Can f(n) equal f(x) in the sum of digits equation?

dapet
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"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.
 
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Can you handle the case where n=1111...1 ?
 
one presumes you mean in base ten as well.
 
Yes, you're right... in base ten.
 
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).
 
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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