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

 Messages: [ Previous | Next ]
 JT Posts: 1,448 Registered: 4/7/12
Re: 7!= 2*5! * 2*3! *7
Posted: Jun 11, 2013 1:09 PM

On 11 Juni, 18:39, JT <jonas.thornv...@gmail.com> wrote:
> On 11 Juni, 18:32, JT <jonas.thornv...@gmail.com> wrote:
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> > On 11 Juni, 18:12, Brian Chandler <imaginator...@despammed.com> wrote:
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> > > 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?
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> > > Well, you can divide both sides by 13 getting:
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> > > 12! = 2*11!* 2*9!*2*7!*2*5!*2*3!
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> > > And if this isn't enough: divide by 12:
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> > > 11! = 2*11!* 2*9!*2*7!*2*5!
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> > > Then 1 = 2*2*9!*2*7!*2*5!
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> > > So even allowing for experimental error, we get 1 > 2^18
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> > > Which means 2 is a lot smaller than previously thought.
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> > > Brian Chandler
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> > 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

Well 5!*3!*7 = 5040 and 7!=5040, tomorrow i give it another try.
And just by chance 9!=362880 and 7!*5!*3!=3628800 so there may be
something, that can be used in the permutation algorithm. I just have

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