Divisibility of n(n^2 − 1) for Odd n > 1 If n is any odd integer greater than 1, what is n(n^2 − 1) = n(n − 1)(n + 1) always divisible by?

Introduction / Context:
This question examines guaranteed divisibility of the product of three consecutive integers centered at an odd n. Recognizing structural factors in consecutive triples is a core technique in number theory problems on aptitude tests.

Given Data / Assumptions:

• n is odd and greater than 1.
• Expression: n(n − 1)(n + 1).
• We must find a fixed integer that always divides this product.

Concept / Approach:
Among three consecutive integers, one is a multiple of 3, so the product is divisible by 3. Because n is odd, both neighbors (n − 1) and (n + 1) are even; they are consecutive even numbers, so one is a multiple of 2 and the other is a multiple of 4. Hence the product has at least a factor of 2 * 4 = 8 from the even terms, and a factor of 3 from the “multiple-of-3” property, totaling 8 * 3 = 24.

Step-by-Step Solution:
Since n is odd, n − 1 and n + 1 are even.One of these consecutive even numbers is divisible by 4, and the other by 2 → combined factor 8.Among n − 1, n, n + 1, one is a multiple of 3 → additional factor 3.Therefore, n(n − 1)(n + 1) is always divisible by 8 * 3 = 24.

Verification / Alternative check:
Try n = 3: product = 3 * 2 * 4 = 24 (divisible by 24). Try n = 5: 5 * 4 * 6 = 120 (24 * 5). The pattern persists for all odd n.

Why Other Options Are Wrong:
Divisibility by 12 is too weak (24 is guaranteed). 48 and 96 require an extra factor of 2 or 2^2 that is not ensured for every odd n. “None of these” is false because 24 always divides the product.

Common Pitfalls:
Assuming both neighbors contribute a 4 each (not guaranteed); overlooking the “multiple-of-3” property of any three consecutive integers.

Final Answer:
divisible by 24 always