If $n$ is any odd number greater than 1, then $n(n^2 - 1)$ is

### Concept & Logic This problem relies on the specific behavior of consecutive even numbers. First, we must factor the algebraic expression. $$ n(n^2 - 1) = (n - 1)n(n + 1) $$ This represents the product of three consecutive integers. ### Step-by-Step Solution * **Given:** $n$ is an *odd* number greater than 1. * **Step 1: Analyze the factors.** We have three consecutive numbers: $(n - 1)$, $n$, and $(n + 1)$. * **Step 2: Apply the odd constraint.** Since $n$ is odd, the numbers immediately before and after it—$(n - 1)$ and $(n + 1)$—must both be *even* numbers. * **Step 3: Use the consecutive even numbers rule.** The product of any two consecutive even numbers is always a multiple of 8. (For example, $2 \times 4 = 8$, or $4 \times 6 = 24$). So, $(n - 1)(n + 1)$ is a multiple of 8. * **Step 4: Use the consecutive integers rule.** In any sequence of three consecutive integers, exactly one of them must be a multiple of 3. * **Conclusion:** The entire product must be divisible by both 8 and 3. Since 8 and 3 are co-prime, the product is always divisible by $8 \times 3 = 24$. ### Exam Strategy & Shortcut **Test the Boundary Condition:** Avoid abstract algebra during a timed exam by picking the smallest number that fits the prompt's rules. The condition says "$n$ is an odd number greater than 1". The smallest valid number is $n = 3$. Substitute $n = 3$ into the expression: $3(3^2 - 1) = 3(9 - 1) = 3 \times 8 = 24$. The result is 24. The only option that 24 is always divisible by is 24 itself. ### Common Pitfall Students often remember that the product of two even numbers is divisible by 4 (since $2 \times 2 = 4$). However, *consecutive* even numbers guarantee that one is a multiple of 2 and the other is a multiple of 4, meaning their product is actually divisible by 8. Underestimating this factor leads to incorrect minimum bounds. ### Final Answer Therefore, the correct answer is **divisible by 24 always**.