If 1 2 · 3 10 + 1 2 2 · 3 9 + … + 1 2 10 · 3 = K 2 10 · 3 10 , then the remainder…

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

$\frac{1}{2·3^{10}}+\frac{1}{2^2·3^9}+\cdots ·\frac{1}{2^9·3^2}+\frac{1}{2^{10}·3}$

$=\frac{2^9+2^8·3+ \cdots 2·3^8+3^9}{2^{10}·3^{16}}=\frac{3^{10}-2^{10}}{2^{10}·3^{10}}$

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

so, $3^{10}=6q_1+3$ and $2^{10}=6q_2+4$
Now $K$ will be of the form $\left(6q_1+3\right)-\left(6q_2+4\right)$
$=6\left(q_1-q_2\right)-1$
Hence, when $K$ is divided by $6$, we get the remainder as $6-1=5$