Let $$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$$. Define $$f : S \to S$$ as $$f(n) = \begin{cases} 2n, & \text{if } n = 1,2,3,4,5 \\ 2n-11 & \text{if } n = 6,7,8,9,10 \end{cases}$$
Let $$g : S \geq S$$ be a function such that $$fog(n) = \begin{cases} n+1, & \text{if } n \text{ is odd} \\ n-1, & \text{if } n \text{ is even} \end{cases}$$, then
$$g(10)(g(1) + g(2) + g(3) + g(4) + g(5))$$ is equal to ______

Given $$S = \{1, 2, \ldots, 10\}$$ and $$f: S \to S$$ defined by:

$$f(n) = \begin{cases} 2n & \text{if } n = 1,2,3,4,5 \\ 2n - 11 & \text{if } n = 6,7,8,9,10 \end{cases}$$

So: $$f(1)=2,\; f(2)=4,\; f(3)=6,\; f(4)=8,\; f(5)=10,\; f(6)=1,\; f(7)=3,\; f(8)=5,\; f(9)=7,\; f(10)=9$$

Note that $$f$$ is a bijection. Its inverse is:

$$f^{-1}(k) = \begin{cases} k/2 & \text{if } k \text{ is even} \\ (k+11)/2 & \text{if } k \text{ is odd} \end{cases}$$

We need $$g: S \to S$$ such that $$f \circ g(n) = \begin{cases} n+1 & \text{if } n \text{ is odd} \\ n-1 & \text{if } n \text{ is even} \end{cases}$$

Since $$f$$ is bijective: $$g(n) = f^{-1}(f \circ g(n))$$.

Computing each value:

$$g(1) = f^{-1}(2) = 1, \quad g(2) = f^{-1}(1) = 6, \quad g(3) = f^{-1}(4) = 2$$

$$g(4) = f^{-1}(3) = 7, \quad g(5) = f^{-1}(6) = 3$$

$$g(6) = f^{-1}(5) = 8, \quad g(7) = f^{-1}(8) = 4, \quad g(8) = f^{-1}(7) = 9$$

$$g(9) = f^{-1}(10) = 5, \quad g(10) = f^{-1}(9) = 10$$

Now compute:

$$g(1) + g(2) + g(3) + g(4) + g(5) = 1 + 6 + 2 + 7 + 3 = 19$$

$$g(10) = 10$$

$$g(10) \cdot (g(1) + g(2) + g(3) + g(4) + g(5)) = 10 \times 19 = 190$$

The correct answer is $$\boxed{190}$$.