Prove that $N - M$ $equiv$ $0$ mod $9$.

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We say that a number $N$ is a "rainbow" number if every digit from $0, 1, .... 9$ shows up exactly once in $N$; for example $1234567890$ and $1029384756$ are both rainbow numbers. Take any two integer rainbow numbers $N$, $M$. Prove that $N - M$ $equiv$ $0$ mod $9$.



I am trying to use direct prove to find an example of $N$ and $M$ in order to prove this question. Are there any two rainbow numbers that can satisfy above condition? I am kinda stuck here.







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    Hint: show that any rainbow number is divisble by 9.
    – Sean Roberson
    Aug 14 at 3:08














up vote
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down vote

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We say that a number $N$ is a "rainbow" number if every digit from $0, 1, .... 9$ shows up exactly once in $N$; for example $1234567890$ and $1029384756$ are both rainbow numbers. Take any two integer rainbow numbers $N$, $M$. Prove that $N - M$ $equiv$ $0$ mod $9$.



I am trying to use direct prove to find an example of $N$ and $M$ in order to prove this question. Are there any two rainbow numbers that can satisfy above condition? I am kinda stuck here.







share|cite|improve this question


















  • 1




    Hint: show that any rainbow number is divisble by 9.
    – Sean Roberson
    Aug 14 at 3:08












up vote
-3
down vote

favorite









up vote
-3
down vote

favorite











We say that a number $N$ is a "rainbow" number if every digit from $0, 1, .... 9$ shows up exactly once in $N$; for example $1234567890$ and $1029384756$ are both rainbow numbers. Take any two integer rainbow numbers $N$, $M$. Prove that $N - M$ $equiv$ $0$ mod $9$.



I am trying to use direct prove to find an example of $N$ and $M$ in order to prove this question. Are there any two rainbow numbers that can satisfy above condition? I am kinda stuck here.







share|cite|improve this question














We say that a number $N$ is a "rainbow" number if every digit from $0, 1, .... 9$ shows up exactly once in $N$; for example $1234567890$ and $1029384756$ are both rainbow numbers. Take any two integer rainbow numbers $N$, $M$. Prove that $N - M$ $equiv$ $0$ mod $9$.



I am trying to use direct prove to find an example of $N$ and $M$ in order to prove this question. Are there any two rainbow numbers that can satisfy above condition? I am kinda stuck here.









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edited Aug 14 at 3:17









Shaun

7,41092972




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asked Aug 14 at 3:02









yoman

1




1







  • 1




    Hint: show that any rainbow number is divisble by 9.
    – Sean Roberson
    Aug 14 at 3:08












  • 1




    Hint: show that any rainbow number is divisble by 9.
    – Sean Roberson
    Aug 14 at 3:08







1




1




Hint: show that any rainbow number is divisble by 9.
– Sean Roberson
Aug 14 at 3:08




Hint: show that any rainbow number is divisble by 9.
– Sean Roberson
Aug 14 at 3:08










1 Answer
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You can't prove it by looking at particular two examples. You have to prove things in general or enumerate things (not a wise move). You can however try to construct some examples (and you should) to help you understand things.



If you want to prove a general statement is false, you can find a counter example, but this is not the case here.



Guide:



  • A number is divisible by $9$ if the sum of each digit is divisible by $9$.

  • Subtraction of two number that is divisible by $m$ is divisible by $m$.





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    up vote
    3
    down vote













    You can't prove it by looking at particular two examples. You have to prove things in general or enumerate things (not a wise move). You can however try to construct some examples (and you should) to help you understand things.



    If you want to prove a general statement is false, you can find a counter example, but this is not the case here.



    Guide:



    • A number is divisible by $9$ if the sum of each digit is divisible by $9$.

    • Subtraction of two number that is divisible by $m$ is divisible by $m$.





    share|cite|improve this answer


























      up vote
      3
      down vote













      You can't prove it by looking at particular two examples. You have to prove things in general or enumerate things (not a wise move). You can however try to construct some examples (and you should) to help you understand things.



      If you want to prove a general statement is false, you can find a counter example, but this is not the case here.



      Guide:



      • A number is divisible by $9$ if the sum of each digit is divisible by $9$.

      • Subtraction of two number that is divisible by $m$ is divisible by $m$.





      share|cite|improve this answer
























        up vote
        3
        down vote










        up vote
        3
        down vote









        You can't prove it by looking at particular two examples. You have to prove things in general or enumerate things (not a wise move). You can however try to construct some examples (and you should) to help you understand things.



        If you want to prove a general statement is false, you can find a counter example, but this is not the case here.



        Guide:



        • A number is divisible by $9$ if the sum of each digit is divisible by $9$.

        • Subtraction of two number that is divisible by $m$ is divisible by $m$.





        share|cite|improve this answer














        You can't prove it by looking at particular two examples. You have to prove things in general or enumerate things (not a wise move). You can however try to construct some examples (and you should) to help you understand things.



        If you want to prove a general statement is false, you can find a counter example, but this is not the case here.



        Guide:



        • A number is divisible by $9$ if the sum of each digit is divisible by $9$.

        • Subtraction of two number that is divisible by $m$ is divisible by $m$.






        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 14 at 3:13

























        answered Aug 14 at 3:07









        Siong Thye Goh

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