Here's the question you clicked on:
Ishaan94
First answer (by Anders) on http://www.quora.com/Mathematics/How-would-I-show-that-49-divides-8-n-7n-1-for-all-n-ge-0 Can anyone explain it to me? Hints'd be appreciated.
@mukushla @KingGeorge @Zarkon
can u type the equation here plz :)
I am not concerned with problem as much as its solution on Quora. And No @nincompoop
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?
I am pretty sure, it is.
oops sorry that should be\[8^n=(1+7)^n=\sum_{k=0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) 7^k\]
but that's not what concerns me. it's the first answer by Anders i am unable to understand.
oh sorry man lol :) im sooo blind
i cant understand it :)
\[ 8^n=(1+7)^n=\sum_{k=0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) 7^n = 1 + 7n + O(7^{(n \geq 2)})\]
Sai Ganesh explanation also no good, right?
\( O(7^{(n \geq 2)}) \) is always divisible by 49
Yes, sorry, I was directing at Ishaan (he wants an "octal" explanation)
Yeah. It didn't help my puny brain much :(
@him1618 maybe you can?
the explanation is good enough
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....
Specifically, I don't understand how did he conclude 'The rest can be divided into equal groups based on their first two nonzero digits.'
... 49* equal groups ...
you got rid of the 0's so you got 7 digits left you can set, right?
So if all the numbers are divisible into 49 groups, then divisibilty by 49 follows. At least, that's what I think it says.
Okay, one silly doubt. Why is he using only first two digits for grouping?
i was acting foolishly yesterday :( thanks.