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re: Solve this math problem and win $1 million dollarsPosted by tiggerthetooth on 6/12/13 at 1:44 am to jmarto1

Solve for what? I dont understand.

ETA: ABC= xyz?

This post was edited on 6/12 at 1:45 am

re: Solve this math problem and win $1 million dollarsPosted by UltimateHog on 6/12/13 at 1:45 am to jmarto1

quote:

= Cz

Looks like it's already solved.

re: Solve this math problem and win $1 million dollarsPosted by FootballNostradamus on 6/12/13 at 1:45 am to tiggerthetooth

That is nowhere near a solvable equation.

Thanks for a LINK.

quote:

Fermat's Last Theorem states that no three positive integers a, b, and c can satisfy the equation a^n + b^n = c^n for any integer value of n greater than two.

Beal's conjecture states that if A^x + B^y = C^z, where A, B, C, x, y, and z are positive integers with x, y, z > 2, then A, B, and C have a common prime factor.

Got all that? Well you'll need to prove or disprove it, and have your work published in a leading mathematics journal to claim your prize.

RockyMtnTigerWDE

New Orleans Saints Fan

14 & Cnting needs my words to live

Member since Oct 2010

59700 posts

New Orleans Saints Fan

14 & Cnting needs my words to live

Member since Oct 2010

59700 posts

re: Solve this math problem and win $1 million dollarsPosted by RockyMtnTigerWDE on 6/12/13 at 1:53 am to jmarto1

180,000.00

re: Solve this math problem and win $1 million dollarsPosted by willymeaux on 6/12/13 at 2:05 am to jmarto1

z = (log(A^x+B^y))/(log(C))

re: Solve this math problem and win $1 million dollarsPosted by tiggerthetooth on 6/12/13 at 2:12 am to Jet12

So in other words the problem in the OP is not even the problem they are attempting to solve.

re: Solve this math problem and win $1 million dollarsPosted by tiggerthetooth on 6/12/13 at 2:13 am to willymeaux

quote:

z = (log(A^x+B^y))/(log(C))

Now plug in that equation into the z variable.

re: Solve this math problem and win $1 million dollarsPosted by JG77056 on 6/12/13 at 2:39 am to tiggerthetooth

Basically, to word it out, find 3 whole numbers that when cubed or greater have one number equal exactly half of the sum total?

re: Solve this math problem and win $1 million dollarsPosted by FootballNostradamus on 6/12/13 at 2:48 am to tiggerthetooth

quote:

So in other words the problem in the OP is not even the problem they are attempting to solve.

Not even remotely close.

re: Solve this math problem and win $1 million dollarsPosted by lsucoonass on 6/12/13 at 3:46 am to UltimateHog

i was thinking the same thing by looking at the letters

re: Solve this math problem and win $1 million dollarsPosted by gthog61 on 6/12/13 at 6:28 am to lsucoonass

Yep, the prize has been out there for 15 years, I am sure some OT baller will knock it out in a day or two.

re: Solve this math problem and win $1 million dollarsPosted by Impotent Waffle on 6/12/13 at 6:37 am to gthog61

True

quote:

Too bad math was never my strong suit.

Obviously, since you didn't state the problem correctly.

Beal's Conjecture is a play on Fermat's Last Theorem. Take the equation A^x + B^y = C^z. Fermat stated that if x, y and z are all the same, then there is no positive integer solution where n > 2.

Beal's Conjecture (which is what the contest is about) states that if x, y and z are all > 2 (but not necessarily the same), then a solution is only possible if A, B and C have a common prime factor.

The Aggies lost two big recruits on Wednesday night.

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