7.4. On the surface of a pentagonal pyramid (see fig.), several gnomes live in pairwise distinct points, and they can live both inside the faces and on the edges or at the vertices. It turned out that on each face (including the vertices and edges that bound it) a different number of gnomes live. What is the minimum number of gnomes living on the pyramid?

![](https://cdn.mathpix.com/cropped/2024_05_06_02a78c455684067eacd3g-2.jpg?height=217&width=171&top_left_y=437&top_left_x=1728)
Answer: 6

Solution: There are 6 faces in total, so at least 5 gnomes live on the most "populated" one. If there are exactly 5 gnomes in total, then all of them live on one face (let's call it face $A$), so the faces where 4 and 3 gnomes live (let's call them $B$ and $C$ respectively) are adjacent to it. Then, on the edge $A-B$ there should be 4, and on the edge $A-C$ there should be 3 gnomes. This means there are already at least 7 gnomes, and we could have counted only one of them twice - the one sitting at the vertex connecting faces $A$, $B$, and $C$. Therefore, there are at least 6 gnomes - a contradiction. An example with 6 gnomes is shown in the figure.

## Criteria:

![](https://cdn.mathpix.com/cropped/2024_05_06_02a78c455684067eacd3g-2.jpg?height=192&width=166&top_left_y=932&top_left_x=1736)

Only the answer - 0 points.

Only a correct example - 2 points.

Only proof that there cannot be fewer than 5 - 1 point.

Only proof that there cannot be fewer than 6 - 3 points.

These points DO NOT add up.

Justification of the example is not required.