Understanding Cube Cutting Constraints

The condition that the pieces are identical after 3 cuts suggests a highly symmetrical cutting pattern. For a solid cube, this typically implies 3 mutually perpendicular planar cuts passing through the center of the cube. These cuts divide the cube into 8 smaller, identical cubes.

• Initial state: 1 piece (the cube).
• Cut 1 (e.g., parallel to XY plane at $z=L/2$): 2 pieces.
• Cut 2 (e.g., parallel to XZ plane at $y=L/2$): 4 pieces.
• Cut 3 (e.g., parallel to YZ plane at $x=L/2$): 8 identical pieces (cubes).

Thus, we establish $x_3 = 8$.

Calculating Possible Values for $x_4$

The number of pieces after $n$ cuts, $x_n$, can be calculated using the relation $x_n = x_{n-1} + k_n$, where $k_n$ is the number of new pieces created by the $n^{th}$ cut. $k_n$ equals the number of regions the $n^{th}$ plane is divided into by its intersections with the previous $n-1$ planes.

For the 4th cut ($n=4$), the plane $P_4$ intersects the previous 3 planes ($P_1, P_2, P_3$). These intersections form 3 lines ($L_1, L_2, L_3$) on the plane $P_4$. The number of regions $k_4$ is determined by how these lines divide $P_4$. The maximum number of regions formed by 3 lines in a plane is 7.

Therefore, $k_4 \le 7$.

The total number of pieces $x_4$ is given by:

Since $x_3 = 8$ and $k_4 \le 7$, the maximum possible value for $x_4$ is:

This means any value of $x_4$ greater than 15 is impossible.

Analyzing the Options

We need to find which option is NOT a possible value for $x_4$. We know $x_4$ must be at least $x_3=8$. Let's examine the possibilities:

• Option A (16): Since the theoretical maximum number of pieces is 15 ($x_4 \le 15$), $x_4=16$ is not possible.
• Option B (12): This value is possible. It occurs if the 4th cut is parallel to one of the first three cuts but does not coincide with it (e.g., $x=L/4$). This cut passes through 4 of the 8 smaller cubes, adding 4 pieces ($k_4=4$). $x_4 = 8 + 4 = 12$.
• Option C (8): This value is possible. It occurs if the 4th cut coincides with one of the previous cuts (e.g., $x=L/2$). This adds no new pieces ($k_4=0$). $x_4 = 8 + 0 = 8$.
• Option D (5): This value is impossible because the number of pieces cannot decrease. We must have $x_4 \ge x_3 = 8$.

Conclusion: The Impossible Value

Both 16 and 5 are mathematically impossible based on the constraints ($x_4 \le 15$ and $x_4 \ge 8$). However, the calculation based on the maximum number of regions created by the intersecting planes yields $x_4 \le 15$. Therefore, 16 is impossible as it exceeds this geometric limit.

The question asks for *a* value that is not possible. Given the options and the derived maximum limit, 16 represents a geometrically impossible outcome.