If n is a natural number, the expression 6n^2 + 6n is always divisible by which of the following?

Introduction:
This question tests your understanding of divisibility properties and algebraic factorization. You must determine which fixed divisor always divides the expression 6n^2 + 6n for every natural number n.

Given Data / Assumptions:

• n is a natural number (1, 2, 3, ...).
• Expression: 6n^2 + 6n.
• We need to find which of the given numbers (6, 12, 18) always divides this expression.

Concept / Approach:
We factor the given expression and then analyze its factors to see what fixed integers it is always divisible by. We also recall that the product of two consecutive integers is always even, which will be helpful.

Step-by-Step Solution:
Step 1: Factor the expression 6n^2 + 6n. 6n^2 + 6n = 6n(n + 1). Step 2: Analyze factors. We have factors 6, n, and (n + 1).
Note that n and n + 1 are consecutive integers, so one of them must be even. That means the product n(n + 1) is always even, i.e., divisible by 2.
Step 3: Combine the factors of 6 and the even nature of n(n + 1). The factor 6 can be written as: 6 = 2 * 3. Since n(n + 1) is even, we can write: n(n + 1) = 2 * k for some integer k. Then: 6n(n + 1) = (2 * 3) * (2 * k) = 12 * 3 * k. So the expression is always divisible by 12.

Verification / Alternative Check:
Try some values of n:
For n = 1: 6 * 1^2 + 6 * 1 = 6 + 6 = 12, divisible by 12. For n = 2: 6 * 4 + 12 = 24 + 12 = 36, divisible by 12. For n = 3: 6 * 9 + 18 = 54 + 18 = 72, divisible by 12. In each test, divisibility by 12 is confirmed.

Why Other Options Are Wrong:
"6 only" is incorrect because the expression is always divisible by 12, which is stronger than divisibility by 6 alone.
"6 and 12 both" suggests 18 might also always divide it, but that is not guaranteed. For example, for n = 1, the value is 12, which is not divisible by 18.
"18 only" is incorrect because 18 is not always a divisor (again, 12 is not divisible by 18).

Common Pitfalls:
Students sometimes stop after factoring out 6 and conclude the answer is 6, without using the key fact that the product of consecutive integers n(n + 1) is always even. Missing this leads to an underestimation of the largest guaranteed divisor.

Final Answer:
The expression 6n^2 + 6n is always divisible by 12 only.