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First answer (by Anders) on http://www.quora.com/Mathematics/HowwouldIshowthat49divides8n7n1forallnge0
Can anyone explain it to me? Hints'd be appreciated.
 one year ago
 one year ago
First answer (by Anders) on http://www.quora.com/Mathematics/HowwouldIshowthat49divides8n7n1forallnge0 Can anyone explain it to me? Hints'd be appreciated.
 one year ago
 one year ago

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Ishaan94Best ResponseYou've already chosen the best response.0
@mukushla @KingGeorge @Zarkon
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
can u type the equation here plz :)
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
I am not concerned with problem as much as its solution on Quora. And No @nincompoop
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
thats not a true statement of binomial theorem\[8^n=(1+7)^n=\sum_{k=0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) 7^n\]right?
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
I am pretty sure, it is.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
oops sorry that should be\[8^n=(1+7)^n=\sum_{k=0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) 7^k\]
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
but that's not what concerns me. it's the first answer by Anders i am unable to understand.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
oh sorry man lol :) im sooo blind
 one year ago

mukushlaBest ResponseYou've already chosen the best response.1
i cant understand it :)
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
\[ 8^n=(1+7)^n=\sum_{k=0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) 7^n = 1 + 7n + O(7^{(n \geq 2)})\]
 one year ago

estudierBest ResponseYou've already chosen the best response.1
Sai Ganesh explanation also no good, right?
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
\( O(7^{(n \geq 2)}) \) is always divisible by 49
 one year ago

estudierBest ResponseYou've already chosen the best response.1
Yes, sorry, I was directing at Ishaan (he wants an "octal" explanation)
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
Yeah. It didn't help my puny brain much :(
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
@him1618 maybe you can?
 one year ago

him1618Best ResponseYou've already chosen the best response.0
the explanation is good enough
 one year ago

estudierBest ResponseYou've already chosen the best response.1
I have to say that the binomial answer is much more natural, it would not occur to me to set out on an octal adventure for that particular problem....
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
Specifically, I don't understand how did he conclude 'The rest can be divided into equal groups based on their first two nonzero digits.'
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
... 49* equal groups ...
 one year ago

estudierBest ResponseYou've already chosen the best response.1
you got rid of the 0's so you got 7 digits left you can set, right?
 one year ago

estudierBest ResponseYou've already chosen the best response.1
So if all the numbers are divisible into 49 groups, then divisibilty by 49 follows. At least, that's what I think it says.
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
Okay, one silly doubt. Why is he using only first two digits for grouping?
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
i was acting foolishly yesterday :( thanks.
 one year ago
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