Chinese Remainder Theorem with 0 mod n I'm trying to get the least x from a system of congruences by applying the Chinese Remainder Theorem. Keep running into issues.
System of congruences: 
$$
x \equiv 0 (_{mod} 7) \\
x \equiv 5 (_{mod} 6) \\
x \equiv 4 (_{mod} 5) \\
x \equiv 3 (_{mod} 4) \\
x \equiv 2 (_{mod} 3) \\
x \equiv 1 (_{mod} 2)
$$
I made the moduli relatively prime by removing 2nd and 6th congruence: 
$1(_{mod} 2)$ is just a special case of $3 (_{mod} 4)$
and the 6th congruence because 
$5 (_{mod} 6)$ splits into 
$2 (_{mod} 3)$ and $1 (_{mod} 2)$, both of which are already represented in the system.  
Product of moduli $m = 7 . 5 . 4 . 3 = 420$ 
Each respective $M_n$
$$
M_1 = 420/7 = 60 \\
M_2 = 420/5 = 84 \\
M_3 = 420/4 = 105 \\
M_3 = 420/3 = 140 \\
$$
Each respective modular inverse $y_n$
$$
y_1 = 0 \\
y_2 = 4 \\
y_3 = 3 \\
y_4 = 2 \\
$$
Trying to find solutions via CRT, I get
$x \equiv a_1  M_1 . y_1 + a_2  M_2 . y_2 + a_3  M_3 . y_3 + a_4  M_4 . y_4$
Plugging in the values:
$$
x \equiv 0 + 4 . 84 . 4 + 3 . 105 . 3 + 2 . 140 . 2 = 2849 \\
\equiv 329 (mod 42)
$$
And the "least" value being #329$ . 
However, 329 doesn't satisfy the equation $329 (_{mod} 4) \equiv 3$. 
What / where am I messing up?

This is very similar to aid78's solution.
\begin{align}
x &\equiv 0 \pmod 7 \\
x &\equiv 5 \pmod 6 \\
x &\equiv 4 \pmod 5 \\
x &\equiv 3 \pmod 4 \\
x &\equiv 2 \pmod 3 \\
x &\equiv 1 \pmod 2
\end{align}
can be rewritten as
\begin{align}
x &\equiv \phantom{-}0 \pmod 7 \\
x &\equiv -1 \pmod 6 \\
x &\equiv -1 \pmod 5 \\
x &\equiv -1 \pmod 4 \\
x &\equiv -1 \pmod 3 \\
x &\equiv -1 \pmod 2
\end{align}
Since $\operatorname{lcm}(2,3,4,5,6)=60$, then the last five congruences are equivalent to $x \equiv -1 \pmod{60}$ So now you have to solve
\begin{align}
x &\equiv \phantom{-}0 \pmod 7 \\
x &\equiv -1 \pmod{60}
\end{align}
$x \equiv -1 \pmod{60}$ is equivalent to saying $x = 60n -1$ for some integer $n$.
This is one way to finish the problem.
\begin{align}
   x &\equiv 0 \pmod{7} \\
   60n - 1 &\equiv 0 \pmod{7} &\text{(Note $60\equiv 4 \pmod 7$)} \\
   4n &\equiv 1 \pmod 7 \\
   8n &\equiv 2 \pmod 7 \\
   n &\equiv 2 \pmod 7 \\
   n &= 7m + 2 \\
   x &= 60n - 1 \\
   x &= 60(7m + 2) - 1 \\
   x &= 420m + 119 \\
   x &\equiv 119 \pmod{420}
\end{align}
