divisibility for numbers like 13,17 and 19 - Compartmentalization method For denominators like 13, 17 i often see my professor use a method to test whether a given number is divisible or not. The method is not the following :
Ex for 17 : subtract 5 times the last digit from the original number, the resultant number should be divisible by 17 etc...
The method is similar to divisibility of 11. He calls it as compartmentalization method. Here it goes.
rule For 17 :
take 8 digits at a time(sun of digits at odd places taken 8 at a time - sum of digits at even places taken 8 at a time)
For Ex : $9876543298765432..... 80$digits - test this is divisible by 17 or not.
There will be equal number of groups (of 8 digits taken at a time) at odd and even places. Therefore the given number is divisible by 17- Explanation.
The number 8 above differs based on the denominator he is considering.
I am not able to understand the method and logic both. Kindly clarify.
Also for other numbers like $13$ and $19$, what is the number of digits i should take at a time? In case my question is not clear, please let me know.

The Similar can also be said for divisibility by $13$
$10^3/13$ gives a remainder of $-1$
$10^6/13$ gievs a remainder of $+1$
So again the rule for $13$, will be same as the rule for $7, 11$.
Group the numbers into triplets as we see alternate changes of $-1$ to $+1$ between 
every three powers of $10$. So the rule for $13$ will also be 
(Sum of triplets at odd places - Sum of triplets at even places)
or 
Sum of digits at odd places taken $3$ at a time - Sum of digits at even places taken three at a time)
As i said, all this can be summarised as:
$10^x/B = +1$ holds for divisors (values of B) like $3,9, 37$ 
$10^x/B = -1$ and $+1$ for divisiors (values of B) like $7, 11, 13, 17$
$10^x/B = 0$ for divisors (values of B) like powers of $2$ and $5$
where $+1,0,-1$ are the remainders of the respective divisions
