adijf0qnf0qi3d01.zulander
Obvious trap question, it's 0 :3
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lim {rounds of rules applied -> inf } PppnRppp where p=pp and ppp=p = 2* 2/3 = 4/3
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it's an undefined number since pppnrppp turns into ppppppnrpppppp which turns into ppnrpp = ppppnrpppp = pppnrppp (pppp is 3 p's + 1 p / 3 turns into 1, 1 turns into 2) = nr = 0 p's
but since there are so many random uninteresting ways to go about this, my answer will be infinite p's :D
(or 42, according to this dutch page on wikipedia it's the answer to everything ^^ https://nl.wikipedia.org/wiki/Het_antwoord_op_de_ultieme_vraag_over_het_Leven,_het_Universum,_en_Alles)
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hahahahaaa, thank you so much for this, can't stop laughing now XD
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When you say infinite p's my heart skipped one beat ^^ that would be a dream come true ^^
But since the first question was kinda uninteresting from the begin with, the only thing I was after was the answer to question number two. ^^
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It wil never be an infinite p's, if you apply the p=pp and ppp=p an infinite times, you'll always come back to 6 which is actually the highest amount of p's OP's nickname can carry according to his rules.
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ppp = p (2 of them) p = pp (2 also)
6/3 = 2 2x2= 4???? Maybe theres infinite amount of p....
Or just read it from ur name XD
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Too bad I was late, I like logic questions like this.
PppnRppp
there are 5 p's in your name, P is not considered because it's capital and P != p.
3 p's will become one p so there are only 3 p's left, now every p becomes 2 p's, that 3x2. So the answer is obviously 6.
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Guardians of Orion
but they changed name to the Orion project
http://store.steampowered.com/app/407840/
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Ended ^^
Just answer one question. Case insensitive!
First one to answer the correct question with the correct answer gets a game
If the first one already have this game, it goes to the second, and so on
if p=pp and ppp=p?
How many "p" is in my nick?
Gabzus was first! ^^
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