If $n$ is a whole number greater than 1, then $n^2(n^2 - 1)$ is always divisible by

### Concept & Logic This problem requires algebraic factorization to reveal the consecutive integer components hidden within the expression. $$ n^2(n^2 - 1) = n \cdot n \cdot (n - 1) \cdot (n + 1) = (n - 1)n(n + 1) \cdot n $$ ### Step-by-Step Solution * **Given:** The expression is $n^2(n^2 - 1)$ where $n$ is a whole number $> 1$. * **Step 1: Factor the expression completely.** As shown in the concept, the expression contains $(n - 1)n(n + 1)$, which represents the product of three consecutive integers. * **Step 2: Apply divisibility rules for consecutive integers.** The product of any three consecutive integers is always divisible by $3! = 6$. So, the expression has a guaranteed factor of 6. * **Step 3: Analyze the remaining $n$.** We have an extra $n$ multiplying our sequence. * If $n$ is even, $n^2$ contains a factor of 4. * If $n$ is odd, then $(n-1)$ and $(n+1)$ are both even. Their product contains a factor of $2 \times 2 = 4$. * In all cases, the entire expression is guaranteed to be divisible by 4. * **Step 4: Combine the factors.** The expression is divisible by 3 (from the consecutive trio) and by 4. Therefore, it is always divisible by $3 \times 4 = 12$. ### Exam Strategy & Shortcut **The Plugin Method:** The quickest way to solve "always divisible by" questions is to plug in the smallest allowed value for the variable. The condition is "$n$ is a whole number greater than 1". The smallest valid number is $n = 2$. Substitute $n = 2$: $2^2(2^2 - 1) = 4(4 - 1) = 4 \times 3 = 12$. Look at the options: 12 is divisible by 12, but not by 8, 10, or 16. You have your answer instantly. ### Common Pitfall A fatal mistake is ignoring the domain constraint "$n$ is greater than 1" and plugging in $n = 1$. This results in $1(0) = 0$. Since 0 is technically divisible by all numbers, it eliminates zero options and wastes your time. ### Final Answer Therefore, the correct answer is **12**.