# Proof a Number is Divisible by 9

1. Feb 5, 2012

### hammonjj

Show that n^3(n+1)^3(n+2)^3 is divisible by 9.

The claim is obvious, so I tried to solve by induction, but I don't see how to simplify it so the two side are equal. You end up with:

(n+1)^3(n+2)^3(n+3)^3

2. Feb 5, 2012

### Curious3141

If f(k+1) - f(k) is a multiple of 9, you've done the inductive step.

So find the difference between those. To help you, use this nifty identity: $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$. Don't expand everything into a big mess. The first term (the a-b part) can be manipulated quite easily into a factor of 3. The second term (the a^2 + ab + b^2 part) can be factorised into 3*something after a bit of manipulation. So the whole expression becomes 9*something, and you're done (apart from showing it for n=1, which is trivial).

3. Feb 5, 2012

### Curious3141

In fact, you don't even need induction. What can you say about three consecutive integers n, (n+1), (n+2)? Would one of them necessarily be a multiple of 3? Every third integer is a multiple of 3, right?

You can immediately see that the cube of one of those is divisible by 27 (and hence by 9).

4. Feb 5, 2012

### hammonjj

I think my reading skills are failing me. Can you explain this a different way? I understand that every third n is divisible by three, just not the second part or how I could write that in proof form.

Thanks!
James

5. Feb 5, 2012

### Curious3141

The fact that out of three consecutive integers, one is a multiple of 3 needs no proof (it's just a consequence of the three times multiplication table!).

So that integer can be expressed as 3k. Its cube is 27k^3. It is therefore divisible by 9 (since 27 = 9*3), and the product of the cubes of those three integers is also obviously divisible by 9.

You can, of course, make an even stronger proof that the product of the cubes is divisible by 27, but the question only asked for divisibility by 9.