The 4$$^{th}$$ term of GP is 500 and its common ratio is $$\frac{1}{m}$$, $$m \in \mathbb{N}$$. Let $$S_n$$ denote the sum of the first $$n$$ terms of this GP. If $$S_6 > S_5 + 1$$ and $$S_7 < S_6 + \frac{1}{2}$$, then the number of possible values of $$m$$ is _____

The 4th term of a GP is 500 and its common ratio is $$\frac{1}{m}$$, where $$m \in \mathbb{N}$$.

The 4th term: $$ar^3 = 500$$, where $$r = \frac{1}{m}$$.

$$a \cdot \frac{1}{m^3} = 500 \implies a = 500m^3$$

Express $$S_n$$.

For a GP with $$|r| < 1$$ (since $$m \geq 2$$ as $$m \in \mathbb{N}$$ and $$r \neq 1$$):

$$S_n = a \cdot \frac{1 - r^n}{1 - r} = 500m^3 \cdot \frac{1 - (1/m)^n}{1 - 1/m}$$

Note that $$S_n - S_{n-1} = ar^{n-1}$$ (the $$n$$th term).

Condition 1: $$S_6 > S_5 + 1$$

$$S_6 - S_5 > 1$$

$$ar^5 > 1$$

$$500m^3 \cdot \frac{1}{m^5} > 1$$

$$\frac{500}{m^2} > 1$$

$$m^2 < 500$$

$$m < \sqrt{500} \approx 22.36$$

So $$m \leq 22$$.

Condition 2: $$S_7 < S_6 + \frac{1}{2}$$

$$S_7 - S_6 < \frac{1}{2}$$

$$ar^6 < \frac{1}{2}$$

$$500m^3 \cdot \frac{1}{m^6} < \frac{1}{2}$$

$$\frac{500}{m^3} < \frac{1}{2}$$

$$m^3 > 1000$$

$$m > 10$$

So $$m \geq 11$$.

We need $$m \in \mathbb{N}$$ with $$11 \leq m \leq 22$$.

The possible values are: $$m = 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22$$.

That gives us $$22 - 11 + 1 = \boxed{12}$$ possible values of $$m$$.