Understanding a proof of Hindman and Strauss involving ternary expansion

Question

I'm trying to understand a proof of lemma 12.18 of Algebra in the Stone-Cech Compactification by Hindman and Strauss. In this proof, the authors slipt the natural numbers into the sets $$C_i = \{3^n(3k+i):n,k\in\omega\}$$ where $i\in\{1,2\}$ and $\omega=\mathbb{N}\cup\{0\}$, saying also that $C_i$ is the set of positive integers whose rightmost nonzero ternary digit is $i$. I didn't understand how this splitting is possible and what does it means that "$C_i$ is the set of positive integers whose rightmost nonzero ternary digit is $i$"?

Answer 1

You have to understand "ternary digits", i.e., integer arithmetic in base $3$ instead of $10$. In base $10$ the corresponding $D_i$ $(1\leq i\leq9)$ would be obtained as follows: Any natural number $N$ has in base $10$ a certain number $n\geq0$ zeros at the end, e.g., the number $304700$ has $n=2$ zeros at the end. When $N$ has $n$ zeros at the end then $$N=10^n\cdot N'$$ with $N'$ a natural number not divisible by $10$, hence with an $i\in[9]$ at the end:$$N'=10k+i,\qquad k\geq0, \quad i\in\{1,2,3,\ldots,9\}\ ,$$ whereby $k$ and $i$ are uniquely determined by $N$. This number $N$ will be put in the part $$D_i=\bigl\{10^n\cdot(10k+i)\bigm| n\geq0, \ k\geq0\bigr\}$$ of our partition $${\mathbb N}_{\geq1}=D_1\cup D_2\cup\ldots\cup D_9\ .$$