General Formula for Divisibility tests

I had given in an earlier blog a procedure for divisibility of any number by the prime 41.I had mentioned about a sequence of numbers to be remembered for this procedure viz 1,10,18,16,37...to be repeated as required.
These numbers are arrived at by finding the remainders when 1,10,100,1000,10000,100000,...etc are divided by 41.For 1 and 10 as they are less than 41,they should be considered as remainders.18 is the remainder when 41 divides 100,...16 the remainder when it divides 1000 etc etc.After 5 steps the same remainders get repeated
The sequence of numbers in some cases go a long way.For instance for 17,you will have 16 numbers before repetition starts,for 19
you will have 18 numbers before repetition etc
I am giving below the sequences for some other primes.Where the sequence contains more than 10 numbers,I am giving only the first 10 numbers of the sequence.
Prime 17- 1,10,15,14,4,6,9,5,/16,7....etc  The mark (/)  shown after the first 8 numbers means that the remaining 8 numbers can be obtained by deducting the first 8 numbers successively from 17.
Prime 19-1,10,5,12,6,3,1115,17,/18......etc....
Prime 73-1,10,27,51,/72,63,46,27 repeated later
Prime 23-1,10,8,11,18,19,6,14,2,20......etc
Prime 29-1,10,13,16,15,5,21,7,12,6......etc
Prime 31-1,10,7,8,18,25,2,20,14,16.....etc
Prime 43-1,10,14,11,24,25,35,6,17,41...etc
Prime 47-1,10,6,13,36,31,28,45,27,35...etc
Prime 79-1,10,21,52,46,65,18,22,62,67,....etc
You could find that all the numbers in each sequence have a connection to the cyclic number generated by the corresponding prime...Viz 17 has 16 numbers in its sequence,19 has 18 numbers,41 has 5 numbers etc
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