Let's arrange six numbers (according to the number of dots on the faces of a cube) in ascending order. We will prove that among them, there are no three consecutive numbers. Suppose such three numbers exist: \(a\), \(a+1\), and \(a+2\). Then the numbers \(a\) and \(a+1\) must be on opposite faces of the cube. No matter where we place the number \(a+2\), it will be adjacent to \(a+1\). This is a contradiction.

Therefore, we need to find a sequence of six numbers with the minimum sum, in which there are no three consecutive numbers. This sequence is \(1, 2, 4, 5, 7, 8\), and the total number of dots is \(1 + 2 + 4 + 5 + 7 + 8 = 27\). An example of their placement on the faces of the cube is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_00e2c6ab04990c8ec18fg-34.jpg?height=369&width=301&top_left_y=90&top_left_x=890)

Fig. 1

## Answer

27 dots.
$\underline{\text { Folklore }}$

Petya wants to make an unusual die, which, as usual, should have the shape of a cube, with dots drawn on its faces (different numbers of dots on different faces), but at the same time, on any two adjacent faces, the number of dots must differ by at least two (at the same time, it is allowed for some faces to have more than six dots). How many dots in total are needed to be drawn for this?

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