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)

$2$

$8^{2n}-(62{)}^{2n+1}$

$={(9-1)}^{2n}-(63-1{)}^{2n+1}=({}^{2n}{C}_0{9}^{2n}{-}^{2n}{C}_1{9}^{2n-1}+....{+}^{2n}{C}_{2n}{9}^0)-({}^{2n+1}{C}_0{\left(63\right)}^{2n}{-}^{2n+1}{C}_1{\left(63\right)}^{2n}+....{+}^{2n+1}{C}_{2n+1}{\left(63\right)}^0)=9k_1+1+63k_2+1=9(k_1+7k_2)+2=9\left(k_3\right)+2$

Hencve, $2$ is the remainder.