Proving Nn1n2 is Always Divisible by 6 Using Mathematical Induction

When dealing with the expression Nn1n2, it is crucial to understand the properties of consecutive integers and their divisibility. Specifically, we aim to prove that this expression is always divisible by 6 for any integer N. This article will delve into both the mathematical induction proof and a more general claim that can be used to establish this result.

Introduction to the Problem

In any set of three consecutive integers, one of them is divisible by 2 and another is divisible by 3. Since 6 is the product of 2 and 3, the product of three consecutive integers is always divisible by 6. We will apply this concept to the expression Nn1n2.

Mathematical Induction

To prove that Nn1n2 is always divisible by 6 for any integer N, we will use mathematical induction. The base case is straightforward, and then we will use the inductive step to show that the statement holds for all integers.

Base Case (n1)

For n1, we have 1 * (1 1) * (1 2) 1 * 2 * 3. Clearly, this product is divisible by 6.

Inductive Step

Assume that the statement is true for some integer k, i.e., kk1k2 is divisible by 6. We need to show that the same is true for (k 1)(k 2)(k 3).

Expression Analysis

Assume the expression for k can be written as:

[ k^2(k 1)(k 2) 6p quad text{where } p in mathbb{Z} ]

Expanding and rearranging, we get:

[ k^3 3k^2 2k 6p ]

Next, consider the expression for (k 1)k(k 2):

[ (k 1)k(k 2) k^3 3k^2 2k ]

Since we know k^3 3k^2 2k 6p, it follows that (k 1)k(k 2) is also divisible by 6.

A More General Claim

A more useful and general claim is that every product of m consecutive integers is divisible by m. Here’s the proof:

Proof: Every Product of m Consecutive Integers is Divisible by m

Let the first of the consecutive integers, denoted as n, have a remainder of k in {0, 1, 2, ldots, m-1} when divided by m. If k0, then n is divisible by m, and thus any product of integers including n is also divisible by m.

Otherwise, k eq 0. Then n-k is divisible by m, and so is n-km. Since n-km is one of the consecutive integers in the product, the product must be divisible by m.

Application to Nn1n2

The claim that every product of m consecutive integers is divisible by m directly applies to the case of three consecutive integers. Therefore, both n(n 1) and (n-1)n(n 1) are divisible by both 2 and 3, making their product divisible by 6.

Final Expression

As shown, we can express the given expression as:

[ Nn(n 1) Nn(n-1)(n 1) n(n-1)n(n 1) n(n-1) ]

Since both terms on the right-hand side are divisible by 6, their sum is also divisible by 6.

Conclusion

In summary, through both mathematical induction and the more general claim, we have proven that the expression Nn1n2 is always divisible by 6. This result is significant in the context of divisibility and provides a thorough understanding of the properties of consecutive integers.