Question

Hii user.....!!!
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1) For any positive integer n, prove that n3 - n is divisible by 6

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Answer 1

n3 - n = n (n2 - 1) = n (n - 1) (n + 1) 

Whenever 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, then n – 1 = 3p + 1 –1 = 3p is divisible by 3.
If n = 3p + 2, then 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 + 1 is always divisible by 3.
⇒ n (n – 1) (n + 1) is divisible by 3.

Answer 2

Hello friend,

Positive Integer n³ - n can be written as n(n² - 1) = (n - 1)(n)(n + 1)

These are three consecutive numbers.

One of every 3 consecutive integers is divisible by 3.

At least one of every 3 consecutive integers is divisible by 2. 


2 and 3 are relatively prime so the product of the 3 integers is divisible by 2 * 3 = 6.

So, Every three consecutive numbers are divisible by 6.

Thank you.