Prove that the product of any three consecutive natural numbers is divisible by 6

Let x denote the smallest of the given three consecutive natural numbers. Then the other two numbers will be x +1 and x + 2.
Let us prove that one of these numbers is necessarily divisible by two.
If the number x is not divisible by 2, then it gives, when divided by 2, the remainder of 1 and the number x can be written as 2k + 1, where k is some integer. In this case, the number x +1 will be equal to 2k + 1 + 1 = 2k + 2 and, therefore, the number x +1 will be divisible by 2.
So, we have shown that one of the numbers x and x +1 will necessarily be divisible by 2.
Let us prove that one of the given three consecutive natural numbers is necessarily divisible by three.
If the number x is not divisible by 3, then it gives a remainder of 1 or a remainder of 2 when divided by 3.
If the number x divided by 3 gives a remainder of 1, then the number x can be represented as 3k + 1, where k is some integer. In this case, the number x + 2 will be equal to 3k + 1 + 2 = 3k + 3 and, therefore, the number x + 2 will be divisible by 3.
If the number x when divided by 3 gives a remainder of 2, then the number x can be represented as 3k + 2, where k is some integer. In this case, the number x + 1 will be equal to 3k + 2 + 1 = 3k + 3 and, therefore, the number x + 1 will be divisible by 3.
So, we have shown that one of the numbers x, x +1 and x +2 will necessarily be divisible by 3. In addition, one of the numbers x and x +1 will necessarily be divisible by 2.
Therefore, the product x * (x +1) * (x +2) will necessarily be divisible by 2 * 3 = 6.