G3.3 Let $k$ be positive integer and $f(k)$ a function that if $\frac{k-1}{k}=0 . k_{1} k_{2} k_{3} \ldots \ldots$, then $f(k)=\overline{k_{1} k_{2} k_{3}}$, for example, $f(3)=666$ because $\frac{3-1}{3}=0.666 \ldots \ldots$, find the value of $D=f(f(f(f(f(112)))))$.
$\begin{array}{l}0.99=1-\frac{1}{100}<\frac{112-1}{112}=1-\frac{1}{112}<1 \Rightarrow f(112)=\overline{99 k_{3}} \\ 0.998=1-\frac{1}{500}<\frac{\overline{99 k_{3}}-1}{\overline{99 k_{3}}}=1-\frac{1}{99 k_{3}}<1-\frac{1}{1000}=0.999 \Rightarrow f(f(112))=998 \\ \Rightarrow f(f(f(112)))=998 \Rightarrow D=f(f(f(f(f(112)))))=998 \\\end{array}$