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Topic: 7!= 2*5! * 2*3! *7
Replies: 14   Last Post: Jun 14, 2013 10:35 PM

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 JT Posts: 1,448 Registered: 4/7/12
Re: 7!= 2*5! * 2*3! *7
Posted: Jun 11, 2013 12:32 PM

On 11 Juni, 18:12, Brian Chandler <imaginator...@despammed.com> wrote:
> JT wrote:
> > Ooops got it wrong there, *2 of course
> > 13!=2*11!*2*9!*2*7!*2*5!*2*3!*13

>
> > Try me, how can you use this?
>
> Well, you can divide both sides by 13 getting:
>
> 12! = 2*11!* 2*9!*2*7!*2*5!*2*3!
>
> And if this isn't enough: divide by 12:
>
> 11! = 2*11!* 2*9!*2*7!*2*5!
>
> Then 1 = 2*2*9!*2*7!*2*5!
>
> So even allowing for experimental error, we get 1 > 2^18
>
> Which means 2 is a lot smaller than previously thought.
>
> Brian Chandler

I think that 2 somehow sneaked in when i was thinking about the
permutation algorithm, bloppers. But i was correct second time.

7!=5!*3!*7
9!=7!*5!*3!*9
11!=9!*7!*5!*3!*11
13!=11!*9!*7!*5!*3!*13

Date Subject Author
6/11/13 JT
6/11/13 JT
6/11/13 JT
6/11/13 Dirk Van de moortel
6/11/13 JT
6/11/13 Brian Chandler
6/11/13 JT
6/11/13 JT
6/11/13 JT
6/11/13 Richard Tobin
6/12/13 JT
6/12/13 JT
6/12/13 JT
6/11/13 Richard Tobin