For any positive integer n, prove that n^3- n divisible by 6.
no spamming
answer in an explained manner
Answers
Answered by
6
Hii There!!
n3 - n = n (n2 - 1) = n (n - 1) (n + 1)
◆Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
=) n = 3p or 3p + 1 or 3p + 2, where p is some integer.
●If n = 3p, then n is divisible by 3.
●If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3
●If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3
So, we can say that one of the numbers among=)
n, n – 1 and n + 1 is always divisible by 3.
=)n (n – 1) (n + 1) is divisible by 3.
Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.
=) n = 2q or 2q + 1, where q is some integer.
●If n = 2q, then n is divisible by 2.
●If n = 2q + 1, so, n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2....
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.
=) n (n – 1) (n + 1) is divisible by 2.
Since, n (n – 1) (n + 1) is divisible by 2 and 3.
=) (n-1) (n+1) = n3 - n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6) .
HOPE IT HELPS..
MARK AS BRAINLIEST..
n3 - n = n (n2 - 1) = n (n - 1) (n + 1)
◆Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
=) n = 3p or 3p + 1 or 3p + 2, where p is some integer.
●If n = 3p, then n is divisible by 3.
●If n = 3p + 1, then n – 1 = 3p + 1 –1 = 3p is divisible by 3
●If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3
So, we can say that one of the numbers among=)
n, n – 1 and n + 1 is always divisible by 3.
=)n (n – 1) (n + 1) is divisible by 3.
Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.
=) n = 2q or 2q + 1, where q is some integer.
●If n = 2q, then n is divisible by 2.
●If n = 2q + 1, so, n – 1 = 2q + 1 – 1 = 2q is divisible by 2 and n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2....
So, we can say that one of the numbers among n, n – 1 and n + 1 is always divisible by 2.
=) n (n – 1) (n + 1) is divisible by 2.
Since, n (n – 1) (n + 1) is divisible by 2 and 3.
=) (n-1) (n+1) = n3 - n is divisible by 6.( If a number is divisible by both 2 and 3 , then it is divisible by 6) .
HOPE IT HELPS..
MARK AS BRAINLIEST..
Answered by
0
The condition for any number to be divisible by 6 is that the number must be individually divisible by 3 and 2.
Check whether n³ −n is divisible by 3.
n³ −n=n(n+1)(n−1)
When a number is divided by 3 then by the remainder theorem, the remainder obtained is either 0 or 1 or 2.
n=3p or n=3p+1 or n=3p+2, where p is some integer.
If n=3p, then the number is divisible by 3.
If n=3p+1, then n−1=3p+1−1=3p. The number is divisible by 3.
If n=3p+2, then n+1=3p+2+1=3(p+1). The number is divisible by 3.
So, any number in the form of n³ −n=n(n+1)(n−1) is divisible by 3.
Check whether n³ −n is divisible by 2.
When a number is divided by 2, the remainder obtained is either 0 or 1 by the remainder theorem.
n=2p or n=2p+1, where p is some integer.
If n=2p, then the number is divisible by 2.
If n=2p+1 then n−1=2p+1−1=2p. The number is divisible by 2.
So, any number in the form of n³ −n=n(n+1)(n−1) is divisible by 2.
Since, the given number n³−n=n(n+1)(n−1) is divisible by both 3 and 2. Therefore, according to the divisibility rule of 6, the given number is divisible by 6.
Hence, n³ − n=n(n+1)(n−1) is divisible by 6.
Similar questions