Sixty-four white $1 \times 1 \times 1$ cubes are used to form a $4 \times 4 \times 4$ cube, which is then painted red on each of its six faces. This large cube is then broken up into its 64 unit cubes. Each unit cube is given a score as follows:

| Exact number of faces painted red |
| :---: |
| 3 |
| 2 |
| 1 |
| 0 |


| Score |
| ---: |
| 3 |
| 2 |
| 1 |
| -7 |

The total score for the $4 \times 4 \times 4$ cube is
(A) 40
(B) 41
(C) 42
(D) 43
(E) 44
Sixty-four white $1 \times 1 \times 1$ cubes are used to form a $4 \times 4 \times 4$ cube, which is then painted red on each of its six faces. This large cube is then broken up into its 64 unit cubes. Each unit cube is given a score as follows:

| Exact number of faces painted red | Score |
| :---: | ---: |
| 2 | 3 |
| 2 | 2 |
| 1 | 1 |
| 0 | -7 |

The total score for the $4 \times 4 \times 4$ cube is
(A) 40
(B) 41
(C) 42
(D) 43
(E) 44

## Solution

First, we deal with the positive points. There is one point assigned for each red face. On each face of the large cube, there will be 16 red faces of $1 \times 1 \times 1$ cubes. This gives $6 \times 16=96$ red faces on $1 \times 1 \times 1$ cubes in total. So there are 96 positive points. However, there will be $2 \times 2 \times 2=8$ unit cubes which have no paint, and these will account for $8 \times(-7)=-56$ points.

Then the point total for the cube is $96+(-56)=40$. (Notice that it was not necessary that we consider cubes with paint on either 3 sides or 2 sides if we use this method.)

ANSWER: (A)