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In regards to repeating decimals and cyclic numbers: I understand that many of them, while multiplying by certain integers will produce a number with the same variant of digits, however my question is in regards to the patterns already occurring within some of them.

Many fractions that convert into a number with repeating decimals, will exhibit a pattern within the repetend. This pattern is made of a consecutive list of numbers growing by the same multiple.

example: $1/31=0.032258064516129...$., the repetend is $032258064516129$ which can be broken into the numbers $032258$, $064516$, $129032$ that are related by multiples of 2: $$032258∗2=064516, \quad 064516∗2=129032.$$

example: $1/13=0.076923....$, the repetend is $076923$ which can be broken into $0769$, $2307$ that are related by multiplying by 3: $$0769∗3=2307$$

example: $1/17=0.0588235294117647....$, the repetend is $0588235294117647$ which can be broken into the numbers $5882352941$, $11764705882$ that are related by multiplying by 2: $$5882352941∗2=11764705882$$

example: $1/19=0.052631578947368421...$, the repetend is $052631578947368421$ which can be broken into the numbers $05263$ , $15789$ , $47367$ , $142101$ (think of the initial 1 in 142101 being removed from the 8 in the preceding 47368) that are related by multiples of 3: $$05263∗3=15789, \quad 15789∗3=47367, \quad 47367∗3=142101$$

example: $1/21=0.047619...$, the repetend is $047619$ which can be broken into the numbers $0476$, $1904$ that are related by multiplying by 4: $$0476∗4=1904$$

What are these multiples? That is, given 1/$n$ with a repeating decimal, what multiple connects the different pieces of the repetend? (And how long are those pieces?)

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    $\begingroup$ I think there is a reasonable question here that can be answered. The basic idea is if you are doing the long division to determine the repeating part of the decimal in all these fractions you start with a remainder of $1$ because that is the numerator. Halfway through the repeat you will have some other remainder. The next set of digits is the set you just finished times this remainder. Voting to reopen $\endgroup$ Jan 31, 2021 at 0:43
  • $\begingroup$ @Ross Millikan Why is it getting closed? I asked the same question on mathoverflow and someone commented to me that this question can be answered on stackexchange by arithmetic modular. $\endgroup$ Jan 31, 2021 at 0:48
  • $\begingroup$ I don't know why it was closed. As I said in my comment, I think it should be reopened. It is not a high enough level question for math overflow, which targets active researchers. $\endgroup$ Jan 31, 2021 at 1:22
  • $\begingroup$ @Brian Hopkins Can’t express the gratitude for the edit, by if it still closed ? $\endgroup$ Jan 31, 2021 at 4:10
  • $\begingroup$ I think this question is related to Doubling sequences of the cyclic decimal parts of the fraction numbers, but you have picked up some longer groups of digits and it's interesting that you found clean doubling/tripling sequences (not overlapping each other) that are longer than the repetends. $\endgroup$
    – David K
    Jan 31, 2021 at 4:17

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The first thing to think about is finding the repeating decimal by long division. Taking $1/17$ for example, your first division starts with a remainder $1$. You append a $0$ to it and try to divide $17$ into $10$. It goes $0$ times, which gives the leading $0$ in the repeat. You then pass a remainder of $10$ to the next digit. You pull down another $0$, getting $100$, divide $17$ into it getting $5$ with a remainder of $15$. The $5$ is the second digit of the repeat and the $15$ goes on for the next digit.

If you have a full length repeat like $17$, you must use each of the remainders from $1$ to $17-1$ once. When you repeat a remainder you have finished the repeat. For $13$, which has only a six digit repeat you only use six remainders. This explains all the relationships you show.

After the $n^{th}$ digit the remainder is $10^n \mod 17$. We can find that $10^{10} \equiv 2 \pmod {17}$ That means after $10$ digits you pass a remainder of $2$ to the next digit. The repeat from there will clearly be twice the repeat starting from the front, where the remainder is $1$.

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