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I Interesting property of idempotent 10-adic number

  1. Jul 6, 2016 #1
    We build a special number from ##5##, by squaring it, appending the next digit of the square to it and repeating the steps.

    ##5^2=25.## The next digit is 2, which is added to 5 to give 25.

    ##25^2=625.## The next digit is 6, which is added to 25 to give 625.

    ##625^2=390625.## The next digit is 0, which is added to 625 to give 0625.

    ##0625^2=390625.## The next digit is 9, which is added to 0625 to give 90625.

    ##90625^2=8212890625.## The next digit is 8, which is added to 90625 to give 890625.

    We end up with the special number ##F=\,...\,4106619977392256259918212890625.##

    Why is it that the number in every step shares the same digits as its square? In other words, prove the existence of this number.

    Source: https://divisbyzero.com/2008/12/29/a-10-adic-number-that-is-a-zero-divisor/
    (The article mentioned that the digit 6 can be used. But I don't think so because ##36\times36=1296##, which doesn't end with 36.)
     
    Last edited: Jul 6, 2016
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  3. Jul 6, 2016 #2

    micromass

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    The number seems to be the limit ##L=\lim_{n\rightarrow +\infty} 5^{2^n}##. We see that
    [tex]L^2 = \lim_{n\rightarrow +\infty} (5^{2^n})^2 = \lim_{n\rightarrow +\infty} 5^{2^{n+1}} = L[/tex]
    So ##L^2 = L## meaning that your number is ##L=0## or ##L = 1##.

    We see that
    [tex]|5^{2^n}|_{10} = 1~\text{and}~|5^{2^n} - 1|_{10} = 1[/tex]
    so the sequence doesn't converge. It does converge in the ##5##-adic numbers:

    [tex]|5^{2^n}|_5 = 2^{-n}[/tex]
    So your number equals ##0## in the ##5##-adic numbers.

    EDIT: the number does exist in the ##10##-adic numbers. The ##10##-adics are not an integral domain so ##L^2 = L~\Rightarrow~L\in \{0,1\}## is not valid. The number doesn't seem to be a rational number though
     
  4. Jul 6, 2016 #3
    I am not familiar with p-adic numbers. Could this be proved without using p-adic numbers?
     
    Last edited: Jul 6, 2016
  5. Jul 6, 2016 #4
    ##625^2=390625## but ##F## ends with ##890625##. So I think ##F## may not be the limit ##L##.
     
  6. Jul 6, 2016 #5

    micromass

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    Yes, you're right. I misunderstood the OP.
     
  7. Jul 6, 2016 #6

    micromass

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    OK, so let ##a_n## be the number formed by the first ##n## digits of ##F##. We know by construction that the ##nth## digit of ##a_n## is the ##n##th digit of ##a_{n-1}^2##. We prove by induction that the first ##n## digits of ##a_n^2## equal ##a_n##.

    We can obviously write ##a_n = x 10^n + a_{n-1}##. Squaring gives us
    [tex]a_n^2 = x^2 10^{2n} + 2x10^n a_{n-1} + a_{n-1}^2[/tex]
    We are interested in the first ##n## digits of this number. The number ##x^2 10^{2n}## has a zero on its first ##2n## digits, so it plays no role. The first ##n-1## digits of ##2x10^n a_{n-1}## are zero. The ##n##th digit of this number is the first digit of ##2xa_{n-1}##. But since the first digit of ##a_{n-1}## is ##5##, we get that this first digit is the first digit of ##2\cdot 5\cdot x##, that is: the first digit is ##0##.

    So the first ##n## digits of ##a_n^2## are the first ##n## digits of ##a_{n-1}^2##, like we wanted.
     
  8. Jul 6, 2016 #7
    ##a_2=25=2(10)+5=2(10^1)+a_1##.

    So ##a_n = x 10^{n-1} + a_{n-1}##.

    Right?
     
  9. Jul 6, 2016 #8

    micromass

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    Right sorry, I should have proof read it. But the proof still goes through modulo these notational errors.
     
  10. Jul 6, 2016 #9

    micromass

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    Using the same proof, I discovered a similar number:

    [tex]2^5 = \mathbf{32},~32^5 = 33554\mathbf{432},~432^5 = 1504591950\mathbf{6432}, ...[/tex]
     
  11. Jul 6, 2016 #10

    mfb

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    The proof needs a slight modification, and relies on the ending number being 5 to make 2x*5 a multiple of 10.
    That also explains why it works with 5 but not with 6.

    The fifth power, starting with 2, also works because then 5x*2 is a multiple of 10. Apart from higher powers (^10, ^15, ... with 2, or ^4, ^6, ... with 5) that should be all we have in the decimal system. Other bases lead to similar results. Start with a divisor in this system, then use a power such that starting number * power is a multiple of the base.
     
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