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Math Help - Multiples of any natural number - only with 1,0 as digits

  1. #1
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    Multiples of any natural number - only with 1,0 as digits

    I read this somewhere recently.

    For any natural number n, there exists a multiple of n, such that the multiple has only 0 and 1 as it's digits.

    For e.g for 2, 3, 4, 5, 6 etc we have 10, 111, 100, 10, 1110 etc

    Any ideas how to go about proving this?
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  2. #2
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    Quote Originally Posted by aman_cc View Post
    I read this somewhere recently.

    For any natural number n, there exists a multiple of n, such that the multiple has only 0 and 1 as it's digits.

    For e.g for 2, 3, 4, 5, 6 etc we have 10, 111, 100, 10, 1110 etc

    Any ideas how to go about proving this?
    OEIS

    id:A004290 - OEIS Search Results

    From OEIS there is a link to this site

    Binary

    Interesting!
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  3. #3
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    Thanks. Infact I was able to work out the first proof given - that of dividing n into 1, 11, 111, 1111, ,,,,, etc etc

    But how to find the smallest such multiple still eludes me. Thanks
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  4. #4
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    Quote Originally Posted by aman_cc View Post
    Thanks. Infact I was able to work out the first proof given - that of dividing n into 1, 11, 111, 1111, ,,,,, etc etc

    But how to find the smallest such multiple still eludes me. Thanks
    If you have time to solve this similar problem

    Problem 303 - Project Euler

    then in the solution forum you will find some nice related discussion. Unfortunately I don't have time to adapt everything to this problem and then hide the source so people can't cheat.

    Problem 303 is pretty easy, but could be hard if you're not already familiar with a programming language.
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