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Thread: Numbers whose decimal digits are 0 and 1

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    MHF Contributor Bruno J.'s Avatar
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    Numbers whose decimal digits are 0 and 1

    Show that for any positive integer $\displaystyle n$, we can find a positive integer $\displaystyle m$ such that the decimal digits of $\displaystyle mn$ are only $\displaystyle 0$'s and $\displaystyle 1$'s.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Show that for any positive integer $\displaystyle n$, we can find a positive integer $\displaystyle m$ such that the decimal digits of $\displaystyle mn$ are only $\displaystyle 0$'s and $\displaystyle 1$'s.
    This is a first impresssion again (not that my last was too succesful) but wouldn't the brute force solution just take $\displaystyle n=a_0+\cdots+a_n10^n$ and let $\displaystyle m=b_0+\cdots+b_m10^m$ multiply the two and then reduce this to a linear algebra question on solving equations?
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    Can we just set $\displaystyle m= \frac{1}{n}\left[ \sum_{j=1}^T 10^{b_j \phi(n) } \right ] $

    where $\displaystyle T $ is a multiple of $\displaystyle n $ and $\displaystyle b_j , j\in\mathbb{N} $ is any sequence (of course positive ) .
    Last edited by simplependulum; May 7th 2010 at 10:47 PM.
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    Quote Originally Posted by simplependulum View Post
    Can we just set $\displaystyle m= \frac{1}{n}\left[ \sum_{j=1}^T 10^{b_j \phi(n) } \right ] $

    where $\displaystyle T $ is a multiple of $\displaystyle n $ and $\displaystyle b_j , j\in\mathbb{N} $ is any sequence (of course positive ) .
    what if $\displaystyle \gcd(n,10) > 1$?
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    Quote Originally Posted by NonCommAlg View Post
    what if $\displaystyle \gcd(n,10) > 1$?
    Perhaps there are some limits in Euler Phi function but if $\displaystyle (n,10) \neq 1 $ then $\displaystyle n $ must be $\displaystyle 2^a 5^b N $ , now $\displaystyle N $ is coprime with $\displaystyle 10 $ . I let $\displaystyle m = 5^a 2^b M $ so $\displaystyle mn = 10^{a+b} MN = MN00000000000000000000000... $ .
    Last edited by simplependulum; May 7th 2010 at 11:40 PM.
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    Quote Originally Posted by simplependulum View Post
    Perhaps there are some limits in Euler Phi function but if $\displaystyle (n,10) \neq 1 $ then $\displaystyle n $ must be $\displaystyle 2^a 5^b N $ , now $\displaystyle N $ is coprime with $\displaystyle 10 $ . I let $\displaystyle m = 5^a 2^b M $ so $\displaystyle mn = 10^{a+b} MN = MN00000000000000000000000... $ .
    now it's a complete proof!
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