Mathematics · 2023
JEE Main · 6 April 2023, Shift 2 · Q6
Among the statements: (S1): 2023^2022-1999^2022 is divisible by 8 (S2): 13(13)^n-11 n-13 is divisible by 144 for infinitely many n ∈ N
Among the statements:
(S1) : $\displaystyle 2023^{2022}-1999^{2022}$ is divisible by $\displaystyle 8$
(S2) : $\displaystyle 13(13)^n-11 n-13$ is divisible by $\displaystyle 144$ for infinitely many $\displaystyle n \in \mathbf{N}$
Official answer
From NTA’s final answer key for this paper.
(4)
both (S1) and (S2) are correct
More from Binomial Theorem
- Suppose Σ_r=0^2023 r^2^2023 C_r=2023 × α × 2^2022. Then the value of α is ____2023
- If Σ_r=0^5 (^11 C_2 r+1)/(2 r+2)=m/n, gcd(m, n)=1, then m-n is equal to ____.2025
- The coefficient of x^18 in the expansion of (x^4-1/x^3)^15 is ____2023
- In the expansion of (1+x)(1-x^2)(1+3/x+3/x^2+1/x^3)^5, x ≠ 0, the sum of the coefficients of x^3 and x^-13 is equal to ____2024
- If the sum of the coefficients of x^7 and x^14 in the expansion of (1/x^3-x^4)^n, x ≠ 0, is zero, then the value of n is ____.2026
- If (1/(^15 C_0)+1/(^15 C_1))(1/(^15 C_1)+1/(^15 C_2)) ⋯(1/(^15 C_12)+1/(^15 C_13))=(α^13)/(^14 C_0^14 C_1 …^14 C_12), then 30 α is equal to ____.2026
- If the coefficients of three consecutive terms in the expansion of (1+x)^n are in the ratio 1: 5: 20, then the coefficient of the fourth term is2023
- The remainder when 428^2024 is divided by 21 is ____.2024
JEE Main 2023 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.