System of congruences and Chinese remainder theorem Find all the integers satisfying this system of congruences
$$\begin{cases}
x \equiv 2 \pmod 5\\
x \equiv 1 \pmod {10}\\
x \equiv 0 \pmod 3
\end{cases}
$$
I think you use Chinese remainder theorem but I'm not sure how to.
 A: As, $5|10,$
$$x\equiv1\pmod{10}\implies x\equiv1\pmod5$$ 
Again we have $x\equiv2\pmod 5$
But $1\not\equiv2\pmod5$
Hence there will be no solution
A: Hint $\ x\equiv 1\pmod{10}\,\Rightarrow\, x\equiv 1\pmod 5\ $ contra $\ x\equiv 2\pmod 5$
Remark $\ $ Generally we can employ the following criterion for existence of a solution
$$\begin{array}{} x\equiv a_1 \pmod{\!m_1 }\\ \quad \vdots \\ x\equiv a_k\pmod{\!m_k} \end{array}\  \text{is solvable}\ \iff\  \color{#c00}{a_i\equiv a_j}\!\!\!\! \pmod{\!\gcd(m_i,m_j)}\ \text{ for all }\ i,j$$
$(\Rightarrow)\ $ has an easy proof: $ $ if $\, d = \gcd(m_i,m_j)\,$ then $\,d\mid m_i,m_j\,$ so  $\ {\rm mod}\ d\!:\ \color{#c00}{a_i\equiv x\equiv a_j}\ $ 
A: Basically the two equivalences
\begin{align}
    x &\equiv a \pmod A \\
    x &\equiv b \pmod B
\end{align}
have a common solution if and only if $$a \equiv b \pmod{\gcd(A,B)}$$
In the case of \begin{align}
x &\equiv 2 \pmod 5\\
x &\equiv 1 \pmod {10}\\
x &\equiv 0 \pmod 3
\end{align}
We notice that $1 \not \equiv 2 \pmod{\gcd(5,10)}$, so there is no common solution to all three equivalences.
Another method is to reduce each equivalence into prime-power congruences and then remove redundant equivalneces. If there are no contradictory congruences , what you are left with is amenable to the regular CRT.
for your problem, $x \equiv 2 \pmod 5$ and $x \equiv 0 \pmod 3$ are already prime-power congruences. Since $10 = 2 \times 5$ we can break $x \equiv 1 \pmod {10}$ into the two prime-power congruences
\begin{align}
    x &\equiv 1 \pmod {2} \\
    x &\equiv 1 \pmod {5}
\end{align}
and we notice that $x \equiv 1 \pmod {5}$ and $x \equiv 2 \pmod {5}$ contradict each other. So, again, we know that the system of congruences has no common solution.
