JEE Main 2023 — Mathematics Algebra

Given,

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

Now we know that $(x-y)$(x−y) divides $({x}^{n}-{y}^{n})\forall n\in \mathbb{N}$(xn−yn)∀n∈N

So, $(2023-1999)$(2023−1999) divides $(2023{)}^{2022}-(1999{)}^{2022}$(2023)2022−(1999)2022

$\Rightarrow 24$⇒24 divides $(2023{)}^{2022}-(1999{)}^{2002}$(2023)2022−(1999)2002

$\Rightarrow 8$⇒8 will divide $(2023{)}^{2022}-(1999{)}^{2002}$(2023)2022−(1999)2002

As $8$8 divides $24$24

Hence, $(S1)$(S1) is correct

Now solving,

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

So using binomial theorem in ${(1+12)}^{n}$(1+12)n we get,

$$13{(1+12)}^{n}-11n-13$$13(1+12)n−11n−13

$$=13({}^{n}{C}_{0}{+}^{n}{C}_{1}12{+}^{n}{C}_{2}{12}^{2}+...{+}^{n}{C}_{n}{12}^{n})-11n-13$$=13(nC0​+nC1​12+nC2​122+...+nCn​12n)−11n−13

$$=12\times 13n-11n+{12}^{2}\lambda$$=12×13n−11n+122λ

$=145n+144\lambda$=145n+144λ which is not divisible by $144$144

Hence, $(S2)$(S2) is incorrect