I'm not going to list all 24 numbers!

To find the sum of divisors, you take the product of each prime factor with its power+1 then divide it by each divisor-1.

In this case, it is

$2^3, 5^2, 7$

$\dfrac{(2^4-1)(5^3-1)(7^2-1)}{(2-1)(5-1)(7-1)}$

$\dfrac{89280}{24}$

$\boxed{3720}$

Suppose we start with powers of 7 namely $7^{0}, 7^{1}$ yielding divisors 1 and 7 whose sum is 8

Now we consider their combination with powers of 5 which are (1, 5, 25) Fetching us two groups of divisors (1, 5, 25) and (7, 35, 175)

This will give an overall sum of all divisors that contain only powers of 7 and 5 as (1 + 7) (1 + 5 + 25)

Finally we throw in the powers of 2 giving the overall sum of divisors as $(1+7)(1+5+25)(1+2+4+8) = (8)(31)(15) = \boxed{3720 }$

The problem can be solved in the following way:

$\frac{p^{a+1} - 1}{p - 1}$ X $\frac{q^{b+1} - 1}{q - 1}$ X $\frac{r^{c+1} - 1}{r - 1}$

where, p,q and r are the first, second and third prime divisor, and a, b and c are the powers of p,q and r respectively.

Thus,

$S_{n}$ = $\frac{2^{3+1} - 1}{2 - 1}$ X $\frac{5^{2+1} - 1}{5 - 1}$ X $\frac{7^{1+1} - 1}{7 - 1}$

$S_{n}$ = $\frac{2^{4} - 1}{1}$ X $\frac{5^{3} - 1}{4}$ X $\frac{7^{2} - 1}{6}$

$S_{n}$ = $\frac{15}{1}$ X $\frac{124}{4}$ X $\frac{48}{6}$

$S_{n}$ = 15 X 31 X 8

$S_{n}$ = $\boxed{3720}$

Simple enough just Put the value in theorem and ans is ready