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

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 $n$.

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