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?