Zero divisors of a ring Let $R = \mathbb Z/25\mathbb Z$ and $S = \{1,7,18,24\}$


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*How many elements of $R$ are non-zero-divisors? How many are invertible?

*Now consider the subring $RS$. How many elements of $RS$ are non-zero, how many are zero-divisors, how many are invertible? 


For (1) I think $R$ has 24 non-zero elements. Also $S$ is set of powers of $7\pmod {25}$.
Thanks for your help!
 A: A zero divisor of a ring $R$ is an element $r \neq 0$ of $R$ which satisfies $rs = 0$ for some nonzero $s$ in $R$. So you haven't answered question (1). Yes, there is only one $0$ in $R$ (this is part of the definition of a ring), but some of the $r \neq 0$ in $R$ might still be zero divisors.
Here's a hint:
Remember that $R = \mathbb{Z}/25\mathbb{Z}$ is a set of equivalence classes $[m]$ of integers modulo 25, i.e., under the equivalence $m_1 \sim m_2 \iff 25 \mid (m_2-m_1)$. So the integers $n \in \mathbb{Z}$ that are in $[0]$ (i.e., $n$ such that $0 \sim n$) are the integers that are divisible by 25.
If $m \in \mathbb{Z}$ and $[m]$ is a zero divisor in $R$, then $[m][n] = [mn] = [0]$ for some nonzero $n \in \mathbb{Z}$. Think about what this says: $[m]$ is a zero divisor in $R$ if and only if there is a nonzero integer $n$ such that $25$ divides $mn$. For which $m \in \mathbb{Z}$ is this possible? How many equivalence classes do these $m$ fall into?
Here's an example of a zero divisor: Consider $[4]$ in the ring $T = \mathbb{Z}/12\mathbb{Z}$. Since $4 \cdot 3 = 12$, we know that $[4][3] = [12] = [0]$. (Of course $[m]$ is a different set of integers for $T$ than it was for $R$, since these rings have different notions of equivalence.)
Hint 2: Once you know the number of zero divisors, you get the number of invertible elements (a.k.a. "units") for free. Why? (How are they related in $\mathbb{Z}/n\mathbb{Z}$?)
Hint 3: For (2), can you find $a \in \mathbb{Z}$ such that $7a \equiv 1 \pmod{25}$? What about finding such an $a$ for other $s \in S$?
Hint 4: Slightly more abstractly, are the elements of $S$ zero divisors, units, or something else? The answer should give you information about $RS$.
A: If $a$ is not a zero divisor then $a^k = 1$ has a solution.  Conversely if $a$ is a unit it is not a zero divisor.  So $a$ is a zero divisor iff it's not a unit.  So take $|R| - |$ units of R $|. \ $  $a$ is a unit iff $\gcd(a, |R|) = 1$.
Since $S$ contains $1$, $RS = R$.  So the problem becomes counting the zero-divisor, non-zero, and unit elements of $R$.
