Amanda has two identical cubes. Each cube has one integer on each face so that the following statements are all true:

- Three adjacent faces of each cube have the numbers 15,9 and 4 as shown.
- The numbers on all pairs of opposite faces have the same sum $s$.
- When both cubes are rolled and the numbers on the top faces are added, the probability that the sum equals 24 is $\frac{1}{12}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_486fbaaa1172337e46fbg-2.jpg?height=320&width=287&top_left_y=1081&top_left_x=973)

Determine the sum of all possible values of $s$.
The sides opposite the sides showing 4,9 , and 15 show the numbers $s-4, s-9$, and $s-15$, respectively.

There are 6 ways that each die can land, hence there are $6 \times 6=36$ ways that the dice can land.

The table below contains the sum of the numbers on the top faces for each of the 36 ways that the dice can land:

|  | 4 | 9 | 15 | $s-4$ | $s-9$ | $s-15$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 4 | 8 | 13 | 19 | $s$ | $s-5$ | $s-11$ |
| 9 | 13 | 18 | 24 | $s+5$ | $s$ | $s-6$ |
| 15 | 19 | 24 | 30 | $s+11$ | $s+6$ | $s$ |
| $s-4$ | $s$ | $s+5$ | $s+11$ | $2 s-8$ | $2 s-13$ | $2 s-19$ |
| $s-9$ | $s-5$ | $s$ | $s+6$ | $2 s-13$ | $2 s-18$ | $2 s-24$ |
| $s-15$ | $s-11$ | $s-6$ | $s$ | $2 s-19$ | $2 s-24$ | $2 s-30$ |

The probability that the total is 24 is given as $\frac{1}{12}=\frac{3}{36}$, which means exactly three of the sums in the table above must equal 24.

There are at least two sums of 24, regardless of the value of $s$.

Therefore, for a value of $s$ to give a probability of $\frac{1}{12}$ of rolling 24 , there needs to be exactly one more sum of 24 .

We will call the sums where the dice show the same value the diagonal sums.

These are the sums in the diagonal of the table above, or $8,18,30,2 s-8,2 s-18$, and $2 s-30$. If the third sum of 24 occurs somewhere other than as a diagonal sum, then there would be at least four possible sums of 24 since the sums on either "side" of the diagonal are equal.

For example, if $s-11=24$, then $s=35$, so one of the values on each die is $35-15=20$. This would lead to a sum of 24 if a 20 and a 4 are rolled, and this can happen in two ways.

Therefore, the only way that the probability of rolling a 24 can be $\frac{1}{12}$ is if $2 s-8=24$, $2 s-18=24$, or $2 s-30=24$.

These equations can be solved for $s$ to get $s=16, s=21$, and $s=27$.

We now need to argue that none of these values of $s$ leads to a probability higher than $\frac{1}{12}$ of rolling a 24 .

The sums that depend on $s$ that are not diagonal sums are

$$
s, s+5, s+11, s-5, s+6,2 s-13, s-11, s-6,2 s-19,2 s-24
$$

Setting each of these values equal to 24 and solving gives

$$
s=24, s=19, s=13, s=29, s=18, s=\frac{37}{2}, s=35, s=30, s=\frac{43}{2}, s=24
$$

and none of $s=16, s=21$, and $s=27$ appear in the list.

The possible values of $s$ are $s=16, s=21$, and $s=27$, so the answer is $16+21+27=64$.

ANSWER: 64