# The MATH thread

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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)

If an integer

Provide the proof that the circumference of a circle

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.

__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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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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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.

The third one, just do your long division. It's straightforward, with no tricks involved.

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I think Falkon was warmer.. call it a hunch.

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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.

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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.

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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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.

(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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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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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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(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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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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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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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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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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*Last edited by jschen; 07-26-07 at 01:18 PM.*

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

*Last edited by jschen; 07-26-07 at 01:18 PM.*

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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?

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

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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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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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I dunno? 6x^2 ? .. but then how do you get the 4 ?

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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.