8.4. On a cubic planet, there live cubic mice, and they live only on the faces of the cube, not on the edges or vertices. It is known that different numbers of mice live on different faces, and the number on any two adjacent faces differs by at least 2. What is the minimum number of cubic mice that can live on this planet, given that there is at least one mouse on each face?
Solution: We will prove that no three consecutive numbers can be the number of mice on the faces. Indeed, if there were $x, x+1$, and $x+2$ mice on some three faces, then $x$ and $x+1$ would have to be on opposite faces. But then $x+2$ mice could not be anywhere.

Consider the first 8 natural numbers. Among the first three, there is at least one missing, and among the next three, there is at least one missing. Therefore, the minimum sum is $1+2+4+5+7+8=27$. It is possible to distribute 27 mice. For this, we will divide all the faces into pairs of opposite faces. On the first pair, there will be 1 and 2 mice, on the second pair - 4 and 5, on the third pair - 7 and 8. Thus, quantities differing by 1 are on opposite faces, and on adjacent faces, the difference is at least 2.

Criteria: Only the answer - 1 point.

Only an example (with or without justification) - 2 points.

Only the estimate - 5 points

Proved that no three consecutive numbers can be the number of mice on the faces - 2 points.