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Old 07-26-07, 11:42 AM   #1
red house
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The MATH thread

Okay, you computer programming savantes and self-proclaimed ''mathemajicians'' of foo, this is your chance to prove your mettle. Now, these mathematical conundrums are not for the faint of heart, infact most of them probably have never been solved.. but, I invite you to try your hand at them anyway.. haha, best of luck (you're going to need it)




Math riddle #1.


If an integer n is greater than 2, provide the 'proof' that the equation a^n + b^n = c^n has no solutions in non-zero integers a, b, and c.




Math riddle #2.


Provide the proof that the circumference of a circle c divided by it's diameter d is infact always equal to a 'nonrational' real integer.




Math riddle #3.


Find a way to factor out (x+4) from the denominator of the trinomial;


(6x^3 + 23x^2 + 6x + 40) / (x+4)








good luck.. and pls show your work.

Last edited by red house; 07-26-07 at 12:34 PM.
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Old 07-26-07, 11:52 AM   #2
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I have a feeling for the 3rd, I'd probably have to factor out the top and multiply by the conjugate.
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Old 07-26-07, 11:55 AM   #3
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I have a feeling for the 3rd, I'd probably have to factor out the top and multiply by the conjugate.

I have the feeling you're probably on to something.. -go with it (!)..
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Old 07-26-07, 11:56 AM   #4
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First two are interesting problems in their own right. Don't expect to come up with an answer yourself. They are classic problems in mathematics that took the mathematics community a REALLY long time to solve.

The third one, just do your long division. It's straightforward, with no tricks involved.
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Old 07-26-07, 12:00 PM   #5
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I think Falkon was warmer.. call it a hunch.
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Old 07-26-07, 12:10 PM   #6
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3) 6x^2-x+10
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Old 07-26-07, 12:26 PM   #7
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3) 6x^2-x+10


ding! ding! ding! ding! ding! ..


Diggidy, you have earned your merit in mathematical excellence.



Okay. . can anyone shed some light on how this feat was accomplished?
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Old 07-26-07, 12:40 PM   #8
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Is this your homework?
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Old 07-26-07, 12:41 PM   #9
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Okay. . can anyone shed some light on how this feat was accomplished?
We already told you. Long division. Do your homework already, before you run out of time to seek assistance like you did last time.
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Old 07-26-07, 12:42 PM   #10
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I remember #1 from long ago...it is impossible to separate any power higher than the second into two like powers. Its a variation of the Pythagorean theorem (a squared + b squared = c squared) ie. (3, 4, 5) ( 5, 12, 13) ( 7, 24, 25) etc.

There are no solutions to this where n as stated above is greater than 2

If you plugged in (3,4,5) into a^n+b^n=c^n where n=3...it wouldn't work, in fact...nothing would work.


EDIT: I just realized I never "proofed it"...I'm too lazy.
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Old 07-26-07, 12:42 PM   #11
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you take the original numerator 6x^3 + 23x^2 6x + 40 and work on reducing it down into this format:
(ax + b)(cx + d)(ex + f)
where a - f are positive or negative integers
You have a denominator thats already reduced in x + 4 so it's a resonable assumptino that ax + b is either going to be x + 4 itself, or a multiple of that y(x + 4) where y is a positive or negative integer, and you just crunch numbers until a sequence hits. I don't remeber if theres a quicker short cut to it, or if I just used to do enough of them that I could see the patterns better than I can now.
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Old 07-26-07, 12:54 PM   #12
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Originally Posted by red house View Post
... Math riddle #2.


Provide the proof that the circumference of a circle c divided by it's diameter d is infact always equal to a 'nonrational' real integer.


Wait a sec, a nonrational real integer? Da hell does that mean? An integer is rational by definition. Or does 'nonrational' have a different definition than the more common term 'irrational'? And by 'real' I assume you mean 'noncomplex' rather than the computer science meaning of 'noninteger'.

Now, it's not difficult to derive the value of pi, but is that what you are asking? Your terminology is corn-fusing.
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Old 07-26-07, 12:57 PM   #13
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Wait a sec, a nonrational real integer? Da hell does that mean? An integer is rational by definition. Or does 'nonrational' have a different definition than the more common term 'irrational'? And by 'real' I assume you mean 'noncomplex' rather than the computer science meaning of 'noninteger'.

Now, it's not difficult to derive the value of pi, but is that what you are asking? Your terminology is corn-fusing.



I think perhaps I meant ''nonrational real number'' .. sorry, my bad.
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Old 07-26-07, 12:58 PM   #14
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Old 07-26-07, 12:58 PM   #15
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Originally Posted by squegeeboo View Post
you take the original numerator 6x^3 + 23x^2 6x + 40 and work on reducing it down into this format:
(ax + b)(cx + d)(ex + f)
where a - f are positive or negative integers
You have a denominator thats already reduced in x + 4 so it's a resonable assumptino that ax + b is either going to be x + 4 itself, or a multiple of that y(x + 4) where y is a positive or negative integer, and you just crunch numbers until a sequence hits. I don't remeber if theres a quicker short cut to it, or if I just used to do enough of them that I could see the patterns better than I can now.


Yeah, I tried that.. unfortunately I didn't have any luck at all.
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Old 07-26-07, 01:03 PM   #16
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We already told you. Long division. Do your homework already, before you run out of time to seek assistance like you did last time.

Who is 'we' ? .. you are the only one who has mentioned long division ? .. haha, 'long division' of a trinomial.. -that's a good one jschen. . .now I've heard it all.

Hey, why don't you show us how to perform this 'magical' long division method of yours for trinomial functions. You are too funny sometimes.. ''long division'' -?
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Old 07-26-07, 01:04 PM   #17
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Yeah, I tried that.. unfortunately I didn't have any luck at all.
Yah it took me a while of trial and error, I started out figuring out with how many ways of getting 40(the last part of the numerator), with one of the numbers being a 4(due to the 4 in the denominator), so in this case 10x1x4 5x2x4 1x10x4 2x5x4 were the only ones I could come up with. tried a bunch of the 5x2 combo's none of those got even close for the 23x^2 part of the original, so I moved to the 10x1 combo, and stumbled into it.
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Old 07-26-07, 01:05 PM   #18
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You don't try to factor the whole thing from scratch. That's a MUCH tougher problem. Set up your division.

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x + 4 | 6 x^3 + 23 x^2 + 6 x + 40


This long division is easy since they already told you what to divide by. Factoring from scratch or solving by trial and error is hard work.
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Old 07-26-07, 01:07 PM   #19
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Old 07-26-07, 01:07 PM   #20
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You don't try to factor the whole thing from scratch. That's a MUCH tougher problem. Set up your division.

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x + 4 | 6 x^3 + 23 x^2 + 6 x + 40


This long division is easy since they already told you what to divide by. Factoring from scratch or solving by trial and error is hard work.


Seriously, wtf is that?

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Old 07-26-07, 01:09 PM   #21
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It's a division problem. Work left to right, just as you would a standard long division problem. I'm trying to in ASCII draw out what I would write on paper to set up the division.

Work with me... What do you have to multiply (x + 4) by to get a 6 x^3 term?
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Old 07-26-07, 01:09 PM   #22
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Originally Posted by red house View Post
ding! ding! ding! ding! ding! ..


Diggidy, you have earned your merit in mathematical excellence.



Okay. . can anyone shed some light on how this feat was accomplished?
If you have the answer and the problem, it shouldn't be too hard to figure out the steps in between. If you are doing this for answers for homework, stop being lazy and do it by yourself. If it is indeed for school, then you will probably see the material again; and you will want to know it then. If it's just for funzies, then thanks; it's been a few months since I got out of high school, and I'm saddened at how hard that was for me.
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Old 07-26-07, 01:10 PM   #23
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Originally Posted by squegeeboo View Post
Yah it took me a while of trial and error, I started out figuring out with how many ways of getting 40(the last part of the numerator), with one of the numbers being a 4(due to the 4 in the denominator), so in this case 10x1x4 5x2x4 1x10x4 2x5x4 were the only ones I could come up with. tried a bunch of the 5x2 combo's none of those got even close for the 23x^2 part of the original, so I moved to the 10x1 combo, and stumbled into it.

Yeah, 23 being a prime was really discouraging and made me think I was going about it all wrong.


So, what does the ( . . )( . . )( . . ) look like?
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Old 07-26-07, 01:12 PM   #24
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It's a division problem. Work left to right, just as you would a standard long division problem. I'm trying to in ASCII draw out what I would write on paper to set up the division.

Work with me... What do you have to multiply (x + 4) by to get a 6 x^3 term?

I dunno? 6x^2 ? .. but then how do you get the 4 ?
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Old 07-26-07, 01:14 PM   #25
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Originally Posted by Diggidy View Post
If you have the answer and the problem, it shouldn't be too hard to figure out the steps in between. If you are doing this for answers for homework, stop being lazy and do it by yourself. If it is indeed for school, then you will probably see the material again; and you will want to know it then. If it's just for funzies, then thanks; it's been a few months since I got out of high school, and I'm saddened at how hard that was for me.


I am not 'lazy' - I am resourceful. And there is no rule saying one must do his or her homework by theirself w/o any help is there? No, I didn't think so, -(unless you happen to reside in North Korea). But thanku much for the help, it's much appreciated.
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