If 1 2 · 3 ^ 10 + 1 2 ^ 2 · 3 ^ 9 + + 1 2 ^ 10 · 3 = K 2 ^ 10 · 3 ^...

We know ${x}^{n} - {y}^{n} = \left(x - y\right)\left({x}^{n - 1} + {x}^{n - 2}y + ⋯ x{y}^{n - 2} + {y}^{n - 1}\right)$

$\dfrac{1}{2 \cdot {3}^{10}} + \dfrac{1}{{2}^{2} \cdot {3}^{9}} + ⋯ \cdot \dfrac{1}{{2}^{9} \cdot {3}^{2}} + \dfrac{1}{{2}^{10} \cdot 3}$

$= \dfrac{{2}^{9} + {2}^{8} \cdot 3 + ⋯ 2 \cdot {3}^{8} + {3}^{9}}{{2}^{10} \cdot {3}^{16}} = \dfrac{{3}^{10} - {2}^{10}}{{2}^{10} \cdot {3}^{10}}$

So, $K = {3}^{10} - {2}^{10}$
Since we need the remainder when $K$ is divided by $6$

so, ${3}^{10} = 6{q}_{1} + 3$ and ${2}^{10} = 6{q}_{2} + 4$
Now $K$ will be of the form \left(6{q}_{1} + 3\right) - \left(6{q}_{2} + 4\right)$<br />$= 6\left({q}_{1} - {q}_{2}\right) - 1$<br />Hence, when$K$is divided by$6$, we get the remainder as$6 - 1 = 5