For any positive integer n3-n is divisible by 6 prove it
Answers
factorise it
n (n^2 - n)
n (n-1)(n+1)
(n-1), (n), (n+1)
they are three consecutive numbers,
as we know that there is a multiple of three withing any three consecutive number,
so one of them have to be divisible by 3
suppose n is a +ve number here,
it could be even , or odd
if it is even
then, one of the number in the expression will become odd
and
if it is odd
then, at least one of them would be even
i.e at least one number will be even and one will be odd here and at least one will be the factor of 3 .
that is all we need for a numbers to be divided by 6,
that number should have an
even number × three
then, only it wil ve completely divisible by 6.
the upper statements show that these requirements are fulfilled by
n (n-1)(n+1) which is n^3 -n
thus, proving that indeed +ve numbers in form of n^3 - n are divisible by 6.
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.