The product of three consecutive integers is divisible by 6. That is, for an integer n, the product n(n+1)(n+1) is always divisible by 6.

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## Prove that the Product of 3 consecutive integers is divisible by 6

By division algorithm, every integer when divided by 3 leaves the remainder 0, 1, or 2. So an integer is always one of the forms:

3k, 3k+1, 3k+2

for some integer k. Now, observe that

- If n=3k, then 3 divides n
- If n=3k+1, then 3 divides n+2
- If n=3k+2, then 3 divides n+1

Thus, for any integer n, the product n(n+1)(n+2) is divisible 3.

On the other hand, 2 divides the product of two consecutive numbers. Hence, 2 divides n(n+1)(n+2).

As both 2 and 3 divides n(n+1)(n+2), we conclude that n(n+1)(n+2) is divisible by 2×3=6.

**Have You Read These?**

Square of an odd integer is of the form 8n+1

## Example

Take n=4.

Then n(n+1)(n+2) = 4×5×6 is divisible by 6.

## FAQs

**Q1: The product of any three consecutive natural number is divisible by 6. True or False?**

Answer: As the product of any 3 consecutive natural number is divisible by 6, the given statement is TRUE.

This article is written by Dr. T. Mandal, Ph.D in Mathematics. On Mathstoon.com you will find Maths from very basic level to advanced level. Thanks for visiting.