<s>[INST] How does mod multiplication work? For example, $10^{10} \equiv 4\pmod{6}$
If I used $\pmod{2}$ and $\pmod{3}$, how does the multiplication process work? 
Since $10^{10} \equiv 0 \pmod{2}$ and $10^{10}\equiv 1\pmod{3}$,
$$
10^{10}\equiv (0,1) \pmod{(2,3)}
$$
how do we get the value $4$ at the end? do we list out the possible values of $0\pmod{2}$ and $1\pmod{3}$?
$$
1\pmod{3} = 1, 4, 7, 10
$$
so on.
Since only $4, 10$ and so on satisfy $\pmod{2}$, only values that satisfy both criteria can be used. 
In general, can we do this for $\pmod{n}$, $n$ being any integer?
 [/INST] This is called the "Chinese Remainder Theorem". You are given two congruent condition and you have to find the answer. I will show an example using slightly bigger numbers.
Suppose we are given
$$ n \equiv 3 \mod 7,  ~~\hbox{and}~~ n\equiv 5\mod 11$$
From the first we can write
$$ n = 7 k + 3$$
Substituting in the second, we get
$$ 7 k + 3\equiv5 \mod 11$$
or 
$$ 7 k \equiv2 \mod 11 \Rightarrow k \equiv (7^{-1}) 2 \mod 11$$
where the inverse is modulo $11$. One way to find $7^{-1} \mod 11$ is to use Euclid's algorithm. For small numbers it is easy to  look at 7,14,21 etc.
Actually since $21=-1 \mod 11$ we can stop and deduce that $$7 \cdot 3\equiv -1 \mod 11$$ or 
$$7 \, (-3) \equiv 1 \mod 11$$ If you prefer positive numbers, you can use $-3 \equiv 11-3 \mod 11$.
Hence
$$ 7^{-1} \equiv 8 \mod 11$$
Hence 
$$ k = 8 \cdot 2 = 16\equiv 5 \mod 11$$
Hence $k = 11 m + 5$ and finally
$$ n =7 \, (11 m + 5) + 3 = 77 m + 38$$
You can easily extend this to any number of congruents.
Added in response to OP's request
We want $$n \equiv 0 \mod 2, ~~~ n \equiv 1 \mod 3$$
From second condition (OP already did it in the other order, so I want to provide a second method)
$$ n = 3 k + 1$$
substituting in the first condition
$$ 3 k + 1 = 0 \mod 2 \Rightarrow k + 1 = 0 \mod 2 ~~~\hbox{since $3 \equiv 1 \mod 2$}$$
So $k = -1 \mod 2$ or if you prefer positive numbers 
$$ k = -1 + 2 \mod 2 \Rightarrow k = 2 m + 1$$
Substituting in the formula for $n$ we have
$$ n = 3 (2 m + 1) + 1 = 6m +4$$
So the possible values of $n$ are 4, 10, 16, 22, -2, -8, -14, -20 etc.
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