Question

The remainder left out when $8^{2n}–{\left(62\right)}^{2n+1}$is divided by $9$, is

• A. $0$
• B. $2$
• C. $7$
• D. $8$

Answer 1

The correct option is B

$2$

Explanation for the correct option:

Find the remainder.

$8^{2n}–{\left(62\right)}^{2n+1}$

$={(9-1)}^{2n}–{(63-1)}^{2n+1}$

$=\left\{{}^{2n}C_0{9}^{2n}-{}^{2n}C_1{9}^{2n-1}+{}^{2n}C_2{9}^{2n-2}.....+{}^{2n}C_{2n}{9}^0\right\}-\left\{{}^{2n+1}C_0{\left(63\right)}^{2n+1}-{}^{2n+1}C_1{\left(63\right)}^{2n}+{}^{2n+1}C_2{\left(63\right)}^{2n-1}.....+{}^{2n+1}C_{2n+1}{\left(63\right)}^0\right\}$

$=9k_1+1+63k_2+1$

$=9(k_1+7k_2)+2$

$=9\left(k_3\right)+2$

Thus, the remainder is $2$.

Hence, option $B$ is the correct answer.