Finding integers whose sum of digits equals the sum of the digits of a multiple of them

For a given integer $$x>0$$, I need to find all integers $$a \in [0, 10^{15}]$$ which have the following property: the digit sum of $$a$$ equals the digit sum of $$x\cdot a$$.

I found this link http://mathworld.wolfram.com/CastingOutNines.html which looks quite relevant to my task, but I can't figure out how to apply it.

Any ideas?

• For this tiny range you may simply program verification of the property. – Wadim Zudilin May 4 '10 at 23:43
• Wouldn't $a_1=1$, $x=10$ work? Also, where do the $a_2,\ldots,a_n$ come in? Presumably, you meant that the digit sum of $a_i$ is equal to the digit sum of $x\cdot a_{i-1}$. However, I'm not sure this is appropriate for MO, please see the FAQ mathoverflow.net/faq#whatnot for a list of some other sites for which your question may be better suited. – Zev Chonoles May 4 '10 at 23:46
• Can't you take $x=10$ for all $a$'s? I guess I do not understand the problem... – Mariano Suárez-Álvarez May 4 '10 at 23:47
• The problem is: For a given integer $x>0$, find all numbers $a$ in the range such that the digit sums of $a$ and $xa$ coincide. – Wadim Zudilin May 4 '10 at 23:49
• Thanks for the clarification Wadim, it makes much more sense now. – Zev Chonoles May 4 '10 at 23:52

If you wish to get the property $S(a)=S(xa)$ (where $S(\cdot)$ denotes the sum of decimal digits) valid for all positive integers $x$, then there are no such numbers $a$. Indeed, if such an $a$ existed, then concantination $\overline{aa}=a\cdot x$ with $x=10^n+1$, $n$ the number of digits in $a$, would give you a number $xa$ with sum of digits twice more than $S(a)$.