Is there a power of $2$ which begins, in base 10, with $999…$? [closed]












1












$begingroup$


Is there a power of $2$ which begins, in base 10, with $999...$?



I don't know how to do it, maybe I must apply Dirichlet principle but in which way?






number-theory







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$endgroup$



closed as off-topic by Henrik, Jyrki Lahtonen, Paul Frost, Cesareo, Andrei Dec 28 '18 at 18:48


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Jyrki Lahtonen, Paul Frost, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    I don't see how the Diichlet principle can be of any value here.
    $endgroup$
    – Henrik
    Dec 28 '18 at 9:19










  • $begingroup$
    @Henrik I'm fairly sure that the OP has Dirichlet's approximation theorem in mind.
    $endgroup$
    – Jyrki Lahtonen
    Dec 28 '18 at 10:17










  • $begingroup$
    Assuming that the string of nines has a finite length (in spite of appearances to the contrary), then this is question is also a duplicate. If that string is not finite, then the question is non-sensical.
    $endgroup$
    – Jyrki Lahtonen
    Dec 28 '18 at 10:20












  • $begingroup$
    @JyrkiLahtonen: That would make more sense.
    $endgroup$
    – Henrik
    Dec 28 '18 at 10:42










  • $begingroup$
    Possible duplicate of Starting digits of $2^n$.
    $endgroup$
    – Andrei
    Dec 28 '18 at 18:48
















1












$begingroup$


Is there a power of $2$ which begins, in base 10, with $999...$?



I don't know how to do it, maybe I must apply Dirichlet principle but in which way?






number-theory







share|cite|improve this question









$endgroup$



closed as off-topic by Henrik, Jyrki Lahtonen, Paul Frost, Cesareo, Andrei Dec 28 '18 at 18:48


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Jyrki Lahtonen, Paul Frost, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.












  • 2




    $begingroup$
    I don't see how the Diichlet principle can be of any value here.
    $endgroup$
    – Henrik
    Dec 28 '18 at 9:19










  • $begingroup$
    @Henrik I'm fairly sure that the OP has Dirichlet's approximation theorem in mind.
    $endgroup$
    – Jyrki Lahtonen
    Dec 28 '18 at 10:17










  • $begingroup$
    Assuming that the string of nines has a finite length (in spite of appearances to the contrary), then this is question is also a duplicate. If that string is not finite, then the question is non-sensical.
    $endgroup$
    – Jyrki Lahtonen
    Dec 28 '18 at 10:20












  • $begingroup$
    @JyrkiLahtonen: That would make more sense.
    $endgroup$
    – Henrik
    Dec 28 '18 at 10:42










  • $begingroup$
    Possible duplicate of Starting digits of $2^n$.
    $endgroup$
    – Andrei
    Dec 28 '18 at 18:48














1












1








1





$begingroup$


Is there a power of $2$ which begins, in base 10, with $999...$?



I don't know how to do it, maybe I must apply Dirichlet principle but in which way?






number-theory







share|cite|improve this question









$endgroup$




Is there a power of $2$ which begins, in base 10, with $999...$?



I don't know how to do it, maybe I must apply Dirichlet principle but in which way?






number-theory




number-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 28 '18 at 9:09









LanceLance

10112




10112




closed as off-topic by Henrik, Jyrki Lahtonen, Paul Frost, Cesareo, Andrei Dec 28 '18 at 18:48


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Jyrki Lahtonen, Paul Frost, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by Henrik, Jyrki Lahtonen, Paul Frost, Cesareo, Andrei Dec 28 '18 at 18:48


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Henrik, Jyrki Lahtonen, Paul Frost, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    $begingroup$
    I don't see how the Diichlet principle can be of any value here.
    $endgroup$
    – Henrik
    Dec 28 '18 at 9:19










  • $begingroup$
    @Henrik I'm fairly sure that the OP has Dirichlet's approximation theorem in mind.
    $endgroup$
    – Jyrki Lahtonen
    Dec 28 '18 at 10:17










  • $begingroup$
    Assuming that the string of nines has a finite length (in spite of appearances to the contrary), then this is question is also a duplicate. If that string is not finite, then the question is non-sensical.
    $endgroup$
    – Jyrki Lahtonen
    Dec 28 '18 at 10:20












  • $begingroup$
    @JyrkiLahtonen: That would make more sense.
    $endgroup$
    – Henrik
    Dec 28 '18 at 10:42










  • $begingroup$
    Possible duplicate of Starting digits of $2^n$.
    $endgroup$
    – Andrei
    Dec 28 '18 at 18:48














  • 2




    $begingroup$
    I don't see how the Diichlet principle can be of any value here.
    $endgroup$
    – Henrik
    Dec 28 '18 at 9:19










  • $begingroup$
    @Henrik I'm fairly sure that the OP has Dirichlet's approximation theorem in mind.
    $endgroup$
    – Jyrki Lahtonen
    Dec 28 '18 at 10:17










  • $begingroup$
    Assuming that the string of nines has a finite length (in spite of appearances to the contrary), then this is question is also a duplicate. If that string is not finite, then the question is non-sensical.
    $endgroup$
    – Jyrki Lahtonen
    Dec 28 '18 at 10:20












  • $begingroup$
    @JyrkiLahtonen: That would make more sense.
    $endgroup$
    – Henrik
    Dec 28 '18 at 10:42










  • $begingroup$
    Possible duplicate of Starting digits of $2^n$.
    $endgroup$
    – Andrei
    Dec 28 '18 at 18:48








2




2




$begingroup$
I don't see how the Diichlet principle can be of any value here.
$endgroup$
– Henrik
Dec 28 '18 at 9:19




$begingroup$
I don't see how the Diichlet principle can be of any value here.
$endgroup$
– Henrik
Dec 28 '18 at 9:19












$begingroup$
@Henrik I'm fairly sure that the OP has Dirichlet's approximation theorem in mind.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 10:17




$begingroup$
@Henrik I'm fairly sure that the OP has Dirichlet's approximation theorem in mind.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 10:17












$begingroup$
Assuming that the string of nines has a finite length (in spite of appearances to the contrary), then this is question is also a duplicate. If that string is not finite, then the question is non-sensical.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 10:20






$begingroup$
Assuming that the string of nines has a finite length (in spite of appearances to the contrary), then this is question is also a duplicate. If that string is not finite, then the question is non-sensical.
$endgroup$
– Jyrki Lahtonen
Dec 28 '18 at 10:20














$begingroup$
@JyrkiLahtonen: That would make more sense.
$endgroup$
– Henrik
Dec 28 '18 at 10:42




$begingroup$
@JyrkiLahtonen: That would make more sense.
$endgroup$
– Henrik
Dec 28 '18 at 10:42












$begingroup$
Possible duplicate of Starting digits of $2^n$.
$endgroup$
– Andrei
Dec 28 '18 at 18:48




$begingroup$
Possible duplicate of Starting digits of $2^n$.
$endgroup$
– Andrei
Dec 28 '18 at 18:48










3 Answers
3






active

oldest

votes


















6












$begingroup$

To find many answers, you can start looking for $n,m$ such that $2^n approx 10^m$, which is equivalent to $nlog2approx mlog10$.



In other words, we look for rational approximations of $frac{log10}{log2}$, in which case $n$ would be the numerator.



Using continued fractions, for



$$frac{log10}{log2} approx 3 + frac1{3 + frac1{9 + frac1{2 + frac1{2 + frac1{4 + frac16}}}}} = frac{13301}{4004}$$



we find $n = 13301$, for which $2^n$ starts with 9999.



Note that since the sequence of continued fractions converges to the value, there is no limit on the number of leading 9's a power of 2 can have.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    If $2^{13301} approx 10^{4004}$ cat it not start with 1000 instead of 9999?
    $endgroup$
    – Todor Markov
    Dec 28 '18 at 9:39








  • 3




    $begingroup$
    With continued fraction approximations you will alternatingly overshoot and undershoot, so you can stop at the right moment to make sure you are below. This doesn't tell you how many 9's you'll have at the beginning. In reality what I did was simply compute the value to check it had at least 3 leading 9's
    $endgroup$
    – doetoe
    Dec 28 '18 at 9:42



















2












$begingroup$

The first power of $2$ that begins with $999$ is $2^{2621}$. The number itself is $789$ digits long.
Done with the following python script:



x,y=1,0

while 1:

    if str(x)[:3]=="999":

        print(x,y)

        break

    x,y=x*2,y+1





share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Just for completeness: $2^{2621}approx9991223times10^{782}$
    $endgroup$
    – Martin Rosenau
    Dec 28 '18 at 9:28





















1












$begingroup$

Let $a = 2 ^ n$ and $b = 10 ^ m$ and $a <b$.



Note: $1024> 1000$ that comes from $128> 125$.



The sequence of powers of 2 is: $$1,2,4,8,16,32,64,128,256, ...$$



Whose first digits are: $$1,2,4,8,1,3,6,1,2,5,1,2,4,8,1,3,6,1,2,5,1, ...$$



At the moment you do not see nines ...



And, in addition, it seems that the sequence is repeated every 10 positions with: $$1,2,4,8,1,3,6,1,2,5$$



But the reality is that that first one and the following digits grow slowly towards the next digit ...



And the 8 becomes 9 with $2 ^ {53} = 2 ^ 3 cdot 2 ^ {50} = 9,007,199,254,740,992$



Everything mentioned above is related to the logarithm in base 10 of 2, which is: $0.3010299956639811952...$



And as you can see it is a real number (transcendent number) that is very close to $3/10$. Hence comes the "almost" repetition of ten digits mentioned above.



With these ingredients we can now propose the equation to solve to find a power number of two that starts with 999 when writing it in decimal:



$$2 ^ n = 10 ^ a cdot b$$



$$b in [0.999, 1); a, n in N$$



Taking base 10 logarithms:



$$n * 0.3010299956639811952... = a + c$$



$$c in [-0.00043451..., 0)$$



Which is easy to solve taking n as 100,000:



$$2 ^ {100,000} = 0.999002093 ... cdot 10 ^ {30103} = 999.002093 ... cdot 10 ^ {30100}$$






share|cite|improve this answer









$endgroup$




















    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    To find many answers, you can start looking for $n,m$ such that $2^n approx 10^m$, which is equivalent to $nlog2approx mlog10$.



    In other words, we look for rational approximations of $frac{log10}{log2}$, in which case $n$ would be the numerator.



    Using continued fractions, for



    $$frac{log10}{log2} approx 3 + frac1{3 + frac1{9 + frac1{2 + frac1{2 + frac1{4 + frac16}}}}} = frac{13301}{4004}$$



    we find $n = 13301$, for which $2^n$ starts with 9999.



    Note that since the sequence of continued fractions converges to the value, there is no limit on the number of leading 9's a power of 2 can have.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      If $2^{13301} approx 10^{4004}$ cat it not start with 1000 instead of 9999?
      $endgroup$
      – Todor Markov
      Dec 28 '18 at 9:39








    • 3




      $begingroup$
      With continued fraction approximations you will alternatingly overshoot and undershoot, so you can stop at the right moment to make sure you are below. This doesn't tell you how many 9's you'll have at the beginning. In reality what I did was simply compute the value to check it had at least 3 leading 9's
      $endgroup$
      – doetoe
      Dec 28 '18 at 9:42
















    6












    $begingroup$

    To find many answers, you can start looking for $n,m$ such that $2^n approx 10^m$, which is equivalent to $nlog2approx mlog10$.



    In other words, we look for rational approximations of $frac{log10}{log2}$, in which case $n$ would be the numerator.



    Using continued fractions, for



    $$frac{log10}{log2} approx 3 + frac1{3 + frac1{9 + frac1{2 + frac1{2 + frac1{4 + frac16}}}}} = frac{13301}{4004}$$



    we find $n = 13301$, for which $2^n$ starts with 9999.



    Note that since the sequence of continued fractions converges to the value, there is no limit on the number of leading 9's a power of 2 can have.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      If $2^{13301} approx 10^{4004}$ cat it not start with 1000 instead of 9999?
      $endgroup$
      – Todor Markov
      Dec 28 '18 at 9:39








    • 3




      $begingroup$
      With continued fraction approximations you will alternatingly overshoot and undershoot, so you can stop at the right moment to make sure you are below. This doesn't tell you how many 9's you'll have at the beginning. In reality what I did was simply compute the value to check it had at least 3 leading 9's
      $endgroup$
      – doetoe
      Dec 28 '18 at 9:42














    6












    6








    6





    $begingroup$

    To find many answers, you can start looking for $n,m$ such that $2^n approx 10^m$, which is equivalent to $nlog2approx mlog10$.



    In other words, we look for rational approximations of $frac{log10}{log2}$, in which case $n$ would be the numerator.



    Using continued fractions, for



    $$frac{log10}{log2} approx 3 + frac1{3 + frac1{9 + frac1{2 + frac1{2 + frac1{4 + frac16}}}}} = frac{13301}{4004}$$



    we find $n = 13301$, for which $2^n$ starts with 9999.



    Note that since the sequence of continued fractions converges to the value, there is no limit on the number of leading 9's a power of 2 can have.






    share|cite|improve this answer











    $endgroup$



    To find many answers, you can start looking for $n,m$ such that $2^n approx 10^m$, which is equivalent to $nlog2approx mlog10$.



    In other words, we look for rational approximations of $frac{log10}{log2}$, in which case $n$ would be the numerator.



    Using continued fractions, for



    $$frac{log10}{log2} approx 3 + frac1{3 + frac1{9 + frac1{2 + frac1{2 + frac1{4 + frac16}}}}} = frac{13301}{4004}$$



    we find $n = 13301$, for which $2^n$ starts with 9999.



    Note that since the sequence of continued fractions converges to the value, there is no limit on the number of leading 9's a power of 2 can have.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 28 '18 at 10:07

























    answered Dec 28 '18 at 9:25









    doetoedoetoe

    1,348811




    1,348811












    • $begingroup$
      If $2^{13301} approx 10^{4004}$ cat it not start with 1000 instead of 9999?
      $endgroup$
      – Todor Markov
      Dec 28 '18 at 9:39








    • 3




      $begingroup$
      With continued fraction approximations you will alternatingly overshoot and undershoot, so you can stop at the right moment to make sure you are below. This doesn't tell you how many 9's you'll have at the beginning. In reality what I did was simply compute the value to check it had at least 3 leading 9's
      $endgroup$
      – doetoe
      Dec 28 '18 at 9:42


















    • $begingroup$
      If $2^{13301} approx 10^{4004}$ cat it not start with 1000 instead of 9999?
      $endgroup$
      – Todor Markov
      Dec 28 '18 at 9:39








    • 3




      $begingroup$
      With continued fraction approximations you will alternatingly overshoot and undershoot, so you can stop at the right moment to make sure you are below. This doesn't tell you how many 9's you'll have at the beginning. In reality what I did was simply compute the value to check it had at least 3 leading 9's
      $endgroup$
      – doetoe
      Dec 28 '18 at 9:42
















    $begingroup$
    If $2^{13301} approx 10^{4004}$ cat it not start with 1000 instead of 9999?
    $endgroup$
    – Todor Markov
    Dec 28 '18 at 9:39






    $begingroup$
    If $2^{13301} approx 10^{4004}$ cat it not start with 1000 instead of 9999?
    $endgroup$
    – Todor Markov
    Dec 28 '18 at 9:39






    3




    3




    $begingroup$
    With continued fraction approximations you will alternatingly overshoot and undershoot, so you can stop at the right moment to make sure you are below. This doesn't tell you how many 9's you'll have at the beginning. In reality what I did was simply compute the value to check it had at least 3 leading 9's
    $endgroup$
    – doetoe
    Dec 28 '18 at 9:42




    $begingroup$
    With continued fraction approximations you will alternatingly overshoot and undershoot, so you can stop at the right moment to make sure you are below. This doesn't tell you how many 9's you'll have at the beginning. In reality what I did was simply compute the value to check it had at least 3 leading 9's
    $endgroup$
    – doetoe
    Dec 28 '18 at 9:42











    2












    $begingroup$

    The first power of $2$ that begins with $999$ is $2^{2621}$. The number itself is $789$ digits long.
    Done with the following python script:



    x,y=1,0
    
while 1:
    
    if str(x)[:3]=="999":
    
        print(x,y)
    
        break
    
    x,y=x*2,y+1





    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Just for completeness: $2^{2621}approx9991223times10^{782}$
      $endgroup$
      – Martin Rosenau
      Dec 28 '18 at 9:28


















    2












    $begingroup$

    The first power of $2$ that begins with $999$ is $2^{2621}$. The number itself is $789$ digits long.
    Done with the following python script:



    x,y=1,0
    
while 1:
    
    if str(x)[:3]=="999":
    
        print(x,y)
    
        break
    
    x,y=x*2,y+1





    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Just for completeness: $2^{2621}approx9991223times10^{782}$
      $endgroup$
      – Martin Rosenau
      Dec 28 '18 at 9:28
















    2












    2








    2





    $begingroup$

    The first power of $2$ that begins with $999$ is $2^{2621}$. The number itself is $789$ digits long.
    Done with the following python script:



    x,y=1,0
    
while 1:
    
    if str(x)[:3]=="999":
    
        print(x,y)
    
        break
    
    x,y=x*2,y+1





    share|cite|improve this answer











    $endgroup$



    The first power of $2$ that begins with $999$ is $2^{2621}$. The number itself is $789$ digits long.
    Done with the following python script:



    x,y=1,0
    
while 1:
    
    if str(x)[:3]=="999":
    
        print(x,y)
    
        break
    
    x,y=x*2,y+1






    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 28 '18 at 9:23

























    answered Dec 28 '18 at 9:18









    FrostFrost

    1776




    1776








    • 1




      $begingroup$
      Just for completeness: $2^{2621}approx9991223times10^{782}$
      $endgroup$
      – Martin Rosenau
      Dec 28 '18 at 9:28
















    • 1




      $begingroup$
      Just for completeness: $2^{2621}approx9991223times10^{782}$
      $endgroup$
      – Martin Rosenau
      Dec 28 '18 at 9:28










    1




    1




    $begingroup$
    Just for completeness: $2^{2621}approx9991223times10^{782}$
    $endgroup$
    – Martin Rosenau
    Dec 28 '18 at 9:28






    $begingroup$
    Just for completeness: $2^{2621}approx9991223times10^{782}$
    $endgroup$
    – Martin Rosenau
    Dec 28 '18 at 9:28













    1












    $begingroup$

    Let $a = 2 ^ n$ and $b = 10 ^ m$ and $a <b$.



    Note: $1024> 1000$ that comes from $128> 125$.



    The sequence of powers of 2 is: $$1,2,4,8,16,32,64,128,256, ...$$



    Whose first digits are: $$1,2,4,8,1,3,6,1,2,5,1,2,4,8,1,3,6,1,2,5,1, ...$$



    At the moment you do not see nines ...



    And, in addition, it seems that the sequence is repeated every 10 positions with: $$1,2,4,8,1,3,6,1,2,5$$



    But the reality is that that first one and the following digits grow slowly towards the next digit ...



    And the 8 becomes 9 with $2 ^ {53} = 2 ^ 3 cdot 2 ^ {50} = 9,007,199,254,740,992$



    Everything mentioned above is related to the logarithm in base 10 of 2, which is: $0.3010299956639811952...$



    And as you can see it is a real number (transcendent number) that is very close to $3/10$. Hence comes the "almost" repetition of ten digits mentioned above.



    With these ingredients we can now propose the equation to solve to find a power number of two that starts with 999 when writing it in decimal:



    $$2 ^ n = 10 ^ a cdot b$$



    $$b in [0.999, 1); a, n in N$$



    Taking base 10 logarithms:



    $$n * 0.3010299956639811952... = a + c$$



    $$c in [-0.00043451..., 0)$$



    Which is easy to solve taking n as 100,000:



    $$2 ^ {100,000} = 0.999002093 ... cdot 10 ^ {30103} = 999.002093 ... cdot 10 ^ {30100}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Let $a = 2 ^ n$ and $b = 10 ^ m$ and $a <b$.



      Note: $1024> 1000$ that comes from $128> 125$.



      The sequence of powers of 2 is: $$1,2,4,8,16,32,64,128,256, ...$$



      Whose first digits are: $$1,2,4,8,1,3,6,1,2,5,1,2,4,8,1,3,6,1,2,5,1, ...$$



      At the moment you do not see nines ...



      And, in addition, it seems that the sequence is repeated every 10 positions with: $$1,2,4,8,1,3,6,1,2,5$$



      But the reality is that that first one and the following digits grow slowly towards the next digit ...



      And the 8 becomes 9 with $2 ^ {53} = 2 ^ 3 cdot 2 ^ {50} = 9,007,199,254,740,992$



      Everything mentioned above is related to the logarithm in base 10 of 2, which is: $0.3010299956639811952...$



      And as you can see it is a real number (transcendent number) that is very close to $3/10$. Hence comes the "almost" repetition of ten digits mentioned above.



      With these ingredients we can now propose the equation to solve to find a power number of two that starts with 999 when writing it in decimal:



      $$2 ^ n = 10 ^ a cdot b$$



      $$b in [0.999, 1); a, n in N$$



      Taking base 10 logarithms:



      $$n * 0.3010299956639811952... = a + c$$



      $$c in [-0.00043451..., 0)$$



      Which is easy to solve taking n as 100,000:



      $$2 ^ {100,000} = 0.999002093 ... cdot 10 ^ {30103} = 999.002093 ... cdot 10 ^ {30100}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $a = 2 ^ n$ and $b = 10 ^ m$ and $a <b$.



        Note: $1024> 1000$ that comes from $128> 125$.



        The sequence of powers of 2 is: $$1,2,4,8,16,32,64,128,256, ...$$



        Whose first digits are: $$1,2,4,8,1,3,6,1,2,5,1,2,4,8,1,3,6,1,2,5,1, ...$$



        At the moment you do not see nines ...



        And, in addition, it seems that the sequence is repeated every 10 positions with: $$1,2,4,8,1,3,6,1,2,5$$



        But the reality is that that first one and the following digits grow slowly towards the next digit ...



        And the 8 becomes 9 with $2 ^ {53} = 2 ^ 3 cdot 2 ^ {50} = 9,007,199,254,740,992$



        Everything mentioned above is related to the logarithm in base 10 of 2, which is: $0.3010299956639811952...$



        And as you can see it is a real number (transcendent number) that is very close to $3/10$. Hence comes the "almost" repetition of ten digits mentioned above.



        With these ingredients we can now propose the equation to solve to find a power number of two that starts with 999 when writing it in decimal:



        $$2 ^ n = 10 ^ a cdot b$$



        $$b in [0.999, 1); a, n in N$$



        Taking base 10 logarithms:



        $$n * 0.3010299956639811952... = a + c$$



        $$c in [-0.00043451..., 0)$$



        Which is easy to solve taking n as 100,000:



        $$2 ^ {100,000} = 0.999002093 ... cdot 10 ^ {30103} = 999.002093 ... cdot 10 ^ {30100}$$






        share|cite|improve this answer









        $endgroup$



        Let $a = 2 ^ n$ and $b = 10 ^ m$ and $a <b$.



        Note: $1024> 1000$ that comes from $128> 125$.



        The sequence of powers of 2 is: $$1,2,4,8,16,32,64,128,256, ...$$



        Whose first digits are: $$1,2,4,8,1,3,6,1,2,5,1,2,4,8,1,3,6,1,2,5,1, ...$$



        At the moment you do not see nines ...



        And, in addition, it seems that the sequence is repeated every 10 positions with: $$1,2,4,8,1,3,6,1,2,5$$



        But the reality is that that first one and the following digits grow slowly towards the next digit ...



        And the 8 becomes 9 with $2 ^ {53} = 2 ^ 3 cdot 2 ^ {50} = 9,007,199,254,740,992$



        Everything mentioned above is related to the logarithm in base 10 of 2, which is: $0.3010299956639811952...$



        And as you can see it is a real number (transcendent number) that is very close to $3/10$. Hence comes the "almost" repetition of ten digits mentioned above.



        With these ingredients we can now propose the equation to solve to find a power number of two that starts with 999 when writing it in decimal:



        $$2 ^ n = 10 ^ a cdot b$$



        $$b in [0.999, 1); a, n in N$$



        Taking base 10 logarithms:



        $$n * 0.3010299956639811952... = a + c$$



        $$c in [-0.00043451..., 0)$$



        Which is easy to solve taking n as 100,000:



        $$2 ^ {100,000} = 0.999002093 ... cdot 10 ^ {30103} = 999.002093 ... cdot 10 ^ {30100}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 28 '18 at 10:51









        Angel MorenoAngel Moreno

        43225




        43225















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