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  • Among the statements : <img alt="\mathrm{(S1) : 2023^{2022} -1999^{2022}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%28S1%29%20%3A%202023%5E%7B2022%7D%20-1999%5E%7B20

Among the statements :
\mathrm{(S1) : 2023^{2022} -1999^{2022}} is divisible by 8
\mathrm{(S2) : 13(13)^{n} -11n -13} is divisible by 144 for infinitely many \mathrm{n\epsilon \mathbb{N}}
 

Option: 1

only (S2) is correct


Option: 2

only (S1) is correct
 


Option: 3

both (S1) and (S2) are incorrect


Option: 4

both (S1) and (S2) are correct


Answers (1)

best_answer

\because \mathrm{x}^{\mathrm{n}}-\mathrm{y}^{\mathrm{n}}=(\mathrm{x}-\mathrm{y})\left[\mathrm{x}^{\mathrm{n}-1}+\mathrm{x}^{\mathrm{n}-2} \mathrm{y}+\mathrm{x}^{\mathrm{n}-3} \mathrm{y}^2+\ldots \ldots+\mathrm{y}^{\mathrm{n}-1}\right] \\
\mathrm{x}^{\mathrm{n}}-\mathrm{y}^{\mathrm{n}} \text { is divisible by } \mathrm{x}-\mathrm{y} \\
\text { Stat } 1 \rightarrow \quad(2023)^{2022}-(1999)^{2022} \\
(2023)-(1999)=24 \\
\therefore \quad(2023)^{2022}-(1999)^{2022} \\
\text { is divisible by } 8 \\
\text { Stat } 2 \rightarrow \quad 13(1+12)^{\mathrm{n}}-11 \mathrm{n}-13 \\
13\left[1+{ }^n C_1,(12)+{ }^n C_2(12)^2+\ldots\right]-11 \mathrm{n}-13 \\
\Rightarrow(156 \mathrm{n}-11 \mathrm{n})+13 \cdot{ }^{\mathrm{n}} \mathrm{C}_2(12)^2+13 \cdot{ }^{\mathrm{n}} \mathrm{C}_3(12)^3+\ldots \\
\Rightarrow 145 \mathrm{n}+13 \cdot{ }^{\mathrm{n}} \mathrm{C}_2(12)^2+13 \cdot{ }^{\mathrm{n}} \mathrm{C}_3(12)^3+\ldots \\

\mathrm{If}(\mathrm{n}=144 \mathrm{~m}, \mathrm{~m} \in \mathrm{N}) then \: it \: is \: divisible\: by \: 144\: for \: infinite \: values \: of \: \mathrm{n}.

Correct Answer is option 4

 

Posted by

jitender.kumar

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