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  #1  
Old 2008-01-22, 10:38
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moothead moothead is offline
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magicball logo

when is the top left magicball logo going to lose his santa hat
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  #2  
Old 2008-01-22, 10:57
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Don't you know that christmas lasts until easter?
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  #3  
Old 2008-01-22, 13:35
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Same discussion as last year, essentially.
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  #4  
Old 2008-01-22, 14:28
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It will be removed on the n'th day of the year. Where n is defined as the x'th prime. And x is defined as y^2 modulo z. y is the number of leap days in the current century, z is x * sqrt(x) * (n - 2)

(have fun)
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Last edited by elmuerte; 2008-01-22 at 14:58.
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  #5  
Old 2008-01-22, 14:45
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y=24
x(z) = 24² mod z
n = x′(z) = d/dz (24² mod z)

Now how to get the derivative of modulo? And how can the nth day be a function?
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  #6  
Old 2008-01-22, 14:47
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elmuerte elmuerte is offline
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oops.. small mistake
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  #7  
Old 2008-01-22, 14:54
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So then:
y=1
x(z) = 1 mod z
-1 ≤ z ≤ 1 <=> x(z) = 0
z < -1 OR z > 1 <=> x(z) = 1
=> x′(z) = 0
n = 0
There is no 0th day of the year.
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  #8  
Old 2008-01-22, 14:57
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ok, corrected it (and yes, it is solvable)
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  #9  
Old 2008-01-22, 15:02
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Quote:
Originally Posted by Kobold View Post
x(z) = 1 mod z
-1 ≤ z ≤ 1 <=> x(z) = 0
z < -1 OR z > 1 <=> x(z) = 1
why "-1 ≤ z ≤ 1" ?

y mod (y+whatever)

is legal, it's simply y
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  #10  
Old 2008-01-22, 15:38
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But
y mod (y+whatever) = y for whatever > 0
is dealed with this:
z < -1 OR z > 1 <=> x(z) = 1
?
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  #11  
Old 2008-01-22, 16:12
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It was never there for the darker themes I think, at least it's not there now.
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  #12  
Old 2008-01-22, 16:12
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y mod (y+c) = 0; if c = 0
y mod (y+c) = y; if c > 0
y mod (y+c) = y - (y * floor(y/(y+c))); if c < 0; which is actually the universal rule, but the above two cases are easier
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  #13  
Old 2008-01-22, 18:38
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17th march :S (random guess)
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  #14  
Old 2008-01-22, 19:04
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Quote:
Originally Posted by elmuerte View Post
Where n is defined as the x'th prime.
The xth prime of which function exactly?
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  #15  
Old 2008-01-22, 19:04
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Quote:
Originally Posted by moothead View Post
17th march :S (random guess)

Maby we should just wait untill it is taken down
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  #16  
Old 2008-01-22, 19:10
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i see no xmas hat! Ha Ha!
(by night theme)
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  #17  
Old 2008-01-22, 19:28
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Quote:
Originally Posted by Kobold View Post
The xth prime of which function exactly?
the x'th prime number, no function, just the x'th prime
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  #18  
Old 2008-01-22, 19:44
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No wonder why I get it wrong - http://en.wikipedia.org/wiki/Prime_%28symbol%29
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  #19  
Old 2008-01-22, 20:02
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x = y² mod z
=> x ≤ y²
For y = 2 (year 2000 isn't part of the 21st century), the possible results for n are:
n ε {2;3;5;7}
All of these dates are in the past.
If you insist in 2000 being part of the 21st century (actually wrong because there was no year 0), the possible results are:
n ε {2;3;5;7;11;13;17;19;23}
23 is the only of these dates not in the past, thus the logo will be removed tomorrow, on the 23rd day of the year.
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  #20  
Old 2008-01-22, 20:05
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there was no year 1 or 2 either.
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  #21  
Old 2008-01-22, 20:09
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Sure.
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  #22  
Old 2008-01-22, 23:52
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Kobold, Elmuerte, WHAT ARE YOU TALKING ABOUT!!!???
All those funny numbers and letters and symbols = WHAT?
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  #23  
Old 2008-01-22, 23:54
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It's called advanced mathematics.
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  #24  
Old 2008-01-23, 11:23
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You what?!
 
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Well the logo is still up so I use advance mathematics to say that it has not been taken down yet.
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  #25  
Old 2008-01-23, 12:36
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Then either my calculation is wrong, the day isn't over yet or El changed his mind. Not that the logo is hurting anyone, I didn't see any snow yet.
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