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

Introduction:

To prove that one of every three consecutive positive integers is divisible by 3, we will use the concept of remainders and the properties of divisibility.

Proof:

Let's assume three consecutive positive integers: x, x+1, and x+2.

Divisibility by 3:

To determine if a number is divisible by 3, we can check if its remainder when divided by 3 is 0.

Case 1: x is divisible by 3
If x is divisible by 3, then the remainder of x divided by 3 is 0. In this case, x is already a multiple of 3.

Case 2: x is not divisible by 3
If x is not divisible by 3, then the remainder of x divided by 3 is either 1 or 2.

Case 2.1: Remainder of x divided by 3 is 1
If the remainder of x divided by 3 is 1, then x+1 has a remainder of 2 when divided by 3. Similarly, x+2 has a remainder of 0 when divided by 3. Therefore, in this case, x+2 is divisible by 3.

Case 2.2: Remainder of x divided by 3 is 2
If the remainder of x divided by 3 is 2, then x+1 has a remainder of 0 when divided by 3. Similarly, x+2 has a remainder of 1 when divided by 3. Therefore, in this case, x+1 is divisible by 3.

Conclusion:

In all possible cases, whether x is divisible by 3 or not, one of the three consecutive positive integers, namely x, x+1, and x+2, will always be divisible by 3. Hence, we have proved that one of every three consecutive positive integers is divisible by 3.