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Sum of digits equation

  1. May 6, 2004 #1
    "sum of digits" equation

    Let f(n) denote the sum of (all) digits of natural number n. Prove that for each natural n we can choose convenient value of natural parameter p such that the equation f(npx)=f(x) has solution in natural numbers x that doesn't contain any "9" in its notation.

    Does anybody have any idea? I don't... but I hope that you do... Actually I can solve a lot of special cases on a lot of pages... but I can't solve it generally. Is there any trick or only hard work? Thank you.
  2. jcsd
  3. May 6, 2004 #2


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    Can you handle the case where n=1111....1 ?
  4. May 6, 2004 #3

    matt grime

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    one presumes you mean in base ten as well.
  5. May 8, 2004 #4
    Yes, you're right... in base ten.
  6. Jun 11, 2004 #5
    If 30 dosen't divide n, the result is true (see Acta Arithmetica 81; Mauduit and Scharkozy's article on sum of digits, theorem 2).
    You can impose lot's of conditions on x (the only thing you really need is that you have an infintly choice on it) as x is in the set {1,11,111,1111,...}

    If 30 divide n, the result is true too, but i don't have reference (it's only a special technical case).
  7. Jun 11, 2004 #6
    Sorry i have done a small confusion : the 2 cases are
    * If 30 is prime with n, then, ...
    * If 30 and n have a commun divisor, then ...
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