9.1. Petya had a large wooden cube with dimensions $30 \times 30 \times 30$ cm $^{3}$. Petya decided to paint the entire cube, which took 100 grams of paint. Later, Petya needed smaller cubes, so he cut the large cube with 6 cuts, parallel to the faces of the cube (2 cuts parallel to each pair of faces), into 27 cubes with dimensions $10 \times 10 \times 10 \mathrm{~cm}^{3}$. How much paint is needed to finish painting all the resulting cubes? Petya does not repaint the already painted surfaces.
Answer: 200 grams.

First solution. Each cut increases the unpainted surface area by an amount equal to twice the area of the cut. It is clear that the area of the cut is the same as the area of a face. Therefore, it is necessary to paint an additional surface area equal to \(6 \cdot 2 = 12\) faces, which is twice the surface area of the cube. Thus, \(100 \cdot 2 = 200\) grams of paint are required.

Second solution. The 27 smaller cubes have a total of \(27 \cdot 6 = 162\) faces. Of these, \(6 \cdot 9 = 54\) faces are already painted. Therefore, \(162 - 54 = 108\) faces remain to be painted, which is twice as many as 54. Thus, painting 108 faces will require \(100 \cdot 2 = 200\) grams of paint.

Third solution. After the cuts, we have 8 corner cubes, each with 3 unpainted faces, 12 non-corner cubes along the edges of the large cube, each with 4 unpainted faces, 6 cubes adjacent to the centers of the faces of the large cube, each with 5 unpainted faces, and one central cube with all faces unpainted. Therefore, we need to paint \(8 \cdot 3 + 12 \cdot 4 + 6 \cdot 5 + 6 = 24 + 48 + 30 + 6 = 108\) faces of the smaller cubes. Initially, we painted \(6 \cdot 9 = 54\) such faces, which required 100 grams of paint. Therefore, painting \(108 = 54 \cdot 2\) faces will require 200 grams of paint.

Comment. Only the answer - 2 points.

The solution follows the first author's solution, but it forgets about the doubling of the unpainted surface area - 2 points.