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Math Help - Divisible by 7

  1. #1
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    Divisible by 7

    The number has nine (not necessarily distinct) decimal digits. The number is such that each of the nine -digit numbers formed by replacing just one of the digits is by the corresponding digit is divisible by . The number is related to in the same way: that is each of the nine numbers formed by replacing one of the by the corresponding is divisible by . Show that for each , is divisible by . (For example, if then may be or , since and are multiples of ).

    I've attempt some experimentation but the numbers are too big to play around with...

    how to do it?
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  2. #2
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    Let's translate each of these sentences into equations: Letting d=d_1 d_2 \ldots d_9 and so on,

    d + 3^{9-i}(e_i-d_i)\equiv 0 \pmod{7}
    e + 3^{9-i}(f_i-e_i)\equiv 0 \pmod{7}

    Summing the first equation over all i we get
    9d + e - d \equiv d+e \equiv 0 \pmod{7}.

    Summing the first two equations gives
    d+e + 3^{9-i}(f_i-d_i) \equiv 3^{9-i}(f_i-d_i) \equiv f_i-d_i \equiv 0 \pmod{7}
    since 3 and 7 are relatively prime.
    Last edited by evouga; January 6th 2010 at 11:47 AM. Reason: incorrect exponent
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  3. #3
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    No idea what you mean... can you show STEP BY STEP working?
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  4. #4
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    I finally get it ... I'm not sure how you got 3^{i-1} though (I'm assuming it's related to the mod) ... I would use 10^{9-i} instead ....

    usagi, the first two congruences are basically the individual replacing of digits. You start with the number d, then you replace one of it's digits d_i by e_i, such that it's divisible by 7. This is the same as adding the difference between d_i and e_i to the appropriate digit. Multiplying the difference by 10^{9-i} (instead of 3^{i-1}) will take care of which digit the difference will be added to. Same thing for converting the number e.

    When you add together the first congruence for all i (i.e. \sum_{i=1}^{9}(d + 10^{9-i}(e_i-d_i)), you get the third congruence.

    Can you figure it out from there?
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  5. #5
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    3^{i-1} \equiv 10^{i-1} \pmod{7} since 3 \equiv 10 \pmod{7}. The exponent in my original post is indeed wrong... I'll fix it.
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  6. #6
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    AH!!! Hahaha ... so obvious ... I thought you used an interesting technique because of the power, but yeah ... makes sense now
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