If 22020+2021 is divided by 9, then the remainder obtained is

Question

If 22020+2021 is divided by 9, then the remainder obtained is (A) 0 (B) 1 (C) 3 (D) 7

Tardigrade Admin · Accepted Answer

22020=(2.2)2019=2.(23)673=2(9 - 1)673 =2(9673 -673 C1 ⋅ 9672 +673 C2 ⋅ 9671 +673 C672 9 - 1) =2 (a multiple of 9-1 ) 22020+2021=9K+2019 9K+2019 when divided by 9 gives remainder 3

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

$0$

$1$

$3$

$7$

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

Q. If 22020+2021 is divided by 9, then the remainder obtained is

0

1

3

7

22020=(2.2)2019=2.(23)673=2(9−1)673 =2(9673−673C1​⋅9672+673C2​⋅9671+673C672​9−1) =2 (a multiple of 9−1 ) 22020+2021=9K+2019 9K+2019 when divided by 9 gives remainder 3

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

$0$

$1$

$3$

$7$

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