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Billionaire Offers $1 Million to Solve Math Problem

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posted on Jun, 7 2013 @ 07:22 PM
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Originally posted by iwontrun
Come on guys the answer is.......purple!
edit on 6/7/2013 by iwontrun because: (no reason given)


I thought the color of The Infinite is blue...?

(We'll see how many Terry Pratchet fans are here I guess.)




posted on Jun, 7 2013 @ 07:23 PM
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Originally posted by Lionhearte

As far as I can tell, I provided a proof for this theorem, but not a counter example. Though it does not state I need a counter example..

Am I missing something? That seemed way too easy.

Someone please correct me if I'm wrong.
And God bless, all you critical thinkers.
edit on 7-6-2013 by Lionhearte because: (no reason given)

edit on 7-6-2013 by Lionhearte because: (no reason given)

That's not a proof. You're just putting numbers into the formula.
A proof would show that the formula/conjecture is true in every instance.



posted on Jun, 7 2013 @ 07:24 PM
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I've always been really good at math. I want to solve it lol.



posted on Jun, 7 2013 @ 07:28 PM
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reply to post by Lionhearte
 


Yeah, you multiplied. Check out an official source and you'll see they're actually exponents. So it should read A^x + B^y = C^z.



posted on Jun, 7 2013 @ 07:32 PM
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Originally posted by Bob Sholtz
reply to post by TeflonBear
 

he wants a numerical solution?


NO



posted on Jun, 7 2013 @ 08:02 PM
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Jeez, no one has made any steps at all in the proof?
I re read the thread just to make sure lol, and to see if there were any hints.

I've done a couple of steps, but I'm kind of stuck.

The thing is ATS has a policy that anything you post becomes theirs, so I'm unwilling to post anything here.

But I will give a hint. There's a first step that I'm fairly certain MUST be done. It's really the only logical direction to prove it.

C'mon ATS. Someone who doesn't mind ATS's policy(although I'd like to think they wouldn't steal someone's million if they came up with the theory here, plus that would be bad for PR), post the first step here so we can talk about lol.

After a couple of steps past the first one, I've come to a problem, where I understand now why proving this would invariably be a major new math idea.

I don't want to say it's impossible, but it very well could be.
edit on 7-6-2013 by Ghost375 because: (no reason given)



posted on Jun, 7 2013 @ 08:35 PM
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reply to post by Ghost375
 


I truly think it may be impossible.

Some tinkering I did was to make the exponents able to be evenly cube rooted then reduced to simplify any of A, B, or C. I also subtracted C^z from both sides to make the total equation = 0 for easier solving.

However, no luck. Then I tried some online variable solvers I remember accidentally coming across a year ago while trying to solve unrelated net ionic equations. After inputting various forms into the generator, every time it told me the solution could not be determined.

In my head it seems logical it would be solvable but I'm not so sure. Maybe that is why it's gone on for hundreds of years.



posted on Jun, 7 2013 @ 08:56 PM
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Originally posted by Ghost375

Originally posted by Bob Sholtz
reply to post by TeflonBear
 

he wants a numerical solution?


NO

calm down. i read the article, yet it didn't provide adequate information.

after reading the real problem i don't think there are any solutions that fit the criteria, and proving that there are no exceptions to the rule will be difficult.



posted on Jun, 7 2013 @ 09:30 PM
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reply to post by Bob Sholtz
 


I was just using the caps emphatically.



posted on Jun, 7 2013 @ 11:40 PM
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reply to post by TeflonBear
 


So how do we contact this guy?
I think I got it lol.

It's rather complex. I''m pleased with myself


edit on 8-6-2013 by Ghost375 because: (no reason given)



posted on Jun, 7 2013 @ 11:46 PM
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The only piece of information I can contribute to disproving this is that A and B cannot both be even numbers, since obviously C would be an even number and all three would have 2 as a common factor.

Aside from that, it would take a much more complex computer than the one I'm typing on to run even a fraction of the infinite number of cases in which this conjecture is true before stumbling across one which disproved it.

In my opinion, with the given restrictions of the conjecture, this is impossible to disprove. In math, you generally find exceptions to the rules using much smaller numbers, like 1 and 2, which is why the restrictions are in place. It's my strong belief that proof of this conjectures validity is the only way this million would ever be awarded.



posted on Jun, 8 2013 @ 12:14 AM
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It seems I have it.
However, if a billionaire wants it and is willing to split with a ...million, it must not be important enough.
I don't even think I'd share it for a billion though, if a person with this much pull wanted it.
Many secrets die and are never found, too bad conquerors burn libraries.

Now back to my book.



posted on Jun, 8 2013 @ 12:52 AM
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Originally posted by FlyersFan
reply to post by boymonkey74
 

OH man .. I was going to say that!


As for the math problem .. I now have dyscalculia (from the sjogrens) so I can't even do the checkbook anymore. Looks like I"m out of the running for the big money. I would have thought that one of the world smarty-pants would have solved it by now.


Thank you, thank you, thank you! I had never heard of dyscalculia before. I have always felt I had some sort of mathematical dyslexia for lack of a better term. Barely squeaked by the most basic math classes in high school. Now I have a starting point for (hopefully) overcoming this!



posted on Jun, 8 2013 @ 02:04 AM
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reply to post by TeflonBear
 


I know the answer












is very hard



posted on Jun, 8 2013 @ 03:12 AM
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reply to post by TeflonBear
 


Ax + By = Cz


I must be missing something. Doesn't this equation mean: A times X plus B times Y equals C times Z?

UGH! I thought it was too easy. The equation is actually:




edit on 6/8/2013 by jiggerj because: (no reason given)



posted on Jun, 8 2013 @ 06:33 AM
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reply to post by Ghost375
 


I'm not sure how to contact this guy. But if you have the answer you just pocketed $1mil and probably secured a future job with an alphabet agency (like it or not)!



posted on Jun, 8 2013 @ 06:51 AM
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a^x + b^y = c^z

I believe the answer is a four dimensional wave with 24 behavioral limits.(original thought).

Actually, it appears to be a four dimensional wave with 6 behavioral limits.
edit on 8-6-2013 by straddlebug because: (no reason given)



posted on Jun, 8 2013 @ 07:58 AM
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Originally posted by straddlebug
a^x + b^y = c^z
Actually, it appears to be a four dimensional wave with 6 behavioral limits.
edit on 8-6-2013 by straddlebug because: (no reason given)


Hmm, I think I am missing something. I believe there should be two additional behavioral limits for a total of 8 limits.

However, I now wonder if there should be additional limits that I have not considered when dealing with a 4 dimensional space.



posted on Jun, 8 2013 @ 08:19 AM
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Variables, by definition, aren't defined so how is it possible to solve something with six unrelated variables?

Seems this belongs in quantum philosophy rather than math...



posted on Jun, 8 2013 @ 09:04 AM
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Originally posted by Thermo Klein
Variables, by definition, aren't defined so how is it possible to solve something with six unrelated variables?
Seems this belongs in quantum philosophy rather than math...


You maybe right. This may be unsolvable, but I think it is possible.

I do have a couple of questions for people who have put some thought into this matter.

Often I have heard that the 4th dimension is time. Is this an accepted definition, and/or does the 4th dimension ALWAYS have to be defined as time? (assuming there are additional dimensions)

Another concern: Is time an absolute where there are both positive and negative values? Plus, is there infinite time, zero time, and NULL time?



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