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

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Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0@mukushla @KingGeorge @Zarkon

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.1can u type the equation here plz :)

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0I am not concerned with problem as much as its solution on Quora. And No @nincompoop

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.1thats 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?

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0I am pretty sure, it is.

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

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0but that's not what concerns me. it's the first answer by Anders i am unable to understand.

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.1oh sorry man lol :) im sooo blind

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.1i cant understand it :)

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1Sai Ganesh explanation also no good, right?

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

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, sorry, I was directing at Ishaan (he wants an "octal" explanation)

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah. It didn't help my puny brain much :(

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0@him1618 maybe you can?

him1618
 2 years ago
Best ResponseYou've already chosen the best response.0the explanation is good enough

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1I 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....

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0Specifically, I don't understand how did he conclude 'The rest can be divided into equal groups based on their first two nonzero digits.'

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0... 49* equal groups ...

estudier
 2 years ago
Best ResponseYou've already chosen the best response.1you got rid of the 0's so you got 7 digits left you can set, right?

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

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0Okay, one silly doubt. Why is he using only first two digits for grouping?

Ishaan94
 2 years ago
Best ResponseYou've already chosen the best response.0i was acting foolishly yesterday :( thanks.
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