A number is divisible by 13 I am studying divisibility and come across this rule. I think the rule is too complicated and hard to understand and remember. What is the best way to judge whether a number is divisible by 13 without having to remember this rule? 

Delete the last digit from the number, and then subtract 9 times the deleted digit from the remaining number. If what is left is divisible by 13, then so is the original number. Repeat the rule if necessary. 


Answer to the OP's Question: 
The best way to judge this divisibility by $13$ would be to use the congruence relation $$1000 \equiv -1\pmod {13}$$
Using this relation, it will be very easy to check for divisibility by $13$ for large-digited numbers (greater than $3$ digits). However, this method does not work for numbers with $1,2,3$ digited numbers. Hence for numbers with less than $4$ digits, it is better to divide and check what happens.
Justification of the Rule:
Say $N=a_0+10a_1+10^2a_2+\ldots +10^na_n$ is any number and $a_0$ is the digit in its units' place.
On deleting the last digit, the number becomes $a_1+10a_2+\ldots +10^{n-1}a_n$
And on subtracting $9$ times the deleted digit, the number becomes $a_1+10a_2+\ldots +10^{n-1}a_n-9a_0$
Then according to the given algorithm, you have to check if
$$a_1+10a_2+\ldots +10^{n-1}a_n-9a_0 \equiv 0 \pmod{13}$$
If the above holds, then it means
$$a_1+10a_2+\ldots +10^{n-1}a_n-9a_0 \equiv 0 \pmod{13}$$
Or $$10a_1+10^2a_2+\ldots +10^na_n-90a_0 \equiv 0 \pmod{13}$$
Or $$a_0+10a_1+10^2a_2+\ldots +10^na_n-91a_0 \equiv 0 \pmod{13}$$
Or $$a_0+10a_1+10^2a_2+\ldots +10^na_n \equiv 91a_0 \pmod{13}$$
And irrespective of $a_0$, $13$ divides $91$. So we get 
$$a_0+10a_1+10^2a_2+\ldots +10^na_n \equiv 91a_0 \equiv 0\pmod{13}$$
that is, $$N \equiv 0\pmod{13}$$
that is, $$13|N$$
