Sequences-2 RATS sequence

Let me now tell you about another sequence.
This also has a peculiar name called  RATS sequence of numbers.
I believe it is based on the procedure of  starting  with a number Reversing it,Adding it to make a Total and Sorting it.
For example if 668 is a number in the sequence,the procedure gives us 668+866=1534,which when sorted to give a number in ascending order gives us 1345,which is the next number in the sequence.Next to this comes from 1345+5431=6776,which when sorted gives us 6677.Similarly the next can be found to be 13444.
The full list of numbers in the sequence is 1,2,4,8,16,77,145,668,1345,6677,13444,55778,  133345
666677,1333444,5567777,1233345,66666677,133333444 and so on.You can see repetition of digits like 3, 6 etc
In the website OEIS referring to the online encyclopedia of integer sequences,this sequence has the number  A004000

Sequences-1 EBAN number sequence

Let me introduce some sequences
The one I first wish to introduce has a funny name.
It is called eban number sequence.
When you write down the names of the numbers in English in this sequence the alphabet E
will be found  banned
The numbers are 2,4,6,30,32,34,36,40,42,44,50,52,54,56,60,62,64,66 and so on.

First attempt to find Kaprekar constants-5 digits

This refers to my earlier blog relating to the 4 digit Kaprekar constant 6174.
To recall the details the number is arrived at by the following process.
Take any 4 digits say 8,3,2,7.
Work out the highest number and smallest number and the difference.
You get 8732-2378=6354
Repeat the procedure now with 6,3,5 and 4 and repeat again as many times as needed.
You will always end up with 6174.
I started similar procedures with 5 digits
I worked out 47 cases.
I got two results 82962 and 74943 in almost equal number of cases
I am trying to work out whether any other result is possible.I feel that no other result could come.
Anyway only a computer programme can show the answer. 

Cyclic number-16 digits- Finding multipliers

In my blog posted on 28 feb 2012,I had mentioned about the following multipliers which make the digits at the end move to the front in the cyclic number  0588235294117647-
1,12,8,11,13,3,2,7...16,5,9,6,4,14,15,10
The second number 12 can be used in another way to find the subsequent multipliers-
The digits 1 and 2 in this number 12 have to be used as follows-
Using any two consecutive numbers in the list say m(1) and m(2),the next number can be found by working out 1*m(1)+2*m(2)
Thus taking m(1)=1 and m(2) =12 we find the next number as 1*1+2*12=25 which reduces to 8 after deducting 17 the generator of the cyclic number.
Again using m(1)=12 and m(2)=8,we find the next number as 1*12+2*8=28 which reduces to 11 after deducting 17 as before.
The next number 13 comes out in a similar way viz 1*8+2*11=30 which reduces to 13 after deducting 17
Next number 1*11+2.13=37 which reduces to 3 after deducting 17 twice
Next number is 1*13+2*3=19 which reduces to 2 as before.
All numbers are found similarly
For example using 4 and 14 the next number will be 1*4+2*14=32 which reduces to 15. 

Remainders -4 Procedure for finding the numbers to get the remainders

Regarding the problem mentioned in my blog Remainder-3 the solution is found as follows-
For remainder 12 in respect of division by 17,the numbers are 29,46,63...
The number 63 satisfies the condition-remainder 11 for division by 13.
For remainder 4 for division by 9,we add  17*13=221 continuously to 63 and test.
The numbers are 63,284,505,726,947,1168,1389,1610,1831..
The number 1831 satisfies remainder 4 for division by 9.
To consider division by 5(Remainder3),we add 17*13*9=1989 continuously to 1831.
The numbers are 1831,3820,5809,7798...
The number 7798 satisfies a remainder 3 for division by 5.
Hence 7798 is the number which satisfies all 4 conditions as required.
The method adopted can be simplified taking into account the fact below-
Instead of the remainder in each case,we consider how much the numbers are short for an even division by the dividing number.
For 13,the remainder should be 11 and for 5 the remainder should be 3-in both case the number should be 2 short for an even division by 13 and 5 and so the smallest such number is 13*5-2=63.
For 17 and 9 ,similarly the number should be short by 5(to give remainders 12 and 4) and so  the smallest such number is 17*9-5=148
This number 148 also gives remainder 3 for division by 5.
We now only have to satisfy the condition of remainder 11 for division by 13.
We hence add 17*9*5=765 continuously to 148 and test.
The numbers we get by adding 765 are 913,1678,2443,3208,3973,4738,5503,6268,7033 and finally7798 which is what we found earlier.

Remainders-3

You can have problems with 4 conditions also-
Example-Remainder 3 on division by 5
                  -do-      4      -do-          9
                  -do-      11     -do-         13
                  -do-      12      -do-        17
Try doing this as per method given by me.
Do you find anything in the problem which will enable you to simplify the method?         

Remainders-2

There has been no response to my previous blog.I desired to know the smallest number which gives remainder 2 when divided by 3,remainder 3 when divided by 5 and  remainder 5 when divided by 7.
There is a method known as chinese remainder theorem.,But I will give you another method
First take the largest divider(7) and the remainder(5).The first number which satisfies the requirement is 7+5=12.,Further numbers can be found by adding 7.They are thus 12,19,26,33,40,47,54,61,68,75,82.....etc
Check these whether you get remainder 3 when divided by 5.Only 68 will satisfy this requirement.
This number 68 surprisingly satisfies the third condition of giving reminder 2 when divided by 3.
So 68 is the required number.
Suppose I had altered condition that you must have a remainder 1 when divided by 3.You then have to add 7*5=35 to 68 repeatedly to check whether you get remainder 1 when divided by 3.The first number you get by adding 35 to 68 is 103.This satisfies the condition that you get remainder 1 when divided by 3
So 103 is the required number which satisfies all the 3 conditions.. .
To get further numbers you have to add 7*5*3=105 repeatedly to 103 .You will have the numbers 208,313,418,523 .....etc.
All these will satisfy all the 3 conditions.
To see whether you have understood the method,I will give you a more difficult problem.
You must find the smallest number which gives remainders as follows-
Remainder 5 when divided by 13
....do-       8     ...do-.             17
...do.....    11.....do-................23
You must get the answer 2150.
Further numbers obtained by adding 13*17*23=5083 repeatedly to 2150.
They are 7233,12316... etc