Problem (Number Theory):
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 $d(n)$.
Suppose that $n$ is a house with $d(n)=1 / 27$. Determine the ten smallest positive integers $m$ (in the standard ordering of the integers) such that $m \in \mathcal{N}(n)$.

Solution:
Here, $\mathcal{N}(n)=\{m \mid m=27 k$, where $3 \nmid k\}$. The ten smallest elements of $\mathcal{N}(n)$ are 27, $54,108,135,189,216,270,297,351$, and 378.

Final answer: 27,54,108,135,189,216,270,297,351,378