4. A net of a cube is shown with one integer on each face. A larger cube is constructed using 27 copies of this cube. What is the minimum possible sum of all of the integers showing on the six faces of the larger cube?
4. Since $3 \times 3 \times 3=27$, the larger cube is made by arranging the 27 smaller cubes in three layers so that each layer consists of 9 small cubes arranged in a $3 \times 3$ square.
This means each face of the larger cube is made up of $3 \times 3=9$ faces of small cubes. A cube has six faces, so the surface of the cube is made up of $6 \times 9=54$ faces of small cubes. Of the 27 small cubes, 1 is completely concealed inside the larger cube, so none of its faces contribute to the surface of the larger cube.

The remaining 26 small cubes each contribute 1,2 , or 3 of their faces to the surface of the larger cube.
There are 6 small cubes that contribute 1 face to the surface of the larger cube, 12 that contribute 2 faces, and 8 that contribute 3 faces.
Observe that $6+12+8=26$, so this accounts for the remaining small cubes.
The 6 small cubes that contribute 1 face are in the centres of the faces of the larger cube.
The 12 small cubes that contribute 2 faces are on the edges of the larger cube.
The 8 small cubes that contribute 3 faces are on the corners of the larger cube.
Observe that $6 \times 1+12 \times 2+8 \times 3=54$, which is the correct total number of faces of small cubes that we expect on the surface of the larger cube.
The goal is to minimize the sum of the integers showing on the surface of the larger cube. This means we want to position each small cube so that the sum of the integers on its visible faces is as small as possible.
Each small cube has a face with the integer 1 on it. This means the 6 small cubes that each contribute 1 face to the surface of the larger cube can be positioned so that the number on this face is 1 .
The 12 small cubes that contribute 2 faces to the surface of the cube contribute 2 faces that share an edge.
The smallest possible total of the integers on 2 faces of a small cube is $1+2=3$.
The faces showing 1 and 2 also share an edge (as indicated in the net given in the problem), so this means these 12 small cubes can be oriented so that they each contribute a total of $1+2=3$ to the total on the surface of the larger cube.
Each of the 8 small cubes that contribute three faces to the surface of the larger cube contribute 3 faces that meet at a vertex.
The smallest possible sum of the integers on 3 faces of a small cube is $1+2+3=6$.
The faces that have the integers 1,2 , and 3 on them meet at a vertex.
This means these 8 cubes can be oriented so that they each contribute $1+2+3=6$ to the total on the surface of the larger cube.
By minimizing the total for each of the 26 cubes that contributes to the surface of the larger cube, we minimize the overall total. Using the arguments above, the minimum possible total of the integers on the surface of the larger cube is
$$
6 \times 1+12 \times 3+8 \times 6=90
$$

Below is an image of what the larger cube could look like when the small cubes are arranged to minimize the total of the visible integers.
ANSWER: 90