First off: I am completely new to modular arithmetic.
I am currently trying to solve the Problem:
$x \equiv3 \pmod{19}$
$x \equiv1 \pmod{20}$
$x \equiv2 \pmod{21}$
$x \equiv1 \pmod{22}$
$x \equiv0 \pmod{23}$
I am using the Chinese remainder theorem, which states that if one knows the remainders of the Euclidean division of an integer n by several integers, then one can determine uniquely the remainder of the division of n by the product of these integers, under the condition that the divisors are pairwise coprime.
I know that $x \equiv0 \pmod{23}$ should be irrelevant to solve this problem, since it cancels itself out to $0$ in the final solution step. Also the $\gcd(20,22) \neq 1$. So I am trying to find a new statement to replace $20$ and $22$. After seeing the first answer of this post, i figured the answer I am looking for might be $x \equiv 1\pmod{220}$. But I don't know how to prove it. Is there some theorem or rule that describes this problem?
In the linked post, they converted
$x \equiv 3 \pmod 4$, and $x \equiv 5 \pmod 6$, to $x \equiv 11 \pmod{12}$
Edit 1:
I know how to find the x value, given that all bases are coprime. But I don't know how to break up modular statements without coprime bases into new statements with coprime bases.
(A) When we have $x \equiv 1 \mod 220$ , it is Equivalent to :
$x=220X+1$
That gives :
$x=22(10X)+1$ which is $x \equiv 1 \mod 22$
$x=20(11X)+1$ which is $x \equiv 1 \mod 20$
(B) When we have $x \equiv 1 \mod 22$ & $x \equiv 1 \mod 20$ , it is Equivalent to :
$x=22Y+1$ & $x=20Z+1$
Which gives :
$x=22Y+1=x=20Z+1$
Hence , $22Y=20Z$ , $11Y=10Z$
Naturally , $Y$ must have factor $10$ , while $Z$ must have factor $11$.
$Y=10y$ & $Z=11z$
Hence , we get :
$x=22(10y)+1$ & $x=20(11z)+1$
Hence $z=220y+1$ & $x=220z+1$
Which is :
$x \equiv 1 \mod 220$
DONE
I am listing out the long calculations , because OP is new to Modular Arithmetic , though it is not necessary : When OP gets comfortable with Modular Arithmetic , it can be simplified.