An $8 \times 8 \times n$ rectangular prism is made up from $1 \times 1 \times 1$ cubes. Suppose that $A$ is the surface area of the prism and $B$ is the combined surface area of the $1 \times 1 \times 1$ cubes that make up the prism. What is the sum of the values of $n$ for which $\frac{B}{A}$ is an integer?
(A) 86
(B) 90
(C) 70
(D) 78
(E) 96
The rectangular prism has two faces whose area is $8 \times 8=64$, and four faces each of whose area is $8 \times n$.

Therefore, $A$ is $2 \times 64+4 \times 8 \times n=128+32 n$.

The prism is made up of $8 \times 8 \times n=64 \times n$ cubes, each of which has dimensions $1 \times 1 \times 1$.

Each $1 \times 1 \times 1$ cube has surface area 6 because each has 6 faces which are all $1 \times 1$ squares.

Therefore, $B=6 \times 64 \times n=384 n$.

Thus, we get

$$
\frac{B}{A}=\frac{384 n}{128+32 n}
$$

This expression can be simplified by recognizing that each of 384,128 and 32 is divisible by 32 . After dividing the numerator and denominator by 32 , we get $\frac{B}{A}=\frac{12 n}{4+n}$. We require that $\frac{B}{A}$ be equal to some integer, so we will determine which integers $\frac{12 n}{4+n}$ can be.

First, note that $n$ is positive, so both $12 n$ and $4+n$ are positive, which means $\frac{12 n}{4+n}$ is positive. This means $\frac{12 n}{4+n}$ is a positive integer, so we determine which positive integers $\frac{12 n}{4+n}$ can equal. If $\frac{12 n}{4+n}=1$, then $12 n=4+n$ which can be rearranged to give $11 n=4$ or $n=\frac{4}{11}$.

Since $n$ must be an integer, we conclude that $\frac{12 n}{4+n}$ cannot be equal to 1 .

What if $\frac{12 n}{4+n}=2$ ? In this case, we need $12 n$ to be twice as large as $4+n$, or $12 n=8+2 n$. This can be rearranged to give $10 n=8$ or $n=\frac{8}{10}$.

Again, this value of $n$ is not an integer, so we conclude that $\frac{12 n}{4+n}$ is not 2 .

Following this reasoning, if $\frac{12 n}{4+n}$ is 3 , we find that $n$ must be $\frac{4}{3}$, which is also not an integer, so $\frac{12 n}{4+n}$ is not equal to 3 .

If $\frac{12 n}{4+n}$ is equal to 4 , we have that $12 n$ is four times $4+n$, or $12 n=16+4 n$.

Rearranging this gives $8 n=16$ which means $n=2$. Therefore, $\frac{12 n}{4+n}$ can be 4 , and it happens when $n=2$.

We continue in this way for all possible positive integer values of $\frac{12 n}{4+n}$ up to and including $\frac{12 n}{4+n}=11$.

The results are summarized in the table below.

| $\frac{12 n}{n+4}$ | $n$ | $n$ is an integer |
| :---: | :---: | :---: |
| 1 | $\frac{4}{11}$ | $\times$ |
| 2 | $\frac{8}{10}$ | $\times$ |
| 3 | $\frac{4}{3}$ | $\times$ |
| 4 | 2 | $\checkmark$ |
| 5 | $\frac{20}{7}$ | $\times$ |
| 6 | 4 | $\checkmark$ |
| 7 | $\frac{28}{5}$ | $\times$ |
| 8 | 8 | $\checkmark$ |
| 9 | 12 | $\checkmark$ |
| 10 | 20 | $\checkmark$ |
| 11 | 44 | $\checkmark$ |

According to the table, $\frac{12 n}{4+n}$ can be any of the integers 4,6,8,9,10, and 11 and these occur when $n$ is equal to $2,4,8,12,20$, and 44 , respectively.

We now consider what happens when $\frac{12 n}{4+n}$ is 12 or greater.

If $\frac{12 n}{4+n}=12$, then $12 n$ is 12 times as large as $4+n$, or $12 n=48+12 n$.

Since $48+12 n$ is always greater than $12 n$, there is no value of $n$ for which $12 n=48+12 n$.

Similarly, since $n$ is a positive integer, there is no value of $n$ for which $\frac{12 n}{4+n}$ is 13 or greater.

We conclude that the only possible positive integer values of $\frac{12 n}{4+n}$ are those in the table, so the only values of $n$ which make $\frac{12 n}{4+n}$ an integer are $2,4,8,12,20$, and 44 .

The sum of these numbers is $2+4+8+12+20+44=90$.