Question

Prove that the product of three consecutive numbers when the first number is even is divisible by 6.

Answer 1

Let three consecutive positive integers be, n, n + 1 and n + 2.were n is an even no
Now,
Case 1:
When a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, ⇒ n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.
If n = 3p + 2, ⇒ n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.
⇒ n (n + 1) (n + 2) is divisible by 3.
Case 2:
Similarly,since the numbers are starting with an even number.
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2.
⇒ n (n + 1) (n + 2) is divisible by 2.
Hence n (n + 1) (n + 2) is divisible by 2 and 3.
∴ n (n + 1) (n + 2) is divisible by 6.