0
$\begingroup$

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 $n$.

$\endgroup$
0
$\begingroup$

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 $x$

$\endgroup$
0
$\begingroup$

\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 & 4 \\ \hline \color{brown}{-9} & 1 & 0 \\ \color{brown}{10} & 0 & 1 \\ \hline \end{array}

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.