Prove n(n+1)(n+2) is divisible by 6

[transmini]

Homework Statement

[/B]
Prove n(n+1)(n+2) is divisible by 6 for all integers n. (WITHOUT using induction, we have yet to get to induction so I figure it would be wise to do this without it.)

Homework Equations

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The section we were given this under primarily talks about the quotient remainder theorem (n = dq+r) though I couldn't figure out how to apply this either.

The Attempt at a Solution

Honestly not sure where to even begin with this one. Nothing even slightly similar has been covered as to give a slight intuition on how to begin this.

I've managed to prove that it's divisible by 2 for all even and odd integers by using n = 2k and n = 2k+1 respectively. That still leaves proving it is divisible by 3 for even and odds though, which is where I get stuck doing that method.

 

[Mark44]

Homework Statement

[/B]
Prove n(n+1)(n+2) is divisible by 6 for all integers n. (WITHOUT using induction, we have yet to get to induction so I figure it would be wise to do this without it.)

Homework Equations

[/B]
The section we were given this under primarily talks about the quotient remainder theorem (n = dq+r) though I couldn't figure out how to apply this either.

The Attempt at a Solution

Honestly not sure where to even begin with this one. Nothing even slightly similar has been covered as to give a slight intuition on how to begin this.

I've managed to prove that it's divisible by 2 for all even and odd integers by using n = 2k and n = 2k+1 respectively. That still leaves proving it is divisible by 3 for even and odds though, which is where I get stuck doing that method.

Any time you have three successive integers as you do here, one of them will be divisible by 3. Of course, this is what you have to prove.

One technique is to break things into three cases, similar to what you did for testing divisibility by 2 with two cases.
Case 1: n = 3k
Case 2: n = 3k + 1
Case 3: n = 3k + 2
What can you conclude from each case? Do all three cases lead you to conclude that n(n + 1)(n + 2) is divisible by 3? Don't forget, though, that one of the factors has to be even so that the overall product has a factor of 6.

 

[transmini]

Any time you have three successive integers as you do here, one of them will be divisible by 3. Of course, this is what you have to prove.

One technique is to break things into three cases, similar to what you did for testing divisibility by 2 with two cases.
Case 1: n = 3k
Case 2: n = 3k + 1
Case 3: n = 3k + 2
What can you conclude from each case? Do all three cases lead you to conclude that n(n + 1)(n + 2) is divisible by 3? Don't forget, though, that one of the factors has to be even so that the overall product has a factor of 6.

So would the proper way to go about Case 1 then be:
n = 3k
3k(3k+1)(3k+2)
Let m = k(3k+1)(3k+2)
3m
3m is divisible by 3, therefore n(n+1)(n+2) is divisble by 3 (only in case 1 currently)
?

And why is it that we set n = 3k and so on, instead of the whole n(n+1)(n+2) = 3k and so on?

 

[Mark44]

So would the proper way to go about Case 1 then be:
n = 3k

If n = 3k, then clearly n(n + 1)(n + 2) is divisible by 3 (since n is divisible by 3). You don't need to say anything else in this case, other than showing that n(n + 1)(n + 2) is even.

3k(3k+1)(3k+2)
Let m = k(3k+1)(3k+2)
3m
3m is divisible by 3, therefore n(n+1)(n+2) is divisble by 3 (only in case 1 currently)
?

And why is it that we set n = 3k and so on, instead of the whole n(n+1)(n+2) = 3k and so on?

If you write this equation, you are automatically assuming that n(n + 1)(n + 2) is divisible by 3. You can't assume this -- you need to prove it.

 

[transmini]

If you write this equation, you are automatically assuming that n(n + 1)(n + 2) is divisible by 3. You can't assume this -- you need to prove it.

Okay I believe that I understand that and it makes sense. Thanks for the quick replies

 

[epenguin]

Counting numbers one by one every how often do you find one divisible by 3? If the first one (n) isn't, how far do you have to go at most before you find one that is?