$a\cdot(b^{-1}\bmod m)$ Can be be solved using modular multiplication

Question

Does $a\cdot(b^{-1}\bmod m) = (a\bmod m) \cdot(b^{-1}\bmod m).$

where $\bmod$ represents remainder left on division with $m$.

$b^{-1} \bmod m$ is multiplicative inverse.

Answer 1

I am assuming that you consider both sides modulo $m$. Then the answer is yes. More generaly, let $a=qm+r$ and $b=km+c$, then $$a\cdot b=(qm+r)(km+c)=m(qkm+cq+kr)+rc.$$ Hence $$ab \equiv rc \hspace{0.2 cm}(mod \hspace{0.2 cm} m).$$ On the other hand, $a\equiv r \hspace{0.2 cm}(mod\hspace{0.2 cm} m)$ and $b \equiv c\hspace{0.2 cm} (mod\hspace{0.2 cm} m)$. Therefore $$(a\equiv (mod \hspace{0.2 cm}m))\cdot (b \equiv (mod\hspace{0.2 cm} m))\equiv r c \hspace{0.2 cm}(\hspace{0.2 cm}mod m).$$