Find integer $x$ such that -2310 ≤ $x$ ≤ 2310...

Question

Find integer $x$ such that -2310 ≤ $x$ ≤ 2310, and

$x$ ≡ 1 (mod 21),

$x$ ≡ 2 (mod 20),

$x$ ≡ 3 (mod 11)

-Currently covering a section on the Chinese remainder theorem and having trouble figuring this problem out or how to start it, any help/clarification is really appreciated.

Answer 1

First find $x$ such that $x \equiv 1 \mod 21$ and $2 \mod 20$, using the Euclidean algorithm on $1 + 21 a = x = 2 + 20 b$. You should get a solution of the form $x \equiv c \mod (20 \times 21)$. Then similarly find $x$ such that $x \equiv c \mod (20 \times 21)$ and $x \equiv 3 \mod 11$.

Answer 2

It is easy to see $x=22$ satisfies the first two congruencies. So, we can merge the first two inequalities to get the system $$x\equiv 22\pmod{420}\qquad x\equiv 3\pmod{11}$$ Thus, $11k+3=420t+22$. Looking at this equality at $\bmod{11}$, we get $11|2t-3$. So, $t\equiv7\pmod{11}$. Then, $$x\equiv 420\cdot7+22\equiv 2962\equiv -1658\pmod{4620}$$