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The number $$\sqrt{308642}$$ has a crazy decimal representation : $$555.5555777777773333333511111102222222719999970133335210666544640008\cdots $$

Is there any mathematical reason for so many repetitions of the digits ?

A long block containing only a single digit would be easier to understand. This could mean that there are extremely good rational approximations. But here we have many long one-digit-blocks , some consecutive, some interrupted by a few digits. I did not calculate the probability of such a "digit-repitition-show", but I think it is extremely small.

Does anyone have an explanation ?

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    $\begingroup$ Hint: $308642=(5000^2+2)/9^2$. $\endgroup$
    – Chen Wang
    Commented Feb 8, 2017 at 13:09
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    $\begingroup$ In interestingly the prime factorization of this number is $ 2 \times 154321 $ I wonder if the 54321 has anything to do with it? $\endgroup$
    – Riemann-bitcoin.
    Commented Feb 8, 2017 at 13:18
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    $\begingroup$ On a related note, see Schizophrenic number $\endgroup$
    – PM 2Ring
    Commented Feb 10, 2017 at 10:29
  • $\begingroup$ Did this come up as an actual problem or just for fun? $\endgroup$
    – Brian Risk
    Commented Feb 10, 2017 at 20:52
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    $\begingroup$ @BrianRisk Just for fun! $\endgroup$
    – Peter
    Commented Feb 11, 2017 at 14:01

2 Answers 2

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The architect's answer, while explaining the absolutely crucial fact that $$\sqrt{308642}\approx 5000/9=555.555\ldots,$$ didn't quite make it clear why we get several runs of repeating decimals. I try to shed additional light to that using a different tool.

I want to emphasize the role of the binomial series. In particular the Taylor expansion $$ \sqrt{1+x}=1+\frac x2-\frac{x^2}8+\frac{x^3}{16}-\frac{5x^4}{128}+\frac{7x^5}{256}-\frac{21x^6}{1024}+\cdots $$ If we plug in $x=2/(5000)^2=8\cdot10^{-8}$, we get $$ M:=\sqrt{1+8\cdot10^{-8}}=1+4\cdot10^{-8}-8\cdot10^{-16}+32\cdot10^{-24}-160\cdot10^{-32}+\cdots. $$ Therefore $$ \begin{aligned} \sqrt{308462}&=\frac{5000}9M=\frac{5000}9+\frac{20000}9\cdot10^{-8}-\frac{40000}9\cdot10^{-16}+\frac{160000}9\cdot10^{-24}+\cdots\\ &=\frac{5}9\cdot10^3+\frac29\cdot10^{-4}-\frac49\cdot10^{-12}+\frac{16}9\cdot10^{-20}+\cdots. \end{aligned} $$ This explains both the runs, their starting points, as well as the origin and location of those extra digits not part of any run. For example, the run of $5+2=7$s begins when the first two terms of the above series are "active". When the third term joins in, we need to subtract a $4$ and a run of $3$s ensues et cetera.

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    $\begingroup$ @Peter It is quite common to unaccept an answer after a better answer appears. Doing so helps guide readers to the best answer, which is often not the highest voted one, due to many factors, e.g. earlier answers usually get more votes, and less technical answers usually get more votes from hot-list activity (as here). This is currently (by far) the best explanation you have. $\endgroup$
    – Bill Dubuque
    Commented Feb 8, 2017 at 20:35
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    $\begingroup$ For the record: The reason I support Peter's decision to accept the architect's answer is that mine is building upon it. Without the observation that $5000/9$ is an extremely good approximation I most likely would not have bothered, and most certainly would not have come up with this refinement. IMHO Math.SE works at its best, when different users add different points of view refining earlier answers. The voters very clearly like both the answers. Sunshine and smiles to all! $\endgroup$
    – Jyrki Lahtonen
    Commented Feb 10, 2017 at 7:36
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    $\begingroup$ @JyrkiLahtonen strong agreement. I love to see answers working in tandem, and the checkmark doesn't give all that many points. Best to have the answer that others build on be the one that people read first :) $\endgroup$
    – hobbs
    Commented Feb 10, 2017 at 8:13
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    $\begingroup$ You may enjoy applying your skills to the Schizophrenic numbers. ;) $\endgroup$
    – PM 2Ring
    Commented Feb 10, 2017 at 10:35
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    $\begingroup$ @JyrkiLahtonen's answer is the correct one. The repeating digits can be inferred from the Taylor expansion of the square root. the_architect's answer is merely an observation. $\endgroup$
    – Klangen
    Commented Mar 12, 2018 at 10:37
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Repeated same numbers in a decimal representation can be converted to repeated zeros by multiplication with $9$. (try it out)

so if we multiply $9 \sqrt{308642} = \sqrt{308642 \times 81} = \sqrt{25 000 002}$ since this number is allmost $5000^2$ it has a lot of zeros in its decimal expansion

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    $\begingroup$ Superb answer! (+1) $\endgroup$
    – Peter
    Commented Feb 8, 2017 at 13:19
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    $\begingroup$ And the underlying reason here is the series expansion $$ \sqrt{a^2+x} = a + \frac{1}{2a}x - \frac{1}{(2a)^3}x^2 + \frac2{(2a)^5}x^3 - \frac{5}{(2a)^7}x^4 + \cdots $$ which can be derived from the generalized binomial theorem. When $2a$ is a large power of $10$, this gives a nice decimal representation of the square root. $\endgroup$
    – hmakholm left over Monica
    Commented Feb 8, 2017 at 14:04
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    $\begingroup$ To check that this is the "right" explanation I'd find it good to have other, similar examples. And here is another one: $\sqrt{1975308642} = 44444.44444472222222222135416666667209201388884650336371564...$ which can be explained by noting that $1975308642 = (400000^2 + 2)/9^2$. $\endgroup$
    – Michael Lugo
    Commented Feb 8, 2017 at 14:37
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    $\begingroup$ It is important to note that $25000002$ is not only almost $5000^2$, but also $>5000^2$. Otherwise the same argument would work for $$\sqrt{30864\color{red}1}\approx 555.55467777708433223768894721... .$$ Does not look so very nice! It is because $81\times 308641<5000^2$, but still close to $5000^2$. $\endgroup$
    – M. Winter
    Commented Jan 25, 2018 at 9:57
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    $\begingroup$ @M.Winter $25000000/81=308641\frac{79}{81}$. So the reason why 308641 produces worse results than 308642 is that its error has larger magnitude, not because it's negative. $\endgroup$
    – Rosie F
    Commented May 7, 2022 at 10:33

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