# Thread: a!b! = a! + b! + c!

1. ## a!b! = a! + b! + c!

This is a BMO 1 question from 2003 that has me completely stumped.

$a!b! = a! + b! + c!$

Any help on basic operations on factorials would be appreciated as I have no idea how they can be manipulated.

Thanks.

EDIT:

Apogies, I forgot to state that all possible values of a b and c are to be found

2. ## Re: a!b! = a! + b! + c!

Originally Posted by jj451
$a!b! = a! + b! + c!$
What is to be done here?
What are the instructions?

3. ## Re: a!b! = a! + b! + c!

Originally Posted by jj451
This is a BMO 1 question from 2003 that has me completely stumped.

$a!b! = a! + b! + c!$
There is one obvious solution (a,b,c) = (3,3,4). I suspect that there are no others, but I don't see how to prove it.

You could start by rewriting the equation as $(a!-1)(b!-1) = c!+1$, but then what?

4. ## Re: a!b! = a! + b! + c!

A quick thought (I don't know if it's useful). You can try independent looking at the possible orderings of {a,b,c}. For example, what happens if $a\leq b\leq c$? We can safely assume that $a\leq b$ because of the symmetry in the equation.

With them being ordered, you will know how much you can "divide out" (sorry, don't know the English word) and where.

5. ## Re: a!b! = a! + b! + c!

If we assume that $a \le b$ say, then we can rewrite $a!b!$ as $(a!)^2 \times \frac{b!}{a!}$ and $a! + b!$ can be rewritten as $2a! + (b! - a!)$ (I know this is obvious but it might help to see something else).

No other ideas at the moment; I'm not too good with factorials.

6. ## Re: a!b! = a! + b! + c!

I found a solution here.

7. ## Re: a!b! = a! + b! + c!

Originally Posted by alexmahone
I found a solution here.
Sehr gut. Marking as solved.