Show that product of any three consecutive integer is divisible by 6 

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.

Discover how to prove that the product of any three consecutive integers is divisible by 6. This mathematical problem explores number theory and divisibility concepts, demonstrating the step-by-step reasoning behind this fundamental mathematical property. Suitable for students in Grades 7-9.

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