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Math Question: (Diophantine equation)
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-dt-
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O.P. Math Question: (Diophantine equation)
;o another math question for you smart math people :P


I've got a diophantine equation

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I've solved part A by doing

14x13x12/ 3! = 364 (which i think it correct :P)

but I'm stuck with part B, i have notes on how to do them if they are >= something but :-/ I've never encountered a <= one so I'm unsure of how to solve it.

so can someone please help the dt :P

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05-22-2008 04:48 PM
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foaly
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RE: Math Question: (Diophantine equation)
I can be completely wrong... but why isn't that 5/12 of the total amount?
05-22-2008 04:54 PM
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Basilis
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RE: Math Question: (Diophantine equation)
Well, what I will say isn't exactly helpful but hasn't anyone of your schoolmates solved it? We haven't done this one at school but if I were you, I would ask the best student of my class. On the other hand, there is Chrono. lol :) I hope someone finds the solution for you.

This post was edited on 05-22-2008 at 05:08 PM by Basilis.
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05-22-2008 05:07 PM
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Adeptus
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RE: Math Question: (Diophantine equation)
quote:
Originally posted by -dt-
i have notes on how to do them if they are >= something but (Smilie) I've never encountered a <= one so I'm unsure of how to solve it.
It seems to me if you know how to solve it for x1 >= 5 (which you are implying you do), then you could subtract the number of those solutions from the total number of solutions and get the answer you want....
05-22-2008 05:40 PM
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-dt-
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O.P. RE: Math Question: (Diophantine equation)
quote:
Originally posted by Adeptus
quote:
Originally posted by -dt-
i have notes on how to do them if they are >= something but (Smilie) I've never encountered a <= one so I'm unsure of how to solve it.
It seems to me if you know how to solve it for x1 >= 5 (which you are implying you do), then you could subtract the number of those solutions from the total number of solutions and get the answer you want....
oh :D thanks I never thought of that
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05-22-2008 11:44 PM
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Jhrono
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RE: Math Question: (Diophantine equation)
quote:
Originally posted by -dt-
quote:
Originally posted by Adeptus
quote:
Originally posted by -dt-
i have notes on how to do them if they are >= something but (Smilie) I've never encountered a <= one so I'm unsure of how to solve it.
It seems to me if you know how to solve it for x1 >= 5 (which you are implying you do), then you could subtract the number of those solutions from the total number of solutions and get the answer you want....
oh :D thanks I never thought of that
Will that work? Remember that the procedure you are aware of also includes the = solutions so, those will be common. By Adeptus method, you'll obtain the < solutions. Or am I looking at this in a wrong way?
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05-22-2008 11:46 PM
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-dt-
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O.P. RE: Math Question: (Diophantine equation)
quote:
Originally posted by Jhrono
quote:
Originally posted by -dt-
quote:
Originally posted by Adeptus
quote:
Originally posted by -dt-
i have notes on how to do them if they are >= something but (Smilie) I've never encountered a <= one so I'm unsure of how to solve it.
It seems to me if you know how to solve it for x1 >= 5 (which you are implying you do), then you could subtract the number of those solutions from the total number of solutions and get the answer you want....
oh :D thanks I never thought of that
Will that work? Remember that the procedure you are aware of also includes the = solutions so, those will be common. By Adeptus method, you'll obtain the < solutions. Or am I looking at this in a wrong way?
if i change it to >= 4 and then subtract it, i should get the same as <= 4, shouldnt i?

edit:

OH i see what you're saying... hm... maybe if i just do > 4 *looks for his notes*

oh wait i would have to subtract > 6 to get the < 4 stuff right? :P or am i just confusing myself more....

edit:

oh *reads Adeptus's post below* *fixes what hes done*

This post was edited on 05-23-2008 at 12:12 AM by -dt-.
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05-22-2008 11:49 PM
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Adeptus
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RE: Math Question: (Diophantine equation)
quote:
Originally posted by Jhrono
Will that work? Remember that the procedure you are aware of also includes the = solutions so, those will be common. By Adeptus method, you'll obtain the < solutions. Or am I looking at this in a wrong way?
The reason why my "method" was fine is because in a diophantine equation, all variables are whole integers.  Thus, if you remove everything >= 5, what remains will be <= 4 -- because there obviously are no whole integers between 4 and 5.
05-23-2008 12:11 AM
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Chrono
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RE: Math Question: (Diophantine equation)
quote:
Originally posted by -dt-
oh wait i would have to subtract > 6
if you know how to do it when it's >=, then you can calulate it when it's >= 5 and substract it to the total, so you will get for how many solutions it is < 5 or in other words <=4 :P

As for the first problem, i dont remember how to solve such kinda problems and i cant figure it out, so i cant verify if your answer is ok :P

i see you solved 14!/(14-3)!*3!, but why? I'd like to know :chrongue:

This post was edited on 05-23-2008 at 12:21 AM by Chrono.
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05-23-2008 12:18 AM
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Jhrono
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RE: Math Question: (Diophantine equation)
quote:
Originally posted by Adeptus
quote:
Originally posted by Jhrono
Will that work? Remember that the procedure you are aware of also includes the = solutions so, those will be common. By Adeptus method, you'll obtain the < solutions. Or am I looking at this in a wrong way?
The reason why my "method" was fine is because in a diophantine equation, all variables are whole integers.  Thus, if you remove everything >= 5, what remains will be <= 4 -- because there obviously are no whole integers between 4 and 5.
Eh, I thought the question was <= 5, not 4 :p Sorry
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05-23-2008 12:38 AM
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