286*. Is it possible to number the vertices of a cube with natural numbers from 1 to 8 so that the sums of the numbers at the ends of each edge of the cube are different?
$\triangle$ The sums of the numbers at the ends of the edges of a cube can take values from $1+2=3$ to $7+8=15$ - a total of 13 different values.

Suppose among the sums there are such:

$$
1+2=3, \quad 1+3=4, \quad 1+4=5
$$

But then there cannot be a sum equal to 6, because the sum $1+5$ cannot be used - since the numbers 1 and 5 are not on the same edge, and the sum $2+4$ - also, for the same reason.

![](https://cdn.mathpix.com/cropped/2024_05_21_0aa0e6ac168e80891953g-093.jpg?height=408&width=394&top_left_y=253&top_left_x=343)

Fig. 35

(Fig. 35). This means that out of the sums $3, 4, 5$ and 6, one, in this case 6, cannot be obtained.

Similarly, it can be proven that out of the sums 12, 13, 14, and 15, one cannot be obtained.

Therefore, there are no more than 11 different sums, while a cube has 12 edges.

Answer: No.