If $$\frac{1}{2 \cdot 3^{10}} + \frac{1}{2^2 \cdot 3^9} + \cdots + \frac{1}{2^{10} \cdot 3} = \frac{K}{2^{10} \cdot 3^{10}}$$, then the remainder when $$K$$ is divided by $$6$$ is

Given,

$$\frac1{2\cdot3^{10}}+\frac1{2^2\cdot3^9}+\cdots+\frac1{2^{10}\cdot3}=\frac{K}{2^{10}\cdot3^{10}}$$

Multiplying both sides by

$$2^{10}\cdot3^{10}$$

we get

$$K=2^9+2^8\cdot3+2^7\cdot3^2+\cdots+3^9$$

Hence,

$$K=\sum_{r=0}^{9}2^{9-r}3^r$$

This is a GP with first term

$$2^9$$

and common ratio

$$\frac32$$

Therefore,

$$K=2^9\cdot\frac{\left(\frac32\right)^{10}-1}{\frac32-1}$$

$$=2^9\cdot\frac{\frac{3^{10}}{2^{10}}-1}{\frac12}$$

$$=2^{10}\cdot\frac{3^{10}-2^{10}}{2^{10}}$$

$$=3^{10}-2^{10}$$

Now, modulo $$6$$,

$$3^{10}\equiv3\pmod6$$

and

$$2^{10}\equiv4\pmod6$$

Hence,

$$K\equiv3-4\equiv-1\equiv5\pmod6$$

Therefore, the remainder when $$K$$ is divided by $$6$$ is $$\boxed{5}$$.