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 $d(m, n)=0$.


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.
Sally (house 32). Prove that $\{n \mid n=3 k+2, k \in \mathbb{Z}\}$ consists exactly of those houses whose distance from Sally's house is at most $1 / 3$.

Solution:
If $n=3 k+2$, then $d(32, n)=d(3 k-30) \leq 1 / 3$. Similarly, if $d(32, n) \leq 1 / 3$, then $3 \mid n-32 \Rightarrow n=3 k+32=3(k+10)+2$.
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.
For any $x$, let $\mathcal{D}_{r}(x)=\{y \mid d(x, y) \leq r\}$, that is, $\mathcal{D}_{r}(x)$ is the disk of radius $r$. Prove that if $d(x, z)=r$, then $\mathcal{D}_{r}(x)=\mathcal{D}_{r}(z) . \quad[2 \mathrm{pts}]$

Solution:
Show that $\mathcal{D}_{r}(x) \subseteq \mathcal{D}_{r}(z)$ and conversely. Suppose that $y \in \mathcal{D}_{r}(x)$. Then $d(x, y) \leq r$. Because $d(x, z)=r$, by $5 \mathrm{~b}, d(y, z) \leq \max \{d(x, y), d(x, z)\}=r$. So $y \in \mathcal{D}_{r}(z)$. Thus $\mathcal{D}_{r}(x) \subseteq \mathcal{D}_{r}(z)$. Similarly, $\mathcal{D}_{r}(z) \subseteq \mathcal{D}_{r}(x)$, so the two sets are equal.
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.
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: $1/3$
Subfield: Algebra


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 $d(n)$.

Later, ARMLopolis finally decides on a drastic expansion plan: now house numbers will be rational numbers. To define $d(p / q)$, with $p$ and $q$ integers such that $p q \neq 0$, write $p / q=3^{k} p^{\prime} / q^{\prime}$, where neither $p^{\prime}$ nor $q^{\prime}$ is divisible by 3 and $k$ is an integer (not necessarily positive); then $d(p / q)=3^{-k}$.
Compute $d(3 / 5), d(5 / 8)$, and $d(7 / 18)$.

Solution:
$\frac{1}{3}, 1, 9$

Final answer: $\frac{1}{3}, 1, 9$
Subfield: Algebra


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 $d(n)$.

Later, ARMLopolis finally decides on a drastic expansion plan: now house numbers will be rational numbers. To define $d(p / q)$, with $p$ and $q$ integers such that $p q \neq 0$, write $p / q=3^{k} p^{\prime} / q^{\prime}$, where neither $p^{\prime}$ nor $q^{\prime}$ is divisible by 3 and $k$ is an integer (not necessarily positive); then $d(p / q)=3^{-k}$.
Determine all pairs of relatively prime integers $p$ and $q$ such that $p / q \in \mathcal{N}(4 / 3)$.

Solution:
Because $d(4 / 3)=1 / 3, \mathcal{N}(4 / 3)=\{p / q$ in lowest terms such that $3 \mid q$ and $9 \nmid q\}$. Thus the set of all possible $(p, q)$ consists precisely of those ordered pairs $\left(p^{\prime} r, q^{\prime} r\right)$, where $r$ is any non-zero integer, $p^{\prime}$ and $q^{\prime}$ are relatively prime integers, $3 \mid q^{\prime}$, and $9 \nmid q^{\prime}$.

Final answer: $(p^{\prime} r, q^{\prime} r)$, where $r$ is any non-zero integer, $p^{\prime}$ and $q^{\prime}$ are relatively prime integers, $3 \mid q^{\prime}$, and $9 \nmid q^{\prime}$
Subfield: Algebra


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 $d(n)$.

Later, ARMLopolis finally decides on a drastic expansion plan: now house numbers will be rational numbers. To define $d(p / q)$, with $p$ and $q$ integers such that $p q \neq 0$, write $p / q=3^{k} p^{\prime} / q^{\prime}$, where neither $p^{\prime}$ nor $q^{\prime}$ is divisible by 3 and $k$ is an integer (not necessarily positive); then $d(p / q)=3^{-k}$.
A longtime resident of IMOpia moves to ARMLopolis and hopes to keep his same address. When asked, he says "My old address was $e$, that is, the sum $\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\ldots$ I'd really like to keep that address, because the addresses of my friends here are all partial sums of this series: Alice's house is $\frac{1}{0 !}$, Bob's house is $\frac{1}{0 !}+\frac{1}{1 !}$, Carol's house is $\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}$, and so on. Just let me know what I have to do in order to be near my friends!" After some head-scratching, the ARMLopolitan planning council announces that this request cannot be satisfied: there is no number, rational or otherwise, that corresponds to the (infinite) sum, or that is arbitrarily close to the houses in this sequence. Prove that the council is correct (and not just bureaucratic).

Solution:
The houses in this sequence actually get arbitrarily far apart from each other, and so there's no single house which they approach. If $H_{n}=\frac{1}{0 !}+\frac{1}{1 !}+\cdots+\frac{1}{n !}$, then $d\left(H_{n}, H_{n-1}\right)=$ $d(1 / n !)=3^{k}$, where $k=\lfloor n / 3\rfloor+\lfloor n / 9\rfloor+\lfloor n / 27\rfloor \ldots$. In particular, if $n !=3^{k} \cdot n_{0}$, where $k>0$ and $3 \nmid n_{0}$, then $d\left(H_{n}, H_{n-1}\right)=3^{k}$, which can be made arbitrarily large as $n$ increases.
Subfield: Geometry


Question: Let $A R M L$ be a trapezoid with bases $\overline{A R}$ and $\overline{M L}$, such that $M R=R A=A L$ and $L R=$ $A M=M L$. Point $P$ lies inside the trapezoid such that $\angle R M P=12^{\circ}$ and $\angle R A P=6^{\circ}$. Diagonals $A M$ and $R L$ intersect at $D$. Compute the measure, in degrees, of angle $A P D$.

Solution:
First, determine the angles of $A R M L$. Let $\mathrm{m} \angle M=x$. Then $\mathrm{m} \angle L R M=x$ because $\triangle L R M$ is isosceles, and $\mathrm{m} \angle R L M=180^{\circ}-2 x$. Because $\overline{A R} \| \overline{L M}, \mathrm{~m} \angle A R M=180^{\circ}-x$ and $\mathrm{m} \angle A R L=180^{\circ}-2 x$, as shown in the diagram below.

<img_3627>

However, $\triangle A R L$ is also isosceles (because $A R=A L$ ), so $\mathrm{m} \angle A L R=180^{\circ}-2 x$, yielding $\mathrm{m} \angle A L M=360^{\circ}-4 x$. Because $\mathrm{m} \angle R M L=\mathrm{m} \angle A L M$, conclude that $360^{\circ}-4 x=x$, so $x=72^{\circ}$. Therefore the base angles $L$ and $M$ have measure $72^{\circ}$ while the other base angles $A$ and $R$ have measure $108^{\circ}$. Finally, the angle formed by diagonals $\overline{A M}$ and $\overline{L R}$ is as follows: $\mathrm{m} \angle R D M=180^{\circ}-\mathrm{m} \angle L R M-\mathrm{m} \angle A M R=180^{\circ}-72^{\circ}-36^{\circ}=72^{\circ}$.

Now construct equilateral $\triangle R O M$ with $O$ on the exterior of the trapezoid, as shown below.

<img_3202>

Because $A R=R M=R O$, triangle $O A R$ is isosceles with base $\overline{A O}$. The measure of $\angle A R O$ is $108^{\circ}+60^{\circ}=168^{\circ}$, so $\mathrm{m} \angle R A O=(180-168)^{\circ} / 2=6^{\circ}$. Thus $P$ lies on $\overline{A O}$. Additionally, $\mathrm{m} \angle P O M=\mathrm{m} \angle A O M=60^{\circ}-6^{\circ}=54^{\circ}$, and $\mathrm{m} \angle P M O=60^{\circ}+12^{\circ}=72^{\circ}$ by construction. Thus $\mathrm{m} \angle M P O=180^{\circ}-72^{\circ}-54^{\circ}=54^{\circ}$, hence $\triangle P M O$ is isosceles with $P M=O M$. But because $O M=R M, \triangle R M P$ is isosceles with $R M=M P$, and $R M=D M$ implies that $\triangle P D M$ is also isosceles. But $\mathrm{m} \angle R M P=12^{\circ}$ implies that $\mathrm{m} \angle P M D=36^{\circ}-12^{\circ}=24^{\circ}$, so $\mathrm{m} \angle D P M=78^{\circ}$. Thus $\mathrm{m} \angle A P D=180^{\circ}-\mathrm{m} \angle O P M-\mathrm{m} \angle D P M=180^{\circ}-54^{\circ}-78^{\circ}=48^{\circ}$.

Final answer: 48
Subfield: Geometry


Question: A regular hexagon has side length 1. Compute the average of the areas of the 20 triangles whose vertices are vertices of the hexagon.

Solution:
There are 6 triangles of side lengths $1,1, \sqrt{3} ; 2$ equilateral triangles of side length $\sqrt{3}$; and 12 triangles of side lengths $1, \sqrt{3}, 2$. One triangle of each type is shown in the diagram below.
<img_3233>

Each triangle in the first set has area $\sqrt{3} / 4$; each triangle in the second set has area $3 \sqrt{3} / 4$; and each triangle in the third set has area $\sqrt{3} / 2$. The average is

$$
\frac{6\left(\frac{\sqrt{3}}{4}\right)+2\left(\frac{3 \sqrt{3}}{4}\right)+12\left(\frac{\sqrt{3}}{2}\right)}{20}=\frac{\frac{6 \sqrt{3}}{4}+\frac{6 \sqrt{3}}{4}+\frac{24 \sqrt{3}}{4}}{20}=\frac{\mathbf{9} \sqrt{\mathbf{3}}}{\mathbf{2 0}} .
$$

Final answer: $\frac{9 \sqrt{3}}{20}$
Subfield: Algebra


Question: Paul was planning to buy 20 items from the ARML shop. He wanted some mugs, which cost $\$ 10$ each, and some shirts, which cost $\$ 6$ each. After checking his wallet he decided to put $40 \%$ of the mugs back. Compute the number of dollars he spent on the remaining items.

Solution:
The problem does not state the number of mugs Paul intended to buy, but the actual number is irrelevant. Suppose Paul plans to buy $M$ mugs and $20-M$ shirts. The total cost is $10 M+6(20-M)$ However, he puts back $40 \%$ of the mugs, so he ends up spending $10(0.6 M)+$ $6(20-M)=6 M+120-6 M=\mathbf{1 2 0}$ dollars.

Final answer: 120
Subfield: Number Theory


Question: Let $x$ be the smallest positive integer such that $1584 \cdot x$ is a perfect cube, and let $y$ be the smallest positive integer such that $x y$ is a multiple of 1584 . Compute $y$.

Solution:
In order for $1584 \cdot x$ to be a perfect cube, all of its prime factors must be raised to powers divisible by 3 . Because $1584=2^{4} \cdot 3^{2} \cdot 11$, $x$ must be of the form $2^{3 k+2} \cdot 3^{3 m+1} \cdot 11^{3 n+2} \cdot r^{3}$, for nonnegative integers $k, m, n, r, r>0$. Thus the least positive value of $x$ is $2^{2} \cdot 3 \cdot 11^{2}=1452$. But in order for $x y$ to be a positive multiple of $1584, x y$ must be of the form $2^{a} \cdot 3^{b} \cdot 11^{c} \cdot d$, where $a \geq 4, b \geq 2, c \geq 1$, and $d \geq 1$. Thus $y$ must equal $2^{2} \cdot 3^{1}=\mathbf{1 2}$.

Final answer: 12
Subfield: Algebra


Question: Emma goes to the store to buy apples and peaches. She buys five of each, hands the shopkeeper one $\$ 5$ bill, but then has to give the shopkeeper another; she gets back some change. Jonah goes to the same store, buys 2 apples and 12 peaches, and tries to pay with a single $\$ 10$ bill. But that's not enough, so Jonah has to give the shopkeeper another $\$ 10$ bill, and also gets some change. Finally, Helen goes to the same store to buy 25 peaches. Assuming that the price in cents of each fruit is an integer, compute the least amount of money, in cents, that Helen can expect to pay.

Solution:
Let $a$ be the price of one apple and $p$ be the price of one peach, in cents. The first transaction shows that $500<5 a+5 p<1000$, hence $100<a+p<200$. The second transaction shows that $1000<2 a+12 p<2000$, so $500<a+6 p<1000$. Subtracting the inequalities yields $300<5 p<900$, so $60<p<180$. Therefore the price of 25 peaches is at least $25 \cdot 61=\mathbf{1 5 2 5}$ cents.

Final answer: 1525
Subfield: Geometry


Question: Circle $O$ has radius 6. Point $P$ lies outside circle $O$, and the shortest distance from $P$ to circle $O$ is 4. Chord $\overline{A B}$ is parallel to $\overleftrightarrow{O P}$, and the distance between $\overline{A B}$ and $\overleftrightarrow{O P}$ is 2 . Compute $P A^{2}+P B^{2}$.

Solution:
Extend $\overline{A B}$ to point $Q$ such that $\overline{P Q} \perp \overline{A Q}$ as shown, and let $M$ be the midpoint of $\overline{A B}$. (The problem does not specify whether $A$ or $B$ is nearer $P$, but $B$ can be assumed to be nearer $P$ without loss of generality.)

<img_3454>

Then $O P=10, P Q=O M=2$, and $O B=6$. Thus $M B=\sqrt{6^{2}-2^{2}}=4 \sqrt{2}$. Because $Q M=O P=10$, it follows that $Q B=10-4 \sqrt{2}$ and $Q A=10+4 \sqrt{2}$. So

$$
\begin{aligned}
P A^{2}+P B^{2} & =\left(Q A^{2}+Q P^{2}\right)+\left(Q B^{2}+Q P^{2}\right) \\
& =(10+4 \sqrt{2})^{2}+2^{2}+(10-4 \sqrt{2})^{2}+2^{2} \\
& =\mathbf{2 7 2}
\end{aligned}
$$

Final answer: 272
Subfield: Number Theory


Question: A palindrome is a positive integer, not ending in 0 , that reads the same forwards and backwards. For example, 35253,171,44, and 2 are all palindromes, but 17 and 1210 are not. Compute the least positive integer greater than 2013 that cannot be written as the sum of two palindromes.

Solution:
If $a+b \geq 2014$, then at least one of $a, b$ must be greater than 1006 . The palindromes greater than 1006 but less than 2014 are, in descending order, 2002, 1991, 1881, ..., 1111. Let a



represent the larger of the two palindromes. Then for $n=2014, a=2002$ is impossible, because $2014-2002=12$. Any value of $a$ between 1111 and 2000 ends in 1 , so if $a+b=2014$, $b$ ends in 3 , and because $b<1000$, it follows that $303 \leq b \leq 393$. Subtracting 303 from 2014 yields 1711, and so $a \leq 1711$. Thus $a=1661$ and $b=353$. A similar analysis shows the following results:

$$
\begin{aligned}
& 2015=1551+464 ; \\
& 2016=1441+575 ; \\
& 2017=1331+686 ; \text { and } \\
& 2018=1221+797
\end{aligned}
$$

But 2019 cannot be expressed as the sum of two palindromes: $b$ would have to end in 8 , so $b=808+10 d$ for some digit $d$. Then $2019-898 \leq a \leq 2019-808$, hence $1121 \leq a \leq 1211$, and there is no palindrome in that interval.

Final answer: 2019
Subfield: Algebra


Question: Positive integers $x, y, z$ satisfy $x y+z=160$. Compute the smallest possible value of $x+y z$.

Solution:
First consider the problem with $x, y, z$ positive real numbers. If $x y+z=160$ and $z$ is constant, then $y=\frac{160-z}{x}$, yielding $x+y z=x+\frac{z(160-z)}{x}$. For $a, x>0$, the quantity $x+\frac{a}{x}$ is minimized when $x=\sqrt{a}$ (proof: use the Arithmetic-Geometric Mean Inequality $\frac{A+B}{2} \geq \sqrt{A B}$ with $A=x$ and $\left.B=\frac{a}{x}\right)$; in this case, $x+\frac{a}{x}=2 \sqrt{a}$. Thus $x+y z \geq 2 \sqrt{z(160-z)}$. Considered as a function of $z$, this lower bound is increasing for $z<80$.

These results suggest the following strategy: begin with small values of $z$, and find a factorization of $160-z$ such that $x$ is close to $\sqrt{z(160-z)}$. (Equivalently, such that $\frac{x}{y}$ is close to $z$.) The chart below contains the triples $(x, y, z)$ with the smallest values of $x+y z$, conditional upon $z$.

| $z$ | $(x, y, z)$ | $x+y z$ |
| :---: | :---: | :---: |
| 1 | $(53,3,1)$ | 56 |
| 2 | $(79,2,2)$ | 83 |
| 3 | $(157,1,3)$ | 160 |
| 4 | $(26,6,4)$ | 50 |
| 5 | $(31,5,5)$ | 56 |
| 6 | $(22,7,6)$ | 64 |

Because $x+y z \geq 2 \sqrt{z(160-z)}$, it follows that $x+y z \geq 64$ for $6 \leq z \leq 80$. And because $x+y z>80$ for $z \geq 80$, the minimal value of $x+y z$ is $\mathbf{5 0}$.

Final answer: 50