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

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



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

Originally Posted by jj451

What is to be done here?
What are the instructions?

Originally Posted by jj451

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

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 , but then what?

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 ? We can safely assume that 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.

If we assume that say, then we can rewrite as and can be rewritten as (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.

I found a solution here.

Originally Posted by alexmahone

I found a solution here.

Sehr gut. Marking as solved.