12. [25] A $3 \times 3 \times 3$ cube is built from 27 unit cubes. Suddenly five of those cubes mysteriously teleport away. What is the minimum possible surface area of the remaining solid? Prove your answer.
Answer: 50
Solution: Orient the cube so that its edges are parallel to the $x-, y$-, and $z$-axes. A set of three unit cubes whose centers differ only in their $x$-coordinate will be termed an " $x$-row"; there are thus nine $x$-rows. Define " $y$-row" and " $z$-row" similarly.
To achieve 50, simply take away one $x$-row and one $y$-row (their union consists of precisely five unit cubes).
To show that 50 is the minimum: Note that there cannot be two $x$-rows that are both completely removed, as that would imply removing six unit cubes. (Similar statements apply for $y$ - and $z$-rows, of course.) It is also impossible for there to be one $x$-row, one $y$-row, and one $z$-row that are all removed, as that would imply removing seven unit cubes. Every $x$-, $y$-, or $z$-row that is not completely removed contributes at least 2 square units to the surface area. Thus, the total surface area is at least $9 \cdot 2+8 \cdot 2+8 \cdot 2=50$.