Q. 43.8( 48 Votes )

# For any positive integer, prove that divisible by 6.

Answer :

n^{3} – n = n(n^{2} – 1) = n(n – 1)(n + 1)

For a number to be divisible by 6, it should be divisible by 2 and 3 both,

**Divisibility by 3:**

n – 1, n and n + 1 are three consecutive whole numbers.

By Euclid’s division lemma

n + 1 = 3q + r, for some integer k and r < 3

As, r < 3 possible values of ‘r’ are 0, 1 and 2.

If r = 0

n + 1 = 3q

⇒ n + 1 is divisible by 3

⇒ n(n – 1)(n + 1) is divisible by 3

⇒ (n^{3} – n) is divisible by 3

If r = 1

⇒ n + 1 = 3q + 1

⇒ n = 3q

⇒ n is divisible by 3

⇒ n(n – 1)(n + 1) is divisible by 3

⇒ (n^{3} – n) is divisible by 3

If r = 2

⇒ n + 1 = 3q + 2

⇒ n + 1 – 2 = 3q

⇒ n – 1 = 3q

⇒ n - 1 is divisible by 3

⇒ n(n – 1)(n + 1) is divisible by 3

⇒ (n^{3} – n) is divisible by 3

**Divisibility by 2:**

If n is even

Clearly, n(n – 1)(n + 1) is divisible by 2

If n is odd

⇒ n + 1 is even

⇒ n + 1 is divisible by 2

⇒ n(n – 1)(n + 1) is divisible by 2

Hence, for any value of n, n^{3} - n is divisible by 2 and 3 both, therefore n^{3} – n is divisible by 6.

Rate this question :

Find the value of p for which the roots of the equation px (x-2) + 6 = 0, are equal.

Mathematics - Board PapersHow many two-digit numbers are divisible by 3?

Mathematics - Board PapersShow that any positive odd integer is of the form (4q + 1) or (4q + 3), where q is a positive integer.

RS Aggarwal - MathematicsWhat is a lemma?

RD Sharma - MathematicsEvery odd integer is of the form 2m - 1, where m is an integer (True/False).

RD Sharma - MathematicsEvery even integer is of the form 2m, where m is an integer (True/False).

RD Sharma - Mathematics