For any $n$ there's a power of $2$ which contains $n$

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So I saw this problem in an Olympiad book,
"Prove that for any natural number $n$, there exists a power of $2$ which contains $n$ in it. "



For example, $n=19$ is in $2^13=8192$, $n=24$ is in $2^10=1024$.



I tried solving it by Pigeonhole principle, but haven't made any progress. Any ideas?




number-theory elementary-number-theory pigeonhole-principle




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    I'd look for a power of two that starts with the digits of $n$.
    – Lord Shark the Unknown
    Aug 19 at 11:53










  • This should be a duplicate. Your claim follows from the fact, that the fractional part of $kcdot log_10(2)$ is equidistributed in the interval $[0,1]$ and therefore $2^k$ can begin with every finite digit string (that does not start with $0$)
    – Peter
    Aug 19 at 12:11















up vote
6
down vote

favorite
3












So I saw this problem in an Olympiad book,
"Prove that for any natural number $n$, there exists a power of $2$ which contains $n$ in it. "



For example, $n=19$ is in $2^13=8192$, $n=24$ is in $2^10=1024$.



I tried solving it by Pigeonhole principle, but haven't made any progress. Any ideas?




number-theory elementary-number-theory pigeonhole-principle




share|cite|improve this question


















  • 2




    I'd look for a power of two that starts with the digits of $n$.
    – Lord Shark the Unknown
    Aug 19 at 11:53










  • This should be a duplicate. Your claim follows from the fact, that the fractional part of $kcdot log_10(2)$ is equidistributed in the interval $[0,1]$ and therefore $2^k$ can begin with every finite digit string (that does not start with $0$)
    – Peter
    Aug 19 at 12:11













up vote
6
down vote

favorite
3









up vote
6
down vote

favorite
3






3





So I saw this problem in an Olympiad book,
"Prove that for any natural number $n$, there exists a power of $2$ which contains $n$ in it. "



For example, $n=19$ is in $2^13=8192$, $n=24$ is in $2^10=1024$.



I tried solving it by Pigeonhole principle, but haven't made any progress. Any ideas?




number-theory elementary-number-theory pigeonhole-principle




share|cite|improve this question














So I saw this problem in an Olympiad book,
"Prove that for any natural number $n$, there exists a power of $2$ which contains $n$ in it. "



For example, $n=19$ is in $2^13=8192$, $n=24$ is in $2^10=1024$.



I tried solving it by Pigeonhole principle, but haven't made any progress. Any ideas?





number-theory elementary-number-theory pigeonhole-principle





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 19 at 11:54

























asked Aug 19 at 11:51









Sudheesh Surendranath

1747




1747







  • 2




    I'd look for a power of two that starts with the digits of $n$.
    – Lord Shark the Unknown
    Aug 19 at 11:53










  • This should be a duplicate. Your claim follows from the fact, that the fractional part of $kcdot log_10(2)$ is equidistributed in the interval $[0,1]$ and therefore $2^k$ can begin with every finite digit string (that does not start with $0$)
    – Peter
    Aug 19 at 12:11













  • 2




    I'd look for a power of two that starts with the digits of $n$.
    – Lord Shark the Unknown
    Aug 19 at 11:53










  • This should be a duplicate. Your claim follows from the fact, that the fractional part of $kcdot log_10(2)$ is equidistributed in the interval $[0,1]$ and therefore $2^k$ can begin with every finite digit string (that does not start with $0$)
    – Peter
    Aug 19 at 12:11








2




2




I'd look for a power of two that starts with the digits of $n$.
– Lord Shark the Unknown
Aug 19 at 11:53




I'd look for a power of two that starts with the digits of $n$.
– Lord Shark the Unknown
Aug 19 at 11:53












This should be a duplicate. Your claim follows from the fact, that the fractional part of $kcdot log_10(2)$ is equidistributed in the interval $[0,1]$ and therefore $2^k$ can begin with every finite digit string (that does not start with $0$)
– Peter
Aug 19 at 12:11





This should be a duplicate. Your claim follows from the fact, that the fractional part of $kcdot log_10(2)$ is equidistributed in the interval $[0,1]$ and therefore $2^k$ can begin with every finite digit string (that does not start with $0$)
– Peter
Aug 19 at 12:11











1 Answer
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Here's a quick sketch of a proof:



  1. Prove $log_10(2)$ is irrational.


  2. Prove that the fractional part of $nalpha$, where $n in mathbbN$ and $alpha$ is irrational, is dense in $(0,1)$


  3. Note that the fractional part of $log(x)$ determines the first few digits, and then use the density of the fractional part of $nalpha$ to prove that the first few digits of the number can be any $m in mathbbN$.


If you need more help, just ask!






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    1 Answer
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    active

    oldest

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    5
    down vote



    accepted










    Here's a quick sketch of a proof:



    1. Prove $log_10(2)$ is irrational.


    2. Prove that the fractional part of $nalpha$, where $n in mathbbN$ and $alpha$ is irrational, is dense in $(0,1)$


    3. Note that the fractional part of $log(x)$ determines the first few digits, and then use the density of the fractional part of $nalpha$ to prove that the first few digits of the number can be any $m in mathbbN$.


    If you need more help, just ask!






    share|cite|improve this answer


























      up vote
      5
      down vote



      accepted










      Here's a quick sketch of a proof:



      1. Prove $log_10(2)$ is irrational.


      2. Prove that the fractional part of $nalpha$, where $n in mathbbN$ and $alpha$ is irrational, is dense in $(0,1)$


      3. Note that the fractional part of $log(x)$ determines the first few digits, and then use the density of the fractional part of $nalpha$ to prove that the first few digits of the number can be any $m in mathbbN$.


      If you need more help, just ask!






      share|cite|improve this answer
























        up vote
        5
        down vote



        accepted







        up vote
        5
        down vote



        accepted






        Here's a quick sketch of a proof:



        1. Prove $log_10(2)$ is irrational.


        2. Prove that the fractional part of $nalpha$, where $n in mathbbN$ and $alpha$ is irrational, is dense in $(0,1)$


        3. Note that the fractional part of $log(x)$ determines the first few digits, and then use the density of the fractional part of $nalpha$ to prove that the first few digits of the number can be any $m in mathbbN$.


        If you need more help, just ask!






        share|cite|improve this answer














        Here's a quick sketch of a proof:



        1. Prove $log_10(2)$ is irrational.


        2. Prove that the fractional part of $nalpha$, where $n in mathbbN$ and $alpha$ is irrational, is dense in $(0,1)$


        3. Note that the fractional part of $log(x)$ determines the first few digits, and then use the density of the fractional part of $nalpha$ to prove that the first few digits of the number can be any $m in mathbbN$.


        If you need more help, just ask!







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 27 at 17:27

























        answered Aug 19 at 12:12









        Isaac Browne

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