This has reference to my previous blog posted some hours ago today.
I was worried the process mentioned would be difficult to understand.
There are two ways of looking at the issue of movement of digits in the cyclic number 142857.
Let us consider the movement of the first digit to the right now(instead of the last digit to the front as considered earlier).Thus we now have 42857l,285714,857142,571428 and 714285.The multipliers required now are 3,2,6,4 and 5(These you can see are the same multipliers considered in the reverse order)
This new set of multipliers are more easier to find.
They are all the successive remainders when you take the reciprocal of 7
While taking the reciprocal of 7 we first add zero to 1 getting us 10 and divide the same by 7.The remainder is 3 which is the first multiplier now.Next we add zero to 3 getting us 30 and divide the same by 7.The remainder is 2 which is the second multiplier now.Next we add zero to 2 getting us 20 and divide the same by 7.The remainder is 6 which is the third multiplier now.The process can be continued to get the next multipliers viz 4 and 5.
You have to understand the necessity of the 2 processes viz.the one in the previous blog and the one now shown.
This will be clear when you consider a cyclic number of too many digits
For instance the cyclic number for the prime 97 has 96 digits.The first 10 of them are 0103092783 and the last 10 of them are 0206185567.If you want to know what multiplier will make the last digit 7 go to the front you have to perform the difficult process of finding the reciprocal right till the end. as per this blog.But if you adopt the process in the previous blog you will see that the value of 'f' is 68 and this multiplier will make the last digit 7 to go to the front.
You may choose in every case whichever process you consider easier and best.
I was worried the process mentioned would be difficult to understand.
There are two ways of looking at the issue of movement of digits in the cyclic number 142857.
Let us consider the movement of the first digit to the right now(instead of the last digit to the front as considered earlier).Thus we now have 42857l,285714,857142,571428 and 714285.The multipliers required now are 3,2,6,4 and 5(These you can see are the same multipliers considered in the reverse order)
This new set of multipliers are more easier to find.
They are all the successive remainders when you take the reciprocal of 7
While taking the reciprocal of 7 we first add zero to 1 getting us 10 and divide the same by 7.The remainder is 3 which is the first multiplier now.Next we add zero to 3 getting us 30 and divide the same by 7.The remainder is 2 which is the second multiplier now.Next we add zero to 2 getting us 20 and divide the same by 7.The remainder is 6 which is the third multiplier now.The process can be continued to get the next multipliers viz 4 and 5.
You have to understand the necessity of the 2 processes viz.the one in the previous blog and the one now shown.
This will be clear when you consider a cyclic number of too many digits
For instance the cyclic number for the prime 97 has 96 digits.The first 10 of them are 0103092783 and the last 10 of them are 0206185567.If you want to know what multiplier will make the last digit 7 go to the front you have to perform the difficult process of finding the reciprocal right till the end. as per this blog.But if you adopt the process in the previous blog you will see that the value of 'f' is 68 and this multiplier will make the last digit 7 to go to the front.
You may choose in every case whichever process you consider easier and best.
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