How to Prove That Only One of Three Consecutive Integers Is Divisible by 3

Divisibility properties of integers are fundamental topics in number theory. One such interesting property is determining how divisibility by 3 behaves in sequences of consecutive numbers. Specifically, if we have three consecutive integers, n, n1, and n2, we want to show that exactly one of them is divisible by 3. This article explores different methods to prove this property, using simple and advanced mathematical concepts such as Euclid's Division Lemma and the product of consecutive integers.

Using General Form and Cases

First, let's consider the general form of an integer. An integer n can be expressed as one of the following forms:

n 3k, for some integer k n 3k 1, for some integer k n 3k 2, for some integer k

Based on these forms, we can examine the divisibility of the consecutive integers n1 and n2 relative to n. Here’s a detailed breakdown:

Case 1: n 3k

In this case, n is divisible by 3. Therefore:

n1 3k - 1 (not divisible by 3) n2 3k 1 (divisible by 3, but since n is already divisible, it’s the exception)

Case 2: n 3k 1

In this case, n is not divisible by 3. Therefore:

n1 3k (divisible by 3) n2 3k 2 (not divisible by 3)

Case 3: n 3k 2

In this case, n is not divisible by 3. Therefore:

n1 3k 1 (not divisible by 3) n2 3k 3 (divisible by 3)

By examining all these cases, we can assert that, for any integer n, exactly one of the three consecutive integers n, n1, and n2 is divisible by 3. This proof covers all possible cases and ensures that the statement holds true.

Using Euclid's Division Lemma

Euclid's Division Lemma states that for any integers a and b, with b > 0, there exist unique integers q and r such that a bq r, where 0 ≤ r If n is divisible by 3, say n 3m, then n1 3m - 1 is not divisible by 3, and n2 3m 1 is not divisible by 3. If n1 is divisible by 3, say n1 3m, then n 3m 1 is not divisible by 3, and n2 3m 2 is not divisible by 3. If n2 is divisible by 3, say n2 3m, then n 3m - 2 is not divisible by 3, and n1 3m - 1 is not divisible by 3.

Thus, we conclude that only one of the numbers n, n1, or n2 is divisible by 3. This method provides a structured approach using division and congruence properties to validate the statement.

Using the Product of Consecutive Integers

Another interesting approach involves the product of consecutive integers. The product of r consecutive integers is divisible by r!. For three consecutive integers n, n1, and n2, their product is divisible by 3! (which is 6). This means that the product of any three consecutive integers is always divisible by 6. Since 6 is composed of 3 and 2, one of the integers must be divisible by 3. This is a more elegant and simpler way to understand the divisibility property without breaking down the integers into different cases.

Additional Insights

1. **Remainder Analysis** - If an integer n leaves a remainder of 1 when divided by 3, then n - 1 is divisible by 3. Therefore, n1 is divisible by 3 in this case. Similarly, if n leaves a remainder of 2, then n - 2 is divisible by 3, and n2 is divisible by 3.

2. **Digital Sum Method** - The digital sum of a number, when reduced modulo 3, can help determine if the number is divisible by 3. Any number that is not divisible by 3 can be made divisible by 3 by adding or subtracting 1. This method provides an intuitive way to understand the divisibility property in practical scenarios.

Scientific Explanation

Integers that are not divisible by 3 can only belong to one of two categories:

Divisible by 2 but not 3 (hence even) Not divisible by 2 or 3 (hence odd)

By examining the divisibility properties and the relationships between integers, we can conclude that exactly one of three consecutive integers is divisible by 3. This property is robust and can be proven through various mathematical methods, each offering deeper insights into the nature of integers and their divisibility.