Question 37

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

Solution:

To prove that one of every three consecutive positive integers is divisible by 3, we can follow these steps:

1. Define the Consecutive Integers:
Let’s denote three consecutive positive integers as $$n$$, $$n+1$$, and $$n+2$$, where $$n$$ is any positive integer.

2. Consider the Remainders When Divided by 3:
Any integer can have a remainder of 0, 1, or 2 when divided by 3. Therefore, we can categorize $$n$$ based on its remainder when divided by 3:
– Case 1: $$n\equiv 0\mathrm{mod}3$$
– Case 2: $$n\equiv 1\mathrm{mod}3$$
– Case 3: $$n\equiv 2\mathrm{mod}3$$

3. Analyze Each Case:
– Case 1: If $$n\equiv 0\mathrm{mod}3$$, then $$n$$ is divisible by 3.
– Case 2: If $$n\equiv 1\mathrm{mod}3$$, then $$n+1\equiv 2\mathrm{mod}3$$ and $$n+2\equiv 0\mathrm{mod}3$$. Thus, $$n+2$$ is divisible by 3.
– Case 3: If $$n\equiv 2\mathrm{mod}3$$, then $$n\equiv 2\mathrm{mod}3$$, $$n+1\equiv 0\mathrm{mod}3$$, and $$n+2\equiv 1\mathrm{mod}3$$. Thus, $$n+1$$ is divisible by 3.

4. Conclusion:
In all three cases, we have shown that at least one of the integers $$n$$, $$n+1$$, or $$n+2$$ is divisible by 3. Therefore, we can conclude that one of every three consecutive positive integers is divisible by 3.

5. Final Statement:
Hence, we have proved that one of every three consecutive positive integers is divisible by 3.