JEE Main 2023 · 6 April · evening · Single correct · +4 / −1
Among the statements :
(S1) : 2023²⁰²²-1999²⁰²² is divisible by 8
(S2) : 13(13)ⁿ-12 n-13 is divisible by 144 for infinitely many n ∈ ℕ
- (a) both (S1) and (S2) are incorrect
- (b) only (S1) is correct
- (c) only (S2) is correct
- (d) both (S1) and (S2) are correct
Answer and solution
Answer (d) both (S1) and (S2) are correct
Solution
Maths sir · 4 steps
Look at both as aᵐ - bᵐ: the factor a - b settles S1, and writing 13 = 1 + 12 settles S2.
- Statement S1. For any natural m, aᵐ - bᵐ is divisible by a - b. Here a - b = 2023 - 1999 = 24 = 8 × 3, so 2023²⁰²² - 1999²⁰²² is a multiple of 24 and therefore of 8. S1 is correct.
- Rewrite S2. 13(13)ⁿ = 13ⁿ⁺¹ and 12n + 13 = 12(n+1) + 1. Put m = n + 1, so the expression becomes 13ᵐ - 12m - 1. Now expand 13ᵐ = (1 + 12)ᵐ by the binomial theorem.
13ᵐ - 1 - 12m = C(m, 2)12² + C(m, 3)12³ + … = 144[C(m, 2) + 12C(m, 3) + …]
- Check S2. Every term left over has at least 12² = 144 in it, so the expression is divisible by 144 for every n ∈ ℕ. For example, n = 1 gives 169 - 12 - 13 = 144. Holding for all n means it holds for infinitely many, so S2 is correct.
- So. Both statements are true, which is option (d).
The idea to keep. For divisibility by a power of k, write the base as 1 + k and expand: the first two binomial terms cancel the extra terms. And aᵐ - bᵐ is always a multiple of a - b.
Answer from the NTA final answer key. Solution by PrepWiser.