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.
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.
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