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.