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While playing on my calculator I accidentally calculated $\frac{1}{9801}$ and the value was $0.00010203040506070809101112131415161718192021222324252627282930\ldots$
We observe that there is a nice pattern in the decimal expansion.
but how long will it continue ie. after how many digits will it repeat?

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    $\begingroup$ Ask your son. $\endgroup$ May 25, 2018 at 9:44
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    $\begingroup$ @CountIblis he couldn't that's why i asked here $\endgroup$ May 25, 2018 at 9:45
  • $\begingroup$ @MohammadZuhairKhan which one? $\endgroup$ May 25, 2018 at 9:46
  • $\begingroup$ It actually repeats all $2$ digit numbers, starting from $00$ to $99$, excluding $98$. $\endgroup$ May 25, 2018 at 9:47
  • $\begingroup$ It may be but it was difficult to find it using the title and the questions are actually different $\endgroup$ May 25, 2018 at 9:49

4 Answers 4

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There is a numberphile video related to this here.

Essentially notice that $9801=99^2$. A similar pattern happens with $\frac{1}{9^2}=\frac{1}{81}$. It's value is $0.\dot{0}1234567\dot{9}$, cycling through all the units except $8$ (To see why, try a long division method). The case with $\frac{1}{99^2}$, it cycles through every two digit number, but excludes $98$ from this list... i.e. it is: $$\frac{1}{99^2}=0.\dot{0}0010203040506\cdots9596979\dot{9}$$ and similar things happen with $\frac{1}{999^2},\frac{1}{9999^2}$, etc.

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Every fraction $\frac{a}{b}$ with integers $a$ and $b$ has a pattern which continues endlessly.

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    $\begingroup$ Discounting zeros, 1/4=0.25 does not continue forever. But if you count zeros then 1/4=0.250000... $\endgroup$
    – pshmath0
    May 25, 2018 at 9:47
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https://www.wolframalpha.com/input/?i=1%2F9801

See for yourself. A fraction when converted to decimal will repeat at some point of times.

Here it repeats till 99 skipping 98 and then start all over again.

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  • $\begingroup$ Incorrect. It skips 98. $\endgroup$ May 25, 2018 at 9:59
  • $\begingroup$ Yup. Sorry. Edited. $\endgroup$ May 25, 2018 at 10:03
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Observe that with $a \in N$

$$ \frac{1}{9801}=\frac{a}{10^n}(1+10^{-n}+10^{-2n}+\cdots) = \frac{a}{10^n}\frac{1}{1-10^{-n}} $$

or

$$ a = \frac{10^n-1}{9801} $$

which gives $n = 198$ and $a = 10203040506070809101112131415161718192021222324252627282930313233343536373\\ 83940414243444546474849505152535455565758596061626364656667686970717273747 5767\\ 7787980818283848586878889909192939495969799$

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