<s>[INST] 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.
 [/INST] 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.
</s>