$$ \begin{array}{|c|l|c|} \hline \alpha-\beta & {\text { Case }} & \mathbf{P} \\ \hline 5 & (6,1) & 1 / 36 \\ \hline 4 & (6,2)(5,1) & 2 / 36 \\ \hline 3 & (6,3)(5,2)(4,1) & 3 / 36 \\ \hline 2 & (6,4)(5,3)(4,3)(3,1) & 4 / 36 \\ \hline 1 & (6,5)(5,4)(4,3)(3,2)(2,1) & 5 / 36 \\ \hline 0 & (6,6)(5,5) \ldots \ldots(1,1) & 6 / 36 \\ \hline-1 & ----- & 5 / 36 \\ \hline-2 & -----& 4 / 36 \\ \hline-3 & ----- & 3 / 36 \\ \hline-4 & (2,6)(1,5) & 2 / 36 \\ \hline-5 & (1,6) & 1 / 36 \\ \hline \end{array} $$

$$E[X^2] = \sum\limits_{i=-5}^{5} (x_i)^2 P(x_i)$$

Substituting the values from table :

$$E[X^2] = 2\left[\frac{25}{36}+\frac{32}{36}+\frac{27}{36}+\frac{16}{36}+\frac{5}{36}\right] = \frac{105}{18} = \frac{35}{6}$$

Next, we calculate the expected value of the differences. The expected value is calculated as the sum of the products of each outcome and its corresponding probability. Given that the table is symmetric around 0, the expected value is 0.

$$E[X] = \sum\limits_{i=-5}^{5} x_i P(x_i) = 0$$

Now, we can calculate the variance, which is the expected value of the squared differences minus the square of the expected value of the differences :

$$Var[X] = E[X^2] - (E[X])^2 = \frac{35}{6} - 0^2 = \frac{35}{6}$$

Here, $$p = 35$$ and $$q = 6$$, and they are co-prime.

The positive divisors of 35 are 1, 5, 7, and 35. The sum of these divisors is $$1 + 5 + 7 + 35 = 48$$.