# Finding all remainders

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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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