Kaprekar numbers-6digits

In my previous post of 26 oct 2012,I had shown 2 series of Kaprekar numbers of 5 digits as follows-
74943,62964,71973,83952,74943(repeat)
61974,82962,75933,63954,61974(repeat)
To enable you to recollect,the process is explained below-
74943.-This consists of digits 3,4,4,7,9
Find the largest and smallest numbers with these and find the difference.You will get the next number in the series viz 62964 and so on.
viz 97443-34479=62964
Now I will show two 6 digit numbers and one series of 6 digit numbers-
1)Number  631764-the process results in the same number..766431-134667=631764
2)Number 549945-behaves like 631764
3) Series-851742,750843,840852,860832,862632,642654,420876,851742(repeat)

Walk the talk-2(Rejoinder)

This is with reference to my blog of 25th April,where I mentioned an imaginary conversation between me and a 10 digit number which explained itself about it's composition-viz the number of times each digit from 0 to 9 occurs in it-viz the first digit shows how many times zeros occur,the second digit how many times 1 occurs,the third digit how many times 2 occurs etc etc,right till the end showing how many times 9 occurs etc
The number is actually 6210001000.You can verify the composition as expressed by itself.

Divisibility tests for all prime numbers upto 100-prime 31

I had dealt with prime 23 in my previous blog.
I now deal with another prime 31,where there is a small difference in the procedure.
Here I choose the number P as 992 which has 31 as a factor.This is of the form (1000m-n) instead of the form (1000m+n) as in the case of prime 23 and hence the need for the change in procedure
No to be tested is the same viz 26105069.(This number was specifically chosen as it is a multiple of 31).
Step 1-We have  the same groups of 3 nos-A 069,B 105 and C 26.
We also now have m=1 and n=8
Step 2 Division by 'm' and multiplication by 'n' of C-
26/1=26.....26*8=208
Step 3-The change in procedure occurs here.We now have to add this result to B instead of deducting as in the case of prime 23.
105+208=313.
Step 4 Process repeated   viz 313/1=313....313*8=2504
Step 5 Result is added to A  viz 069+2504=2573
Step 6-The result in step 5 is the final result to be tested for divisibility by 31.
As 2573 is divisible by 31....(2573=31*83) we conclude the number under test 26105069 is divisible by 31.Actually the quotient is 842099.
In all cases where P is of the form (1000m-n) the change in procedure in step 3 viz addition as done above is to be followed.

Divisibility tests for all primes upto 100

I have devised another general method for testing divisibility of any number by all the primes below 100.
I will cover cases of each prime one after another.The method is based on a procedure similar to that for 17,19 and 59 shown in my earlier blogs.
In each case a suitable number -call it P- for which the prime considered is a factor should first be selected of the form (1000m+n) or(1000m-n)
Let be show the procedure for the next prime after 19 viz 23
The number P chosen for 23 is 2001 which has 23 as a factor.In this case m=2 and n=1
Let us consider the number 26105069 for test.
Step 1-This is broken up into groups of 3 digits starting from right calling them A,B and C.
We thus have A=069,B=105, and C=26.
Step2-We divide C by m and then multiply by n
We get 26/2=13 and 13*1=13
Step3-The result is subtracted from B .We get 105-13=92
Step4-Process in Step 2 is now repeated with the result in previous step.
We get 92/2=46 and 46*1=46
Step5-The result is subtracted from A.We get 069-46=23
Step6-The final result is tested for divisibility by 23.If it is divisible, the number under test can be considered as divisible
In our case the final result is itself 23-divisible obviously by 23 and hence the number under test viz 26105069 is so divisible
The values of P ,m and n vary for each prime otherwise the process is more or less similar.