We placed a cube in the center of a potter's wheel, with each of its faces inscribed with a natural number. Just before we spun the wheel, we could see three faces of the cube and thus only three numbers from our position. Their sum was 42. After rotating the potter's wheel by $90^{\circ}$, we observed from the same position three faces with numbers summing to 34, and after another $90^{\circ}$ rotation, we saw three numbers summing to 53 from the same spot.

1. Determine the sum of the three numbers we will see from our position after the wheel is rotated by another $90^{\circ}$.
2. The cube was lying on a face with the number 6 throughout the entire process. Determine the maximum possible sum of all six numbers on the cube.

(L. Šimůnek)

Hint. Focus on the relationship between the numbers on mutually parallel side faces.
Let the numbers we see before spinning the circle be denoted as $a, b, c$, where $c$ is the number on the top face. After a $90^{\circ}$ rotation, we lose the face with the number $a$ from our view and a face parallel to it becomes visible. According to the problem, the sum of the visible numbers changes from 42 to 34, which means it decreases by 8. Therefore, the newly visible number is 8 less than $a$, i.e., $a-8$.

Similarly, consider the next $90^{\circ}$ rotation. During this rotation, we lose the face with the number $b$ from our view, and the sum of the visible numbers changes from 34 to 53, which means it increases by 19. Therefore, the number on the remaining side face is $b+19$.

![](https://cdn.mathpix.com/cropped/2024_04_17_39f221c1ba6e0b54eafbg-4.jpg?height=494&width=880&top_left_y=1318&top_left_x=588)

After another $90^{\circ}$ rotation, we see the faces with numbers $b+19, a, c$. From the problem, we know that $a+b+c=42$, so $a+b+19+c=42+19=61$. The sum 61 is the solution to the first problem.

Now we solve the second problem. Before spinning the circle, we see three faces with a sum of 42. After a $180^{\circ}$ rotation, we see two different side faces and the same top face with the number $c$, which now has a sum of 53. Therefore, the sum of the numbers on these five faces is $42+53-c$. The number 6 is written on the bottom of the cube, so the sum of all its numbers is $6+42+53-c$, which is 101 - $c$. To find the maximum possible value of this expression, we substitute the smallest permissible value for $c$, which is 1. Thus, the sum of the numbers on the cube can be at most 100.

Another solution to the second problem. From the above solution, we use the diagram and its description. The sum of all the numbers on the cube is

$$
a+b+(a-8)+(b+19)+c+6=2a+2b+c+17.
$$

All unknowns are natural numbers, and $a \geq 9$ to ensure that $a-8$ is a natural number. The last condition is $a+b+c=42$. To obtain the maximum value of the expression $2a+2b+c+17$ for a given sum $a+b+c$, we choose the smallest permissible value for $c$, which is 1, because the other unknowns are represented in their multiples in the expression. The sum $a+b$ then takes the value 41, and the sum of all six numbers on the cube can be at most

$$
2a+2b+c+17=2 \cdot 41+1+17=100.
$$