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Math Help - Divisible by 9

  1. #1
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    Divisible by 9

    hey guys,

    I was wondering how to prove this
    Problem:
    Let m be a positive integer and let n be an integer obtained from m by rearranging the digits of m in some way. (for example 381762 reaaranging 378126) prove m-n is divisble 9

    thanks
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  2. #2
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    Re: Divisible by 9

    If the digits of any number sum to a multiple of nine then it is also divisible by nine. Rearranging them won't change this.
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  3. #3
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    Re: Divisible by 9

    Hello,

    Well a simple way to do it is to remember that, S(.) being the sum of the digits of a number, m\equiv S(m) \bmod 9 (this can be easily proved with congruences, writing the number as powers of 10). And since S(m)=S(n) (rearranging the digits won't change their sum !), then m-n \equiv S(m)-S(n) \equiv 0 \bmod 9
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    Re: Divisible by 9

    Quote Originally Posted by pickslides View Post
    If the digits of any number sum to a multiple of nine then it is also divisible by nine. Rearranging them won't change this.
    Yes, but it isn't stated that m is divisible by 9
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    Re: Divisible by 9

    what if the sum of its digits is not multiple to nine like 721 and 172
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    Re: Divisible by 9

    Quote Originally Posted by Moo View Post
    Yes, but it isn't stated that m is divisible by 9
    Well I did say if..
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  7. #7
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    Re: Divisible by 9

    More generally, if the remainder of the sum of digits of a number, divided by 9, is n, then the remainder of the number, divided by 9, is also n. Since rearranging the digits in a number does not change the sum of digits, it also does not change the remainder when divided by n. Let A be the original number, and let n be its remainder when divided by 9: A= 9k+ n for some integer k. If B is any rearrangement of the digits of A, then B= 9j+ n for some integer j. A-B= (9k+ n)- (9j+ n)= 9(k- j).

    You can prove that statement about the remainders by looking at the numbers expansion in powers of 10:
    A= a_0+ 10a_1+ 100a_2+ \cdot\cdot\cdot+ 10^ma_m
    = a_0+ (9+ 1)a_1+ (99+ 1)a_2+ \cdot\cdot\cdot+ (9(1+ 10+\cdot\cdot\cdot+ 10^{m-1})a_m
    = (a_0+ a_1+ a_2+ \cdot\cdot\cdot+ a_m)+ 9(a_1+ 11a_2+ \cdot\cdot\cdot (1+ 10^{k-1})a_m)
    Last edited by HallsofIvy; August 14th 2011 at 10:47 AM.
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  8. #8
    Moo
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    Re: Divisible by 9

    Quote Originally Posted by HallsofIvy View Post
    More generally, if the remainder of the sum of digits of a number, divided by 9, is n, then the remainder the number, divided by 9, is also n. Since rearranging the digits in a number does not change the sum of digits, it also does not change the remainder when divided by n. Let A be the original number, and let n be its remainder when divided by 9: A= 9k+ n for some integer k. If B is any rearrangement of the digits of A, then B= 9j+ n for some integer j. A-B= (9k+ n)- (9j+ n)= 9(k- j).

    You can prove that statement about the remainders by looking at the numbers expansion in powers of 10:
    A= a_0+ 10a_1+ 100a_2+ \cdot\cdot\cdot+ 10^ma_m
    = a_0+ (9+ 1)a_1+ (99+ 1)a_2+ \cdot\cdot\cdot+ (9(1+ 10+\cdot\cdot\cdot+ 10^{m-1})a_m
    = (a_0+ a_1+ a_2+ \cdot\cdot\cdot+ a_m)+ 9(a_1+ 11a_2+ \cdot\cdot\cdot (1+ 10^{k-1})a_m)
    Hm well, isn't that equivalent to what I wrote in #3 ? (I'm quite disturbed by your "more generally", not by the fact that you're giving another way to see through the answer)
    The congruence gives the remainder, so...

    And the OP seems to have understood the solution, given the private messages he's sent me...
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  9. #9
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    Re: Divisible by 9

    Yes, you are completely correct! I just did not read your post properly. The "more generally" was referring to PickSlides' post #2.
    Last edited by HallsofIvy; August 17th 2011 at 04:41 AM.
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  10. #10
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    Re: Divisible by 9

    Quote Originally Posted by Moo View Post
    Hm well, isn't that equivalent to what I wrote in #3 ? (I'm quite disturbed by your "more generally", not by the fact that you're giving another way to see through the answer)
    The congruence gives the remainder, so...

    And the OP seems to have understood the solution, given the private messages he's sent me...
    Yes, Moo. Your post, #3, was ignored earlier, particularly by OP.
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  11. #11
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    Re: Divisible by 9

    Quote Originally Posted by SammyS View Post
    Yes, Moo. Your post, #3, was ignored earlier, particularly by OP.
    Yes, I'm very sad that it was ignored !!

    And the OP has thanked it and sent me a pm about what I wrote in it, so I guess he didn't ignore it !
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  12. #12
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    Re: Divisible by 9

    well sum adds up to 27 so the number is divisble by 9 and 3 and even if you rearrange it adds to same
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