Question on recurring decimal digits In my discrete maths class, I have come across an interesting

Brooklynn Hubbard 2022-04-10 Answered
Question on recurring decimal digits
In my discrete maths class, I have come across an interesting phenomenon for which I can't find an explanation!
If we divide 1 by 13 we obtain 0.07692307
If we divide 3 by 13 we obtain 0.23076923
If we divide 4 by 13 we obtain 0.30769230
As you can see, the digits are recurring in the same order but starting at a different point in the sequence.
Can someone explain this to me? What exactly is happening here?
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Answers (1)

Eden Bradshaw
Answered 2022-04-11 Author has 21 answers
Well, let me start by writing some equations which are boring:
1 13 = 0.0769230769230
10 13 = 0.769230769230
100 13 = 7.69230769230
1000 13 = 76.9230769230
10000 13 = 769.230769230
100000 13 = 7692.30769230
Obviously, multiplying by 10 just shifts the sequence over. However, suppose we subtract out the integer part of each of these equations. Then they're suddenly interesting as they each repeat the same pattern starting from a different place:
1 13 0 = 1 13 = 0.0769230769230
10 13 0 = 10 13 = 0.769230769230
100 13 7 = 9 13 = 0.69230769230
1000 13 76 = 12 13 = 0.9230769230
10000 13 769 = 3 13 = 0.230769230
100000 13 7692 = 4 13 = 0.30769230
Hey, those equations are all pretty neat- they have the same property that the digits appear in a shifted order, and if you just wrote out the fraction without seeing where it comes from, that's not terribly obvious.
Notice that this only works for the numerators 1 , 3 , 4 , 9 , 10 , and 12. The reason these numbers are special is that they are the powers of 10 mod 13. More generally, we could replace 13 by any number n coprime to 10 and say that this is true of the powers of 10 mod n. This is particularly interesting when every integer between 1 and n 1 is a power of 10 mod n, meaning that every non-integer fraction would have this property (which is true of n = 7 and other numbers)
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