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.

Question

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.

anonymous · Answer

@mukushla @KingGeorge @Zarkon

anonymous · Answer

can u type the equation here plz :)

anonymous · Answer

@mukushla try again.

anonymous · Answer

I am not concerned with problem as much as its solution on Quora. And No @nincompoop

anonymous · Answer

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?

anonymous · Answer

I am pretty sure, it is.

anonymous · Answer

oops sorry that should be$$8^n=(1+7)^n=\sum_{k=0}^{n}   \left(\begin{matrix}n \ k\end{matrix}\right) 7^k$$

anonymous · Answer

oh yeah. k. lol

anonymous · Answer

but that's not what concerns me. it's the first answer by Anders i am unable to understand.

anonymous · Answer

oh sorry man lol :) im sooo blind

anonymous · Answer

i cant understand it :)

experimentx · Answer

$$ 8^n=(1+7)^n=\sum_{k=0}^{n}   \left(\begin{matrix}n \ k\end{matrix}\right) 7^n = 1 + 7n + O(7^{(n \geq 2)})$$

anonymous · Answer

Sai Ganesh explanation also no good, right?

experimentx · Answer

$ O(7^{(n \geq 2)}) $ is always divisible by 49

anonymous · Answer

Yes, sorry, I was directing at Ishaan (he wants an "octal" explanation)

anonymous · Answer

Yeah. It didn't help my puny brain much :(

anonymous · Answer

@him1618 maybe you can?

anonymous · Answer

the explanation is good enough

anonymous · Answer

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....

anonymous · Answer

true

anonymous · Answer

Specifically, I don't understand how did he conclude  'The rest can be divided into  equal groups based on their first two nonzero digits.'

anonymous · Answer

... 49* equal groups ...

anonymous · Answer

7*7

anonymous · Answer

you got rid of the 0's so you got 7 digits left you can set, right?

anonymous · Answer

right.

anonymous · Answer

So if all the numbers are divisible into 49 groups, then divisibilty by 49 follows.

At least, that's what I think it says.

anonymous · Answer

Okay, one silly doubt. Why is he using only first two digits for grouping?

anonymous · Answer

because 7*7  = 49

anonymous · Answer

i was acting foolishly yesterday :(

thanks.