Finding U13 Cyclic Numbers: A Faster Way?

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Homework Statement



Is U13 cyclic?

The Attempt at a Solution



I know the elements are
{1,2,3,4,5,6,7,8,9,10,11,12}. I have eliminated 1,2,3,4,5 and I am working on 6. I am doing it this way:

60=1
61=6
62=10
63=8
64=9
65=2

..and so on, but I did, for example, 62=36-13=23=10 and that is how I found 10. But when I get into larger numbers it is very time consuming to type, for example, 64=1296-13=1283-13=1270-13...and so on. Does anyone know a faster way to solve these?
 
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If you're looking for a remainder, you could use the long division algorithm. Also, you can make use of what you've already found.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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