Q:

Finding a generator for the intersection of two subgroups

In $\mathbb{Z}_{24}$ find a generator for $\langle 21\rangle\cap \langle 10\rangle$.

My attempt:
$\langle 21\rangle = \{21, 18, 15, 12, 9, 6, 3 ,0 \}$ from adding multiples of $21\pmod{24}$
$\langle 10\rangle = \{10,20,6, 16, 2, 12, 22, 8, 18, 4 , 14, 0\}$ from adding multiples of $10\pmod{24}$ 
Then I take the intersection:
$ \langle 21\rangle \cap \langle 10\rangle = \{0, 6, 12, 18\}$
I think I have the intersection of subgroups right but now what do I add to each of them to check which if any are a generator? Did I do it correctly up to this point, also?

A:

So we have:
$$\langle 21\rangle=\{21,18,15,12,9,6,3,0\}$$
and
$$\langle 10\rangle=\{10,20,6,16,2,12,22,8,18,4,14,0\}$$
and therefore 
$$ \langle 21\rangle \cap \langle 10\rangle = \{ 18,12,6,0\}$$
Since this is a pretty small set, you can check each member if they do or do not generate it. Then, you will find that $6$ certainly does.

