One more cyclic number-Rejoinder 2

The link mentioned in the just preceding blog occurs as follows-

The numbers used for multiplication in the reverse order are 1,....621517,...269324,...441277,....251300,..25130,...2513,...1864802,..1429514,..557296,..1713108.

1000000,100000,10000,1000,100,10..

The number 2513 can be worked from the previous 6 numbers by successively multiplying them with 6,2,1,5,1,7 forming part of 621517 and adding the results.Similarly the number 1864802 can be worked from the previous 6 numbers etcetc-see below-

(1*6)+(621517*2)+(269324*1)+(441277*5)+(251300*1)+(25130*7)=

6+1243034+269324+2206385+251300+175910=4145959.From this result the maximum possible multiple of 2071723 viz 4143446 is deducted producing 2513.

Similarly (621517*6)+(269324*2)+(441277*1)+(251300*5)+(25130*1)+(2513*7)=6008248 which after subtracting 4143446 gives us 1864802.etc etc.

You can similarly get all the subsequent numbers.

One more cyclic number-Rejoinder

This is with reference to my post of 22 june 2012,where I had introduced a cyclic number of 17 digits with 6 zeros viz 00000048269001213-Call this 'N'.
I had mentioned that the 6 zeros occur due to the action of finding the reciprocal of  2071723-You can call this as'g'.
I had also mentioned that by multiplying the number 'N, with  621517, you will get a new number as follows-
30000004826900121-viz the 3 at the end of the original number 'N' moving to the left with all other digits remaining as before in the same order
I had not mentioned how this number for multiplication could be found.It is found as follows-
You have to multiply g=2071723 by a digit 'd'which gets a result of the form (10k-1)  and you will find that k=621517.
We have d=3 and the multiplication gives us 6215169 and so k=621517.
The subsequent numbers which on multiplication with 'N" produce results like 13000000482690012..
21300000048269001 etc etc-showing the last digits moving to the left successively are all the remainders at each stage while taking the reciprocal of g=2071723.
The procees of finding the reciprical gives these numbers as follows-
The first stage for getting a quotient while dividing 1 by g occurs after adding 7 zeroes to 1,making it
10000000-The division by 'g' gives a quotient 4 and remainder 1713108.
At the step dividing 17131080 by 'g' gives the quotient 8 and the remainder 557296.
Proceeding further we get quotients 2,6.9 etc forming part of 4826900121 .
All the remainders are 1713108,557296,1429514,1864802,2513,25130,251300,441277,269324 and finally 621517 and 1-The cycle repeats from here again.
These remainders when used in the reverse order for multiplication of 'N' will produce results showing movement of the last digits successively to the left as shown below-
621517 will produce 30000004826900121
269324 will produce 13000000482690012
441277 will produce 21300000048269001 etc .etc.
All these numbers used for multiplication also have a link in another way which I will show in my next post.  

Ramanujan number-2

In my previous blog of 18th feb,I had mentioned about the number 1729-called Ramanujan number-the only number which could be expressed as a sum of 2 cubes in 2 different ways-viz 1000+729=1728+1=1729.
If we consider differences between 2 cubes (this means the same thing as considerig the cube of one negative number),we can write the equations differently....12^3+{(-)10}^3=1729-1000=728 also 9^3+{(-1)}^3=729-1=728.
I can now show a case where a number can be expressed as a difference of 2 cubes in 3 different ways
34^3-33^3=39304-35937=3367
15^3-2^3 = 3375-8 =3367
16^3-9^3 = 4096-729=3367
There could be many more such cases.-also numbers which could be expressed in more than 3 different ways.

Sums of 2 square numbers-rejoinder

With reference to my 2 prvious blogs on the subject there are only 4 cases relating to numbers between 2001 and 3000.The following are those cases.

484+1784=2248,400+1849=2249,225+2025=2250
1296+1296=2592,2304+289=2593,1225+1369=2594
784+1600=2384,81+2304=2385,361+2025=2386
196+2116=2312,9+2304=2313,289+2025=2314.

I am yet to work about cases of numbers above 3000.

Sum of 2 square numbers-2

With ref to my blog of 9 march,no one has come with a response.The following are the 8 cases of numbers between 1001 and 2000
196+900=1096,256+841=1097,9+1089=1098
576+576=1152,64+1089=1153,529+625=1154
16+1296=1312,289+1024=1313,225+1089=1314
64+1600=1664,576+1089=1665,441+1225=1666
144+1600=1744,784+961=1745,225+1521=1746
900+900=1800,576+1225=1801,121+1681=1802
576+1296=1872,784+1089=1873,25+1849=1874
196+1764=1960,361+1600=1961,441+1521=1962.
I will show the cases of numbers between 2001 and 3000 in my next blog