Answer to Question #273407 in Discrete Mathematics for haji

Question

Prove that the product of any three consecutive integers is a multiple of 3. 

Answer 1

Let the integers be $n, n + 1$ and $n+2, n\in \Z.$

If $n$ is divisible by $3,$ then the product $n(n + 1)(n + 2)$ is divisible by $3.$

If $n$ is not divisible by $3,$ then either $n = 3l + 1, l\in \Z$ or $n = 3m + 2,$

$m\in \Z.$

By the division algorithm

In the first case, $n + 2 = 3l + 3 = 3(l + 1), l\in \Z$ is divisible by 3 and therefore the product

$$n(n + 1)(n + 2) = 3(3l+1)(3l + 2)(l+1)$$

is divisible by 3.

In the second case, $n + 1 = 3m +3 = 3(m + 1), m\in \Z$ is divisible by 3 and therefore the product

$$n(n + 1)(n + 2) = 3(3m+2)(m+1)(3m+4)$$

is divisible by 3.

This shows that the product of any three consecutive integers is a multiple of $3.$