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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 $$n \in \mathbb{N}$$

JEE · Math · previous-year question

  1. A.both (S1) and (S2) are incorrect
  2. B.only (S1) is correct
  3. C.only (S2) is correct
  4. D.both (S1) and (S2) are correctcorrect

Answer

D. both (S1) and (S2) are correct

Explanation

We have, $S_1$ : $(2023)^{2022}-(1999)^{2022}$ $$ \begin{aligned} & =(1999+24)^{2022}-(1999)^{2022}={ }^{2022} C_0(1999)^{2022}(24)^0 \\\\ & +{ }^{2022} C_1(1999)^{2021}(24)^1+{ }^{2022} C_2(1999)^{2020}(24)^2 \\\\ & +\ldots-(1999)^{2022} \\\\ & ={ }^{2022} C_1(1999)^{2021}(24)+{ }^{2022} C_2(1999)^{2022}(24)^2 \\\\ & =24\left({ }^{2022} C_1(1999)^{2021}+{ }^{2022} C_2(1999)^{2022}(24)+\ldots+\ldots\right) \\\\ & \Rightarrow S_1 \text { is divisible by } 24 \end{aligned} $$ $$ \begin{aligned} & \text { Now, } S_2: 13(13)^n-12 n-13 \\\\ & \text { Here, } 13^n=(1+12)^n \\\\ & \quad=1+{ }^n C_1 12+{ }^n C_2(12)^2+{ }^n C_3(12)^3 \\\\ & \begin{aligned} \therefore S_2: & 13\left(1+{ }^n C_1(12)+{ }^n C_2(12)^2+{ }^n C_3(12)^3+\ldots\right)-12 n-13 \\\\ \quad= & 13+156 n+13\left({ }^n C_2(12)^2+{ }^n C_3(12)^3+\ldots\right)-12 n-13 \\\\ \quad= & 144 \times 13\left({ }^n C_2+{ }^n C_3(12)+\ldots\right) \end{aligned} \end{aligned} $$ $\Rightarrow S_2$ is divisible by 144 for infinitely many $n \in N$

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