Question

prove that the product of the three consecutive integers is divisible by 6.​

Answer 1

considering 4,5,6.

multiplying....

we get 120....

its divisible by 6...

now we take....

x, x+1 ,x+2

mltiplying we get...

a product which is divisible by 6..

so product of the three consecutive integers is divisible by 6.

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Answer 2

Let us three consecutive  integers be, x, x + 1 and x + 2.

Whenever a number is divided by 3 the remainder obtained is either 0 or 1 or 2.

let x = 3p or 3p + 1 or 3p + 2, where p is some integer.

If x = 3p, then n is divisible by 3.

If x = 3p + 1, then x + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.

If x = 3p + 2, then x + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.

So that x, x + 1 and x + 2 is always divisible by 3.

⇒ x (x + 1) (x + 2) is divisible by 3.

 

Similarly, whenever a number is divided 2 we will get the remainder is 0 or 1.

∴ x = 2q or 2q + 1, where q is some integer.

If x = 2q, then x and x + 2 = 2q + 2 = 2(q + 1) are divisible by 2.

If x = 2q + 1, then x + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.

So that x, x + 1 and x + 2 is always divisible by 2.

⇒ x (x + 1) (x + 2) is divisible by 2.

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Hope it helps!!

 

But n (n + 1) (n + 2) is divisible by 2 and 3.

 

∴ n (n + 1) (n + 2) is divisible by 6