solving a modular equation $6x=3\operatorname{mod}27, 6x=2\operatorname{mod}10$

Question

Determine the kernel of the following group homomorphism: $$ \phi\colon\mathbb Z/270\mathbb Z\to\mathbb Z/270\mathbb Z\colon\overline x\mapsto\overline{6x}. $$ Then find the solutions of the following system of equations in $\mathbb Z/270\mathbb Z$: \begin{align} 6x=3\mod 27\\ 6x=2\mod 10 \end{align}

Since $6*45=270$, the kernel is $\{\overline{45}, \overline{90}, \overline{135}, \overline{180}, \overline{225}, \overline{270}\}$. Instead of working with $6x$, I solved the following equations: \begin{align} a=3\mod 27\\ a=2\mod 10, \end{align} using the Chinese remainder theorem. I found: $a=192=32*6$. So I would guess they are looking for this answer: $$ \left\{\overline{32+k45}:k\in\{1,2,3,4,5,6\}\right\}. $$ But I'm a bit confused by their phrasing, because we initially solved for $x\in\{0,1,\dots,269\}$, and not for $\overline a\in\mathbb Z/270\mathbb Z$... So I could only sort of guess what I had to do, but could someone clarify their wording? Why can it be interpreted as: find $\overline a\in\mathbb Z/270\mathbb Z$, such that for $$ \phi'=\theta\circ\phi, $$ where $$ \theta\colon\mathbb Z/270\mathbb Z\to\mathbb Z/27\mathbb Z\times\mathbb Z/10\mathbb Z\colon a\mapsto (a,a), $$ we have $\phi'(\overline a)=(3,2)$.

edit

I think I got it: to find solutions in $\mathbb Z/n\mathbb Z$ just means to find solutions modulo $n$.

Answer 1

we can write $$6x=3+27m$$ and $$6x=2+10n$$ where $m,n$ are integers, from here we get $$1=10n-27m$$ solving this Diophantine equation we get $$m=7+10k,n=19+27k$$ and from here you will get $x$

Answer 2

\begin{align} 6x &\equiv 3 \pmod {27}\\ 2x &\equiv 1 \pmod{9} \\ \color{red}{x} &\color{red}{\equiv} \color{red}{5 \pmod{9}} \\ \hline 6x &\equiv 2 \pmod{10} \\ 3x &\equiv 1 \pmod{5} \\ \color{red}x &\color{red}{\equiv} \color{red}{2 \pmod{5}} \\ \end{align}

\begin{array}{c|cc} &\mod 5 &\mod 9 \\ \hline 9 & 4 & 0 \\ 5 & 0 & 5 \\ \hline \color{brown}{-9} & 1 & 0 \\ \color{brown}{10} & 0 & 1 \\ \hline \end{array}

$$ \left. \begin{array}{c} x \equiv 5 \pmod 9 \\ x \equiv 2 \pmod 5 \end{array} \right\} \implies x \equiv \color{brown}{-9}(\color{red}2) +\color{brown}{10}(\color{red}5) \equiv 32 \pmod{45} $$