Exams › JEE Main › Maths

If 49^λ + 16ⁿ + λ is divisible by 64 for every n ∈ N, then the greatest negative value of λ is

1. −2
2. −1
3. −3
4. −4

Correct answer: −1

Solution

The expression 49^λ + 16ⁿ + λ must be evaluated under modulo 64 conditions. For n being a natural number, 16ⁿ is always divisible by 64 when n ≥ 2, thus the focus is on the terms 49^λ and λ. The term 49^λ can be simplified as (49 mod 64)^λ, which is 49, and we find that for λ = -1, the expression becomes 1 + 0, satisfying the divisibility condition. Hence, the greatest negative value of λ that maintains this divisibility is -1.

Related JEE Main Maths questions

• If Sₙ = cosⁿ θ + sinⁿ θ, then the value of 3S₄ - 2S₆ is
• For a natural number n, the inequality 2^(n - 1) < nⁿ holds when
• For any natural number n, the product n(n + 1) is always
• For which positive integers n does the inequality 3ⁿ exceed n³?
• For which values of n does the inequality 4ⁿ/(n+1) < (2n)!/(n!)² hold, if P(n) denotes this statement?
• For a positive integer n, define a(n) = 1 + 1/2 + 1/3 + 1/4 +... + 1/(2ⁿ - 1). Which of the following is true?

⚔️ Practice JEE Main Maths free + battle 1v1 →