How does divisibility test using congruence work? In the book, it said:
Let $n = a_{k}10^{k} + a_{k-1}10^{k-1} + a_{k-2}10^{k-2} + ... + a_110 + a_0$
Then, because $10 \equiv 0 \pmod{2}$ it follows that $10^j \equiv 0 \pmod{2^j}$ 
What congruence property did they use in this case? Is that:
If $a \equiv b \pmod{k_1}$ and $c \equiv d \pmod{k_2}$ then,
$ab \equiv cd \pmod{k_1k_2}$ ?
I saw one property in the book, which is:
$a \equiv b \pmod{k}$ and $c \equiv d \pmod{k}$then,
$ab \equiv cd \pmod{k}$
But I really don't understand how this property relates to the one above it. Any idea?  

You you need to use this property $j$ times, since:
$10 \equiv 0 \mod{2} $ and $10 \equiv 0 \mod{2}$, then $10 \cdot 10 \equiv 10^2 \equiv 0 \mod{2^2}$
You know that $2|10$ so it must be that $2^2 | 10^2$ (factorization). Repeat again:
$10 \equiv 0   \mod{2}$ and $ 10^2 \equiv 0  \mod{2^2}$, then $10 \cdot 10^2 \equiv 10^3 \equiv 0  \mod{2^3}$
So in the end you get:
$10 \equiv 0  \mod{2}$ and $ 10^{j-1} \equiv 0 \mod{2^{j-1}}$, then $10 \cdot 10^{j-1} \equiv 10^j \equiv 0  \mod{2^j}$
