Guaranteed divisibility — The product of three consecutive even natural numbers is always divisible by what largest natural number? Justify using factor patterns.

Introduction / Context:
Problems about guaranteed divisibility rely on number patterns. For three consecutive even numbers, certain prime powers and factors must appear in their product regardless of where the sequence starts. The question asks for the largest natural number that always divides such a product.

Given Data / Assumptions:

• Let the consecutive even numbers be 2n, 2n + 2, 2n + 4 for some integer n.
• We analyze the product P = (2n)(2n + 2)(2n + 4).
• We seek the largest fixed divisor valid for all integers n.

Concept / Approach:
Factor out common elements and inspect guaranteed multiples. Note that P = 2 * 2 * 2 * n * (n + 1) * (n + 2) = 8 * n * (n + 1) * (n + 2). Among n, n + 1, and n + 2, there is always exactly one multiple of 3. Also, at least one of these three consecutive integers is even, contributing at least one extra factor 2 beyond the 8 already factored out. Therefore, the product is always divisible by 8 * 2 * 3 = 48.

Step-by-Step Solution:
Write P = 8 * n * (n + 1) * (n + 2).Among three consecutive integers, one is a multiple of 3 → factor 3 present.Among these three, at least one is even → at least one extra factor 2 beyond the initial 8.Therefore guaranteed factor = 8 * 2 * 3 = 48.

Verification / Alternative check:
Test small cases: (2, 4, 6) product = 48; (4, 6, 8) product = 192, which is 48 * 4; (6, 8, 10) product = 480 = 48 * 10. Each case is divisible by 48, supporting the general proof.

Why Other Options Are Wrong:

• 16 / 24: Too small; they always divide the product, but larger guaranteed factors exist.
• 96: Not always guaranteed (counterexample 2 * 4 * 6 = 48).
• 12: Also divides, but not the largest guaranteed divisor.

Common Pitfalls:
Overlooking the extra factor of 2 from n(n + 1)(n + 2); assuming a fixed multiple of 4 inside n(n + 1)(n + 2) when it is not guaranteed for all n.

Final Answer:
48