Among the statements : ( S1 ) : 2023 ^ 2022 - 1999 ^ 2022 is...

Among the statements :

$( S1 ) : {2023}^{2022} - {1999}^{2022}$ is divisible by $8$ .

$( S2 ) : 13 ( 13{)}^{n} - 11n - 13$ is divisible by $144$ for infinitely many $n \in \mathbb{N}$

Options

AOnly $\left(S2\right)$ is correct
BOnly $( S1 )$ is correct
CBoth $\left(S1\right)$ and $\left(S2\right)$ are correct
DBoth $\left(S1\right)$ and $\left(S2\right)$ are incorrect

Correct answer

B. Only $( S1 )$ is correct

Step-by-step solution

Given,
$( S1 ) : ( 2023{)}^{2022} - ( 1999{)}^{2022} is divisible by$ 8 $Now we know that$ ( x - y ) $divides \left({x}^{n} - {y}^{n}\right) \forall n \in \mathbb{N}$
So, \left(2023 - 1999\right) $divides$ ( 2023{)}^{2022} - ( 1999{)}^{2022} $$ ⇒ 24 $divides$ ( 2023{)}^{2022} - ( 1999{)}^{2002} $$ ⇒ 8 $will divide$ ( 2023{)}^{2022} - ( 1999{)}^{2002} $As$ 8 $divides$ 24 $Hence, \left(S1\right)$ is correct
Now solving,
\left(S2\right): 13{\left(13\right)}^{n} - 11n - 13 $is divisible by$ 144 $for$ n \in \mathbb{N} $So using binomial theorem in$ {\left(1 + 12\right)}^{n} $we get,$ 13{\left(1 + 12\right)}^{n} - 11n - 13 $$ = 13\left({}^{n}{C}_{0}{+}^{n}{C}_{1}12{+}^{n}{C}_{2}{12}^{2} + . . . {+}^{n}{C}_{n}{12}^{n}\right) - 11n - 13 $$ = 12 \times 13n - 11n + {12}^{2}\lambda $$ = 145n + 144\lambda $which is not divisible by$ 144 $Hence, \left(S2\right)$ is incorrect

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