The number $6n^2 + 6n$ for natural number $n$ is always divisible by

### Concept & Logic To find the universal divisors of an algebraic expression, factor it completely to reveal its core components. The product of any two consecutive integers is always an even number (a multiple of 2). ### Step-by-Step Solution * **Given:** The algebraic expression is $6n^2 + 6n$, where $n$ is a natural number ($n \ge 1$). * **Step 1: Factor the expression.** Pull out the greatest common factor, which is $6n$. * $6n^2 + 6n = 6n(n + 1)$ * **Step 2: Analyze the components.** We have a constant $6$ multiplied by $n(n + 1)$. * **Step 3: Apply number properties.** For any natural number $n$, the numbers $n$ and $(n + 1)$ are consecutive integers. Exactly one of them must be even. Therefore, their product $n(n + 1)$ is always a multiple of 2. * Let $n(n + 1) = 2k$ (where $k$ is an integer). * **Step 4: Substitute back.** * $6n(n + 1) = 6(2k) = 12k$. * **Conclusion:** The expression is always a multiple of 12. If a number is divisible by 12, it is fundamentally also divisible by its factors, including 6. ### Exam Strategy & Shortcut **The Plugin Method:** Test the first two natural numbers to find the common divisors rapidly. Test $n = 1$: $6(1)^2 + 6(1) = 6 + 6 = 12$. * 12 is divisible by 6 and 12, but NOT 18. (Eliminates option d). Test $n = 2$: $6(2)^2 + 6(2) = 24 + 12 = 36$. * 36 is divisible by both 6 and 12. Since it works for both 6 and 12 across different test cases, option (b) is the robust answer. ### Common Pitfall The word "only" in the options is a trap. A student might factor out the 6, see $6 \times n(n+1)$, and immediately select "6 only" without realizing that the variable portion $n(n+1)$ guarantees an additional factor of 2, bumping the true universal divisor up to 12. ### Final Answer Therefore, the correct answer is **6 and 12**.