JEE Main 2022 — Mathematics Algebra

We know ${x}^{n}-{y}^{n}=(x-y)({x}^{n-1}+{x}^{n-2}y+\cdots x{y}^{n-2}+{y}^{n-1})$xn−yn=(x−y)(xn−1+xn−2y+⋯xyn−2+yn−1)

$$\frac{1}{2\cdot {3}^{10}}+\frac{1}{{2}^{2}\cdot {3}^{9}}+\cdots \cdot \frac{1}{{2}^{9}\cdot {3}^{2}}+\frac{1}{{2}^{10}\cdot 3}$$2⋅3101​+22⋅391​+⋯⋅29⋅321​+210⋅31​

$$=\frac{{2}^{9}+{2}^{8}\cdot 3+\cdots 2\cdot {3}^{8}+{3}^{9}}{{2}^{10}\cdot {3}^{16}}=\frac{{3}^{10}-{2}^{10}}{{2}^{10}\cdot {3}^{10}}$$=210⋅31629+28⋅3+⋯2⋅38+39​=210⋅310310−210​

So, $K={3}^{10}-{2}^{10}$K=310−210

Since we need the remainder when $K$K is divided by $6$6

so, ${3}^{10}=6{q}_{1}+3$310=6q1​+3 and ${2}^{10}=6{q}_{2}+4$210=6q2​+4

Now $K$K will be of the form $(6{q}_{1}+3)-(6{q}_{2}+4)$(6q1​+3)−(6q2​+4)

$$=6({q}_{1}-{q}_{2})-1$$=6(q1​−q2​)−1

Hence, when $K$K is divided by $6$6, we get the remainder as $6-1=5$6−1=5