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Math Help - Proof required

  1. #1
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    Proof required

    I have observed that if one takes a whole number, between 10 and 9999, then re-arranges the digits in any way. Then, one subtracts the lesser number from the greater, the answer is always divisible by three. Eg let 5124 be original number. Re-arranged to be 4512. 5124 - 4512 = 612, which is divisible by three.

    Is there a proof as to why this is so?

    Many thanks if anyone can help

    c0unsel
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  2. #2
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    Not only is the number divisible by 3 its divisible by 9!

    Say n=A_1A_2...A_N be a number. And let m=B_1B_2...B_N be a permutation of its digits.

    Then, n=10^nA_1+10^{n-1}A_{n-1}+...+A_N and m=10^nB_1+10^{n-1}B_2+...+B_N

    Thus,
    n-m = 10^n(A_1-B_1)+10^{n-1}(A_2-B_2)+...+(A_N-B_N) \equiv (A_1-B_1)+(A_2-B_2)+...+(A_N-B_N)(\bmod 9)
    Since 10\equiv 1(\bmod 9)\implies 10^k \equiv 1(\bmod 9).
    But,
    (A_1-B_1)+...(A_N-B_N)=(A_1+...+A_N)-(B_1+...+B_N)=0.
    Q.E.D.
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  3. #3
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    Thankyou for the proof which i confess, is too complicated for me to follow. Is there any prospects of breaking down the steps further. If not, thankyou anyway and i will try and decipher.

    c0unsel
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  4. #4
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    Quote Originally Posted by c0unsel View Post
    Thankyou for the proof which i confess, is too complicated for me to follow. Is there any prospects of breaking down the steps further. If not, thankyou anyway and i will try and decipher.

    c0unsel
    Are you familar with the basic laws of modular arithmetic?
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  5. #5
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    Unfortunately not. However, in the event that the same proves too complex to explain a suitable link would be gratefully digested.
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  6. #6
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    Maybe this is a little too advanced.
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