As I remmember , I had seen this problem before 2002,Maybe I am wrong. I have a simple solution: use induction: if $x+d(x)=m_k$have exactly k solution, then there exist $0\leq t\leq 8$so that $2+m_k+2t$is a multipy of 9 let $l=\frac{2+m_k+2t}{9}$ let $m_{k+1}=m_k+10^l+1$ then $x+d(x)=m_{k+1}$have exactly {k+1} solution.

Sorry,I didn't see how you apply induction and why there exist exactly $ k$ solutions but what about more than $ k$? The solution below was rather long,because we should concentrate on exactly $ k$ solutions... By the way,soory for this bad-writing solution,because both the idea and construction are extraordinary complicated

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This is the official solution...But I still missed the part for exactly $ k$ solutions... Is this part trivial to solve for I just missed some simple facts???

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Doc1.pdf (202kb)

Hawk is right.

123eeee wrote: Hawk is right. Probably,but could anyone help me to explain?Thanks in advance

Answer in chinese.

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2003tst.pdf (58kb)

Thank you very much!I finally found the thing I missed,that is,if $ n<10^k,s(n)=9k-s(10^k-1-n)$

Thank you so much for your explaination,I have rewritten it into English

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First, let's make the observation that we are essentially tasked with finding $m$ so that the number of ways to represent it as a sum of at most $9$ $11$'s, at most $9$ $101$'s, at most $9$ $1001$'s, etc. We will actually show the stronger statement that there exists such an $m$ which is $7$ (mod $9$). We'll show this by induction on $k$. When $k = 1$, $m = 88$ suffices. Suppose it holds when $k = t.$ When $k = t+1$, let $9\ell - 2$ be the $m$ that works when $k = t.$ We claim that $8 \cdot (10^{\ell} + 1)$ works when $k = t+1.$ It's clear that we cannot use things $\ge 10^{\ell + 1}.$ If we use $8 \cdot (10^{\ell}+1)$, we cannot use anything else. Now, observe that we must otherwise use $7$ $(10^{\ell} + 1)$'s. In this case, observe that $10^{\ell} + 1 - 9(10 ^{\ell - 1} +1 + 10^{\ell - 2} + 1 + \cdot + 10 + 1 + 1 + 1) = 2 - 9 \ell.$ Hence, we must use $9$ of each of $10^{\ell - 1} + 1, 10^{\ell - 2} +1, \cdots, 10 + 1, 1+1$ except for some which sum to a total of $9 \ell -2.$ However, from our inductive hypothesis, there are exactly $t$ ways to exclude numbers with sum $9 \ell - 2.$ Hence, combining with the $1$ way from earlier ($8$ copies of $10^{\ell} + 1$), we have completed the induction. $\square$