Finding U13 Cyclic Numbers: A Faster Way?

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SUMMARY

The discussion focuses on finding U13 cyclic numbers using a more efficient method than manual calculations. The user has identified elements {1,2,3,4,5,6,7,8,9,10,11,12} and has begun eliminating numbers through a method involving modular arithmetic. They suggest using the long division algorithm to expedite the process of finding remainders, which is crucial for determining cyclicity. The conversation emphasizes the need for a systematic approach to handle larger numbers effectively.

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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.
 

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