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


Question: Compute the smallest positive integer $n$ such that

$$
n,\lfloor\sqrt{n}\rfloor,\lfloor\sqrt[3]{n}\rfloor,\lfloor\sqrt[4]{n}\rfloor,\lfloor\sqrt[5]{n}\rfloor,\lfloor\sqrt[6]{n}\rfloor,\lfloor\sqrt[7]{n}\rfloor, \text { and }\lfloor\sqrt[8]{n}\rfloor
$$

are distinct.

Solution:
Inverting the problem, the goal is to find seven positive integers $a<b<c<d<e<f<g$ and a positive integer $n$ such that $a^{8}, b^{7}, c^{6}, \ldots, g^{2} \leq n$ and $n<(a+1)^{8},(b+1)^{7}, \ldots,(g+1)^{2}$. Proceed by cases starting with small values of $a$.

If $a=1$, then because $n<(a+1)^{8}, n<256$. But because $n \geq(a+3)^{5}, n \geq 4^{5}=1024$. So it is impossible for $a$ to be 1 .

If $a=2$, then $a^{8}=256$ and $(a+1)^{8}=6561$, so $256 \leq n<6561$. Then $b \geq 3 \Rightarrow b^{7} \geq 2187$ and $c \geq 4 \Rightarrow c^{6} \geq 4096$. So $n \geq 4096$. Because $(3+1)^{7}=16384$ and $(4+1)^{6}=15625$, the condition $n<6561$ found previously guarantees that $\lfloor\sqrt[7]{n}\rfloor=3$ and $\lfloor\sqrt[6]{n}\rfloor=4$. Notice that if $4096 \leq n<6561$, then $\lfloor\sqrt[5]{n}\rfloor=5,\lfloor\sqrt[4]{n}\rfloor=8$, and $\lfloor\sqrt[3]{n}\rfloor \geq 16$. In fact, $\lfloor\sqrt[3]{4096}\rfloor=2^{4}=16$ and $\lfloor\sqrt{4096}\rfloor=2^{6}=64$. So the desired value of $n$ is 4096 .

Final answer: 4096
Subfield: Number Theory


Question: If $n$ is a positive integer, then $n$ !! is defined to be $n(n-2)(n-4) \cdots 2$ if $n$ is even and $n(n-2)(n-4) \cdots 1$ if $n$ is odd. For example, $8 ! !=8 \cdot 6 \cdot 4 \cdot 2=384$ and $9 ! !=9 \cdot 7 \cdot 5 \cdot 3 \cdot 1=945$. Compute the number of positive integers $n$ such that $n !$ ! divides 2012!!.

Solution:
If $n$ is even and $n \leq 2012$, then $n$ !! $\mid 2012$ !! trivially, while if $n>2012,2012$ !! $<n$ !!, so $n$ !! cannot divide 2012!!. Thus there are a total of 1006 even values of $n$ such that $n$ !! | 2012!!. If $n$ is odd and $n<1006$, then $n$ !! | 2012!!. To show this, rearrange the terms of 2012!! and factor:

$$
\begin{aligned}
2012 ! ! & =2 \cdot 4 \cdot 6 \cdots 2010 \cdot 2012 \\
& =(2 \cdot 6 \cdot 10 \cdots 2010)(4 \cdot 8 \cdot 12 \cdots 2012) \\
& =2^{503}(1 \cdot 3 \cdot 5 \cdots 1005)(4 \cdot 8 \cdot 12 \cdots 2012)
\end{aligned}
$$

However, the condition $n<1006$ is not necessary, only sufficient, because $n$ !! also divides 2012 if $1007 \cdot 1009 \cdots n \mid(4 \cdot 8 \cdot 12 \cdots 2012)$. (The factor of $2^{503}$ is irrelevant because all the factors on the left side are odd.) The expression $(4 \cdot 8 \cdot 12 \cdots 2012)$ can be factored as $4^{503}(1 \cdot 2 \cdot 3 \cdot \cdots 503)=4^{503} \cdot 503$ !. Examining the numbers $1007,1009, \ldots$ in sequence shows that 1007 is satisfactory, because $1007=19 \cdot 53$. On the other hand, 1009 is prime, so it cannot be a factor of $4^{503} \cdot 503$ !. Thus the largest possible odd value of $n$ is 1007 , and there are 504 odd values of $n$ altogether. The total is $1006+504=\mathbf{1 5 1 0}$.

Final answer: 1510
Subfield: Combinatorics


Question: One face of a $2 \times 2 \times 2$ cube is painted (not the entire cube), and the cube is cut into eight $1 \times 1 \times 1$ cubes. The small cubes are reassembled randomly into a $2 \times 2 \times 2$ cube. Compute the probability that no paint is showing.

Solution:
Call each $1 \times 1 \times 1$ cube a cubelet. Then four cubelets are each painted on one face, and the other four cubelets are completely unpainted and can be ignored. For each painted cubelet, the painted face can occur in six positions, of which three are hidden from the outside, so the probability that a particular painted cubelet has no paint showing is $3 / 6=1 / 2$. Thus the probability that all four painted cubelets have no paint showing is $(1 / 2)^{4}=\frac{1}{\mathbf{1 6}}$.

Final answer: $\frac{1}{16}$
Subfield: Geometry


Question: In triangle $A B C, A B=B C$. A trisector of $\angle B$ intersects $\overline{A C}$ at $D$. If $A B, A C$, and $B D$ are integers and $A B-B D=7$, compute $A C$.

Solution:
Let $E$ be the point where the other trisector of $\angle B$ intersects side $\overline{A C}$. Let $A B=B C=a$, and let $B D=B E=d$. Draw $X$ on $\overline{B C}$ so that $B X=d$. Then $C X=7$.

<img_3688>

The placement of point $X$ guarantees that $\triangle B E X \cong \triangle B D E$ by Side-Angle-Side. Therefore $\angle B X E \cong \angle B E X \cong \angle B D E$, and so $\angle C X E \cong \angle A D B \cong \angle C E B$. By Angle-Angle, $\triangle C E X \sim \triangle C B E$. Let $E X=c$ and $E C=x$. Then comparing ratios of corresponding sides yields

$$
\frac{c}{d}=\frac{7}{x}=\frac{x}{d+7}
$$

Using the right proportion, $x^{2}=7(d+7)$. Because $d$ is an integer, $x^{2}$ is an integer, so either $x$ is an integer or irrational. The following argument shows that $x$ cannot be irrational. Applying the Angle Bisector Theorem to $\triangle B C D$ yields $D E=c=\frac{d}{d+7} \cdot x$. Then $A C=2 x+c=$ $x\left(2+\frac{d}{d+7}\right)$. Because the expression $\left(2+\frac{d}{d+7}\right)$ is rational, $A C$ will not be an integer if $x$ is irrational.

Hence $x$ is an integer, and because $x^{2}$ is divisible by $7, x$ must also be divisible by 7 . Let $x=7 k$ so that $d=c k$. Rewrite the original proportion using $7 k$ for $x$ and $c k$ for $d$ :

$$
\begin{aligned}
\frac{c}{d} & =\frac{x}{d+7} \\
\frac{c}{c k} & =\frac{7 k}{c k+7} \\
7 k^{2} & =c k+7 \\
7 k & =c+\frac{7}{k} .
\end{aligned}
$$



Because the left side of this last equation represents an integer, $7 / k$ must be an integer, so either $k=1$ or $k=7$. The value $k=1$ gives the extraneous solution $c=0$. So $k=7$, from which $c=48$. Then $d=336$ and $A C=2 x+c=2 \cdot 49+48=\mathbf{1 4 6}$.

Final answer: 146
Subfield: Algebra


Question: The rational number $r$ is the largest number less than 1 whose base-7 expansion consists of two distinct repeating digits, $r=0 . \underline{A} \underline{B} \underline{A} \underline{B} \underline{A} \underline{B} \ldots$ Written as a reduced fraction, $r=\frac{p}{q}$. Compute $p+q$ (in base 10).

Solution:
In base 7, the value of $r$ must be $0.656565 \ldots=0 . \overline{65}_{7}$. Then $100_{7} \cdot r=65 . \overline{65}_{7}$, and $\left(100_{7}-1\right) r=$ $65_{7}$. In base $10,65_{7}=6 \cdot 7+5=47_{10}$ and $100_{7}-1=7^{2}-1=48_{10}$. Thus $r=47 / 48$, and $p+q=95$.

Final answer: 95
Subfield: Geometry


Question: Let $T=95$. Triangle $A B C$ has $A B=A C$. Points $M$ and $N$ lie on $\overline{B C}$ such that $\overline{A M}$ and $\overline{A N}$ trisect $\angle B A C$, with $M$ closer to $C$. If $\mathrm{m} \angle A M C=T^{\circ}$, then $\mathrm{m} \angle A C B=U^{\circ}$. Compute $U$.

Solution:
Because $\triangle A B C$ is isosceles with $A B=A C, \mathrm{~m} \angle A B C=U^{\circ}$ and $\mathrm{m} \angle B A C=(180-2 U)^{\circ}$. Therefore $\mathrm{m} \angle M A C=\left(\frac{180-2 U}{3}\right)^{\circ}=\left(60-\frac{2}{3} U\right)^{\circ}$. Then $\left(60-\frac{2}{3} U\right)+U+T=180$, so $\frac{1}{3} U=$ $120-T$ and $U=3(120-T)$. Substituting $T=95$ yields $U=\mathbf{7 5}$.

Final answer: 75
Subfield: Combinatorics


Question: Let $T=75$. At Wash College of Higher Education (Wash Ed.), the entering class has $n$ students. Each day, two of these students are selected to oil the slide rules. If the entering class had two more students, there would be $T$ more ways of selecting the two slide rule oilers. Compute $n$.

Solution:
With $n$ students, Wash Ed. can choose slide-rule oilers in $\left(\begin{array}{l}n \\ 2\end{array}\right)=\frac{n(n-1)}{2}$ ways. With $n+2$ students, there would be $\left(\begin{array}{c}n+2 \\ 2\end{array}\right)=\frac{(n+2)(n+1)}{2}$ ways of choosing the oilers. The difference is $\frac{(n+2)(n+1)}{2}-\frac{n(n-1)}{2}=T$. Simplifying yields $\frac{\left(n^{2}+3 n+2\right)-\left(n^{2}-n\right)}{2}=2 n+1=T$, so $n=\frac{T-1}{2}$. Because $T=75, n=37$.

Final answer: 37
Subfield: Number Theory


Question: Compute the least positive integer $n$ such that the set of angles

$$
\left\{123^{\circ}, 246^{\circ}, \ldots, n \cdot 123^{\circ}\right\}
$$

contains at least one angle in each of the four quadrants.

Solution:
The first angle is $123^{\circ}$, which is in Quadrant II, the second $\left(246^{\circ}\right)$ is in Quadrant III, and the third is in Quadrant I, because $3 \cdot 123^{\circ}=369^{\circ} \equiv 9^{\circ} \bmod 360^{\circ}$. The missing quadrant is IV, which is $270^{\circ}-246^{\circ}=24^{\circ}$ away from the second angle in the sequence. Because $3 \cdot 123^{\circ} \equiv 9^{\circ} \bmod 360^{\circ}$, the terminal ray of the $(n+3)^{\mathrm{rd}}$ angle is rotated $9^{\circ}$ counterclockwise from the $n^{\text {th }}$ angle. Thus three full cycles are needed to reach Quadrant IV starting from the second angle: the fifth angle is $255^{\circ}$, the eighth angle is $264^{\circ}$, and the eleventh angle is $273^{\circ}$. So $n=11$.

Final answer: 11
Subfield: Number Theory


Question: Let $T=11$. In ARMLvania, license plates use only the digits 1-9, and each license plate contains exactly $T-3$ digits. On each plate, all digits are distinct, and for all $k \leq T-3$, the $k^{\text {th }}$ digit is at least $k$. Compute the number of valid ARMLvanian license plates.

Solution:
There are 9 valid one-digit plates. For a two-digit plate to be valid, it has to be of the form $\underline{A} \underline{B}$, where $B \in\{2, \ldots, 9\}$, and either $A \in\{2, \ldots, 9\}$ with $A \neq B$ or $A=1$. So there are 8 ways to choose $B$ and $8-1+1=8$ ways to choose $A$, for a total of $8 \cdot 8=64$ plates. In general, moving from the last digit to the first, if there are $k$ ways to choose digit $n$, then there are $k-1$ ways to choose digit $n-1$ from the same set of possibilities as digit $n$ had, plus one additional way, for a total of $k-1+1=k$ choices for digit $n-1$. So if a license plate has $d$ digits, there are $10-d$ choices for the last digit and for each digit before it, yielding $(10-d)^{d}$ possible $d$-digit plates. Using $d=T-3=8$, there are $2^{8}=\mathbf{2 5 6}$ plates.

Final answer: 256
Subfield: Algebra


Question: Let $T=256$. Let $\mathcal{R}$ be the region in the plane defined by the inequalities $x^{2}+y^{2} \geq T$ and $|x|+|y| \leq \sqrt{2 T}$. Compute the area of region $\mathcal{R}$.

Solution:
The first inequality states that the point $(x, y)$ is outside the circle centered at the origin with radius $\sqrt{T}$, while the second inequality states that $(x, y)$ is inside the tilted square centered at the origin with diagonal $2 \sqrt{2 T}$. The area of the square is $4 \cdot \frac{1}{2}(\sqrt{2 T})^{2}=4 T$, while the area of the circle is simply $\pi T$, so the area of $\mathcal{R}$ is $4 T-\pi T=\mathbf{1 0 2 4}-\mathbf{2 5 6 \pi}$.

Final answer: $1024-256 \pi$
Subfield: Combinatorics


Question: This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

<img_3859>

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
Consider the hideout map $M$ below.

<img_3592>

Show that one Cop can always catch the Robber.

Solution:
Have the Cop stay at $A$ for 2 days. If the Robber is not at $A$ the first day, he must be at one of $B_{1}-B_{6}$, and because the Robber must move along an edge every night, he will be forced to go to $A$ on day 2 .
Subfield: Combinatorics


Question: This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

![](https://cdn.mathpix.com/cropped/2023_12_21_279781de8be6584eef7dg-1.jpg?height=336&width=447&top_left_y=889&top_left_x=836)

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
Consider the hideout map $M$ below.

![](https://cdn.mathpix.com/cropped/2023_12_21_c8468e5f43f43636bd4ag-1.jpg?height=350&width=401&top_left_y=324&top_left_x=908)

Show that one Cop can always catch the Robber.

Solution:
Have the Cop stay at $A$ for 2 days. If the Robber is not at $A$ the first day, he must be at one of $B_{1}-B_{6}$, and because the Robber must move along an edge every night, he will be forced to go to $A$ on day 2 .

Final answer: 证明题，略
Subfield: Combinatorics


Question: This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

<img_3859>

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
The map shown below is $\mathcal{C}_{6}$, the cyclic graph with six hideouts. Show that three Cops are sufficient to catch the Robber on $\mathcal{C}_{6}$, so that $C\left(\mathcal{C}_{6}\right) \leq 3$.

<img_3546>

Solution:
The Cops should stay at $\left\{A_{1}, A_{3}, A_{5}\right\}$ for 2 days. If the Robber evades capture the first day, he must have been at an even-numbered hideout. Because he must move, he will be at an odd-numbered hideout the second day. Equivalently, the Cops could stay at $\left\{A_{2}, A_{4}, A_{6}\right\}$ for 2 days.
Subfield: Combinatorics


Question: This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.

<img_3859>

Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .

The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.

Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.

The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
Show that for all maps $M, C(M)<h(M)$.

Solution:
Let $n=h(M)$. The following strategy will always catch a Robber within two days using $n-1$ Cops, which proves that $C(M) \leq n-1$. Choose any subset $\mathcal{S}$ of $n-1$ hideouts and position $n-1$ Cops at the hideouts of $\mathcal{S}$ for 2 days. If the Robber is not caught on the first day, he must have been at the hideout not in $\mathcal{S}$, and therefore must move to a hideout in $\mathcal{S}$ on the following day.