Senderov V.A.

At the vertices of a cube, numbers from 1 to 8 were written, and on each edge - the absolute difference of the numbers at its ends. What is the smallest number of different numbers that can be written on the edges?

#
Three numbers must be necessarily present: for this, it is sufficient to consider the three edges of the cube coming out of the vertex where the number 1 (or 8) is written. Let's prove that there will be an arrangement of numbers for which exactly three numbers will be required. Consider 2 squares. In the vertices of the first, place the numbers $1,2,3,4$ clockwise, and in the vertices of the second, also clockwise, place the numbers $5,6,7,8$. So far, we have used two different numbers: 1 and 3. Now place the first square below the second: 1 below 5, 2 below 6, and so on.

## Answer

$$
3.00
$$

Send a comment