Riley has 64 cubes with dimensions $1 \times 1 \times 1$. Each cube has its six faces labelled with a 2 on two opposite faces and a 1 on each of its other four faces. The 64 cubes are arranged to build a $4 \times 4 \times 4$ cube. Riley determines the total of the numbers on the outside of the $4 \times 4 \times 4$ cube. How many different possibilities are there for this total?
We can categorize the 64 small cubes into four groups: 8 cubes that are completely in the interior of the larger cube (and are completely invisible), $4 \times 6=24$ "face cubes" that have exactly one of their faces showing, $2 \times 12=24$ "edge cubes" that have exactly two faces showing, and 8 "corner cubes" that have 3 faces showing.

Notice that $8+24+24+8=64$, so this indeed accounts for all 64 small cubes.

The diagram below has all visible faces of the smaller cubes labelled so that faces of edge cubes are labelled with an $\mathbf{E}$, faces of face cubes are labelled with an $\mathbf{F}$, and faces of corner cubes are labelled with a $\mathbf{C}$ :

![](https://cdn.mathpix.com/cropped/2024_04_17_54c374478de6843636d1g-16.jpg?height=905&width=984&top_left_y=163&top_left_x=625)

Each of the 8 corner cubes has 3 of its faces exposed.

Because of the way the faces are labelled on the small cubes, each of the 8 corner cubes will always contribute a total of $1+1+2=4$ to the sum of the numbers on the outside.

Regardless of how the small cubes are arranged, the corner cubes contribute a total of $8 \times 4=32$ to the total of all numbers on the outside of the larger cube.

The edge cubes can each show either a 1 and a 1 or a 1 and a 2 , for a total of either 2 or 3 .

Thus, each edge cube contributes a total of either 2 or 3 to the total on the outside of the larger cube.

This means the smallest possible total that the edge cubes contribute is $24 \times 2=48$, and the largest possible total that they contribute is $24 \times 3=72$.

Notice that it is possible to arrange the edge cubes to show any total from 48 to 72 inclusive.

This is because if we want the total to be $48+k$ where $k$ is any integer from 0 to 24 inclusive, we simply need to arrange exactly $k$ of the edge cubes to show a total of 3 .

Each of the face cubes shows a total of 1 or a total of 2 . There are 24 of them, so the total showing on the face cubes is at least 24 and at most $24 \times 2=48$.

Similar to the reasoning for the edge cubes, every integer from 24 to 48 inclusive is a possibility for the total showing on the face cubes.

Therefore, the smallest possible total is $32+48+24=104$, and the largest possible total is $32+72+48=152$.

By earlier reasoning, every integer between 104 and 152 inclusive is a possibility, so the number of possibilities is $152-103=49$.

ANSWER: 49