Prove that x^n + y^n = z^n has no solution in integers when the integer n > 2. :D :D :DQuote:
So any possible questions out there?
Fierytycoon
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Prove that x^n + y^n = z^n has no solution in integers when the integer n > 2. :D :D :DQuote:
So any possible questions out there?
Fierytycoon
Just recently proved, but still was considered a conjecture for over a hundred years (two hundred maybe?). Anyway, the proof is *probably* (hope I'm not offending anyone) beyond anyone's ability at CodeGuru.
Another question please...
Ok, Fermat's last theorem remained unproven for over 300 years.
I thought Kevin would never come back and play...:)
I'm ready...
:cool:
I guess you have just finished your dinner...:)Quote:
Originally posted by KevinHall
I'm ready...
:cool:
I have a question that I asked once in another thread but no one wants to answer, now I ask again...:)
Given a series of number from one to 9 ,use some operators to turn everything into 111111111.
This means A=111111111; find A ?
The answer is obviously A=111111111.
Next question.
As the question is so poorly stated almost anything would be a valid answer.
Yes, please take care (whomever is asking the question) to state the question as clearly as you possibly can, as failure to do so has been the main cause of disturbance in this thread's history. If people can't understand the question, it turns into a debate over what the question was instead of what the answer should be.
Thank you for your answer but unfortunately incorrect....It is not like that...Quote:
Originally posted by Tom Frohman
The answer is obviously A=111111111.
Next question.
As the question is so poorly stated almost anything would be a valid answer.
I call this problem MY LONELY KNIGHTS *
Okay, please calm down....I will explain more on the problem...
A can be 12x4+5x9, 3^12^4, etc.
So what is A inorder that A=111111111
A hint: each number is used only once but there is one digit is used twice... What is it ? and how can you arrange them in what order so that they can be those nine lonely knights trying to fight all night:)
Yes, answer please....
* Ideas borrowed from Mick-2002(TM), all right reserved.
I just wanted to declare one important thing to all people who are reading this thread that THIS POST OF MINE IS EXACTLY 1000th.:) Really nice....
So Could you give me the answer to the question above?
How about 12345679*9 = 111111111.
If you have a simple calculator and want to impress people ask them their favorite number between 1 and 9. If they say 3 then type 12345679*3*9 and you get 3.33333333 and the little e symbol. It's a great trick.
-Ben
:( I am sad because you were correct about the problem though my answer is slightly different from yours....But generally, the same, yes, same...:)Quote:
Originally posted by bmacri
How about 12345679*9 = 111111111.
If you have a simple calculator and want to impress people ask them their favorite number between 1 and 9. If they say 3 then type 12345679*3*9 and you get 3.33333333 and the little e symbol. It's a great trick.
-Ben
Mr Ben your up now, you have any question ?:)
I hate to burst your bubble so to speak, but the post before was in fact the thousandth. (What did you expect, "I hate to burst your bubble, so I won't"? :rolleyes: )Quote:
Originally posted by hometown
I just wanted to declare one important thing to all people who are reading this thread that THIS POST OF MINE IS EXACTLY 1000th.:) Really nice....
So Could you give me the answer to the question above?
One of these proofs are not correct. Which one?
Proof 1:
Assume a=b where a,b are nonzero integers.
a^2= ab
a^2 - b^2 = ab - b^2
(a + b)(a - b) = b(a - b)
(a + b)(a - b)/(a - b) = b(a - b)/(a - b)
a + b = b
but since a=b then
2b = b
2 = 1
Proof 2:
Posting on CodeGuru takes both time and money (for the computer, electricity, ...) so
Posting = Time * Money
We all know time is money so
Posting = Money * Money = Money^2
But money is the root of all evil so
Posting = (sqrt(Evil))^2
Which becomes
Posting = Evil.
So posting on CodeGuru is Evil.
One of these is not true but which one????
-Ben
What are you crazy ;).Quote:
Originally posted by bmacri
(a + b)(a - b)/(a - b) = b(a - b)/(a - b)
A bit sketchy, but if you say it's true...Quote:
Posting on CodeGuru takes both time and money (for the computer, electricity, ...) so
Posting = Time * Money
We all know time is money so
Posting = Money * Money = Money^2
But money is the root of all evil so
Posting = (sqrt(Evil))^2
Which becomes
Posting = Evil.
So posting on CodeGuru is Evil.