Question

The remainder left out when $8^{2n}- (62)^{2n+1}$ − is divided by $9$ is

• A. $0$
• B. $2$
• C. $7$
• D. $8$

Answer 1

Answer: (B)

$8^{2n}-\left(62\right)^{2n+1}=\left(1+63\right)^{n}-\left(63-1\right)^{2n+1}$
$=\left(1+63\right)^{n}+\left(1-63\right)^{2n+1}=\left(1+^{n}c_{1}63+^{n}c_{2}\left(63\right)^{2}+....+\left(63\right)^{n}\right)+\left(1-^{\left(2n+1\right)}c_{1}63+^{\left(2n+1\right)}c_{2}\left(63\right)^{2}+....+\left(-1\right)\left(63\right)^{\left(2n+1\right)}\right)$
$=2+63\left(^{n}c_{1}+^{n}c_{2}\left(63\right)+....+\left(63\right)^{n-1}-^{\left(2n+1\right)}c_{1}+^{\left(2n+1\right)}c_{2}\left(63\right)+....-\left(63\right)^{\left(2n\right)}\right)$
$\therefore $ Reminder is 2