If n is a natural number, then the expression 6n² + 6n is always divisible by which of the following?

Introduction / Context:
This question focuses on divisibility and factorisation. You are given an expression 6n² + 6n where n is a natural number and asked which numbers always divide the value of this expression. Instead of plugging in many values of n, the correct strategy is to factor the expression and reason about the divisibility properties of the factors, especially for evenness and multiples of 3.

Given Data / Assumptions:

• n is a natural number (1, 2, 3, and so on).
• Expression: 6n² + 6n.
• We need to determine whether it is always divisible by 6, by 12, or by 18.

Concept / Approach:
Factor out common terms from the expression. The greatest common factor is 6n, so 6n² + 6n = 6n(n + 1). Because n and n + 1 are consecutive integers, their product has predictable divisibility properties. Specifically, among two consecutive integers one must be even, and among n and n + 1, at least one of them may be a multiple of 3 depending on n. By examining these two factors together with the constant factor 6, we can determine divisibility by 6, 12 and 18.

Step-by-Step Solution:
Factor the expression: 6n² + 6n = 6n(n + 1). Because n and n + 1 are consecutive integers, one of them is even. Thus n(n + 1) is always even and has at least one factor of 2. The factor 6 contributes 2 * 3. Therefore, 6n(n + 1) always contains at least 2 * 2 * 3 = 12 as a factor. So the expression is always divisible by 12. It is obviously always divisible by 6 as well because 6 is a factor of 12. To check 18, we need the expression to have at least 2 * 3² as factors. For n = 1, the expression is 6 * 1² + 6 * 1 = 6 + 6 = 12. 12 is not divisible by 18, so the expression is not always divisible by 18.

Verification / Alternative check:
Test a few values of n. For n = 1, the expression is 12, which is divisible by both 6 and 12 but not by 18. For n = 2, we get 6 * 4 + 6 * 2 = 24 + 12 = 36, which is divisible by 6, 12 and 18, but this is just a special case. For n = 3, 6 * 9 + 6 * 3 = 54 + 18 = 72, again divisible by 6 and 12, and also by 18. However, since n = 1 already shows that divisibility by 18 fails in at least one case, 18 cannot be an always divisor. The pattern of factorisation confirms that 6 and 12 are guaranteed divisors for all natural n.

Why Other Options Are Wrong:

• 6 only and 12 only: These options ignore the fact that if the expression is always divisible by 12, it is automatically always divisible by 6. So both 6 and 12 are valid always divisors.
• 18 only: The example n = 1 gives 12, which is not divisible by 18, so this is incorrect.
• Neither 6 nor 12: This contradicts the clear factorisation 6n(n + 1), where 6 is explicitly a factor and n(n + 1) is always an integer.

Common Pitfalls:
Some learners do not notice the factorisation and try to reason directly from the original quadratic expression, which is harder. Others test only a couple of values of n and draw incorrect general conclusions. Always factor when possible, and remember that consecutive integers have very useful divisibility properties. Look for guaranteed factors that hold for all allowed values of the variable, not just special cases.

Final Answer:
The expression 6n² + 6n is always divisible by both 6 and 12 for any natural number n.