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