In an earlier blog I had shown a cyclic number of 16 digits reproduced below-
0588235294117647
I had shown that multiplication of this number successively by 12,8,11,13,3,2,7,16,5,9.6,4,14,15 and 10 makes the last digit at right move to the left side successively.
Let me now tell you how this sequence of multipliers could be found.
The cyclic number is actually found by working out the reciprocal of 17,which could be treated as a generator of the same.The first number in the sequence viz 12 is the most important multiplier.Every subsequent multiplier is found by repeatedly multiplying the previous one by 12.From every result of multiplication you have to however deduct 17 as many times as necessary.
Thus the second multiplier is 12*12-17*8= 144-136=8
Third multiplier is 8*12-17*5= 96-85= 11
Fourth 11*12-17*7= 132-119=13
Fifth 13*12-17*9= 156-153=3 etc etc
You can go upto the multiplier 7 like this .The remaining multipliers are those found by deducting successively those found earlier from 17...Viz17-1=16, 17-12=5, 17-8=9 17-11=6 etc etc
I will inform you in another blog how the first number 12 can be used to find the subsequent multipliers in another way
0588235294117647
I had shown that multiplication of this number successively by 12,8,11,13,3,2,7,16,5,9.6,4,14,15 and 10 makes the last digit at right move to the left side successively.
Let me now tell you how this sequence of multipliers could be found.
The cyclic number is actually found by working out the reciprocal of 17,which could be treated as a generator of the same.The first number in the sequence viz 12 is the most important multiplier.Every subsequent multiplier is found by repeatedly multiplying the previous one by 12.From every result of multiplication you have to however deduct 17 as many times as necessary.
Thus the second multiplier is 12*12-17*8= 144-136=8
Third multiplier is 8*12-17*5= 96-85= 11
Fourth 11*12-17*7= 132-119=13
Fifth 13*12-17*9= 156-153=3 etc etc
You can go upto the multiplier 7 like this .The remaining multipliers are those found by deducting successively those found earlier from 17...Viz17-1=16, 17-12=5, 17-8=9 17-11=6 etc etc
I will inform you in another blog how the first number 12 can be used to find the subsequent multipliers in another way
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