Among the statements S1: 2023²⁰²²−1999²⁰²²... | Mathematical Reasoning

JEE MAIN 2023

08-04-2023 S2

Question

Among the statements: (S1): $2023^{2022}-1999^{2022}$ is divisible by 8 . $(\mathrm{S} 2): 13(13)^n-11 \mathrm{n}-13$ is divisible by 144 for infinitely many $\mathrm{n} \in \mathrm{N}$.

Select the correct option:

both (S1) and (S2) are incorrect

only (S2) is correct

both (S1) and (S2) are correct

only (S1) is correct

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Sol. $$ \begin{aligned} & \mathrm{S}_1=(1999+24)^{2022}-(1999)^{2022} \\ & \Rightarrow \mathrm{C}_1(1999)^{2021}(24)+{ }^{2022} \mathrm{C}_2(1999)^{2020}(24)^2+\ldots . s o \text { on } \end{aligned} $$ $\mathrm{S}_1$ is divisible by 8 $$ \begin{aligned} & S_{2 n}: 13\left(13^n\right)-11 n-13 \\ & 13^n=(1+12)^n=1+12 n+{ }^n C_2 12^2+{ }^n C_3 12 \ldots \ldots \\ & 13\left(13^n\right)-11 n-13=145 n+{ }^n C_2 12+{ }^n C_3 12^3 \ldots \ldots \end{aligned} $$ If ( $n=144 m, m \in N$ ), then it is divisible by 144 For infinite value of $n$.

Question Tags

JEE Main

Mathematics

Medium

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