Answer: 75.

Solution. There are 9 ways to cut off a "brick" consisting of two $1 \times 1 \times 1$ cubes from our figure. In each such "brick," there are two opposite faces $1 \times 1$, the distance between which is 2. Let's correspond these two faces to each other.

Consider one such pair of faces: on one of them, the dots were not erased, while on the other, they were erased. It is not difficult to understand that originally there were the same number of dots on these faces.

Let's calculate the total number of dots that were originally on all 9 pairs of faces. We get

$$
2 \cdot(1+1+6+2+2+5+3+3+4)=54
$$

There are 6 faces $1 \times 1$ left, about which we still know nothing. However, it can be noticed that they are divided into pairs according to the following principle: in one pair, there will be faces from one $1 \times 1 \times 1$ cube. In each such pair, the sum of the numbers was 7. Then we get that the number of dots on the surface of the figure originally was

$$
54+3 \cdot 7=75
$$
Problem 8.5. There are 7 completely identical cubes, each of which has 1 dot marked on one face, 2 dots on another, ..., and 6 dots on the sixth face. Moreover, on any two opposite faces, the total number of dots is 7.

These 7 cubes were used to form the figure shown in the diagram, such that on each pair of glued faces, the same number of dots is marked. All dots were erased from all faces except nine, as shown in the diagram. What is the total number of dots that were initially marked on the surface of the figure?

![](https://cdn.mathpix.com/cropped/2024_05_06_d3a039ae044fd2dec0bcg-27.jpg?height=513&width=553&top_left_y=266&top_left_x=450)