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LolWolf
Group Title
Could anyone help out on Number theory? A proof of the following:
 one year ago
 one year ago
LolWolf Group Title
Could anyone help out on Number theory? A proof of the following:
 one year ago
 one year ago

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LolWolf Group TitleBest ResponseYou've already chosen the best response.0
\[\binom{pa}{pb}\equiv\binom{a}{b}\mod p^3\]Where p is prime and\[p>2\]
 one year ago

Lethal Group TitleBest ResponseYou've already chosen the best response.0
what class?
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
Well, it'd be a grad level number theory class, but it's a problem I was given in a summer camp as part of a worksheet.
 one year ago

Lethal Group TitleBest ResponseYou've already chosen the best response.0
haha well idk anything have you tried looking up similar problems or proofs like this http://f2.org/maths/nthproof.html
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
Eeyup, haven't found anything.
 one year ago

Lethal Group TitleBest ResponseYou've already chosen the best response.0
oh, well i'm sorry i won't be of any help then. I just barged in here due to the fact no one else was trying to help. good luck.
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
It's fine, thanks. Yeah, I haven't been able to solve this problem since quite the time.
 one year ago

Lethal Group TitleBest ResponseYou've already chosen the best response.0
and you could try posting this on Yahoo! answers.
 one year ago

Herp_Derp Group TitleBest ResponseYou've already chosen the best response.0
Have you tried: http://math.stackexchange.com/ http://mathoverflow.net/ http://www.mathworld.wolfram.com Wikipedia??? (Actually, if you post your question on the talk page of a user who seems knowledgeable, you can occasionally get an answer.)
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.0
I've tried the latter of the three, but I was going to post it on Math Overflow/Stack Exchange, haven't yet... thanks, though...
 one year ago
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