Is n(n-1)(2n-1) Always Divisible by 6?

[cragar]

Homework Statement

show that n(n-1)(2n-1) is always divisible by 6.

The Attempt at a Solution

I see how to get the factor of 2 if n is even then n is divisible by 2.
is n is odd then n-1 is even.
Here is my idea to get the factor of 3.
we know that n and n-1 are consecutive integers.
and we know that if n is divisible by 3 then so is 2n and the same goes for
n-1 but 2n-1 comes right before 2n, so we could some how make an argument as to why
n or n-1 or 2n-1 is divisible by 3. I need to think about it more.

 

[rollingstein]

If n or (n-1) are divisible by 3 then your problem is solved.

If not, n+1 surely is divisible by 3.

Hence

n+1 = 3k, where k is some integer

Hence

2n -1 = 2 (3k-1) - 1

2n-1=3(2k-1)

Hence (2n-1) has to be divisible by 3.

QED

 

[SteamKing]

It's not clear what is meant by 'always divisible by 6'.

By inspection, n(n-1)(2n-1) is divisible by 6 IFF n >= 2

 

[rs1n]

I would break this up into cases, some of which you already addressed. Consider the possible choices for n. The three cases are

n = 3k
n=3k+1
n=3k+2

where k is an integer. Consider

\frac{n(n-1)(2n-1)}{6}

What happens when you substitute each case in for n into the fraction above?

 

[rs1n]

It's not clear what is meant by 'always divisible by 6'.

By inspection, n(n-1)(2n-1) is divisible by 6 IFF n >= 2

I think the issue should actually be that n must itself be an integer. However, there is nothing wrong with n = 0, n = 1, (in both cases, the product is 0, which is divisible by 6), or even if n is a negative integer.