Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
Starting digits of 2^n

While I was randomly working with number patterns, I came along with some interesting pattern which seems to turn to a conjecture in fact.

My conjecture can be stated as follows : For any number $n$ there exists any number $p$ such that the number $np$ can be a power of two or $2^k$ ( here the $np$ doesn't represent product of $n$ an $p$ but the number $p$ is placed to the right of $n$ , for eg: if the number $n$ is $11$ then if $p$ is $12$ then $np$ is $1112$ not $11 \cdot 12$ ). For a practical demonstration let us consider a number $11$ there exists any number you choose for example $80591620717411303424$ ( I have selected this number deliberately and I need to find an effective way of choosing numbers ) such that if the number is placed to right of $11$ it becomes $1180591620717411303424$ which in turn makes the complete entity as $2^{70}$ .

For a practical formulation, it took some time for me in the mathematical representation.
My conjecture can be stated as " Consider a number $N$, then there exist a number $P$ such that $$((N\cdot 10^{\lceil\,\log_{10} P\,\rceil})+P) \equiv 0\pmod{2^k} $$for some $K$. "

Does this already exists or it is new? , if it exists please give me a reference.

P.S : I have been trying for a proof, and I have a proof with me. Is it worth of publishing ?

Thank you.

share|improve this question

marked as duplicate by Andres Caicedo, Asaf Karagila, Jyrki Lahtonen, Henning Makholm, J. M. Apr 25 '12 at 12:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

10  
This is an immediate consequence of the fact that the fractional parts of $\log_{10} 2^k$ are dense in $(0,1)$ because $\log_{10} 2$ is irrational. –  Henning Makholm Apr 13 '12 at 11:49
1  
It should be $$ ((N\cdot 10^{\lceil\,\log_{10} P\,\rceil})+P) \equiv 0\pmod{2^k} $$ –  example Apr 13 '12 at 11:59
    
@Iyengar if $P=1235$ you would create 1235 zeroes to the right of $N$ with your formula instead of the correct 4. –  example Apr 13 '12 at 12:04
    
Oh Yes, now realized my mistake, Ya it should be the length of the P. Thank you. @example –  Iyengar Apr 13 '12 at 12:05
    
@Zander : But the answers are different and also informative answer is present here. –  Iyengar Apr 14 '12 at 16:11
show 3 more comments

1 Answer 1

up vote 9 down vote accepted

If I understood it correctly you are asking, whether any finite sequence $S$ of decimals (that we can also treat as an integer) forms a sequence of the most significant digits of an integer power of two?

This is, indeed, the case. It follows from Kronecker's density theorem stating that if $\theta$ is an irrational real number, $\alpha$ is a real number, and $\varepsilon>0$ is any positive real number, then there exist integers $h,k$ with $0<k$ such that $$ |k\theta-h-\alpha|<\varepsilon. $$

Let $\theta=\log_{10}2$. This is irrational by virtue of $\mathbf{Z}$ being a UFD. Let $\alpha=\log_{10}S$. What the above density result means is that there exist an integer $k>0$ such that $\log_{10}(2^k)$ differs from $\log_{10}S$ by something that can be made arbitrarily close to an integer with an appropriate choice of $k$. This means that $2^k$ is very close to $S\cdot 10^h$, which is exactly what you (hopefully) want. To make this rigorous we would actually need to select $\alpha=\log_{10}(S+0.5)$, and then select $\varepsilon$ to be small enough such that both $\log_{10}(S+1)$ and $\log_{10}S$ are further than $\varepsilon$ away from $\alpha.$

For a proof of Kronecker's density theorem, see e.g. Apostol's GTM series book on Modular functions. Probably Kronecker's theorem is in all the books that touch the basics of Diophantine approximation. I believe I have seen this particular problem as an exercise in one such book, but unfortunately cannot name one at this time.

share|improve this answer
    
Oh really ? . Which book ? –  Iyengar Apr 13 '12 at 12:03
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.