Question

for any positive integer n use euclid's division lemma to prove that n3-n is divisible by 6​

Answer 1

Hola mate..

Here's your answer..

Let a = n3 – n

⇒ a = n × (n2 -1)

⇒ a = n × (n-1) × (n + 1) [Using (a2 –b2) = (a−b) × (a + b)]

⇒ a = (n-1) × n × (n + 1)

We know that

I If a number is completely divisible by 2 and 3, then it is also divisible by 6.

II If the sum of digits of any number is divisible by 3, then it is also divisible by 3.

III. If one of the factors of any number is an even number, then it is also divisible by 2.

Since, a = (n-1) × n × (n + 1)

Sum of the digits = n−1 + n + n + 1 = 3n which is a 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 be even.

Hence, by condition I the number n3-n is always divisible by 6, where n is any

positive integer.

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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.