In this problem, we prove a basic property of consecutive positive integers. We will show that among any three consecutive positive integers, one integer is always divisible by 3.

Question

Prove that one of every three consecutive positive integers is divisible by 3.

Solution

Let the three consecutive positive integers be

$$n,\; n+1,\; n+2$$

where n is a positive integer.

Every positive integer can be written in one of the following three forms:

$$3q,\; 3q+1,\; 3q+2$$

where q is an integer.

Now, we consider the possible cases.

Case 1: $n = 3q$

Then n is divisible by 3.

Case 2: $n = 3q + 1$

Then,

$$n + 2 = 3q + 3 = 3(q + 1)$$

So, $n + 2$ is divisible by 3.

Case 3: $n = 3q + 2$

Then,

$$n + 1 = 3q + 3 = 3(q + 1)$$

So, $n + 1$ is divisible by 3.

Thus, in all possible cases, one of the three consecutive integers n, n + 1, or n + 2 is divisible by 3.

Conclusion

Therefore, one of every three consecutive positive integers is always divisible by 3.

$$\boxed{\text{Hence proved.}}$$

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