$${1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}$$

$$ \Rightarrow {1 \over {2\,.\,{3^{10}}}}\left[ {{{{{\left( {{3 \over 2}} \right)}^{10}} - 1} \over {{3 \over 2} - 1}}} \right] = {K \over {{2^{10}}\,.\,{3^{10}}}}$$

$$ = {{{3^{10}} - {2^{10}}} \over {{2^{10}}\,.\,{3^{10}}}} = {K \over {{2^{10}}\,.\,{3^{10}}}} \Rightarrow K = {3^{10}} - {2^{10}}$$

Now $$K = {(1 + 2)^{10}} - {2^{10}}$$

$$ = {}^{10}{C_0} + {}^{10}{C_1}2 + {}^{10}{C_2}{2^3} + \,\,....\,\, + \,\,{}^{10}{C_{10}}{2^{10}} - {2^{10}}$$

$$ = {}^{10}{C_0} + {}^{10}{C_1}2 + 6\lambda + {}^{10}{C_9}\,.\,{2^9}$$

$$ = 1 + 20 + 5120 + 6\lambda $$

$$ = 5136 + 6\lambda + 5$$

$$ = 6\mu + 5$$

$$\lambda ,\,\mu \in N$$

$$\therefore$$ remainder = 5