Question

For all $n \in \mathbb{N}$, if $n(n^2+3)$ is divisible by $k$, then the maximum value of $k$ is

• A. 4
• B. 6
• C. 8
• D. 2

Hint:

For divisibility problems, compute the expression for small $n$ to find the GCD. Check odd and even cases to confirm the maximum common divisor.

Answer 1

The Correct Option is D

We need to determine the maximum value of k for which n(n²+3) is divisible by k for all n ∈ ℕ.

Step 1: Consider the expression n(n²+3). Expanding this gives us:

n³ + 3n

Step 2: Analyze divisibility for a few small values of n.

• If n = 1, then 1³ + 3×1 = 4.
• If n = 2, then 2³ + 3×2 = 14.
• If n = 3, then 3³ + 3×3 = 36.

Step 3: Determine the greatest common divisor (GCD) of these values.

• GCD(4, 14) = 2
• GCD(2, 36) = 2

Observing these results, the expressions consistently give a GCD of 2. Thus, n(n²+3) is divisible by 2 for any natural number n, but not necessarily by any larger number for all n.

Conclusion: The maximum value of k such that n(n²+3) is divisible by k for all n ∈ ℕ is 2.