We are tasked with proving that the sum of any five consecutive integers is divisible by 5. The input note suggests using $n$ as the first integer and expressing the sum in a form that shows it's divisible by 5.

Let's break it down step by step.

Step-by-Step Proof:

1. Let the five consecutive integers be:
   $n, n+1, n+2, n+3, n+4$

2. Sum of these integers: $S = n + (n+1) + (n+2) + (n+3) + (n+4)$ Simplifying the sum: $S = n + n + 1 + n + 2 + n + 3 + n + 4$ $S = 5n + 10$

3. Factor out the common factor of 5: $S = 5(n + 2)$

4. Conclusion: Since the sum can be factored as $5(n + 2)$, it is clearly divisible by 5, for any integer value of $n$.

Thus, the sum of any five consecutive integers is always divisible by 5.

Input for the box:

The expression that goes into the box is: $n + 2$ The full expression becomes: $5(n + 2)$

Let me know if you'd like further details or clarifications.

Here are some related questions to explore this concept further:

1. What happens if we take the sum of 3 consecutive integers? Is it divisible by 3?
2. Can you generalize the sum of any odd number of consecutive integers?
3. What is the sum of 7 consecutive integers? Is it divisible by 7?
4. Can this method be used to prove divisibility for other types of number sets?
5. How would you prove the sum of consecutive even integers is divisible by 2?

Tip: Whenever working with consecutive integers, always look for patterns or relationships between the numbers to factor expressions easily.