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Oh dear god no...Quote:
Originally posted by ih8mylife
you have 12 ball bearings
1 of them is either heavier or lighter
you have 1 balance that you can use 3 times
how do you find out which one is the odd ball (and you must know if its heavier or lighter)
nice work
uhateurlife, :D
problem is we had this one before... post something new
so there's a new one
suppose you have 3 numbers X and Y, both having equal number of digits, and N (no digit limitation)
and two other numbers X..X and Y..Y that are formed by puttin N times Xs and Ys together (simply next to each-other)
(ex X=45 Y=98 N=3 therefore X..X = 454545 and Y..Y = 989898)
which is bigger
X/Y or X..X/Y..Y
and why?
DO you mean to say positive whole numbers?
X...X/Y...Y = X/Y^N so It depends on whether X >= Y or X < Y
look at the example :pQuote:
Originally posted by me
X..X and Y..Y are formed by puttin N times Xs and Ys together (simply next to each-other)
(ex X=45 Y=98 N=3 therefore X..X = 454545 and Y..Y = 989898)
although they must be whole and for more convenience positive ;):rolleyes:
What about the example? Just because your example uses positive whole numbers does not mean it is a constraint on the problem. :)
Anyway, Did I not answer the question?:)
What... is the Airspeed Velocity of an Unladen Swallow?
FAST
no you did notQuote:
Originally posted by souldog
What about the example? Just because your example uses positive whole numbers does not mean it is a constraint on the problem. :)
Anyway, Did I not answer the question?:)
It does not depend on anything
look at my post
there are NO powers! (no X^N)
OOPS:rolleyes:
They are always equal.
e.g.
454545 = 45 (10000 + 100 + 1)
989898 = 98 (10000 + 100 + 1)
In general if The number of digits in the numbers X and Y is k
then
X..X = X (10^(k*(N-1) + 10^(k*(N-2) +...+10^0)
A european swallow or an african swallow?Quote:
Originally posted by Mick_2002
What... is the Airspeed Velocity of an Unladen Swallow?
that is correct
:D
heh heh...well I don't know...ahhhhhhhhhhhhhhhQuote:
Originally posted by Elrond
A european swallow or an african swallow?
Even if you don't know, you have to be firm and just make a decision. Life is full of arbitrary choices. H_E_L_L life is a friggin arbitrary choice. SO come on Mick.... Be a man for god's sack.
Ohh come on now soul..you do realize that's from monty phyton's holy grail....Quote:
Originally posted by souldog
Even if you don't know, you have to be firm and just make a decision. Life is full of arbitrary choices. H_E_L_L life is a friggin arbitrary choice. SO come on Mick.... Be a man for god's sack.
No, I didn't know that, but you still haven't made a choice.:)
Ok here is a question. Are there an infinite number of prime numbers p such that p+2 is also prime.
e.g. 5 and 7;)
I will choose freewillQuote:
Originally posted by souldog
No, I didn't know that, but you still haven't made a choice.:)
I don't know why you guys/gals post your little riddle questions...the answer is 3, it will always be 3...
Well my last question is kinda funny
Yes. I have a wonderful proof, but it's too big to fit in this small margin :DQuote:
Originally posted by souldog
Ok here is a question. Are there an infinite number of prime numbers p such that p+2 is also prime.
e.g. 5 and 7;)
[Ups ^^]
African, or European? (When you're a king you have to know these things)Quote:
Originally posted by Mick_2002
What... is the Airspeed Velocity of an Unladen Swallow?
And souldog, most certainly yes.
well i dont have a proof like Yves but it really sounds as if they are infinite :)Quote:
Originally posted by souldog
Ok here is a question. Are there an infinite number of prime numbers p such that p+2 is also prime.
e.g. 5 and 7;)
I heard this in my class some time ago...:)Quote:
Originally posted by souldog
Ok here is a question. Are there an infinite number of prime numbers p such that p+2 is also prime.
e.g. 5 and 7 :o
when we were introduced about properties of Primes and
Diophantine equations...:p , Moreover,
your question I assume is taken from one of many FAMOUS open problems relating to Prime Number Theorem...:o :p
Yeah.... Hometown is Nina, Nani is Nina....Yes.
And another thing to pay back is...
cut that out!
In case I am allowed to make a question now to Souldog, I would like to ask you to prove:[lim(n*log(n))]/pn=1 Knowing that n->inf, p is prime.
:)
Seventh Star...Quote:
Originally posted by SeventhStar
cut that out!
I am trying to solve that problem, I thought that you would agree to tell me whether my answer to that question was correct or not, which is also what I am expecting from you now....
I think that your answer to that question was just like you were finishing only half of your little loaf of bread, and what I posted was the other half but because you asked me to stop, so I deleted it, I hope you wont remember any word of it....
:confused: :confused: But I am still :cool:
Souldog willnot get angry at me Okay ? And SeventhStar willnot either, hah.....
:) :):)
Here's one:
Prove that every integer n > 5 is the sum of three primes. :D
Fierytycoon
False.Quote:
Originally posted by hometown
Souldog willnot get angry at me Okay ? And SeventhStar willnot either, hah.....
That's what you get for posting that in the Question and Answer thread! ;):D
It's been a while with all this proof stuff but here goes. Here's a proof by induction.
I can show this works for
6 = 3+2+1
7 = 3+3+1
8 = 5+2+1
9 = 5+3+1
Now let x be a number that's the sum of three primes. If I can show that x+1 will be the sum of three primes then I'm done.
Assume x is even. Then one of the three numbers in the sum must be 2 because 2 is the only even prime and to add three numbers together and get an even number you can either add three even numbers or two odd numbers and one even. Since we can't do three even numbers then one of three numbers has to be 2 ( 2+2+2 will work but that's as high as it will go). So x can be written like
x = a + b + 2 where a and b are odd primes.
Change the 2 to 3 and there we go.
x + 1 = a + b + 3.
That's if x is even. What about if x is odd? Well it's not quite as clear but here goes.
Assume x = a + b + c and x is an odd number and a,b,c are primes. Since x+1 is an even number we know x+1 = r + s + t where r,s,t are primes. One of them must be 2 from the argument above. Let's say t=2 so x+1 = r + s + 2.
That means x = r + s + 1.
And we are done. Let me know if you find any problems with this proof or have any questions. It might seem a little backwards but I think it works.
-Ben
1 is not prime. You can do it like this, though:Quote:
Originally posted by bmacri
6 = 3+2+1
7 = 3+3+1
8 = 5+2+1
9 = 5+3+1
6 = 2 + 2 + 2
7 = 3 + 2 + 2
8 = 3 + 3 + 2
9 = 3 + 3 + 3
You're using what you're supposed to prove to prove what you're supposed to prove.Quote:
Assume x = a + b + c and x is an odd number and a,b,c are primes. Since x+1 is an even number we know x+1 = r + s + t where r,s,t are primes. One of them must be 2 from the argument above. Let's say t=2 so x+1 = r + s + 2.
That means x = r + s + 1.
You're on the right track though. But who am I to criticize when I myself am too lazy to do it? ;):rolleyes:
Your right 1 isn't a prime numnber. So much for that.
quote:
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You're using what you're supposed to prove to prove what you're supposed to prove.
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Not really although it seems like that. I had shown at that point that any even number could be written as the sum of three primes. So from that point forward I could use that information. It seems wierd and a think it can be better written but at that point I think I can use that piece of information. I don't have time now to redo the proof with out using 1 but I'll try to put something up tomorrow.
bmacri,
Nice try, but even the mathemeticians who have studied this topic have not been able to prove it. Here is a link concerning this topic:
http://www.utm.edu/research/primes/notes/conjectures/
- Kevin
P.S. Can we have some riddles which are not impossible (or virtually impossible) to solve?
Thanks Kevin. At least I won't waste anymore time trying to do it. If only one was a prime number then it just might be possible to prove.
So any possible questions out there?
-Ben
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?