Among the statements:(S1): 2023^{2022} - 1999 | vedclass

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(C) For $(S1):$ Let $a = 2023$ and $b = 1999$. Note that $a \equiv 7 \pmod{8}$ and $b \equiv 7 \pmod{8}$.Then $a^{2022} \equiv 7^{2022} \pmod{8}$ and

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both $(S1)$ and $(S2)$ are correct

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(C) For $(S1):$ Let $a = 2023$ and $b = 1999$. Note that $a \equiv 7 \pmod{8}$ and $b \equiv 7 \pmod{8}$.Then $a^{2022} \equiv 7^{2022} \pmod{8}$ and $b^{2022} \equiv 7^{2022} \pmod{8}$.Since $7 \equiv -1 \pmod{8}$,$7^{2022} \equiv (-1)^{2022} \equiv 1 \pmod{8}$.Thus,$2023^{2022} - 1999^{2022} \equiv 1 - 1 \equiv 0 \pmod{8}$. So,$(S1)$ is correct.For $(S2):$ Let $f(n) = 13(13^{n}) - 11n - 13 = 13^{n+1} - 11n - 13$.Using binomial expansion,$13^{n+1} = (1 + 12)^{n+1} = 1 + (n+1)12 + \binom{n+1}{2}12^2 + \dots = 1 + 12n + 12 + 144 \binom{n+1}{2} + \dots$So,$f(n) = 1 + 12n + 12 + 144 \binom{n+1}{2} + \dots - 11n - 13 = n + 144 \binom{n+1}{2} + \dots$For $f(n)$ to be divisible by $144$,we need $n$ to be a multiple of $144$. Since there are infinitely many such $n \in N$,$(S2)$ is correct.

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

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