QJEEM23Mac8da: JEE Main 2023 Mathematics - Class 11 JEEM23Mac8da Solution: Among the statements : (S1) : is divisible by 8 (S2) : is…

Question

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Among the statements :

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

(S2) : $13(13)^{n}-12 n-13$ is divisible by 144 for infinitely many $n \in \mathbb{N}$

Accepted Answer

We have,

S_1

:

(2023)^{2022}-(1999)^{2022}

\begin{aligned}&amp; =(1999+24)^{2022}-(1999)^{2022}={ }^{2022} C_0(1999)^{2022}(24)^0 \\\\&amp; +{ }^{2022} C_1(1999)^{2021}(24)^1+{ }^{2022} C_2(1999)^{2020}(24)^2 \\\\&amp; +\ldots-(1999)^{2022} \\\\&amp; ={ }^{2022} C_1(1999)^{2021}(24)+{ }^{2022} C_2(1999)^{2022}(24)^2 \\\\&amp; =24\left({ }^{2022} C_1(1999)^{2021}+{ }^{2022} C_2(1999)^{2022}(24)+\ldots+\ldots\right) \\\\&amp; \Rightarrow S_1 \text { is divisible by } 24\end{aligned}

\begin{aligned}&amp; \text { Now, } S_2: 13(13)^n-12 n-13 \\\\&amp; \text { Here, } 13^n=(1+12)^n \\\\&amp; \quad=1+{ }^n C_1 12+{ }^n C_2(12)^2+{ }^n C_3(12)^3 \\\\&amp; \begin{aligned}\therefore S_2: &amp; 13\left(1+{ }^n C_1(12)+{ }^n C_2(12)^2+{ }^n C_3(12)^3+\ldots\right)-12 n-13 \\\\\quad= &amp; 13+156 n+13\left({ }^n C_2(12)^2+{ }^n C_3(12)^3+\ldots\right)-12 n-13 \\\\\quad= &amp; 144 \times 13\left({ }^n C_2+{ }^n C_3(12)+\ldots\right)\end{aligned}\end{aligned}

\Rightarrow S_2

is divisible by 144 for infinitely many

n \in N

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