I have the following congruence system:
$x \equiv 1 \mod 5 \\ x \equiv 2 \mod 7 \\ x \equiv 0 \mod 8 \\ x \equiv 3 \mod 11$
I used the Chinese Remainder Theorem to get a solution, but it only seems to work for the last three congruences, but not for $x \equiv 1 \mod 5$.
Here's my solution:
$x = 1\cdot (7\cdot 8\cdot 11)\cdot x_{1}+2\cdot (5\cdot 8\cdot 11)\cdot x_{2}+0\cdot (5\cdot 7\cdot 11)\cdot x_{3}+3\cdot (5\cdot 7\cdot 8)\cdot x_{4}$
$(7\cdot 8\cdot 11)\cdot x_{1} \equiv 1 \mod 5 \Rightarrow x_{1} = 1\\ (5\cdot 8\cdot 11)\cdot x_{2} \equiv 1 \mod 7 \Rightarrow x_{2} = 6\\ (5\cdot 7\cdot 11)\cdot x_{3} \equiv 1 \mod 8 \Rightarrow x_{3} = 1\\ (3\cdot 5\cdot 7)\cdot x_{4} \equiv 1 \mod 11 \Rightarrow x_{4} = 4\\$
$\Rightarrow x = 7912\\ m = 5\cdot7\cdot8\cdot11 = 3080\\ 7912 \mod 3080 = 1752$
Formula for all x:
$x=1752 + z\cdot3080, z \in \mathbb{Z}$
But if I use 1752 or any combination with 3080 for x, it only works for
$x \equiv 2 \mod 7 \\
x \equiv 0 \mod 8 \\
x \equiv 3 \mod 11 \\$
But not for $x \equiv 1 \mod 5$ because $1752 \mod 5 = 2$.
What did I do wrong? Can anyone spot my error?