Quotient over modulo

Question

Suppose $a$ is divisible by $b$, and their quotient $(\frac ab)$ is $k$. Does it hold that for any prime number $p$, $ k \bmod p= (a \bmod p * (b^{-1})\bmod p) \mod p$? Note that $b^{-1}$ is the modular inverse of $b$ with respect to $p$.

I am aware this statement does not hold for the quotient of $a$ by $b$ when $a \bmod b$ is not $0$. But in the case where $b$ does divide $a$, is this expansion valid?

Answer 1

Yes.

If $a=bk$ in $\Bbb Z$, then $a=bk$ in $\Bbb Z/p\Bbb Z$.

If furthermore $b$ is invertible then you can multiply both sides by $b^{-1}$ to get $$ab^{-1}=k$$ in $\Bbb Z/p\Bbb Z$