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Topic: Factorial Product (Read 383 times) |
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Sir Col
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impudens simia et macrologus profundus fabulae
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Factorial Product
« on: Mar 7th, 2004, 4:26am » |
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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! ?
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Barukh
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Re: Factorial Product
« Reply #1 on: Mar 7th, 2004, 7:49am » |
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on Mar 7th, 2004, 4:26am, Sir Col wrote:Given that a,b,c,d[in][bbn], and a<b<c, solve a!*b! = c!. |
| I think a, b = a! - 1, c = a! would do. Quote:Prove that the solution is unique. |
| ??
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towr
wu::riddles Moderator Uberpuzzler
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Re: Factorial Product
« Reply #2 on: Mar 7th, 2004, 8:00am » |
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I was gonna say that.. but I'm too slow today just to contribute at least something, :: a > 2, else a<b<c won't hold if c=a! and b=c-1 ::
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« Last Edit: Mar 7th, 2004, 8:04am by towr » |
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Sir Col
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Re: Factorial Product
« Reply #3 on: Mar 7th, 2004, 8:32am » |
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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...
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towr
wu::riddles Moderator Uberpuzzler
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Re: Factorial Product
« Reply #4 on: Mar 7th, 2004, 3:08pm » |
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I found one solution, whether it's the only one, I can't prove yet.. :: 6!*7!=10! ::
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Barukh
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Re: Factorial Product
« Reply #6 on: Mar 8th, 2004, 12:01am » |
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on Mar 7th, 2004, 8:32am, Sir Col wrote:Okay, a new restriction... a<b<c–1. Now go prove the existence of the unique solution... |
| 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. So, Sir Col, if you've got the proof I would definitely like to see it!
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