Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math.independent

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
JT

Posts: 1,148
Registered: 4/7/12
Re: 7!= 2*5! * 2*3! *7
Posted: Jun 11, 2013 12:39 PM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

On 11 Juni, 18:32, JT <jonas.thornv...@gmail.com> wrote:
> 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


No i was not 9! = 362880



Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.