Find the elements $[x]_{50}$ such that $[15]_{50}\cdot[x]_{50} = [0]_{50}$

Question

I'm given the integer residue ring $R = \Bbb Z/50\Bbb Z$

I'm wanting to find all elements $[x]_{50}$ such that $[15]_{50}\cdot[x]_{50} = [0]_{50}$.

Because the prime factorizations of $15$ and $50$ are $15:3,5$ and $50:2,5,5$, wouldn't any multiple of $10$ work for $[15]_{50}\cdot[x]_{50} = [0]_{50}$, making $x=\{[10]_{50},[20]_{50},[30]_{50},[40]_{50}\}$? Do I only want to only include numbers in the complete canonical residue system? Did I write out the solutions correctly? Am I supposed to include $[0]_{50}$? I thought that because it's not a zero divisor that it doesn't get included.

Answer 1

Hint:

What integer(s) $\;n\;$ can you come up with such that $\;15n\;$ is a multiple of $\;50\;$ ...?