Let's investigate whether the product of three consecutive even numbers is always a multiple of 24.

Step 1: Represent the numbers

Let the three consecutive even numbers be:

• $2n$
• $2n + 2$
• $2n + 4$

where $n$ is an integer.

Step 2: Expression for the product

The product of these three numbers is:

$P = 2n \times (2n + 2) \times (2n + 4)$

Step 3: Check for divisibility by 24

We need to determine if this product is always divisible by 24.

Divisibility by 8:

• Since all three numbers are even, at least one of them is divisible by 4, and another one is divisible by 2. This guarantees that the product of these three numbers includes at least $2 \times 2 \times 2 = 8$, which ensures divisibility by 8.

Divisibility by 3:

• Among any three consecutive integers (in this case, $2n, 2n+2, 2n+4$), one of them must be divisible by 3. This is a property of consecutive numbers.

Thus, the product $P = 2n \times (2n + 2) \times (2n + 4)$ is divisible by both 8 and 3, meaning it is divisible by $8 \times 3 = 24$.

Conclusion:

The product of three consecutive even numbers is always a multiple of 24 because it is divisible by both 8 and 3.

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Would you like further details or clarification on any part of this explanation?

Here are 5 related questions for further exploration:

1. Why is the product of any three consecutive integers divisible by 6?
2. How can you prove that the product of five consecutive integers is divisible by 120?
3. What is the role of the least common multiple (LCM) when considering divisibility?
4. How does the property of divisibility by 3 work in sets of consecutive numbers?
5. What are the divisibility rules for numbers like 16, 18, or 36?

Tip: When working with products of consecutive integers, it’s helpful to break the problem down by checking divisibility for small prime numbers (e.g., 2, 3, 5).