## Task A-4.7.

Given are 27 identical standard dice $1 \times 1 \times 1$ with numbers 1 to 6, as shown in the top image (opposite 1 is 6, opposite 2 is 5, and opposite 3 is 4).

From them, a cube $3 \times 3 \times 3$ is constructed such that the dice always touch sides with the same number. We observe the sums of the numbers written on the faces of the large cube. These sums are denoted by the letters $A, B, C, D, E, F$, as shown in the image below (pairs on opposite faces are $A$ and $B, C$ and $D$ and $E$ and $F$).

If $A=9$ and $C=36$, determine $B, D, E$ and $F$.

![](https://cdn.mathpix.com/cropped/2024_05_30_a897c7e1109e2574c8d7g-27.jpg?height=525&width=372&top_left_y=1231&top_left_x=1296)
## Solution.

If on one side of a large cube at some place the number $n \in\{1,2,3,4,5,6\}$ is written, then on the opposite side at the corresponding place the number $7-n$ is written.

Indeed, this directly follows from the rule of joining small cubes. Thus, we immediately have that

$$
B=9 \cdot 7-A=63-9=54 \quad \text { and } \quad D=9 \cdot 7-C=63-36=27
$$

Since $A=9$ and this sum involves 9 numbers, each of which is at least 1, we conclude that all numbers written on the front side are equal to 1.

Consider the upper right cube on the front side and the first column of the right side. There are exactly 4 possibilities for the position of the observed cube such that 1 is on the front side. Each position uniquely determines the observed column:

- if the number on the cube on the right side is 2, then the column must contain the numbers $2,5,2$ in order;
- if the number on the cube on the right side is 5, then the column must contain the numbers $5,2,5$ in order;
- if the number on the cube on the right side is 3, then the column must contain the numbers $3,4,3$ in order;
- if the number on the cube on the right side is 4, then the column must contain the numbers $4,3,4$ in order.

![](https://cdn.mathpix.com/cropped/2024_05_30_a897c7e1109e2574c8d7g-28.jpg?height=665&width=762&top_left_y=581&top_left_x=567)

Since $C=36$, and the maximum possible sum of a column is 12, we conclude that the sum in each column must be 12 (the numbers must be $5,2,5$).

Due to the rule of joining cubes, we conclude (using the same logic as at the beginning of the solution) that the numbers on the sides with sums $C$ and $E$ are as follows.

| 5 | 5 | 5 |
| :--- | :--- | :--- |
| 2 | 2 | 2 |
| 5 | 5 | 5 |


| 4 | 3 | 4 |
| :--- | :--- | :--- |
| 3 | 4 | 3 |
| 4 | 3 | 4 |

Finally, $E=32$ and $F=9 \cdot 7-E=63-32=31$. 1 point

Note: A student who only calculates the value of $B$ without any further progress deserves 0 points.

Note: If a student constructs an example of a cube that satisfies all the conditions of the problem and based on that calculates the required numbers, but does not show deductively that this is the only possibility, they should receive 3 points.