Question: I have two coins $A$ and $B$. The diameter of coins A is $d\_A=18$ mm. The smaller coin $B$ rolls around $A$ without sliding. Both coins have marks on the edge. At the beginning of the movement, the coins touch the marks (see the figure). 

[![enter image description here](https://i.stack.imgur.com/5T9GY.jpg)](https://i.stack.imgur.com/5T9GY.jpg)

To make the labels match again the first time, coin B should make two turns around $A$. 

**Question.** What can be the diameter of coin $B$, $d\_B$?

**My attempt.** The length of the circumference $C\_A$ is related to the diameter $d\_A$ by: $C\_A=\pi d\_A$, and the length of the circumference $C\_B$ is $C\_B=\pi d\_B$. We know that $2 C\_B = C\_A$, therefore, $C\_B =\frac{ C\_A}{2}$, next $\pi d\_B = \pi \frac{d\_A}{2}$, $d\_B = \frac{d\_A}{2}=18/2=9$mm. 

But correct answer is $12$ mm. Where is my error?

Answer: 
$12$mm is just one possible answer. What we need is that $B$ makes an odd number of revolutions while circling $A$ twice. So this would mean that twice the circumference of $A$ will be $k$ times the circumference of $B$, where $k$ is odd (to avoid matching on the first time around).

So the options are that $B$ has $\frac 23, \frac 25, \frac 27, \frac 29$ etc. of the circumference of $A$, and hence also the same proportion of the diameter.

For an integer mm diameter for $B$, we have the options of $\frac 23\times 18=12$mm and $\frac 29\times 18=4$mm.
If $2C\_B=C\_A$, then $B$ would have to go around $A$ just one time to 'match up'!