Among the statements ( S 1 ) : 2023 2022 - 1999 2022 i s d i v i s i b l e b y 8 ( S 2 ) : 13 ( 13 ) n - 11 n - 13 i s d i v i s i b l e b y 144 f o r i n f i n i t e l y m a n y n ? [2023]

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Solution

Q.
Among the statements

$(S 1) : 2023^{2022} - 1999^{2022} i s d i v i s i b l e b y 8$

$(S 2) : 13 {(13)}^{n} - 11 n - 13 i s d i v i s i b l e b y 144 f o r i n f i n i t e l y m a n y n \in ℕ$ [2023]

1 only (S2) is correct

2 both (S1) and (S2) are incorrect

3 both (S1) and (S2) are correct

4 only (S1) is correct

Ans.
(3)

As we know that $x^{n} - y^{n}$ is always divisible by $x - y$

$∴$ $2023^{2022} - 1999^{2022}$ is divisible by 2023 - 1999 = 24, which is divisible by 8. $∴$ $S 1 is true.$

Also, $13 {(13)}^{n} - 11 n - 13 = 13 {(12 + 1)}^{n} - 11 n - 13$

$= 13 [{}^{n}{C_{0}} 12^{n} + {}^{n}{C_{1}} 12^{n - 1} + {}^{n}{C_{2}} 12^{n - 2} + \dots + {}^{n}{C_{n}}] - 11 n - 13$

$= 13 [{}^{n}{C_{0}} 12^{n} + {}^{n}{C_{1}} 12^{n - 1} + \dots + {}^{n}{C_{n - 2}} 12^{2}] + 13 \cdot 12 \cdot n + 13 - 11 n - 13$

$= 13 \times 12^{2} [{}^{n}{C_{0}} 12^{n - 2} + {}^{n}{C_{1}} 12^{n - 3} + \dots + {}^{n}{C_{n - 2}}] + 145 n$

Which is divisible by 144, for all $n \geq 144$ .

$i.e., It is divisible by 144 for infinitely many n \in N .$

$∴$ $S 2 is also true.$

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