Question

For any positive integer n, prove that n3-n is divisible by6

Answer 1

solution :

To prove : any positive integer n cube minus n is divisible by 6

By using euclid's division Lemma,

Let n be any positive integer such that

n = 6q+r , where b= 6 and r is greater than or equal to zero less than 6

=> r=0,1,2,3 ... 5.

If , r=0

n=6q +0=> 6q

n^3 -n (substitute value)

(6q)^3 -6q

216q^3 - 6q

=> 6(36q^2 - q)

Thus , it is divisible by 6

Similarly you can pro for all value of ' r'

Thank you

Answer 2

Answer:

n3 – n = n(n2 – 1) = n(n+1)(n – 1) = (n – 1)n(n+1) = product of three consecutive positive integers.

Now, we have to show that the product of three consecutive positive integers is divisible by 6.

We know that any positive integer n is of the form 3q, 3q + 1 or 3q + 2 for some positive integer q.

Now three consecutive positive integers are n, n + 1, n + 2.

Case I. If n = 3q.

n(n + 1) (n + 2) = 3q(3q + 1) (3q + 2)

But we know that the product of two consecutive integers is an even integer.

∴ (3q + 1) (3q + 2) is an even integer, say 2r.

⟹ n(n + 1) (n + 2) = 3q × 2r = 6qr, which is divisible by 6.

Case II. If n = 3n + 1.

∴ n(n + 1) (n + 2) = (3q + 1) (3q + 2) (3q + 3)

= (even number say 2r) (3) (q + 1)

= 6r (q + 1),

which is divisible by 6.

Case III. If n = 3q + 2.

∴ n(n + 1) (n + 2) = (3q + 2) (3q + 3) (3q + 4)

= multiple of 6 for every q

= 6r (say),

which is divisible by 6.

Hence, the product of three consecutive integers is divisible by 6.