Pedro spends $1 \mathrm{~mL}$ of gray ink to paint $100 \mathrm{~cm}^{2}$ of surface.

a) The solid in the figure below was made by gluing one face of a cube with an edge of $10 \mathrm{~cm}$ to one face of a cube with an edge of $20 \mathrm{~cm}$. How many milliliters of ink does Pedro need to paint this solid?

![](https://cdn.mathpix.com/cropped/2024_05_01_1c2853cb1112cfb2ee2dg-12.jpg?height=234&width=192&top_left_y=1436&top_left_x=835)

b) Pedro used $54 \mathrm{~mL}$ of ink to paint a cube and then divided this painted cube into two identical rectangular blocks, as shown in the next figure below. How many more milliliters of ink will he need to finish painting these two blocks?

![](https://cdn.mathpix.com/cropped/2024_05_01_1c2853cb1112cfb2ee2dg-12.jpg?height=165&width=508&top_left_y=1905&top_left_x=680)

c) Pedro used $54 \mathrm{~mL}$ of ink to paint another cube. After painting, this cube was divided into identical smaller cubes, and Pedro used an additional $216 \mathrm{~mL}$ of ink to paint all the faces of the smaller cubes that were not painted. Into how many smaller cubes did he divide the cube?
Solution

a) $1^{\text{st}}$ solution: The surface area of the solid is equal to the sum of the surface areas of the cubes minus the "lost" area where they touch, which is equal to twice the area of one face of the smaller cube. Thus, the area of the solid obtained, in square centimeters, is equal to $6 \times 20 \times 20 + 6 \times 10 \times 10 - 2 \times 10 \times 10 = 2400 + 600 - 200 = 2800$. Since Pedro uses $1 \text{ mL}$ of paint to paint $100 \text{ cm}^2$, he will use $\frac{2800}{100} = 28 \text{ mL}$ of paint to paint the surface of the solid.

$2^{\text{nd}}$ solution: Each face of the larger cube has an area of $20 \text{ cm} \times 20 \text{ cm} = 400 \text{ cm}^2$. Thus, Pedro will use $\frac{400}{100} = 4 \text{ mL}$ of paint to paint each face of the larger cube; analogously, he will use $1 \text{ mL}$ of paint to paint each face of the smaller cube. Therefore, he will use $6 \times 4 + 6 \times 1 - 2 \times 1 = 28 \text{ mL}$ of paint to paint the entire solid.

b) To paint one of the faces of the cube, Pedro used $\frac{54}{6} = 9 \text{ mL}$ of paint. The cut created two new surfaces, each with an area equal to that of one of the cube's faces; to paint these two surfaces, Pedro must use $2 \times 9 = 18 \text{ mL}$ of paint.

c) $1^{\text{st}}$ solution: To divide the cube into equal smaller cubes, cuts must be made parallel to the faces and equally spaced. As we saw in part b), each of these cuts creates $1800 \text{ cm}^2$ of unpainted surface. Therefore, the number of cuts was $\frac{21600}{1800} = 12$. Since the smaller cubes are equal, the horizontal, vertical, and longitudinal cuts must all be of the same number, that is, in a number of $\frac{12}{3} = 4$. These cuts result in 5 horizontal, vertical, and longitudinal layers of smaller cubes, and it follows that the original cube was divided into $5 \times 5 \times 5 = 125$ smaller cubes.

$2^{\text{nd}}$ solution: As in part b), we concluded that Pedro used $9 \text{ mL}$ of paint to paint each face of the cube, so each face has an area of $900 \text{ cm}^2$. We conclude that the edge of the cube measures $30 \text{ cm}$. Pedro used $54 + 216 = 270 \text{ mL}$ of paint in total, so he painted $27000 \text{ cm}^2$. It is intuitive that the number of horizontal, vertical, and longitudinal layers is the same. We call this number $n$. The number of smaller cubes is then $n^3$ and the edge of each of the smaller cubes measures $\frac{30 \text{ cm}}{n}$. Thus, the area of one face of a smaller cube is $\frac{30 \text{ cm}}{n} \times \frac{30 \text{ cm}}{n} = \frac{900}{n^2} \text{ cm}^2$, and we have:

$$
27000 = \underbrace{6 \times \frac{900}{n^2}}_{\begin{array}{l}
\text{total area of} \\
\text{one face of a} \\
\text{smaller cube}
\end{array}} \times n^3 = 5400 \times n
$$

It follows that $n = 5$ and thus the number of smaller cubes is $5^3 = 125$.