16.
For any positive integer n, use Euclid's division lemma to prove that
n3-n is divisible by 6.
Answers
a=n3-n a=n*(n2-1)
a=n*(n-1)*(n+1) [a2-b2=(a+b) (a-b)
a=(n-1)*n*(n+1)
We know that
¹ if a number is completely divisible by 2 and 3 it is also divisible by 6
² if the sum of any digit is divisible by 3 then number is also divisible by 3
³ is one of the factor of any number is any number then it is also divisible by 2
a=(n-1)*n*(n+1)
=n-1+n+n+1=3n
Multiple of 3 where n is a 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 since the condition ² and ³ are satisfied
Hence by conditions ¹ n3-n is 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.