di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Geometry


Question: 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 $d(m, n)=0$.

A visitor to ARMLopolis is surprised by the ARMLopolitan distance formula, and starts to wonder about ARMLopolitan geometry. An ARMLopolitan triangle is a triangle, all of whose vertices are ARMLopolitan houses.
Show that $d(17,51) \leq d(17,34)+d(34,51)$ and that $d(17,95) \leq d(17,68)+d(68,95) . \quad[1 \mathrm{pt}]$

Solution:
Because $d(17,34)=d(17)=1, d(34,51)=d(17)=1$, and $d(17,51)=d(34)=1$, it follows that $d(17,51) \leq d(17,34)+d(34,51)$. Because $d(17,68)=d(51)=1 / 3$, $d(68,95)=d(27)=1 / 27$, and $d(17,95)=d(78)=1 / 3$, it follows that $d(17,95) \leq$ $d(17,68)+d(68,95)$.
Subfield: Geometry


Question: 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 $d(m, n)=0$.

A visitor to ARMLopolis is surprised by the ARMLopolitan distance formula, and starts to wonder about ARMLopolitan geometry. An ARMLopolitan triangle is a triangle, all of whose vertices are ARMLopolitan houses.
If $a=17$ and $b=68$, determine all values of $c$ such that $d(a, c)=d(a, b)$.

Solution:
Because $d(17,68)=1 / 3$, the condition implies that $3 \mid(c-17)$ but $9 \nmid(c-17)$. Thus either $c-17=9 k+3$ or $c-17=9 k+6$ for some $k \in \mathbb{Z}$. Solving for $c$ yields $c=9 k+20$ or $c=9 k+23$, equivalently $c=9 k+2$ or $c=9 k+5, k \in \mathbb{Z}, k \geq 0$.

Final answer: $c=9 k+2$ or $c=9 k+5, k \in \mathbb{Z}, k \geq 0$
Subfield: Geometry


Question: 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 $d(m, n)=0$.

A visitor to ARMLopolis is surprised by the ARMLopolitan distance formula, and starts to wonder about ARMLopolitan geometry. An ARMLopolitan triangle is a triangle, all of whose vertices are ARMLopolitan houses.
Prove that, for all $a$ and $b, d(a, b) \leq \max \{d(a), d(b)\}$.

Solution:
Write $a=3^{\alpha} a_{0}$ and $b=3^{\beta} b_{0}$, where $3 \nmid a_{0}$ and $3 \nmid b_{0}$. First consider the case $\alpha=\beta$. Then $a-b=3^{\alpha}\left(a_{0}-b_{0}\right)$. In this case, if $3 \mid\left(a_{0}-b_{0}\right)$, then $3^{\alpha}\left(a_{0}-b_{0}\right)=3^{\gamma} c$, where $3 \nmid c$ and $\alpha<\gamma$; so $d(a-b)=3^{-\gamma}<3^{-\alpha}=d(a)=d(b)$. If $3 \nmid\left(a_{0}-b_{0}\right)$, then $d(a, b)=3^{-\alpha}=d(a)=d(b)$. If $\alpha \neq \beta$, suppose, without loss of generality, that $\alpha<\beta$ (so that $d(a)>d(b))$. Then $a-b=3^{\alpha}\left(a_{0}-3^{\beta-\alpha} b_{0}\right)$. In this second factor, notice that the second term, $3^{\beta-\alpha} b_{0}$ is divisible by 3 but the first term $a_{0}$ is not, so their difference is not divisible by 3 . Thus $d(a, b)=3^{-\alpha}=d(a)$. Therefore $d(a, b)=d(a)$ when $d(a)>d(b)$, and similarly $d(a, b)=d(b)$ when $d(b)>d(a)$. Hence $d(a, b) \leq \max \{d(a), d(b)\}$.
Subfield: Geometry


Question: 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 $d(m, n)=0$.

A visitor to ARMLopolis is surprised by the ARMLopolitan distance formula, and starts to wonder about ARMLopolitan geometry. An ARMLopolitan triangle is a triangle, all of whose vertices are ARMLopolitan houses.
Prove that, for all $a, b$, and $c, d(a, c) \leq \max \{d(a, b), d(b, c)\}$.

Solution:
Note that $d(a, c)=d(a-c)=d((a-b)-(c-b))=d(a-b, c-b)$. Then $d(a, c) \leq$ $\max \{d(a, b), d(c, b)\}$ by part 5 a, and the fact that $d(c, b)=d(b, c)$.
Subfield: Geometry


Question: 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 $d(m, n)=0$.

A visitor to ARMLopolis is surprised by the ARMLopolitan distance formula, and starts to wonder about ARMLopolitan geometry. An ARMLopolitan triangle is a triangle, all of whose vertices are ARMLopolitan houses.
Prove that, for all $a, b$, and $c, d(a, c) \leq d(a, b)+d(b, c)$.

Solution:
Because $d(x, y)>0$ for all $x \neq y, \max \{d(a, b), d(b, c)\}<d(a, b)+d(b, c)$ whenever $a \neq b \neq c$. Thus $d(a, c)<d(a, b)+d(b, c)$ when $a, b, c$ are all different. If $a=b \neq c$, then $d(a, b)=0, d(a, c)=d(b, c)$, so $d(a, c) \leq d(a, b)+d(b, c)$. And if $a=c \neq b$, then $d(a, c)=0$ while $d(a, b)=d(b, c)>0$, so $d(a, c)<d(a, b)+d(b, c)$.
Subfield: Geometry


Question: 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 $d(m, n)=0$.

A visitor to ARMLopolis is surprised by the ARMLopolitan distance formula, and starts to wonder about ARMLopolitan geometry. An ARMLopolitan triangle is a triangle, all of whose vertices are ARMLopolitan houses.
After thinking about it some more, the visitor announces that all ARMLopolitan triangles have a special property. What is it? Justify your answer.

Solution:
The foregoing shows that all ARMLopolitan triangles are isosceles! Examining the proof in 5a, note that if $d(a) \neq d(b)$, then $d(a, b)=\max \{d(a), d(b)\}$. Applying that observation to the proof in $5 \mathrm{~b}$, if $d(a, b) \neq d(b, c)$, then $d(a, c)=\max \{d(a, b), d(b, c)\}$. So either $d(a, b)=d(b, c)$ or, if not, then either $d(a, c)=d(a, b)$ or $d(a, c)=d(b, c)$. Thus all ARMLopolitan triangles are isosceles.
Subfield: Geometry


Question: 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.
Prove that the distance between houses with consecutive numbers does not change after the move.

Solution:
Note that $d(m+18, n+18)=d((m+18)-(n+18))=d(m-n)=d(m, n)$. In particular, if $m=n+1$, then $d(m, n)=1$, thus $d(m+18, n+18)$ is also 1 .
Subfield: Geometry


Question: 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.
The residents of $\mathcal{N}(1)$ value their suburban location and protest the move, claiming it will make them closer to City Hall. Will it? Justify your answer.

Solution:
No, they won't move. If $a \in \mathcal{N}(1)$, then $3 \nmid a$. Thus $3 \nmid(a+18)$, and so $d(a+18)=1$.
Subfield: Geometry


Question: 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.
Some residents of $\mathcal{N}(9)$ claim that their tightly-knit community will be disrupted by the move: they will be scattered between different neighborhoods after the change. Will they? Justify your answer.

Solution:
Yes, they will be split apart. If $a \in \mathcal{N}(9)$, then $9 \mid a$ but $27 \nmid a$, so $a=27 k+9$ or $a=27 k+18$. Adding 18 has different effects on these different sets of houses: $27 k+18 \mapsto 27 k+36$, which is still in $\mathcal{N}(9)$, while $27 k+9 \mapsto 27 k+27$, which is in $\mathcal{N}(27 r)$ for some $r \in \mathbb{Z}^{+}$, depending on the exact value of $a$. For example, $a=9 \mapsto 27$, while $a=63 \mapsto 81$, and $d(81)=3^{-4}$. That is, $d(a+18) \leq 3^{-3}$ whenever $a=27 k+9$, with equality unless $a \equiv 63 \bmod 81$.
Subfield: Geometry


Question: 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.
Other residents of $\mathcal{N}(9)$ claim that their community will be disrupted by newcomers: they say that after the move, their new neighborhood will also contain residents previously from several different old neighborhoods (not just the new arrivals to ARMLopolis). Will it? Justify your answer

Solution:
The question is whether it is possible that $x \notin \mathcal{N}(9)$ but $x+18 \in \mathcal{N}(9)$. Because $d(9)=1 / 9$, there are two cases to consider: $d(x)>1 / 9$ and $d(x)<1 / 9$. In the first case, $9 \nmid x$ (either $x$ is not divisible by 3 , or $x$ is divisible by 3 and not 9 ). These houses do not move into $\mathcal{N}(9): x \not \equiv 0 \bmod 9 \Rightarrow x+18 \not \equiv 0 \bmod 9$. On the other hand, if $d(x)<1 / 9$, then $27 \mid x$, i.e., $x=27 k$. Then $x+18=27 k+18$ which is divisible by 9 but not by 27 , so $d(x+18)=9$. So in fact every house $x$ of ARMLopolis with $d(x)<1 / 9$ will move into $\mathcal{N}(9)$ !
Subfield: Geometry


Question: 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.
One day, Paul (house 23) and Sally (house 32), longtime residents of $\mathcal{N}(1)$, are discussing the " 2 side" of the neighborhood, that is, the set of houses $\{n \mid n=3 k+2, k \in \mathbb{Z}\}$. Paul says "I feel like I'm at the center of the " 2 side": when I looked out my window, I realized that the "2 side" consists of exactly those houses whose distance from me is at most $1 / 3 . "$ Prove that Paul is correct.

Solution:
If $n=3 k+2$, then $d(23, n)=d(3 k-21)$, and because $3 k-21=3(k-7), d(3 k-21) \leq$ $1 / 3$. On the other hand, if $d(23, n) \leq 1 / 3$, then $3 \mid(n-23)$, so $n-23=3 k$, and $n=3 k+23=3(k+7)+2$.