Shoutbox

;o another math question for you smart math people


I've got a diophantine equation



I've solved part A by doing

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

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

I can be completely wrong... but why isn't that 5/12 of the total amount?

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.

quote:

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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.

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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....

quote:

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Originally posted by Adeptus

quote:

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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.

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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....

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oh thanks I never thought of that

quote:

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Originally posted by -dt-

quote:

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Originally posted by Adeptus

quote:

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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....

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oh thanks I never thought of that

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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?

quote:

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Originally posted by Jhrono

quote:

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Originally posted by -dt-

quote:

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Originally posted by Adeptus

quote:

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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....

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oh thanks I never thought of that

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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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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? or am i just confusing myself more....

edit:

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

quote:

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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?

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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.

quote:

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Originally posted by -dt-
oh wait i would have to subtract > 6

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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

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

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

quote:

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Originally posted by Adeptus

quote:

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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?

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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.

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Eh, I thought the question was <= 5, not 4 Sorry