4. The numbers from 1 to 8 are arranged at the vertices of a cube such that the sum of the numbers in any three vertices lying on the same face is at least 10. What is the smallest possible sum of the numbers at the vertices of one face?
Answer: 16.

Solution. Each face has a vertex where a number not less than 6 is placed. Indeed, otherwise, one of the triples of the remaining largest numbers $2,3,4,5$ would give a sum less than 10 (specifically, the triple $2,3,4$ with a sum of 9).

Consider a face containing the vertex where the number 6 is placed. Since the sum of the numbers in the other three vertices is not less than 10, the sum of all numbers in the vertices of this face is not less than 16.

![](https://cdn.mathpix.com/cropped/2024_05_06_73b5fdf426b8c910f128g-1.jpg?height=432&width=465&top_left_y=2371&top_left_x=1435)

An example of an arrangement where the smallest sum of numbers in the vertices of one face is 16 is shown in the figure: the sum of the numbers in the front face is $2+3+5+6=16$.