Question

for any positive integer n,prove that n3-n is divisible by 6​

Answer 1

Answer:

Step-by-step explanation:

Let n be any positive integer and b=6. Using division algorithm on n and 6,there exists unique integers q and r such that

n=6q+r (0≤r≤6)

Therefore, r=1,2,3,4,5

Now, n=6q, 6q+1,6q+2,6q+3,6q+4,6q+5.

If n =6q

n³-n

=216q³-6q

=6(36q³-q)

Therefore, n³-n/6

Similarly taking n=6q+1,6q+2,6q+3,6q+4,6q+5 it can be prove that n³-n is divisible by 6

Hence, for any integer n, n³-n is divisible by 6.

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.