Let three consecutive positive integers be n, n + 1 and n + 2.#Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.Therefore :n = 3p or 3p + 1 or 3p + 2, where p is some integer.If n = 3p, then n is divisible by 3.1.If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.2.If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.

Introduction:

In this proof, we will demonstrate that among any three consecutive integers, at least one of them is divisible by 3. This property can be observed by analyzing the patterns of integers and their remainders when divided by 3.

Proof:

1. Division by 3:
When an integer is divided by 3, there are only three possible remainders: 0, 1, or 2.

2. Analyzing consecutive integers:
Consider three consecutive integers: n, n+1, and n+2. We will analyze the possible remainders when each of these integers is divided by 3.

3. Case 1: n is divisible by 3 (remainder 0):
If n is divisible by 3, then n/3 has a remainder of 0. In this case, n itself is divisible by 3.

4. Case 2: n+1 is divisible by 3 (remainder 0):
If n+1 is divisible by 3, then (n+1)/3 has a remainder of 0. Rearranging the equation, we have n = 3k - 1, where k is an integer. In this case, n is divisible by 3.

5. Case 3: n+2 is divisible by 3 (remainder 0):
If n+2 is divisible by 3, then (n+2)/3 has a remainder of 0. Rearranging the equation, we have n = 3k - 2, where k is an integer. In this case, n is divisible by 3.

6. Conclusion:
From the above analysis, we can observe that in each of the three cases, one of the three consecutive integers (n, n+1, or n+2) is divisible by 3.

Summary:
Hence, we have proven that among any three consecutive integers, at least one of them is divisible by 3. This property is a result of the nature of division and the limited number of possible remainders (0, 1, and 2) when dividing an integer by 3.