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: $x \equiv 2 \pmod{11}$?
 A: 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?
A: It just jumped out at me that if I define $y=x+1$ the first two become 
$$3y \equiv 0 \pmod 5\\3y \equiv 0 \pmod 7$$
which clearly calls for $y$ to be a multiple of $35$.  Now we have $$3y \equiv 9 \pmod {11}\\y \equiv 3 \pmod {11}$$
We note that $35 \equiv 2 \pmod {11}$ so $7 \cdot 35 = 245 \equiv 7\cdot 2 \equiv 14\equiv 3 \pmod{11}$ and $y=245, x=244$ is a solution.
A: This is an old-fashioned solution.
$$3x \equiv 2 \mod 5 \qquad 3x \equiv 4 \mod 7 \qquad 3x \equiv 6 \mod{11}$$
\begin{array}{|r|rrr|}
\hline
 385 &  5 &  7 & 11 \\
\hline
  77 &  2 &  0 &  0 \\
  55 &  0 &  6 &  0 \\
  35 &  0 &  0 &  2 \\
\hline
 231 &  1 &  0 &  0 \\
 -55 &  0 &  1 &  0 \\
-175 &  0 &  0 &  1 \\
\hline
\end{array}
$$3x \equiv 2(231) -4(55) -6(175) \equiv -808 + 385^{\#} \equiv -423 \pmod {385}$$
$$x \equiv -141 \equiv 244 \pmod {385}$$
($\#:$ Adding $385$ makes the answer, $-423$, a multiple of $3$.)
