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If $\operatorname{ord}_ma=10$, find $\operatorname{ord}_ma^6$. Justify you're answer.

I'm not sure what I should say for my answer to be justified. However, I expect $\operatorname{ord}_ma^6=5$, because $5\cdot 6=30$ and $10\mid 30$.

Like I said, no idea if this is how to go about solving this problem.

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    I expect that this, together with "No lower positive multiple of $6$ is divisible by $10$" should do the trick.2012-11-04

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If $ord_ma=d, ord_m(a^k)=\frac{d}{(d,k)}$ (Proof @Page#95)

Here $d=10,k=6\implies ord_m(a^6)=\frac{10}{(10,6)}=\frac{10}2=5$

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    The linked proof is also on our site, e.g. see [this simpler one-line proof.](http://math.stackexchange.com/a/134139/242)2012-11-05
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Hint $\ $ If $\rm\ ord\ a\, = 10\:$ then $\rm\:a^{6n}\equiv 1\iff 10\mid 6n\iff 5\mid 3n\iff 5\mid n.$

Remark $\ $ Above we invoked: $\ $ if $\rm\:ord\ a\, = k\:$ then $\rm\:a^n\equiv 1\iff k\mid n.\:$ This can be viewed much more conceptually: the set $\cal O$ of integers $\rm\,n\,$ with $\rm\:a^n\equiv 1\:$ is closed under subtraction, hence, by a fundamental theorem, every element of $\cal O$ is a multiple of the least positive element of $\cal O,\,$ which in our case is called the order.