<s>[INST] Simultaneous congruences $3x \equiv 2 \pmod{5}$, $3x \equiv 4 \pmod{7}$, $3x \equiv 6 \pmod{11}$ I am stuck in a simultaneous linear congruence problem:
\begin{cases}
3x \equiv 2 \pmod{5} \\[4px]
3x \equiv 4 \pmod{7} \\[4px]
3x \equiv 6 \pmod{11}
\end{cases}
Using the Chinese remainder theorem, 
I started with the 'highest' divisor: $11$. Since $(3, 11) =1$ there is a unique solution. $x= 6 \cdot 3 ^{\phi (11) -1} \equiv 6 \cdot 3^4\pmod{11}$ But to be honest, I have no clue how to continue.
Perhaps, cancel out the last equation to: $x \equiv 2 \pmod{11}$?
 [/INST] First, you could observe that
\begin{cases}
3x \equiv 2 \pmod{5} \Leftrightarrow 3x \equiv 2+10 \equiv 12=3 \cdot 4 \pmod{5} \Leftrightarrow x \equiv \color{red}{4} \pmod{5} \\[4px]
3x \equiv 4 \pmod{7} \Leftrightarrow 3x \equiv 4+14 \equiv 18=3 \cdot 6\pmod{7} \Leftrightarrow x \equiv \color{lime}{6} \pmod{6} \\[4px]
3x \equiv 6 \pmod{11} \Leftrightarrow x \equiv \color{blue}{2} \pmod{11} \\[4px]
\end{cases}
Next, you could find integers $a, b, c$ with
\begin{cases}
a \cdot 7 \cdot 11 \equiv 1 \pmod{5} \Leftrightarrow 2a \equiv 1 \pmod{5} \Leftrightarrow a \equiv 3 \pmod{5}\\[4px]
b \cdot 5 \cdot 11 \equiv 1 \pmod{7} \Leftrightarrow 6b \equiv 1 \pmod{7} \Leftrightarrow b \equiv 6 \pmod{7} \\[4px]
c \cdot 5 \cdot 7 \equiv 1 \pmod{11} \Leftrightarrow 2c \equiv 1 \pmod{11} \Leftrightarrow c \equiv 6 \pmod{11} \\[4px]
\end{cases}
Then, all solutions of your simultaneous congruences are $$x \equiv a \cdot 7 \cdot 11 \cdot \color{red}{4} + b \cdot 5 \cdot 11 \cdot \color{lime}{6} + c \cdot 5 \cdot 7 \cdot \color{blue}{2} \pmod{5 \cdot 7 \cdot 11}$$ so $x \equiv 244 \pmod{385}$.
Btw: Did anybody notice that this way of constructing a solution is exactly the same as finding an interpolation polynomial using Lagrange polynomials?
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