SOLUTION: For any positive integer 'n' prove that 'n3-n' is divisible by 6.

Question 747469: For any positive integer 'n' prove that 'n3-n' is divisible by 6.

Answer by tommyt3rd(5050) (Show Source): You can put this solution on YOUR website!
not true:
n=1
n^3-n=0
Maybe you mean if n is 2 or greater?

now for

this product will include 3 consecutive numbers and that means at least one of them will be even and exactly one will be a multiple of 3. Therefore each will be a multiple of 6.

:)
RELATED QUESTIONS
Prove that {{{ n^3 + 5n }}} is divisible by 6, where n is any positive... (answered by ikleyn)

Use mathematical induction to prove the statement is true for all positive integers n. (answered by ikleyn)

Prove that for every positive integer n, n3 + n is... (answered by jim_thompson5910)

prove that for every positive integer n, n3 + n is even (answered by solver91311)

Prove by induction and through divisibility algorithm that 11^n - 6 is divisible by 5... (answered by ikleyn)

Prove that {{{ 2^n + 5^n }}} is divisible by 7, where n is any odd, positive... (answered by Alan3354)

prove by contradiction. for any integer n, n^2-2 is not divisible by... (answered by Edwin McCravy)

Prove that for any positive integer n, the value of the expression 3^(2n+2)-8n-9 is... (answered by richard1234)

Prove that n^5 − n is divisible by 5 for any natural... (answered by richard1234)