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Congruence sum proof

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that every positive integer is congruent to the sum of its digits (mod 9). (for example 38 is congruent to 11(mod 9))

    2. Relevant equations

    If a is congruent to b (mod n), then n divides (a-b)

    3. The attempt at a solution

    let a= {a1, a2..... a9} be digits
    where 0 is less than or equal to a which is less than 10

    That is where i don't know where to go, i have these digits and I don't know how to show that when added together they = b(mod 9)
    maybe i should do something about 9 divides some digits, but i have no clue.

    Any suggestions or tips would be great =)
     
  2. jcsd
  3. Feb 9, 2009 #2
    Let anan-1...a0 represent your number, where the ai are the digits. Another way of representing your number is [tex] 10^n a_n + \cdots + 10^0 a_0 [/tex]. So what you want to prove is that 9 divides [tex] (10^n a_n + \cdots + 10^0 a_0) - (a_n + \cdots + a_0) [/tex]
     
    Last edited by a moderator: Feb 9, 2009
  4. Feb 9, 2009 #3
    I know this may sound stupid, but how am I supposed to show that. Right now I have:

    9 divides a_n (10^(n)-1 )+ a_(n-1) (10^(n-1)-1)+⋯a_1 (10-1).
    How do i get that 9 divides that?
     
  5. Feb 9, 2009 #4
    Every term in [tex] a_n (10^n - 1) + a_{n-1} (10^{n-1} - 1) + \cdots + a_1 (10-1) [/tex] is of the form [tex] a_i (10^i -1) [/tex]. If you can prove that 9 divides any such term, then you are done.

    Can you figure out how to prove that 9 divides [tex] a_i (10^i -1) [/tex] for all positive integers i?
     
  6. Feb 9, 2009 #5
    well a_1(10-1)= a_1(9), and 9 divides a_1(9). So does that mean that it divides all i? I thought that only would apply for n=1
     
  7. Feb 9, 2009 #6

    Dick

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    10 is congruent to 1 mod 9. Hence 10^n is congruent to 1 mod 9.
     
  8. Feb 10, 2009 #7
    Where does the a_n go? wouldn't (a_1)10 be congruent to 1 mod 9, meaning that a_n would be congruent to 1 mod 9?
     
  9. Feb 10, 2009 #8
    ok, nevermind, i just figured out that that doesn't matter because 9 divides both a_n10^n and also 9 divides just 10^n
     
  10. Feb 10, 2009 #9
    well (10^n -1) that is
     
  11. Feb 10, 2009 #10

    Dick

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    a_n doesn't have to be congruent to anything. The point is that (10^n-1) is divisible by nine, as HitMan-2 pointed out. So a_n*10^n is congruent to a_n mod 9.
     
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