Question

If $2^{2020}+2021$ is divided by $9,$ then the remainder obtained is

• A. $0$
• B. $1$
• C. $3$
• D. $7$

Answer 1

Answer: (C)

$2^{2020}=\left(2.2\right)^{2019}=2.\left(2^{3}\right)^{673}=2\left(9 - 1\right)^{673}$
$=2\left(9^{673} -^{673} C_{1} \cdot 9^{672} +^{673} C_{2} \cdot 9^{671} +^{673} C_{672} 9 - 1\right)$
$=2$ (a multiple of $9-1$ )
$2^{2020}+2021=9K+2019$
$9K+2019$ when divided by $9$ gives remainder $3$