The maximum sum is $6 \cdot 6 + 12 \cdot 11 + 8 \cdot 15 = 288$.

We will show that the total sum can take any intermediate value between 90 and 288. For this, we will start from the cube where the minimum sum is achieved and rotate the dice, step by step, so as to increase the sum by 1 at each step.

By rotating a type I die, we can increase the total sum on the faces of the large cube by one unit at each step, until a 6 appears on this die. We continue this procedure so that on all type I dice, the face 6 becomes visible. ... $1 p$

Now we rotate a type II die from $1+2$ to $5+6$ and, at the same time, rotate two of the type I dice to 1 and 4; thus, we have increased the total sum by 1. In the next steps, we rotate those two type I dice on which we acted, increasing the total sum by one unit each time; in the end, the number 6 will appear on them again. We continue this procedure so that on all type II dice, the faces 5 and 6 become visible. ... $2 p$

Then, we rotate a type III die from $1+2+3$ to $4+5+6$, while at the same time rotating two of the type I dice to 2; thus, we have increased the total sum by 1. In the next steps, we rotate those two type I dice on which we acted, increasing the total sum by one unit each time; in the end, the number 6 will appear on them again. We continue this procedure so that on all type III dice, the faces 4, 5, and 6 become visible.

We have obtained a cube of edge 3 with the sum of the numbers on the faces being 288. In conclusion, the sum on the faces can take all values from 90 to 288.