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I need yours help one more time.
I have got very hard task to do but i don' know how to do it :)
I need to find all possible remainders of $6^n \bmod 9$.
It's very important to me :)
Thanks for help,
John

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1  
I dont know how old you are , but - with respect - this is a trivial question and i read the tag homework. So im not sure if i should answer and even less if you are suppose to ask this. I think it is best to think about this yourself. Its not really difficult. I dont know the exact intentions of your book or teacher so i dont know what i should say more. Try some things on paper. I cant say another more usefull thing considering the other replies that are already given. Good luck. –  mick Sep 2 '12 at 21:19
    
Adding on Mick's words: you could at the very least show some self effort, give some ideas, what have you read in your book, in the web...something to show you're not expecting others to do your homeweokr for you. -1 –  DonAntonio Sep 3 '12 at 12:05

3 Answers 3

Try n=0, 1, 2, 3, 4 ... can you spot a pattern?

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The remainder is $0$ if $n \ge 2$, for then $36$ divides $6^n$. You can compute separately the remainders when $n=0$ and $n=1$. Not a hard task!

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Simpler:

$$6^n=2^n 3^n$$

For $n\geq 2$, we hav a factor of $9$ so the remainder is $0$. Else, you get $1$ and $6$, which should be fairly easy to compute.


Well, using the binomial theorem, $$6^n=(9-3)^n=9^n+\text{many multiples of 9}+(-3)^n $$

then, $\mod 9$, we must have

$$6^n \mod 9\equiv (-3)^n\mod 9$$

If $n\geq 2$, we have the remainder is $0$, since we have $3^2=9$, and from there on we always have a factor of $9$. And for $n=0$ and $n=1$, the computation is straightforward.

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@AndréNicolas I'm lost on that one. Where? –  Pedro Tamaroff Sep 4 '12 at 23:02
    
Something about the limit of $(1+n)^{1/n}$. A while ago someone got upset, after editing one of my posts, that I edited further, hence erasing his or her name from the edit credit. There was an accusation that I was trying to erase the name by re-editing. So I am trying to avoid that. –  André Nicolas Sep 5 '12 at 1:19

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