[Algebra]
Problem: 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 $d(m, n)=0$.
Of the houses with positive numbers less than 100, find, with proof, the house or houses which is (are) closest to City Hall.

Solution: If $n=3^{k} m$ where $3 \nmid m$, then $d(n)=1 / 3^{k}$. So the smallest values of $d(n)$ occur when $k$ is largest. The largest power of 3 less than 100 is $3^{4}=81$, so $d(81)=1 / 3^{4}=1 / 81$ is minimal.

Final answer: 81