Marachev A.

A cube $3 \times 3 \times 3$ is made from $1 \times 1 \times 1$ cubes. What is the maximum number of cubes that can be removed from it so that the remaining figure has the following two properties:

- from the side of each face of the original cube, the figure looks like a $3 \times 3$ square (looking perpendicularly at this face, we do not see any gaps - 9 cubes of the figure are visible);
- by moving from cube to cube through their common face, you can get from any cube to any other cube in the figure


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Example. The layers from bottom to top are shown in the figure. The 13 remaining cubes are marked in black. Each layer has cubes in all columns and rows. When the layers are overlaid, a black $3 \times 3$ square is formed. Therefore, the first condition is met. The bottom and middle layers are connected and glued by the central cubes. Each cube in the top layer is glued to the corresponding cube in the middle layer. Thus, the second condition is also met.

![](https://cdn.mathpix.com/cropped/2024_05_06_f68ab0b4ba9161e79ed2g-25.jpg?height=209&width=1144&top_left_y=2211&top_left_x=480)

Estimate. Let there be $n$ cubes remaining. We see $6 \cdot 9$ of their faces. For connectivity, at least $n-1$ glues are necessary. Therefore, at least $2 n-2$ faces are not visible. Thus, $2 n-2+54 \leq 6 n$, from which $n \geq 13$.

## Answer

14 cubes.

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