Question

For any positive ineger n, prove that n^3-n is divisible by 6

Answer 1

Let a = n3 - n ⇒ a = n.(n2 - 1)
⇒ a = n.(n-1).(n+1)  [∵(a2 - b2) = (a-b)(a+b)]
⇒ a = (n-1).n.(n+1)  ...(i)
We know that,
1. If a number is completely divisible by 2 and 3, then it is also divisible by 6.
2. If the sum of digits of any number is divisible by 3, then it is also divisible by 3.
3. If one of the factor of any number is an even number,then it is also divisible by 2.
∴ a = (n-1).n.(n+1)  [from Eq.(i)]
Now,sum of the digits = n-1+n+n+1 = 3n
= multiple of 3,where n is any positive integer,
and (n-1)-n-(n+1) will always be even, as one out of (n-1) or n or (n+1) must of even. Since,
conditions II and III is completely satify the Eq.(i).
Hence, by condition I the number n3 - n is always divisible by 6, where n is any positive integer.
Hence proved.       

Answer 2

Step-by-step explanation:

▶ 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.

Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.

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

If n = 2q, then n is divisible by 2.

If n = 2q + 1, then n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.

So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.

⇒ n (n – 1) (n + 1) is divisible by 2.

Since, n (n – 1) (n + 1) is divisible by 2 and 3.

∴ n ( n - 1 ) ( n + 1 ) = n³ - n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6)

✔✔ Hence, it is solved ✅✅.