Let $$[t]$$ denote the greatest integer less than or equal to $$t$$. Let $$f : [0, \infty) \rightarrow \mathbb{R}$$ be a function defined by $$f(x) = \left[\frac{x}{2} + 3\right] - [\sqrt{x}]$$. Let $$S$$ be the set of all points in the interval $$[0, 8]$$ at which $$f$$ is not continuous. Then $$\sum_{a \in S} a$$ is equal to ___________

We need to find the sum of all points in $$[0, 8]$$ where $$f(x) = [x/2 + 3] - [\sqrt{x}]$$ is discontinuous, where $$[t]$$ is the greatest integer function.

The function $$[x/2 + 3]$$ is discontinuous when $$x/2 + 3$$ is an integer, i.e., $$x/2$$ is an integer, which gives $$x = 0, 2, 4, 6, 8$$.

The function $$[\sqrt{x}]$$ is discontinuous when $$\sqrt{x}$$ is an integer, i.e., $$x = 0, 1, 4$$ (in $$[0,8]$$, we have $$\sqrt{x} = 0, 1, 2$$ at $$x = 0, 1, 4$$; at $$9$$, $$\sqrt{x} = 3$$ but $$9 > 8$$).

Potential discontinuity points of $$f = [x/2 + 3] - [\sqrt{x}]$$ are the union $$\{0, 1, 2, 4, 6, 8\}$$, though some discontinuities may cancel.

At $$x = 0$$ (the left endpoint), we check right continuity. We have $$f(0) = [3] - [0] = 3$$, and as $$x \to 0^+$$, $$f = [x/2 + 3] - [\sqrt{x}] = 3 - 0 = 3$$, so $$f$$ is continuous from the right at 0.

At $$x = 1$$, near the left we get $$f(1^-) = [3 + \epsilon] - [1 - \epsilon] = 3 - 0 = 3$$. At $$x = 1$$, $$f(1) = [3.5] - [1] = 3 - 1 = 2$$, so a jump occurs and $$f$$ is discontinuous at 1.

At $$x = 2$$, $$f(2^-) = [4 - \epsilon] - [\sqrt{2 - \epsilon}] = 3 - 1 = 2$$, while $$f(2) = [4] - [\sqrt{2}] = 4 - 1 = 3$$, so there is a jump and $$f$$ is discontinuous at 2.

At $$x = 4$$, we calculate carefully: $$f(4) = [4/2 + 3] - [\sqrt{4}] = [5] - [2] = 5 - 2 = 3$$. The left-hand limit is $$f(4^-) = [2 - \epsilon + 3] - [\sqrt{4 - \epsilon}] = [5 - \epsilon] - [2 - \delta] = 4 - 1 = 3$$, and the right-hand limit is $$f(4^+) = [2 + \epsilon + 3] - [\sqrt{4 + \epsilon}] = [5 + \epsilon] - [2 + \delta] = 5 - 2 = 3$$. Since both one-sided limits equal $$f(4)$$, the discontinuities cancel and $$f$$ is continuous at 4.

At $$x = 6$$, $$f(6^-) = [6 - \epsilon] - [\sqrt{6 - \epsilon}] = 5 - 2 = 3$$, while $$f(6) = [6] - [\sqrt{6}] = 6 - 2 = 4$$, so a jump occurs and $$f$$ is discontinuous at 6.

At $$x = 8$$, $$f(8^-) = [7 - \epsilon] - [\sqrt{8 - \epsilon}] = 6 - 2 = 4$$ and $$f(8) = [7] - [2\sqrt{2}] = 7 - 2 = 5$$, so $$f$$ is discontinuous at 8.

Thus the discontinuity points in $$[0,8]$$ are $$\{1,2,6,8\}$$, and their sum is $$1 + 2 + 6 + 8 = 17$$, which is the final answer.