Question

Prove that n3-n is divisible by 3 for any
natural number
n.​

Answer 1

Answer:

hence proved

Step-by-step explanation:

Hint – Simplify the given equation. Check for the divisibility of the equation with 3 and 2 as 6 = 2 × 3.

Complete step by step answer:

Simplify n3–n

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)

Note: A number is divisible by 6 if it is divisible by both 2 and 3. I.e. if a number is divisible by two distinct numbers m and n, then that number should be divisible by m × n, which is a product of both numbers.