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Math Help - Numbers whose decimal digits are 0 and 1

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

    Show that for any positive integer n, we can find a positive integer m such that the decimal digits of mn are only 0's and 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 n, we can find a positive integer m such that the decimal digits of mn are only 0's and 1's.
    This is a first impresssion again (not that my last was too succesful) but wouldn't the brute force solution just take n=a_0+\cdots+a_n10^n and let 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  m= \frac{1}{n}\left[ \sum_{j=1}^T 10^{b_j \phi(n) } \right ]

    where  T is a multiple of  n and  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  m= \frac{1}{n}\left[ \sum_{j=1}^T 10^{b_j \phi(n) } \right ]

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