In a previous blog I had mentioned that by using the multipliers 5,4,6,2,and 3 on the cyclic number 142857,one can see the last digit at right moving successively to the front viz producing the results 714285,571428,857142,285714 and 428571,the other digits remaining in the same order in each case.
The multipliers 5,4,6,2 and 3 can be found as follows-
I had mentioned that the cyclic number 142857 was found by taking the reciprocal of prime number 7.We can call 7 as the 'generator'(g) of the cyclic number.
The first multiplier (f) was shown by me as 5
The relation between 'f' and 'g' is that (10f-1) should be a multiple of 'g'
In the case of g=7,we have f=5 as 50-1=49 is a multiple of 7.
For the subsequent multipliers you have to repeatedly multiply 5 by itself and find modulus 7 of the result.
(I now have to explain the term modulus.The modulus of any number say 45 with reference to another number say 7 is the remainder when 45 is divided by 7.In this case the remainder is 3 and so modulus 7 of 45 is 3)
For our case we have 5 multiplied by 5 gives us 25 and modulus 7 of 25 is 4 which is the second multiplier.
Next 4 multiplied by 5 gives us 20 and modulus 7 of 20 is 6,which is the third multiplier.
Next 6 multiplied by 5 gives us 30 and modulus 7 of 30 is 2 which is the fourth multiplier.
Next 2 multiplied by 5 gives us 10 and modulus 7 of 10 is 3 which is the fifth multiplier.
The process stops here as 3 multiplied by 5 gives us 15 and the modulus 7of 15 is 1.
You may find the processes a bit difficult to understand but read again you will understand the same .
Now let me give you the problem -whether you will be able to work out how the multipliers 4,3,12,9 and 10 can be found for the cyclic number 076923 relating to g=13(The cyclic number 076923 can be found by taking the reciprocal of 13).
The multipliers 5,4,6,2 and 3 can be found as follows-
I had mentioned that the cyclic number 142857 was found by taking the reciprocal of prime number 7.We can call 7 as the 'generator'(g) of the cyclic number.
The first multiplier (f) was shown by me as 5
The relation between 'f' and 'g' is that (10f-1) should be a multiple of 'g'
In the case of g=7,we have f=5 as 50-1=49 is a multiple of 7.
For the subsequent multipliers you have to repeatedly multiply 5 by itself and find modulus 7 of the result.
(I now have to explain the term modulus.The modulus of any number say 45 with reference to another number say 7 is the remainder when 45 is divided by 7.In this case the remainder is 3 and so modulus 7 of 45 is 3)
For our case we have 5 multiplied by 5 gives us 25 and modulus 7 of 25 is 4 which is the second multiplier.
Next 4 multiplied by 5 gives us 20 and modulus 7 of 20 is 6,which is the third multiplier.
Next 6 multiplied by 5 gives us 30 and modulus 7 of 30 is 2 which is the fourth multiplier.
Next 2 multiplied by 5 gives us 10 and modulus 7 of 10 is 3 which is the fifth multiplier.
The process stops here as 3 multiplied by 5 gives us 15 and the modulus 7of 15 is 1.
You may find the processes a bit difficult to understand but read again you will understand the same .
Now let me give you the problem -whether you will be able to work out how the multipliers 4,3,12,9 and 10 can be found for the cyclic number 076923 relating to g=13(The cyclic number 076923 can be found by taking the reciprocal of 13).
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