Problem (Geometry):
In ARMLopolis, every house number is a positive integer, and City Hall's address is 0. However, due to the curved nature of the cowpaths that eventually became the streets of ARMLopolis, the distance $d(n)$ between house $n$ and City Hall is not simply the value of $n$. Instead, if $n=3^{k} n^{\prime}$, where $k \geq 0$ is an integer and $n^{\prime}$ is an integer not divisible by 3 , then $d(n)=3^{-k}$. For example, $d(18)=1 / 9$ and $d(17)=1$. Notice that even though no houses have negative numbers, $d(n)$ is well-defined for negative values of $n$. For example, $d(-33)=1 / 3$ because $-33=3^{1} \cdot-11$. By definition, $d(0)=0$. Following the dictum "location, location, location," this Power Question will refer to "houses" and "house numbers" interchangeably.

Curiously, the arrangement of the houses is such that the distance from house $n$ to house $m$, written $d(m, n)$, is simply $d(m-n)$. For example, $d(3,4)=d(-1)=1$ because $-1=3^{0} \cdot-1$. In particular, if $m=n$, then $d(m, n)=0$.


The neighborhood of a house $n$, written $\mathcal{N}(n)$, is the set of all houses that are the same distance from City Hall as $n$. In symbols, $\mathcal{N}(n)=\{m \mid d(m)=d(n)\}$. Geometrically, it may be helpful to think of $\mathcal{N}(n)$ as a circle centered at City Hall with radius $d(n)$.


Unfortunately for new development, ARMLopolis is full: every nonnegative integer corresponds to (exactly one) house (or City Hall, in the case of 0). However, eighteen families arrive and are looking to move in. After much debate, the connotations of using negative house numbers are deemed unacceptable, and the city decides on an alternative plan. On July 17, Shewad Movers arrive and relocate every family from house $n$ to house $n+18$, for all positive $n$ (so that City Hall does not move). For example, the family in house number 17 moves to house number 35.
Ross takes a walk starting at his house, which is number 34 . He first visits house $n_{1}$, such that $d\left(n_{1}, 34\right)=1 / 3$. He then goes to another house, $n_{2}$, such that $d\left(n_{1}, n_{2}\right)=1 / 3$. Continuing in that way, he visits houses $n_{3}, n_{4}, \ldots$, and each time, $d\left(n_{i}, n_{i+1}\right)=1 / 3$. At the end of the day, what is his maximum possible distance from his original house? Justify your answer.

Solution:
The maximum possible distance $d\left(34, n_{k}\right)$ is $1 / 3$. This can be proved by induction on $k: d\left(n_{1}, 34\right) \leq 1 / 3$, and if both $d\left(n_{k-1}, 34\right) \leq 1 / 3$ and $d\left(n_{k-1}, n_{k}\right) \leq 1 / 3$, then $\max \left\{d\left(n_{k-1}, 34\right), d\left(n_{k-1}, n_{k}\right)\right\} \leq 1 / 3$ so by 5 b, $d\left(34, n_{k}\right) \leq 1 / 3$.

Final answer: $1/3$