Cesko-Slovacko-Poljsko

Obviously, the sequence $(a_n{)}_{n=1}^{\infty }$ is increasing until the first term with the zero digit occurs and is constant following this term. Our aim is to find such values $a$ and $k$ that the zero digit first occurs in $a_{2009}$. We will solve the problem in general – given an integer $m>4$, we present $a$ and $k$ such that the zero digit first occurs in $a_m$. Set $$a=\frac{10^{2m-5}-1}{9}=\underset{2m-5\text{ ones}}{\underbrace{11\dots 1}},k=10^{m-3}+4=1\underset{m-4\text{ zeros}}{\underbrace{00\dots 0}}4.$$

Then we have $$a_1=a=\underset{2m-5}{\underbrace{11\dots 1}}$$ $$\varrho (a_1)=1.$$ $$a_2=a_1+k=a_1+\underset{m-1}{\underbrace{100\dots 04}}=\underset{m-3}{\underbrace{11\dots 12}}\underset{m-4}{\underbrace{11\dots 15}}$$ $$\varrho (a_2)=10.$$ $$a_3=a_2+10k=a_2+\underset{m-4}{\underbrace{100\dots 040}}=\underset{m-4}{\underbrace{11\dots 122}}\underset{m-5}{\underbrace{11\dots 155}}$$ $$\varrho (a_3)=100.$$ $$a_i=a_{i-1}+10^{i-2}k=\underset{m-1}{\underbrace{11\dots 122\dots 2}}\underset{m-1}{\underbrace{11\dots 155\dots 5}}$$ $$\varrho (a_i)=10^{i-1}.$$ $$a_{m-2}=a_{m-3}+10^{m-4}k=\underset{m-3}{\underbrace{122\dots 255\dots 5}}$$ $$\varrho (a_{m-2})=10^{m-3}.$$ $$a_{m-1}-a_{m-2}+10^{m-3}k=\underset{m-4}{\underbrace{100\dots 0}}\underset{m-3}{\underbrace{100\dots 0}}\underset{m-3}{\underbrace{22\dots 2655\dots 5}}$$ $$\varrho (a_{m-1})=6\cdot 10^{m-3}.$$ $$a_m=a_{m-1}+6\cdot 10^{m-3}k=\underset{m-5}{\underbrace{600\dots 0}}\underset{m-3}{\underbrace{2400\dots 0}}\underset{m-5}{\underbrace{822\dots 25055\dots 5}}$$ $$\varrho (a_m)=0.$$

Second solution. Put $$a=\underset{2007}{\underbrace{611\dots 1}}\text{and}k=\underset{2007}{\underbrace{33\dots 34}}=\frac{1}{6}\cdot \underset{2007}{\underbrace{200\dots 04}}$$

Then we have $$\begin{array}{rlrlrl}{a}_1& =\underset{2007}{\underbrace{611\dots 1}},& \varrho (a_1)& =6,& k\varrho (a_1)& =\underset{2007}{\underbrace{200\dots 01}}\\ a_2& =\underset{2006}{\underbrace{2611\dots 15}},& \varrho (a_2)& =60,& k\varrho (a_2)& =\underset{2007}{\underbrace{200\dots 040}}\\ a_3& =\underset{2005}{\underbrace{22611\dots 155}},& \varrho (a_3)& =600,& k\varrho (a_3)& =\underset{2007}{\underbrace{200\dots 0400}}\\ a_4& =\underset{2004}{\underbrace{222611\dots 1555}},& \varrho (a_4)& =6000,& k\varrho (a_4)& =\underset{2007}{\underbrace{200\dots 04000}}\\ & & & & & \\ a_{t+1}=\underset{2007-t}{\underbrace{22\dots 2611\dots 155\dots 5}},& \varrho (a_{t+1})& =\underset{t}{\underbrace{60\dots 0}},& k\varrho (a_{t+1})& =\underset{2007}{\underbrace{200\dots 0400\dots 0}}& \\ & & & & & \\ a_{2007}=\underset{2006}{\underbrace{22\dots 2615\dots 5}}& \varrho (a_{2007})& =\underset{2006}{\underbrace{60\dots 0}},& k\varrho (a_{2007})& =\underset{2007}{\underbrace{200\dots 0400\dots 0}}& \\ a_{2008}=\underset{2007}{\underbrace{22\dots 2655\dots 5}}& \varrho (a_{2008})& =\underset{2007}{\underbrace{60\dots 0}},& k\varrho (a_{2008})& =\underset{2007}{\underbrace{200\dots 0400\dots 0}}& \\ a_{2009}=\underset{2007}{\underbrace{22\dots 23055\dots 5}}& \varrho (a_{2009})& =0,& k\varrho (a_{2009})& =0& \end{array}$$ and obviously $a_{2009},a_{2010},a_{2011},\dots$