Solution system $3x \equiv 6\,\textrm{mod}\,\, 12$, $2x \equiv 5\,\textrm{mod}\,\, 7$ , $3x \equiv 1\,\textrm{mod}\,\, 5$ Have solution the following congruence system?
$$\begin{array}{ccl}
      3x & \equiv & 6\,\textrm{mod}\,\, 12\\
      2x & \equiv & 5\,\textrm{mod}\,\, 7\\
      3x & \equiv & 1\,\textrm{mod}\,\, 5
    \end{array}$$
Point of Interest: This question requires some special handling due to the mixture of factors among the moduli. This is more than the run of the mill Chinese Remainder Theorem problem

The congruences simplify to 
$$\left\{ 
\begin{array}{ccc}
x&\equiv& 2\pmod{4} \\
x&\equiv& 6 \pmod{7}\\
x&\equiv& 2 \pmod{5}
\end{array}
\right.$$
The first and third congruences combine to $x\equiv 2 \pmod{20}$.
Then adding the modulus, we have
$\pmod{20}$: $x\equiv 2\equiv 22\equiv 42\equiv 62$
And $62$ satisfies the remaining congruence.  So the solution is $x\equiv 62\pmod{140}$
