To prove that the product of three consecutive even numbers is always a multiple of 24, let us break this into logical steps.

1. Representation of Consecutive Even Numbers

Let the three consecutive even numbers be represented as: $n, \; n+2, \; n+4$ where $n$ is an even integer.

2. Expression for the Product

The product of these three numbers is: $P = n(n+2)(n+4)$

3. Multiple of 8

Since $n$ is even, we can write $n = 2k$, where $k$ is an integer. Substituting $n = 2k$ into the product: $P = (2k)(2k+2)(2k+4)$ Factor out 2 from each term: $P = 2 \cdot 2 \cdot 2 \cdot k(k+1)(k+2) = 8 \cdot k(k+1)(k+2)$ Thus, $P$ is always divisible by $8$.

4. Multiple of 3

Among $k$, $k+1$, and $k+2$, one of these numbers is always divisible by $3$. This is because any three consecutive integers contain exactly one multiple of $3$. Hence, $k(k+1)(k+2)$ is divisible by $3$.

5. Conclusion: Multiple of 24

Since $P = 8 \cdot k(k+1)(k+2)$ is divisible by both $8$ and $3$, it is divisible by $8 \times 3 = 24$.

Therefore, the product of three consecutive even numbers is always a multiple of $24$.

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Would you like further examples or clarifications? Here are some related questions to explore:

1. Can we extend this proof to three consecutive odd numbers for other divisors?
2. How would this proof change if we used consecutive integers instead of even numbers?
3. What are the divisors of the product of four consecutive even numbers?
4. How can we generalize this proof to $n$-consecutive numbers and their product?
5. Can we use modular arithmetic to prove the divisibility in this case?

Tip: When proving divisibility, always check both prime factors and their powers to verify the result fully!