Problem 995

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 $2\times64+4\times8\times n=128+32n$.

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, $B=6\times64\times n=384n$.

Thus, we get $$\frac{B}{A}=\frac{384n}{128+32n}.$$ 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 $\frac{B}{A}=\frac{12n}{4+n}$. We require that $\dfrac{B}{A}$ be equal to some integer, so we will determine which integers $\dfrac{12n}{4+n}$ can be.

First, note that $n$ is positive, so both $12n$ and $4+n$ are positive, which means $\dfrac{12n}{4+n}$ is positive.

This means $\dfrac{12n}{4+n}$ is a positive integer, so we determine which positive integers $\dfrac{12n}{4+n}$ can equal.

If $\dfrac{12n}{4+n}=1$, then $12n=4+n$ which can be rearranged to give $11n=4$ or $n=\frac{4}{11}$.

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

What if $\dfrac{12n}{4+n}=2$? In this case, we need $12n$ to be twice as large as $4+n$, or $12n=8+2n$.

This can be rearranged to give $10n=8$ or $n=\frac{8}{10}$.

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

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

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

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

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

The results are summarized in the table below.

According to the table, $\dfrac{12n}{4+n}$ 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 $\dfrac{12n}{4+n}$ is 12 or greater.

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

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

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

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

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