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

$$\begin{gathered} \frac{1}{2·3^{10}}+\frac{1}{2^2·3^9}+\dots \frac{1}{2^{10}·3}=\frac{K}{2^{10}·3^{10}} \\ \mathrm{K}=2^9+2^8·3+2^7·3^2+\dots \dots +3^9 \\ =\frac{2^9\left({\left(\frac{3}{2}\right)}^{10}-1\right)}{\frac{3}{2}-1}=3^{10}-2^{10} \\ \text{ Now, }{3}^{10}-2^{10}=\left(3^5-2^5\right)\left(3^5+2^5\right) \\ =\left(211\right)\left(275\right) \\ =(35\times 6+1)(45\times 6+5) \\ =6\lambda +5 \end{gathered}$$

Remainder is 5.