Spring Mathematical Tournament

Since the sum of digits of an integer divisible by $k-1$, is divisible by $k-1$, too, $(k-1{)}^3$ divides all the numbers obtained after the second step. $(k-1{)}^3$. On the other hand, if $a=\overline{a_n{a}_{n-1}\dots a_0(k)}$, $n\ge 4$ or $a_3\ge 2$, $n=3$, then

$$\begin{array}{rl}a-(k-1{)}^2(a_n+a_{n-1}+\cdots +a_0)& \ge 2(k^3-(k-1{)}^2)-a_1((k-1{)}^2-k)-a_0((k-1{)}^2-1))\\ & \ge 2(k^3-(k-1{)}^2)-(k-1)(2(k-1{)}^2-k-1)=5k^2-2k-1>0.\end{array}$$ This shows that we shall get a number of the form $a=\overline{a_3{a}_2{a}_1{a}_0(k)}$, where $a_3=0,1$. The next number is $$(k-1{)}^2(a_3+a_2+a_1+a_0)\le (k-1{)}^2(1+3(k-1))<4(k-1{)}^3.$$

Hence this number is $(k-1{)}^3$, $2(k-1{)}^3$ or $3(k-1{)}^3$. Note that $$(k-1{)}^3={\overline{k-3,2,k-1}}_{(k)}\to 2(k-1{)}^3,k>2,$$ $$2(k-1{)}^3={\overline{1,k-6,5,k-2}}_{(k)}\to 2(k-1{)}^3,k>5,$$ $$3(k-1{)}^3={\overline{2,k-9,8,k-3}}_{(k)}\to 2(k-1{)}^3,k>8.$$ Since $3.5^3={\overline{1423}}_{(6)}\to 10.5^2=2.5^3$, $3.6^3={\overline{1614}}_{(7)}\to 12.6^2=2.6^3$ and $3.7^3={\overline{2005}}_{(8)}\to 7.7^2=7^3\to 2.7^3$, we conclude that if $k>5$, then the numbers are equal to $2(k-1{)}^3$ from some point onwards.