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I was just playing around with dividing random numbers when I noticed that any $x\in\mathbb{N}$ when divided by an integer that is some series of $1$s, such as $34/111$, ends up being $0.\overline{9\cdot x}$, or in this case $0.\overline{306}$. This acts similar to how dividing $x$ by a series of $9$s becomes $0.\overline{x}$.

I know this isn't anything revolutionary or anything but I noticed this works in other bases as well. I wanted to extend the formula for any number $x\in\mathbb{N}$ in any base $b\ge2$ because I can't seem to find it anywhere online, and ended up coming up with this: $$\frac{x}{\sum_{i=0}^{\left\lceil \log_{b+1}(x) \right\rceil}b^i}=0.\overline{(b-1)\cdot x}$$

Since I am still a high school student who doesn't know too much advanced math, I am looking for feedback on errors/oversights I made in this identity or ideas to improve and/or make this more readable if necessary. Additionally, if this is already a known thing, I would love for someone to help me find it so I can read more into this.

Bill Dubuque
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swivet
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    Your formula can be used to prove $0.999…=\frac{111}{111}=1$ – d ds Jun 26 '25 at 02:52
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    $x/111=9x/999.$ – Thomas Andrews Jun 26 '25 at 03:09
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    It is worth thinking of this as a question about base $b=10^3.$ A number written in base $b^n$ can be written easily in base $b$ by converting each base $b^n$ digit into $n$ base $b.$ Then in base $b=10^3,$ $$\frac{34}{111}=\frac{34\cdot9}{10^3-1}=0.\overline{(34\cdot9)}$$ – Thomas Andrews Jun 26 '25 at 03:20
  • More generally, if $q$ is a denominator of a fraction not divisible by $2$ or $5,$ then the repetition length for a reduced fraction $p/q$ is the smallest $n$ such that $10^n-1$ is divisible by $q.$ For example, $3$ is a factor of $10^1-1,$ so $1/3$ has a single digit repeating. $7$ is a factor of $10^6-1,$ so $1/7=0.\overline{142857}$ has a repetition length $6.$ This explains why $1/37=0.\overline{027}$ has a repetition length $3,$ since $37$ is a factor of $10^3-1.$ – Thomas Andrews Jun 26 '25 at 03:30
  • And that, of course, works for any base, similarly. So, in base $11,$ $1/7$ has repetition length $3,$ and $11^3-1$ is divisible by $7.$ – Thomas Andrews Jun 26 '25 at 03:34

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