To show that the product of any three consecutive integers is divisible by 6, let's consider three consecutive integers. We can denote them as $n$, $n+1$, and $n+2$, where $n$ is any integer.

Step 1: Understanding the product

The product of these three consecutive integers is: $n(n+1)(n+2)$

Step 2: Divisibility by 2

• Among any three consecutive integers, one of them must be even.
• Since an even number is divisible by 2, the product $n(n+1)(n+2)$ is divisible by 2.

Step 3: Divisibility by 3

• Among any three consecutive integers, one of them must be divisible by 3.
• This is because every third integer is divisible by 3.

Step 4: Divisibility by 6

• Since the product $n(n+1)(n+2)$ is divisible by both 2 and 3, it is divisible by their least common multiple, which is 6.

Conclusion

Therefore, the product of any three consecutive integers $n(n+1)(n+2)$ is divisible by 6.

Would you like more details on this proof or have any questions? Here are some related questions to consider:

1. What are the properties of divisibility by other numbers like 4 or 5?
2. How can we generalize this proof to the product of four consecutive integers?
3. What happens if we take the product of two consecutive integers?
4. Can we apply a similar approach to prove divisibility by 12 for four consecutive integers?
5. How does this result connect to factorials and their divisibility properties?

Tip: When proving divisibility, consider the properties of smaller factors (like 2 and 3) and how they combine.