# “Multi-facets” rope puzzle

I've done the following, can you tell me if it's correct?

If $n$ is the number of sides of the rope and $k$ is the number of rotation, e.g. $k=0$ for glue each side to itself then I think the number of colours needed is $$\# \tt{colours} = \gcd (n,k)$$

I think I can view the rope as $G = (\mathbb Z / n \mathbb Z, +)$ and $k$ as an element of $G$. Then the order of $k$ determines how many sides we can reach. In particular, we can reach $n/\gcd(n,k)$ sides with one colour, the size of the subgroup generated by $k$ (which equals the size of the subgroup generated by $\gcd(n,k)$).

Is this right? Thanks for help!

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Yes. Good thought. – Ross Millikan Oct 11 '12 at 16:14
@RossMillikan Thank you. Is it all correct as is? – Rudy the Reindeer Oct 11 '12 at 16:15
Yes it is........ – Ross Millikan Oct 11 '12 at 16:17
Spot on, I like it. +1 – Old John Oct 11 '12 at 16:18
Basically, if $G=\mathbb Z/n\mathbb Z$ then $G/kG \cong \mathbb Z/(k,n)\mathbb Z$ – Thomas Andrews Oct 11 '12 at 16:24

We can indeed view an $n$-faceted rope as $\mathbb Z / n \mathbb Z$ with addition. If $k$ denotes the number of sides we rotate by then the number of sides we colour with one colour is $n / \gcd (n,k)$ which is the size of the subgroup generated by $k$.
As pointed out in the comments by Thomas Andrews, we have $\mathbb Z / k \mathbb Z \cong \mathbb Z / \gcd (n,k) \mathbb Z$. If $\langle k \rangle$ denotes the subgroup generated by $k$ then $\langle k \rangle \cong \mathbb Z / (n / \gcd (n,k)) \mathbb Z$.