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.
Best Answer
You quote two different rules with different results. When testing for divisibility by 17 by subtracting 5 times the last digit from the orignal number without its last digit, you are using the fact that $51$ is divisible by $17$, so $10a+b \equiv 10a-50b \pmod {17}$, then the fact that $10(a-5b)$ is a multiple of $17$ if and only if $(a-5b)$ is. Unless you do further computation, you lose the remainder if the original number is not a multiple.
When you take blocks of 8 digits, you use the fact that $10^8+1 \equiv 0 \pmod {17}$, so $10^8a+b \equiv b-a \pmod {17}$ You retain the remainder in this case. For 13, you need half the period of its repeating decimal, which is 6, so you use blocks of 3. Note that $10^3+1=1001 \equiv 0 \pmod {13}$