JEE Main 2023 (Online) 6th April Evening ShiftQuestion 1
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 $n\in\mathbb{N}$
Binomial Theorem · Remainder and Divisibility
Solution
Concept. (S1) uses $(a-b)\mid(a^n-b^n)$. (S2) expands $13^{n+1}=(1+12)^{n+1}$ modulo $144$. Formulas used. (1) $(a-b)\mid\left(a^n-b^n\right)$(2) $(1+12)^{N}=1+12N+(\text{multiple of }144)$Step 1 - (S1). By (1), $2023^{2022}-1999^{2022}$ is divisible by $2023-1999=24$, hence by $8$. True. Step 2 - (S2). $13(13)^n=13^{n+1}$. By (2), $$13^{n+1}-11n-13\equiv1+12(n+1)-11n-13=n\pmod{144}.$$ So the number is divisible by $144$ exactly when $144\mid n$, which happens for infinitely many $n$. True. Step 3 - Conclude. Both statements are correct. Answer: (D)