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        riddles >> easy >> Factorial Product
        
(Message started by: Sir Col on Mar 7^th, 2004, 4:26am)

Title: Factorial Product
Post by Sir Col on Mar 7^th, 2004, 4:26am

Given that a,b,c,d[in][bbn], and a<b<c, solve a!*b! = c!.
Prove that the solution is unique.

What about a!*b!*c! = d! ?

Title: Re: Factorial Product
Post by Barukh on Mar 7^th, 2004, 7:49am

on 03/07/04 at 04:26:38, Sir Col wrote:

Given that a,b,c,d[in][bbn], and a<b<c, solve a!*b! = c!.

I think [hide]a, b = a! - 1, c = a! [/hide] would do.

Quote:

Prove that the solution is unique.

?? ::) ::)

Title: Re: Factorial Product
Post by towr on Mar 7^th, 2004, 8:00am

I was gonna say that.. but I'm too slow today

just to contribute at least something, ::[hide] a > 2, else a<b<c won't hold if c=a! and b=c-1 [/hide]::

Title: Re: Factorial Product
Post by Sir Col on Mar 7^th, 2004, 8:32am

Dang! I missed that trivial case, but good spot you two.
Obviously if b=c–1, a!=c!/b!=c.

Okay, a new restriction... a<b<c–1.

Now go prove the existence of the unique solution... :P

Title: Re: Factorial Product
Post by towr on Mar 7^th, 2004, 3:08pm

I found one solution, whether it's the only one, I can't prove yet..
::[hide] 6!*7!=10! [/hide]::

Title: Re: Factorial Product
Post by KarmaBandit on Mar 7^th, 2004, 4:39pm

Here's Mathworld's input: http://mathworld.wolfram.com/FactorialProducts.html

16! = 14! 5! 2!
10! = 7! 5! 3! = 7! 6!

It looks like mathworld doesn't know of any uniqueness proof, either. That's my signal to give up!

Title: Re: Factorial Product
Post by Barukh on Mar 8^th, 2004, 12:01am

on 03/07/04 at 08:32:37, Sir Col wrote:

Okay, a new restriction... a<b<c–1.

Now go prove the existence of the unique solution... :P

It seems that towr's solution may be the unique one. However, credible sources (including KarmaBandit's link) tell me that the proof is far beyond the scope of the easy section. ;D

So, Sir Col, if you've got the proof I would definitely like to see it!