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\frac{1}{2} | AI-MO/NuminaMath-1.5/olympiads | Final Answer: The final answer is $\boxed{1/2}$ | Mathematical logic and foundations | Given a quadrilateral $A B C D$ with an area of 1. From its internal point O, perpendiculars OK, OL, OM, and ON are dropped to the sides $A B, B C, C D$, and $D A$ respectively. It is known that $A K \geq K B, B L \geq L C, C M \geq M D$, and $D N \geq N A$. Find the area of the quadrilateral $K L M N$. | From two inclined lines drawn from one point, the one with the larger projection is larger. Therefore, from the inequalities given in the problem, it follows that $O A \geq O B \geq O C \geq O D \geq O A$. Hence, $O A=O B=O C=O D$, which means $O$ is the center of the circle circumscribed around quadrilateral $A B C D$... | 710648c7e0789888 | true | 2 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
36 | AI-MO/NuminaMath-1.5/olympiads | Let $ABC$ be a triangle with sides $AB=26$, $BC=30$, and $AC=28$. Let $BH$ be the altitude from $B$ to $AC$, and $BD$ be the angle bisector of $\angle ABC$, where $H$ and $D$ are on $AC$. We want to find the area of triangle $BHD$.
First, we find the position of $H$ on $AC$. Let $AH = x$. Then $CH = 28-x$.
By Pythagor... | Geometry | [ [ Relationship,
In triangle $ABC$, the three sides are given: $AB=26$, $BC=30$, and $AC=28$. Find the part of the area of this triangle enclosed between the height and the bisector drawn from vertex $B$.
# | Apply Heron's formula and the property of the angle bisector of a triangle.
## Solution
Let $B P$ and $B Q$ be the height and the angle bisector of the given triangle $A B C$. By Heron's formula,
On the other hand, $S=1 / 2$ AC$\cdot$BP.
Therefore, $B P=\frac{2 S_{\triangle A B C}}{A C}=$
$\frac{2 \cdot 336}{28}=2... | c8ca584a8d928f91 | true | 3.75 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
12 | AI-MO/NuminaMath-1.5/amc_aime | We are asked to find the sum of all integer solutions to the inequality $1 < (x-2)^2 < 25$.
We can split this compound inequality into two separate inequalities:
\begin{enumerate}
\item $(x-2)^2 > 1$
\item $(x-2)^2 < 25$
\end{enumerate}
For the first inequality, $(x-2)^2 > 1$, we take the square root of both s... | Number theory | What is the sum of all integer solutions to $1<(x-2)^2<25$?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 25$ | $(x-2)^2$ = perfect square.
1 < perfect square < 25
Perfect square can equal: 4, 9, or 16
Solve for $x$:
$(x-2)^2=4$
$x=4,0$
and
$(x-2)^2=9$
$x=5,-1$
and
$(x-2)^2=16$
$x=6,-2$
The sum of all integer solutions is
$4+5+6+0+(-1)+(-2)=\boxed{\textbf{(B)} 12}$ | 5bae26dcf0c3215d | true | 1.25 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
6 | AI-MO/NuminaMath-1.5/olympiads | We are given $a^{2}+4 b^{2}+9 c^{2}=4 b+12 c-2$, which can be rewritten as $a^{2}+4 b^{2}-4 b+9 c^{2}-12 c+2=0$.
Completing the square, we have $a^{2}+(2 b-1)^{2}-1+(3 c-2)^{2}-4+2=0$, so $a^{2}+(2 b-1)^{2}+(3 c-2)^{2}=3$.
Let $x=a, y=2 b-1, z=3 c-2$. Then $a=x, b=\frac{y+1}{2}, c=\frac{z+2}{3}$.
The constraint becomes... | Real functions | 10. Let positive real numbers $a, b, c$ satisfy $a^{2}+4 b^{2}+9 c^{2}=4 b+12 c-2$, find the minimum value of $\frac{1}{a}+\frac{2}{b}+\frac{3}{c}$. | $$
a^{2}+4 b^{2}+9 c^{2}=4 b+12 c-2 \Rightarrow a^{2}+(2 b-1)^{2}+(3 c-2)^{2}=3,
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(1+1+1)\left[a^{2}+(2 b-1)^{2}+(3 c-2)^{2}\right] \geqslant(a+2 b+3 c-3)^{2} \\
\Rightarrow(a+2 b+3 c-3)^{2} \leqslant 9 \Rightarrow 0<a+2 b+3 c \leqslant 6 .
\end{array}
$$... | bcc57c0c72d3fca2 | true | 6.5 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
66 | AI-MO/NuminaMath-1.5/amc_aime | We consider two types of equilateral triangles: those with exactly two vertices in $\{A_1, \dots, A_9\}$ and those with exactly three vertices in $\{A_1, \dots, A_9\}$.
Case 1: Equilateral triangles with exactly three vertices in $\{A_1, \dots, A_9\}$.
We found that there are 3 such triangles: $\{A_1, A_4, A_7\}, \{A_... | Combinatorics | Given the nine-sided regular polygon $A_1 A_2 A_3 A_4 A_5 A_6 A_7 A_8 A_9$, how many distinct equilateral triangles in the plane of the polygon have at least two vertices in the set $\{A_1,A_2,\dots,A_9\}$?
$\text{(A) }30 \qquad \text{(B) }36 \qquad \text{(C) }63 \qquad \text{(D) }66 \qquad \text{(E) }72$ | Each of the $\binom{9}{2} = 36$ pairs of vertices determines two equilateral triangles, for a total of 72 triangles. However, the three triangles $A_1A_4A_7$, $A_2A_5A_8$, and $A_3A_6A_9$ are each counted 3 times, resulting in an overcount of 6. Thus, there are $\boxed{66}$ distinct equilateral triangles. | ff388ce341b2c497 | true | 5.25 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{5}{7} | AI-MO/NuminaMath-1.5/olympiads | Let $A_0 = 1$ be the area of the initial square.
In the first step, we cut off 4 triangles, each with area $\frac{1}{18}$. Total area cut off is $C_1 = 4 \times \frac{1}{18} = \frac{2}{9}$.
Area of $P_1$ is $A_1 = A_0 - C_1 = 1 - \frac{2}{9} = \frac{7}{9}$.
In the second step, we cut off 8 triangles, each with area $\f... | Geometry | Konyagin S.V.
Given a square with side length 1. From it, four corners are cut off - four triangles, each of which has two sides along the sides of the square and these sides are $1 / 3$ of their lengths. With the resulting 8-sided polygon, the same operation is performed: from each vertex, a triangle is cut off, two ... | Let $M_{0}$ be the original square, and $M_{1}, M_{2}, M_{3}, \ldots$ be the polygons obtained from $M_{0}$ by sequentially cutting off corners. It is convenient to also consider the polygon $N_{k}$, whose vertices are the midpoints of the sides of $M_{k} (k=0,1,2, \ldots)$.
Let $A$ be an arbitrary vertex of the polyg... | e88a0f014bb545ff | true | 5 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
105:44 | AI-MO/NuminaMath-1.5/olympiads | Let $A$ and $B$ be the areas of plot A and plot B, respectively. Let $g$ be the growth rate of grass per unit area per day, and let $G$ be the initial amount of grass per unit area. Let $C$ be the rate at which the herd eats grass.
The initial amount of grass on plot A is $AG$. In 7 days, the total grass grown on plot ... | Mathematical logic and foundations | 6. A uniformly growing uniform grassland is divided into two unequal parts, $\mathrm{A}$ and $\mathrm{B}$. A herd of cows first eats the grass on plot A, finishing it in exactly 7 days. Then the herd continues to eat the grass on plot B at the same speed, finishing it in 4 days, at which point plot A has just recovered... | 【Answer】105:44
【Analysis】A The grassland from being eaten clean to growing back to the original amount of grass takes 4 days. Assuming A grassland grows 1 unit of grass per day, then in 4 days it grows 4 units. The original amount of grass in A grassland is 4 units. The speed at which the cow eats grass is $(4+7) \div ... | fe6ad07e9fb9c823 | true | 7.6 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\text{85 kg} | AI-MO/NuminaMath-1.5/olympiads | Let $m_1$ be the mass of first-grade apples in kilograms, and $m_2$ be the mass of second-grade apples in kilograms.
Let $P_1$ be the price per kilogram of first-grade apples in rubles, and $P_2$ be the price per kilogram of second-grade apples in rubles.
We are given that the total cost of first-grade apples is 228 r... | General algebraic systems | 3.11 Apples of the first grade for a total of 228 rubles and apples of the second grade for a total of 180 rubles were delivered to the store. During unloading, the delivered apples were accidentally mixed. It was found that if all the apples are now sold at one price - 90 kopecks lower than the price per kilogram of f... | 3.11 Let's construct the following table:
| Apple Variety | Cost, rub | Quantity, tons | Price, rub/kg |
| :---: | :---: | :---: | :---: |
| First | 228 | $x$ | $\frac{228}{x}$ |
| Second | 180 | $x+5$ | $\frac{180}{x+5}$ |
According to the problem, we have the equation
$\left(\frac{228}{x}-0.9\right)(2 x+5)=408$,
... | f20bc40e5e1ede97 | true | 3 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
0.6 | AI-MO/NuminaMath-1.5/olympiads | Let the scale division value of the amperemeter be $d = 0.1 \mathrm{~A}$.
The readings of the amperemeter are rounded to the nearest whole division, which means the readings are rounded to the nearest multiple of $0.1 \mathrm{~A}$.
Let $I$ be the true value of the current.
Let $R$ be the reading of the amperemeter.
Wh... | Probability theory and stochastic processes | 308. The scale division value of the amperemeter is 0.1 A. The readings of the amperemeter are rounded to the nearest whole division. Find the probability that an error exceeding $0.02 \mathrm{~A}$ will be made during the reading. | Solution. The rounding error of the measurement can be considered as a random variable $X$, which is uniformly distributed in the interval between two adjacent integer divisions. The density of the uniform distribution $f(x)=1 /(b-a)$, where $(b-a)$ is the length of the interval in which the possible values of $X$ are ... | d0e773c9bc2f9a5b | true | 2 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\((2,4,4), (2,3,6), (3,3,3)\) | AI-MO/NuminaMath-1.5/olympiads | Let the three positive integers be x, y, and z.
We are given the equation:
$$ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 $$
Without loss of generality, assume $x \le y \le z$.
Since x, y, z are positive integers, we must have $x > 1$, otherwise if $x=1$, then $\frac{1}{1} + \frac{1}{y} + \frac{1}{z} = 1$, which implie... | Number theory | 3-ча 1. The sum of the reciprocals of three positive integers is equal to 1. What are these numbers? Find all solutions. | Solve 1. Answer: \((2,4,4),(2,3,6)\) or \((3,3,3)\).
Let \(x \leqslant y \leqslant z\) be natural numbers, and \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\). Then \(x1000^{10}\), so the number \(2^{100}\) has at least 31 digits. On the other hand,
\[
\frac{1024^{10}}{1000^{10}}<\left(\frac{1025}{1000}\right)^{10}=\left(\... | 5c5ca9d90e4ec4e6 | true | 2 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
3323 | AI-MO/NuminaMath-1.5/olympiads | The length of a Gregorian year is, on average, $365 + \frac{97}{400} = \frac{146097}{400} = 365.2425$ days.
The length of the precise year is given as 365 days 5 hours 48 minutes 46 seconds.
Convert 5 hours 48 minutes 46 seconds to days:
5 hours = $5/24$ days
48 minutes = $48/60$ hours = $48/(60 \times 24)$ days = $48/... | Numerical analysis | According to the Gregorian calendar, 97 leap days must be inserted over 400 consecutive years. After how many years will there be a 1-day difference between the Gregorian calendar and the "precise" calendar, if the length of a year is 365 days 5 hours 48 minutes 46 seconds? | Regardless of the number of years, the Gregorian calendar offers 97 days every 400 years, which is $97 \cdot 24=2328$ hours, compared to 5 hours, 48 minutes, and 46 seconds 400 times over. It is reasonable to count the minutes as $\frac{4}{5}$ of an hour, thus in 400 years, $400 \cdot\left(5+\frac{4}{5}\right)=$ 2320 h... | 4c6ae5502cd56a0f | true | 5.25 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{7}{8} | AI-MO/NuminaMath-1.5/olympiads | Final Answer:
We are given that $S_1 = \text{Area}(\triangle CDE) = 1$ and $DE = 2AE$.
Since $\triangle ACE$ and $\triangle CDE$ have the same altitude from $C$ to $AD$, the ratio of their areas is equal to the ratio of their bases $AE$ and $DE$.
$\frac{S_{ACE}}{S_{CDE}} = \frac{AE}{DE} = \frac{AE}{2AE} = \frac{1}{2}$.... | Geometry | Example 8 As shown in Figure 1.4.9, in trapezoid $A B C D$, $A B / / D C, C E$ is the bisector of $\angle B C D$, $C E \perp A D$ at $E, D E=2 A E, C E$ divides the trapezoid into two parts with areas $S_{1}$ and $S_{2}$, if $S_{1}=1$, find $S_{2}$. | Extend $C B$ and $D A$ to intersect at $F$, let $S_{\triangle A B F}=S_{3}$. Since $C E$ bisects $\angle B C D$ and $C E \perp A D$, we know that $\triangle C D F$ is an isosceles triangle, thus $S_{1} = S_{2} + S_{3}$.
Also, because $D E = 2 A E$ and $A B \parallel C D$, we know that $A F = \frac{1}{4} D F$. Therefore... | 47543e316b137c4a | true | 3.4 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
320^{\circ} | AI-MO/NuminaMath-1.5/olympiads | We are asked to find the sum of all solutions to the equation $1+\frac{\sin x}{\sin 4 x}=\frac{\sin 3 x}{\sin 2 x}$ for $0^{\circ}<x<180^{\circ}$.
We rewrite the equation as $\cos \frac{9x}{2} (\cos \frac{5x}{2} - \cos \frac{3x}{2}) = 0$.
Case 1: $\cos \frac{9x}{2} = 0$.
$\frac{9x}{2} = 90^{\circ} + k \cdot 180^{\circ}... | General algebraic systems | 20. [10] There exist several solutions to the equation
$$
1+\frac{\sin x}{\sin 4 x}=\frac{\sin 3 x}{\sin 2 x},
$$
where $x$ is expressed in degrees and $0^{\circ}<x<180^{\circ}$. Find the sum of all such solutions. | Answer: $320^{\circ}$
Solution: We first apply sum-to-product and product-to-sum:
$$
\begin{array}{c}
\frac{\sin 4 x+\sin x}{\sin 4 x}=\frac{\sin 3 x}{\sin 2 x} \\
2 \sin (2.5 x) \cos (1.5 x) \sin (2 x)=\sin (4 x) \sin (3 x)
\end{array}
$$
Factoring out $\sin (2 x)=0$,
$$
\sin (2.5 x) \cos (1.5 x)=\cos (2 x) \sin (3 x... | 9cca75520f6d37fe | true | 10 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{(x-2)^{2}}{2 x+2} | AI-MO/NuminaMath-1.5/olympiads | Rationale:
We analyze the given graph and compare it with the properties of the functions in the options.
The graph has a vertical asymptote at $x = -1$ and as $x \to -1^+$ the function tends to $-\infty$. The function is increasing.
Let's check each option:
(A) $f(x)=\log _{3}(x+1)-1$
Domain: $x+1 > 0 \Rightarrow x >... | Real functions | A2. The graph of the function on the image is:
(A) $f(x)=\log _{3}(x+1)-1$
(B) $f(x)=\frac{(x-2)^{2}}{2 x+2}$
(C) $f(x)=2^{x+1}+3$
(D) $f(x)=\frac{2}{3} x-1$
(E) None of the above. | A2. In the graph, there is a rational function with a pole at $x=-1$, a double zero at $x=2$, and an initial value of 2, that is, the function $f(x)=\frac{(x-2)^{2}}{2 x+2}$. | bcac71a51263e3e9 | true | 2 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
3:1 | AI-MO/NuminaMath-1.5/olympiads | Let $O$ be the origin. Let $\vec{OE} = \mathbf{e}$ and $\vec{OG} = \mathbf{g}$. Since $O$ lies on side $EG$, $\mathbf{e}$ and $\mathbf{g}$ are collinear and in opposite directions. Let $OE = x$ and $OG = y$. Then $\mathbf{e} = -x \mathbf{u}$ and $\mathbf{g} = y \mathbf{u}$ for some unit vector $\mathbf{u}$.
Let $\vec{O... | Geometry | When triangle $E F G$ is rotated by an angle $\arccos _{1 / 3}$ around point $O$, which lies on side $E G$, vertex $F$ moves to vertex $E$, and vertex $G$ moves to point $H$, which lies on side $F G$. Find the ratio in which point $O$ divides side $E G$. | Isosceles triangles $E O F, G O H$ and $E F G$ are similar.
## Otвет
$3: 1$.
## Answer
$3: 1$. | d0c0cd67f88b221b | true | 3 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{1}{5} | AI-MO/NuminaMath-1.5/amc_aime | Let $E$ be the event that team B wins the second game.
Let $F$ be the event that team A wins the series.
Let $G$ be the event that team B wins the first game.
We want to find $P(G | E \cap F) = \frac{P(G \cap E \cap F)}{P(E \cap F)}$.
We need to find the sequences of games where team B wins the second game and team A ... | Geometry | Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first gam... | There are at most $5$ games played.
If team $B$ won the first two games, team $A$ would need to win the next three games. So the only possible order of wins is $BBAAA$.
If team $A$ won the first game, and team $B$ won the second game, the possible order of wins are: $ABBAA, ABABA,$ and $ABAAX$, where $X$ denotes that... | 3ace2f546c0e4578 | true | 3 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{\sqrt{6}}{4} | AI-MO/NuminaMath-1.5/amc_aime | Let $H$ be the origin. Let $\vec{HD} = (0,0,z)$, $\vec{HG} = (x,0,x)$. Then $\cos(\angle DHG) = \frac{\vec{HD} \cdot \vec{HG}}{||\vec{HD}|| \cdot ||\vec{HG}||} = \frac{x z}{\sqrt{z^2} \sqrt{2x^2}} = \frac{1}{\sqrt{2}}$, so $\angle DHG = 45^\circ$.
Let $\vec{HE} = (0,y,0)$. Then $\vec{HF} = \vec{HG} + \vec{HE} = (x,y,x)... | Geometry | In the adjoining figure of a rectangular solid, $\angle DHG=45^\circ$ and $\angle FHB=60^\circ$. Find the cosine of $\angle BHD$.
$\text {(A)} \frac{\sqrt{3}}{6} \qquad \text {(B)} \frac{\sqrt{2}}{6} \qquad \text {(C)} \frac{\sqrt{6}}{3} \qquad \text{(D)}\frac{\sqrt{6}}{4}\qquad \text{(E)}\frac{\sqrt{6}-\sqrt{2}}{4... | WLOG, let $CD=1$.
Looking at square GHDC, we see that $\angle DHC=45$, which implies that $DC=CH=1$ and $DH=\sqrt{2}$
Taking each cross-section one at a time, we look at square DHFB. We obviously know that CHB is a $30$ degree angle, giving $BH=\frac{2\sqrt{3}}{3}$, and $BC=\frac{\sqrt{3}}{3}$.
Looking at square ABCD,... | 9f060aab2f7d7203 | true | 4 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{2+\sqrt{2}}{4} | AI-MO/NuminaMath-1.5/olympiads | Let $A = (0, 0, 0)$, $B = (a, 0, 0)$, $C = (a, a, 0)$, $D = (0, a, 0)$.
The center of the square $ABCD$ is $O = (\frac{a}{2}, \frac{a}{2}, 0)$. The height of the pyramid is $PO = \frac{\sqrt{2}a}{2}$.
So $P = (\frac{a}{2}, \frac{a}{2}, \frac{\sqrt{2}a}{2})$.
The cube $ABCD-EFGH$ is constructed on the other side of the ... | Linear and multilinear algebra; matrix theory | 4. Given a regular quadrilateral pyramid $P-A B C D$ with all edges of equal length. Taking $A B C D$ as one face, construct a cube $A B C D-E F G H$ on the other side of the pyramid. Then, the cosine value of the angle formed by the skew lines $P A$ and $C F$ is | 4. $\frac{2+\sqrt{2}}{4}$.
Assume the edge lengths of the quadrilateral pyramid $P-ABCD$ are all 2. Taking the center $O$ of the square $ABCD$ as the origin, and the directions of $\overrightarrow{DA}$, $\overrightarrow{DC}$, and $\overrightarrow{OP}$ as the positive directions of the $x$-axis, $y$-axis, and $z$-axis,... | 8af90ab84c548e84 | true | 4 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
13 | AI-MO/NuminaMath-1.5/olympiads | We are given the equation $3 x^{2}+5 y^{2}=345$ and we need to find the maximum value of $x+y$ where $x, y$ are integer solutions.
First, we can simplify the equation by dividing by the greatest common divisor if possible.
We can see that $345 = 3 \times 115 = 3 \times 5 \times 23$.
Divide the equation by 3: $x^{2} + ... | Number theory | 1. Find the maximum value of the expression $x+y$, where $x, y-$ are integer solutions of the equation $3 x^{2}+5 y^{2}=345$ | # Solution
Notice that 345 and $5 y^{2}$ are divisible by 5, so $3 x^{2}$ must also be divisible by 5. Therefore, $\quad x=5 t, t \in Z$. Similarly, $y=3 n, n \in Z$. After simplification, the equation becomes $5 t^{2}+3 n^{2}=23$. Therefore, $t^{2} \leq \frac{23}{5}$, $n^{2} \leq \frac{23}{3}$ or $|t| \leq 2,|n| \leq... | ca34d2811a28a389 | true | 3 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$n=7 \text{ or } n=14$ | AI-MO/NuminaMath-1.5/olympiads | Let $n+2$ be the number of students. Two students scored a total of 8, and the other $n$ students scored equal scores $x$. The sum of scores is $8+nx$. The total number of games played is $\binom{n+2}{2} = \frac{(n+2)(n+1)}{2}$.
Thus, $8+nx = \frac{(n+2)(n+1)}{2}$.
$16+2nx = (n+2)(n+1) = n^2+3n+2$.
$2nx = n^2+3n-14$.
$... | Combinatorics | 8th CanMO 1976 Problem 3 n+2 students played a tournament. Each pair played each other once. A player scored 1 for a win, 1/2 for a draw and nil for a loss. Two students scored a total of 8 and the other players all had equal total scores. Find n. | There are (n+2)(n+1)/2 matches, so the total score is (n+2)(n+1)/2. Let the other players score k each. Then 8 + nk = (n+2)(n+1)/2, so n 2 - (2k-3) - 14 = 0. We know this equation has one root which is a positive integer. The product of the roots is -14, so the possibilities for the roots are: 1, -14; 2, -7; 7, -2; 14,... | 84c0ad44e41a1698 | true | 6 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\left[0, \frac{\sqrt{6}}{6}\right] | AI-MO/NuminaMath-1.5/olympiads | The domain of the function $f(x)=\sqrt{\frac{x+1}{x^{2}+4 x+7}}$ is determined by the condition $\frac{x+1}{x^{2}+4 x+7} \ge 0$.
Since $x^2+4x+7 = (x+2)^2+3 > 0$ for all real $x$, we need $x+1 \ge 0$, which means $x \ge -1$.
Let $g(x) = \frac{x+1}{x^2+4x+7}$. We want to find the range of $f(x) = \sqrt{g(x)}$ for $x \ge... | Real functions | (1) The range of the function $f(x)=\sqrt{\frac{x+1}{x^{2}+4 x+7}}$ is $\qquad$ | (1) $\left[0, \frac{\sqrt{6}}{6}\right]$ | 2e3cd9da5d2f3982 | true | 3 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{5}{2} | AI-MO/NuminaMath-1.5/amc_aime | Rationale:
Let $S$ be the number of students who preferred spaghetti and $M$ be the number of students who preferred manicotti. We are asked to find the ratio $\frac{S}{M}$. We need to read the number of students who preferred each type of pasta from the bar graph (which is not provided here). However, we can test each... | Mathematics education | $650$ students were surveyed about their pasta preferences. The choices were lasagna, manicotti, ravioli and spaghetti. The results of the survey are displayed in the bar graph. What is the ratio of the number of students who preferred spaghetti to the number of students who preferred manicotti?
$\mathrm{(A)} \frac{2}... | The answer is $\dfrac{\text{number of students who preferred spaghetti}}{\text{number of students who preferred manicotti}}$
So,
$\frac{250}{100}$
Simplify,
$\frac{5}{2}$
The answer is $\boxed{\textbf{(E)}\ \dfrac{5}{2}}$ | da4a7ae24e4bf1c0 | true | 3.2 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
382 | AI-MO/NuminaMath-1.5/olympiads | Let the six odd numbers be $x_1, x_2, x_3, x_4, x_5, x_6$. We are given that $\sum_{i=1}^6 x_i = 42$. Since the sum of six odd numbers is even, this is possible.
Let $x_i = 7 + \delta_i$, where $\sum_{i=1}^6 \delta_i = 0$. Since $x_i$ are odd, $\delta_i = x_i - 7$ must be even integers. Let $\delta_i = 2k_i$, where $k_... | Number theory | $30 \cdot 35$ Six odd numbers, their sum is 42, their sum of squares can only be
(A) 280.
(B) 368.
(C) 382.
(D) 423.
(China Jiangxi Nanchang Junior High School Mathematics Competition, 1990) | [Solution] The sum of the squares of six odd numbers must be even, so we exclude $(D)$.
Let these six odd numbers be $2 k_{1}+1, 2 k_{2}+1, \cdots 2 k_{6}+1$. Note that
$$
\begin{aligned}
& \left(2 k_{1}+1\right)^{2}+\left(2 k_{2}+1\right)^{2}+\cdots+\left(2 k_{6}+1\right)^{2} \\
= & 4\left(k_{1}^{2}+k_{2}^{2}+\cdots+k... | 22098cad61a4eab0 | true | 2.67 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
9801 | AI-MO/NuminaMath-1.5/olympiads | Let H be the height of the pyramid, and A be the area of the base.
Let the lateral edge of the pyramid be divided into 100 equal parts by 99 division points.
Planes parallel to the base are drawn through these 99 division points. This results in 99 cross-sections.
Let the heights of the smaller pyramids formed by these... | Geometry | [ [ Cross-sectional area $]$
The lateral edge of the pyramid is divided into 100 equal parts, and planes parallel to the base are drawn through the points of division. Find the ratio of the areas of the largest and smallest of the resulting cross-sections.
# | Let $S$ be the area of the base of the pyramid, $S_1$ and $S_2$ be the areas of the largest and smallest sections, respectively. The largest and smallest sections are polygons similar to the polygon of the base of the pyramid with coefficients $\frac{99}{100}$ and $\frac{1}{100}$, respectively. Therefore,
$$
S_1 = \le... | 4e50a84d2fe4fa46 | true | 2 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
1347 | AI-MO/NuminaMath-1.5/olympiads | We want to find the number of natural numbers $n$ in the range $1 \le n \le 2020$ that can be expressed in the form $n = [x] + [2x] + [3x]$ for some real number $x$.
Let $f(x) = [x] + [2x] + [3x]$. We analyzed the values of $f(x)$ for $x \in [k, k+1)$ for integer $k \ge 0$.
For $x \in [k, k+1)$, $f(x)$ takes values $6k... | Real functions | 7. Let $[a]$ denote the greatest integer not exceeding $a$, for example: $[8]=8, [3.6]=3$.
Some natural numbers can be expressed in the form $[x]+[2 x]+[3 x]$, such as 6 and 3:
$$
\begin{array}{c}
6=\left[\frac{5}{4}\right]+\left[2 \times \frac{5}{4}\right]+\left[3 \times \frac{5}{4}\right], \\
3=[0.8]+[2 \times 0.8]+[... | $1347$ | d283b9d7a2bbac85 | true | 6 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$30 \mathrm{~cm}^{2}$ | AI-MO/NuminaMath-1.5/olympiads | Let $Area(ABC)$ denote the area of triangle $A B C$. We are given $Area(ABC) = 60 \mathrm{~cm}^{2}$.
Let $s$ be the side length of equilateral triangle $A B C$. Then $Area(ABC) = \frac{\sqrt{3}}{4} s^2 = 60$.
Since $K, M, O$ are midpoints of sides $A B, B C, C A$, triangle $KMO$ is equilateral with side length $\frac{s... | Geometry | For an equilateral triangle $A B C$, a regular hexagon $K L M N O P$ is inscribed such that points $K, M, O$ lie sequentially at the midpoints of sides $A B, B C$, and $A C$. Calculate the area of the hexagon $K L M N O P$ if the area of triangle $A B C$ is $60 \mathrm{~cm}^{2}$. (K. Pazourek) | Let's inscribe a hexagon $K L M N O P$ into the triangle $A B C$ in the prescribed manner.
Given that both figures as a whole are symmetric with respect to three axes of symmetry, the centro... | d2a78d72e8bce47b | true | 3 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
999 | AI-MO/NuminaMath-1.5/olympiads | Final Answer: The final answer is $\boxed{999}$ | Number theory | 5. Let $\Sigma(n)$ denote the sum of the digits of the number $n$. Find the smallest three-digit $n$ such that $\Sigma(n)=\Sigma(2 n)=\Sigma(3 n)=\ldots=\Sigma\left(n^{2}\right)$ | Answer: 999.
Solution: Let the desired number be $\overline{a b c}$. Note that this number is not less than 101 (since 100 does not work). Therefore, $101 \cdot \overline{a b c}=\overline{a b c 00}+\overline{a b c}$ also has the same sum of digits. But the last digits of this number are obviously $b$ and $c$, so the s... | ccc3ed17d9261dd9 | true | 6 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
58 | AI-MO/NuminaMath-1.5/amc_aime | Let $\widehat{AB} = \widehat{BC} = \widehat{CD} = \widehat{DE} = x$ and $\widehat{EF} = \widehat{FG} = \widehat{GH} = \widehat{HI} = \widehat{IA} = y$.
We are given that $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$, so $\angle ABD - \angle AHG = 12^\circ$.
$\angle ABD = \frac{1}{2} \widehat{AD} = \frac{1}{2} (\wideh... | General algebraic systems | Point $A,B,C,D,$ and $E$ are equally spaced on a minor arc of a circle. Points $E,F,G,H,I$ and $A$ are equally spaced on a minor arc of a second circle with center $C$ as shown in the figure below. The angle $\angle ABD$ exceeds $\angle AHG$ by $12^\circ$. Find the degree measure of $\angle BAG$. | Let $O$ be the center of the circle with $ABCDE$ on it.
Let $x$ be the degree measurement of $\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}$ in circle $O$
and $y$ be the degree measurement of $\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}$ in circle $C$.
$\angle ECA$ is, therefore, $5y$ by way... | 67f8fe346b09317f | true | 4.2 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
4 | AI-MO/NuminaMath-1.5/olympiads | I2.1 Let $P$ be a two-digit number. The $50^{\text {th }}$ power of $P$, $P^{50}$, is a 69-digit number.
This means that $10^{68} \le P^{50} < 10^{69}$.
Taking logarithm base 10 on all sides, we get
$\log_{10}(10^{68}) \le \log_{10}(P^{50}) < \log_{10}(10^{69})$
$68 \le 50 \log_{10} P < 69$
Divide by 50:
$\frac{68}{50}... | Field theory and polynomials | I2.1 If the $50^{\text {th }}$ power of a two-digit number $P$ is a 69-digit number, find the value of $P$.
(Given that $\log 2=0.3010, \log 3=0.4771, \log 11=1.0414$.)
I2.2 The roots of the equation $x^{2}+a x-P+7=0$ are $\alpha$ and $\beta$, whereas the roots of the equation $x^{2}+b x-r=0$ are $-\alpha$ and $-\beta... | Reference: 1995 HG5 . . $37^{100} \ldots$ 157-digit number, $37^{15} \ldots n$-digit ....
$$
\begin{array}{l}
P^{50}=y, 10<P \leq 99,10^{68} \leq y<10^{69} \\
P=y^{\frac{1}{50}} ; 10^{68 \div 50}<P<10^{69 \div 50} \\
1.34<\log P<1.38 \\
\log 22=\log 2+\log 11=1.3424 ; \log 24=3 \log 2+\log 3=1.3801 \\
\log 22<\log P<\l... | 07fbf42f4e0b4aef | true | 3 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
225 | AI-MO/NuminaMath-1.5/amc_aime | Let $k$ be the number of distinct values in the list.
Let one of these values be the unique mode, which occurs exactly $10$ times.
The remaining $k-1$ distinct values must each occur at most $9$ times, since the mode is unique.
To minimize $k$, we want to maximize the frequencies of these $k-1$ distinct values.
So we a... | Combinatorics | A list of $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list?
$\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234$ | To minimize the number of distinct values, we want to maximize the number of times a number appears. So, we could have $223$ numbers appear $9$ times, $1$ number appear once, and the mode appear $10$ times, giving us a total of $223 + 1 + 1 = \boxed{\textbf{(D)}\ 225}.$ | 92778e98834aa268 | true | 4 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
59 | AI-MO/NuminaMath-1.5/amc_aime | Let $n_A, n_B, n_C$ be the number of rocks in piles $A, B, C$ respectively, and let $\mu_A, \mu_B, \mu_{AB}, \mu_{AC}$ be the mean weights given. We are given $\mu_A = 40$, $\mu_B = 50$, $\mu_{AB} = 43$, $\mu_{AC} = 44$. Let $\mu_{BC}$ be the mean weight of rocks in combined piles $B$ and $C$. We want to find the great... | Mathematical logic and foundations | $A$, $B$, $C$ are three piles of rocks. The mean weight of the rocks in $A$ is $40$ pounds, the mean weight of the rocks in $B$ is $50$ pounds, the mean weight of the rocks in the combined piles $A$ and $B$ is $43$ pounds, and the mean weight of the rocks in the combined piles $A$ and $C$ is $44$ pounds. What is the gr... | Let pile $A$ have $A$ rocks, and so on.
The total weight of $A$ and $C$ can be expressed as $44(A + C)$.
To get the total weight of $B$ and $C$, we add the weight of $B$ and subtract the weight of $A$: $44(A + C) + 50B - 40A = 4A + 44C + 50B$
Therefore, the mean of $B$ and $C$ is $\frac{4A + 44C + 50B}{B + C}$, which i... | 2280828ca5f9d7b3 | true | 3.4 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$f(x) = 0 \text{ or } f(x) = x$ | AI-MO/NuminaMath-1.5/olympiads | Final Answer: The solutions are $\boxed{f(x) = 0, f(x) = x}$ | Difference and functional equations | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying:
$$
f(x(1+y))=f(x)(1+f(y))
$$ | The zero constant function is a solution, but we exclude this case. By setting $x=0$, we get $f(0)=0$, and by setting $y=-1$, we get $f(-1)=-1$. By setting $x=-1$ and $y=-\frac{1}{2}$, we get $f\left(-\frac{1}{2}\right)=-\frac{1}{2}$. Finally, by setting $x=-\frac{1}{2}$ and $y=1$, we get $f(1)=1$. Then, by setting $x=... | 604c3458242cf84a | true | 5.75 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
17 \text{ cm} | AI-MO/NuminaMath-1.5/olympiads | Let $A B C F$ be a rectangle with $A B=30 \mathrm{~cm}$ and $A F=14 \mathrm{~cm}$.
The area of rectangle $A B C F$ is $A B \times A F = 30 \times 14 = 420 \mathrm{~cm}^{2}$.
We are given that $F E=5 \mathrm{~cm}$ and points $E$ and $D$ are on $F C$. Let $D C = x$.
Since $F C = A B = 30 \mathrm{~cm}$, we have $F D = F C... | Game theory, economics, social and behavioral sciences | In the diagram, $A B C F$ is a rectangle with $A B=30 \mathrm{~cm}$ and $A F=14 \mathrm{~cm}$. Points $E$ and $D$ are on $F C$ so that $F E=5 \mathrm{~cm}$ and the area of quadrilateral $A B D E$ is $266 \mathrm{~cm}^{2}$. The length of $D C$ is
(A) $17 \mathrm{~cm}$
(B) $16 \mathrm{~cm}$
(C) $19 \mathrm{~cm}$
(D) $20 ... | Solution 1
Since $A B$ and $E D$ are parallel, quadrilateral $A B D E$ is a trapezoid.
We know that $A B=30 \mathrm{~cm}$.
Since $A B C F$ is a rectangle, then $F C=A B=30 \mathrm{~cm}$.
Suppose that $D C=x \mathrm{~cm}$.
Then $E D=F C-F E-D C=(30 \mathrm{~cm})-(5 \mathrm{~cm})-(x \mathrm{~cm})=(25-x) \mathrm{cm}$... | c54660382eb88b03 | true | 2.4 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{1}{3} | AI-MO/NuminaMath-1.5/olympiads | Let $B$ be the value of Ben's prize and $J$ be the value of Jamie's prize.
Ben's prize can be $\$ 5, \$ 10$ or $\$ 20$, each with probability $\frac{1}{3}$.
Jamie's prize can be $\$ 30$ or $\$ 40$, each with probability $\frac{1}{2}$.
We want to find the probability that $B+J = 50$.
We consider the possible values of $... | Probability theory and stochastic processes | Ben participates in a prize draw. He receives one prize that is equally likely to be worth $\$ 5, \$ 10$ or $\$ 20$. Jamie participates in a different prize draw. She receives one prize that is equally likely to be worth $\$ 30$ or $\$ 40$. What is the probability that the total value of their prizes is exactly $\$ 50$... | Since there are two possible prizes that Jamie can win and each is equally likely, then the probability that Jamie wins $\$ 30$ is $\frac{1}{2}$ and the probability that Jamie wins $\$ 40$ is $\frac{1}{2}$. If Jamie wins $\$ 30$, then for the total value of the prizes to $\$ 50$, Ben must win $\$ 20$. The probability t... | b87ec707839457ba | true | 1 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\sqrt{6} | AI-MO/NuminaMath-1.5/olympiads | Let the isosceles triangle be ABC, with AB = AC. Let CH be the height dropped from vertex C to the lateral side AB.
The height CH divides the side AB into segments AH and HB. We are given that these segments are of lengths 2 and 1, counting from the vertex of the triangle. Since CH is dropped to the lateral side AB, th... | Geometry | [Pythagorean Theorem (direct and inverse).]
The height of an isosceles triangle, dropped to the lateral side, divides it into segments of 2 and 1, counting from the vertex of the triangle. Find the base of the triangle. | Let $CH$ be the height of the isosceles triangle $ABC$ dropped to the lateral side $AB$, with $AH=2$, $BH=1$. Then $AC=AB=3$.
By the Pythagorean theorem from the right triangles $ACH$ and $BCH$, we find that
$$
CH^2=AC^2-AH^2=9-4=5, BC^2=BH^2+CH^2=1+5=6 \text{. }
$$
Therefore, $BC=\sqrt{6}$.
## Answer
$\sqrt{6}$. | c268418b704d8f8d | true | 2 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
(43, 257), (100, 200), (98, 202), (125, 175) | AI-MO/NuminaMath-1.5/olympiads | To solve this task, we need to pair the given numbers so that the sum of each pair is the same.
First, let's calculate the sum of all the given numbers:
$43 + 202 + 100 + 175 + 98 + 257 + 125 + 200$
To make the addition easier, let's rearrange and group the numbers:
$(43 + 98) + (100 + 200) + (125 + 175) + (202 + 257... | Mathematical logic and foundations | ## Task 4
Pair the following numbers so that they always add up to the same sum.
$\begin{array}{llllllll}43 & 202 & 100 & 175 & 98 & 257 & 125 & 200\end{array}$ | The sum is always 300. Pairs are (43, 257), (100, 200), (98, 202), (125, 175) | 0da8ae73f72e3733 | true | 1.4 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{1}{58} | KbsdJames/Omni-MATH/HMMT_2 | Let $N$ be the number of rolls that are 20 in five rolls of fair icosahedral dice.
We are given that at least half of the rolls are 20, which means $N \ge 3$. We want to find the probability that all three remaining concealed dice show 20. We interpret this as the probability that all five rolls are 20, given that at l... | Combinatorics | Your math friend Steven rolls five fair icosahedral dice (each of which is labelled $1,2, \ldots, 20$ on its sides). He conceals the results but tells you that at least half of the rolls are 20. Assuming that Steven is truthful, what is the probability that all three remaining concealed dice show $20 ?$ | The given information is equivalent to the first two dice being 20 and 19 and there being at least two 20's among the last three dice. Thus, we need to find the probability that given at least two of the last three dice are 20's, all three are. Since there is only one way to get all three 20's and $3 \cdot 19=57$ ways ... | a4457562dde4bba7 | true | 2 | {'domain': ['Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations'], 'difficulty': 4.0} |
$90^{\circ}$ | AI-MO/NuminaMath-1.5/olympiads | Let $M$ be the midpoint of $AB$. We want to find the angle $\angle GMH$. We calculated $\vec{MG} \cdot \vec{MH} = 0$ for any point $P$ on $BC$. This means that the vectors $\vec{MG}$ and $\vec{MH}$ are orthogonal, so the angle $\angle GMH = 90^\circ$, unless one of the vectors is zero.
We found that $\vec{MG} = 0$ and ... | Geometry | In the isosceles right triangle $ABC$, the right angle is at $C$. Let $P$ be any point on the side $BC$, and let $G$ be the orthogonal projection of point $C$ onto $AP$. Let $H$ be the point on segment $AP$ such that $AH = CG$. At what angle is the segment $GH$ seen from the midpoint of $AB$? | Solution. Let $F$ be the midpoint of segment $AB$.
Since triangle $ABC$ is half of a square, $CF = AF$, and triangles $ACF$ and $BCF$ are isosceles right triangles. $\angle CAP = \angle PCG$,... | 81aa2b100e4230cd | true | 4.25 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
400 \% | AI-MO/NuminaMath-1.5/olympiads | The price of filling the tank in 1972 was $P_{1972} = \$ 29.90$.
The price of filling the tank in 1992 was $P_{1992} = \$ 149.70$.
The increase in price is $P_{1992} - P_{1972} = 149.70 - 29.90 = \$ 119.80$.
The percentage increase in the price of gasoline is given by:
Percentage increase $= \frac{\text{Increase in pri... | Mathematics education | In 1972, filling the tank of a small car cost $R \$ 29.90$, and in 1992, it cost $\$ 149.70$ to fill the same tank. Which of the following values best approximates the percentage increase in the price of gasoline over this 20-year period?
(a) $20 \%$
(b) $125 \%$
(d) $300 \%$
(d) $400 \%$
(e) $500 \%$ | The increase in value was
$$
149.70-29.90=119.80 \text { reais }
$$
which corresponds to:
$$
\frac{119.80}{29.90} \times 100 \% = 400.66 \%
$$
The correct option is (d). | 34a69a3a5da3136b | true | 2.4 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
13 | AI-MO/NuminaMath-1.5/olympiads | We are given the system of linear equations
$$\left\{\begin{array}{l}7 x-5 y=11 \\ 5 x+7 y=k\end{array}\right.$$
We solved for $x$ and $y$ in terms of $k$:
$$x = \frac{77 + 5k}{74}$$
$$y = \frac{7k - 55}{74}$$
For integer solutions $(x, y)$, we need $77 + 5k$ to be divisible by 74 and $7k - 55$ to be divisible by 74.
W... | Geometry | 3. Given an integer $k$ satisfying $1000<k<2020$, and such that the system of linear equations in two variables $\left\{\begin{array}{l}7 x-5 y=11 \\ 5 x+7 y=k\end{array}\right.$ has integer solutions, then the number of possible values for $k$ is $\qquad$. | $13$ | 83658df674e1a7ab | true | 3 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
5392 | AI-MO/NuminaMath-1.5/olympiads | Let $p(x) = Ax+B$.
For any integers $a, b, c$ such that $a+b+c=2022$, we have $p(a)+p(b)+p(c) = (Aa+B)+(Ab+B)+(Ac+B) = A(a+b+c)+3B = 2022A+3B$.
We are given that $p(a)+p(b)+p(c) = p(674)$.
$p(674) = A(674)+B$.
Thus, $2022A+3B = 674A+B$.
$2022A - 674A + 3B - B = 0$.
$1348A + 2B = 0$.
$2B = -1348A$.
$B = -674A$.
So $p(x)... | Commutative algebra | 4. The polynomial $p(x)$ has the following property: for every triplet of integers $a, b, c$ such that $a+b+c=2022$ we have that $p(a)+p(b)+p(c)=p(674)$. It is also known that $p(0)=-2696$. What is the value of $p(2022)$?
(A) -2696
(B) 674
(C) 5392
(D) 8088
(E) It is not possible to determine it with the given data. | 4. The answer is $\mathbf{( C )}$. Substituting $a=b=c=674$ (integers that indeed satisfy $a+b+c=$ 2022) we get $3 p(674)=p(674)$, which means $p(674)=0$. Substituting then $a=b=0$ and $c=2022$ we obtain
$$
2 p(0)+p(2022)=p(674)=0 \Rightarrow p(2022)=-2 p(0)=5392 \text {. }
$$ | 1319b26beaf63d6d | true | 4 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
7 | AI-MO/NuminaMath-1.5/olympiads | Let $E = \sqrt{16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9}$. We rewrite the expression inside the square root as
$16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9 = (4a-2b)^2 - 3(4a-2b) + 9$.
Let $x = 4a-2b$. Then $E = \sqrt{x^2 - 3x + 9}$.
We want to find the maximum value of $E$ subject to $17\left(a^{2}+b^{2}\right)-30 a b-16=0$.
We need... | Algebraic geometry | 20. Let $a$ and $b$ be real numbers such that $17\left(a^{2}+b^{2}\right)-30 a b-16=0$. Find the maximum value of $\sqrt{16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9}$.
(2 marks)
設 $a$ 、 $b$ 為實數 , 使得 $17\left(a^{2}+b^{2}\right)-30 a b-16=0$ 。 求 $\sqrt{16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9}$ 的最大值。 | 20. 7
20. Rewrite the given equation as $(a+b)^{2}+16(a-b)^{2}=16$. Hence we may let $a+b=4 \cos x$ and $a-b=\sin x$. Note that
$$
\sqrt{16 a^{2}+4 b^{2}-16 a b-12 a+6 b+9}=\sqrt{(4 a-2 b)^{2}-3(4 a-2 b)+9}=\sqrt{\left(4 a-2 b-\frac{3}{2}\right)^{2}+\frac{27}{4}} .
$$
Since $4 a-2 b=(a+b)+3(a-b)=4 \cos x+3 \sin x$, wh... | 94f0c9001cc9bf03 | true | 7 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{2}{3} | AI-MO/NuminaMath-1.5/olympiads | Let $W$ be the event that a white ball is added, and $B$ be the event that a black ball is added. We are given $P(W) = P(B) = 1/2$.
Let $D$ be the event that a white ball is drawn from the urn.
We want to find the conditional probability that the remaining ball in the urn is also white, given that a white ball was draw... | Probability theory and stochastic processes | 14. A ball was added to an urn containing one white ball - either white or black (with equal probabilities of selection). After this, one ball was randomly drawn from the urn. It turned out to be white. What is the conditional probability that the remaining ball in the urn is also white? | Solution. Let
$$
\begin{aligned}
A & =\{\text { a white ball was added to the urn }\}, \\
C & =\{\text { a white ball was drawn }\}, \\
D & =\{\text { a white ball remains }\} .
\end{aligned}
$$
Note that $A=C D$. Then
$$
\begin{aligned}
\mathrm{P}(D \mid C)=\frac{\mathrm{P}(D C)}{\mathrm{P}(C)}=\frac{\mathrm{P}(A)}... | 16f94773a3a8c5c4 | true | 2 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
(1, 4) | AI-MO/NuminaMath-1.5/olympiads | We are given the equation $37+27^{m}=n^{3}$, where $m$ and $n$ are positive integers.
We rewrite the equation as $n^{3}-27^{m}=37$, or $n^{3}-(3^{m})^{3}=37$.
Let $x = 3^{m}$. Then the equation becomes $n^{3}-x^{3}=37$.
Factoring the difference of cubes, we have $(n-x)(n^{2}+nx+x^{2})=37$.
Since $m$ and $n$ are positiv... | Number theory | 3. Determine all pairs of positive integers $m$ and $n$ for which $37+27^{m}=n^{3}$.
The written part of the school round in category B takes place
## on Thursday, January 21, 2010
so that it starts in the morning and the participants have 4 hours of pure time to solve the problems. For each problem, participants ca... | 3. We rewrite the equation in the form $37=n^{3}-27^{m}$ and factor the difference of cubes:
$$
37=\left(n-3^{m}\right)\left(n^{2}+n \cdot 3^{m}+9^{m}\right) .
$$
The number 37 is a prime number, and on the right side of the equation, there is a product of two integers, with the second factor being greater than 1. Th... | ae2fd1d2abb14500 | true | 3 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\{(1, 2), (9, 22), (8, 9)\} | AI-MO/NuminaMath-1.5/olympiads | We are given that $k^2+4m = a^2$ and $m^2+5k = b^2$ for some positive integers $a, b$.
From the first equation, $4m = a^2-k^2$. Since $m>0$, $a>k$. Let $a = k+x$ for some positive integer $x$. Then $4m = (k+x)^2-k^2 = 2kx+x^2$, so $m = \frac{x(2k+x)}{4}$.
From the second equation, $5k = b^2-m^2$. Since $k>0$, $b>m$. Le... | Number theory | LVIII OM - I - Problem 2
Determine all pairs of positive integers $ k $, $ m $, for which each of the numbers $ {k^2+4m} $, $ {m^2+5k} $ is a perfect square. | Suppose the pair $ (k,m) $ satisfies the conditions of the problem.
If the inequality $ m\geq k $ holds, then
$$(m+3)^2=m^2+6m+9>m^2+5m\geq m^2+5k>m^2,$$
and since $ m^2+5k $ is a square of an integer, it follows that one of the equalities $ m^2+5k=(m+1)^2 $ or $ m^2+5k=(m+2)^2 $ must hold.
If $ m^2+5k=(m+1)^2=m^2+2... | 2a9114fe3d1f7eb1 | true | 5 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\sqrt{2007} | AI-MO/NuminaMath-1.5/olympiads | Let $a = \sqrt{2006}$ and $b = \sqrt{2007}$. Then $f(x) = x^6 - 2ax^5 - x^4 + x^3 - 2bx^2 + 2x - a$. We want to find $f(a+b)$.
Let $x = a+b$. Then $x-a = b$ and $x-b = a$.
From $x-a = b$, we have $(x-a)^2 = b^2 = 2007$. So $x^2 - 2ax + a^2 = 2007$. $x^2 - 2ax + 2006 = 2007$. $x^2 - 2ax = 1$.
From $x-b = a$, we have $(x... | Field theory and polynomials | $$
\begin{array}{l}
\text { 7. If } f(x)=x^{6}-2 \sqrt{2006} x^{5}-x^{4}+x^{3}-2 \sqrt{2007} x^{2}+2 x-\sqrt{2006} \text {, then } f(\sqrt{2006}+ \\
\sqrt{2007})= \\
\end{array}
$$ | 7. $\sqrt{2007}$ Hint: $f(x)=x^{4}\left(x^{2}-2 \sqrt{2006} x-1\right)+x\left(x^{2}-2 \sqrt{2007} x+1\right)+x-\sqrt{2006}-\sqrt{2007}$ $+\sqrt{2007}, f(\sqrt{2006}+\sqrt{2007})=\sqrt{2007}$. | a9d469aee7388520 | true | 5.4 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
3 \sqrt{4+\sqrt{3}} | AI-MO/NuminaMath-1.5/olympiads | Final Answer: The final answer is $\boxed{(A)}$ | Difference and functional equations | 16. On the sides of the equilateral triangle $A B C$, which have length 1, three squares are constructed, as shown in the figure. What is the perimeter of the triangle $A^{\prime} B^{\prime} C^{\prime}$?
(A) $3 \sqrt{4+\sqrt{3}}$
(D) $\frac{3}{2} \sqrt{3}$
(E) $\frac{3}{2}+\frac{3}{4} \sqrt{3}$
(C) $3 \sqrt{1+2 \sqrt{3... | (16) The correct answer is $(A)$.
Triangle $A^{\prime} B^{\prime} C^{\prime}$ is equilateral for evident reasons of symmetry (or, if you prefer, by observing the congruence of triangles $A^{\prime} B^{\prime} A, B^{\prime} C^{\prime} B, C^{\prime} A^{\prime} C$).
Let $D$ be the orthogonal projection of $A^{\prime}$ o... | d7633ed40d738e57 | true | 3.4 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{41}{45} | AI-MO/NuminaMath-1.5/olympiads | Let $V_{ABCD}$ be the volume of the pyramid $ABCD$. We are given $V_{ABCD} = 1$.
We are given points $K, L, M$ on edges $AD, BD, CD$ such that $AK = \frac{1}{3} AD, KD = \frac{2}{3} AD$, $BL = \frac{2}{3} BD, LD = \frac{1}{3} BD$, $CM = \frac{3}{5} CD, MD = \frac{2}{5} CD$.
The volume of the pyramid $KLMD$ is given by ... | Geometry | The volume of the pyramid $ABCD$ is 1. Points $K, L$, and $M$ are taken on the edges $AD, BD, CD$ respectively, such that $2AK = KD, BL = 2LD$, and $2CM = 3MD$. Find the volume of the polyhedron $ABCKLM$.
# | $$
V_{K L M D}=\frac{D K}{D A} \cdot \frac{D L}{D B} \cdot \frac{D M}{D C} \cdot V_{A B C D}=\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{2}{5} \cdot 1=\frac{4}{45}
$$
Therefore,
$$
V_{A B C K L M}=V_{A B C D}-V_{K L M D}=1-\frac{4}{45}=\frac{41}{45}
$$
## Answer
$\frac{41}{45}$ | e5df6d985c2919a1 | true | 3 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
10 | AI-MO/NuminaMath-1.5/olympiads | We want to find the number of triples of distinct natural numbers $(a, b, c)$ such that $a|b$, $b|c$, and $c|12$.
The divisors of 12 are $\{1, 2, 3, 4, 6, 12\}$. Thus, $c \in \{1, 2, 3, 4, 6, 12\}$.
Since $b|c$ and $b \neq c$, $b$ must be a proper divisor of $c$.
Since $a|b$ and $a \neq b$, $a$ must be a proper divisor... | Combinatorics | 10. How many triples of distinct natural numbers $(a, b, c)$ are there such that the number $a$ is a divisor of $b$, the number $b$ is a divisor of $c$, and the number $c$ is a divisor of 12?
(A) 6
$\begin{array}{ll}\text { (B) } 7 & \text { (C) } 8\end{array}$
(D) 10
(E) 9 | Problem 10. The correct answer is (D).
The possible triples $(a, b, c)$ are: $\quad(1,2,4), \quad(1,2,6), \quad(1,2,12), \quad(1,3,6), \quad(1,3,12), \quad(1,4,12), \quad(1,6,12)$, $(2,4,12), \quad(2,6,12), \quad(3,6,12)$. In total, there are 10. | fc82239ac0387506 | true | 3.25 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
8 | AI-MO/NuminaMath-1.5/olympiads | The color of a point with integer coordinate $x$ depends only on $x \pmod 7$. Let $C(x)$ denote the color of the point with coordinate $x$. We are given that $C(x) = C(y)$ if $x \equiv y \pmod 7$. We need to determine the colors $C(0), C(1), C(2), C(3), C(4), C(5), C(6)$.
From rule b), we have:
$C(20) = R$. Since $20 \... | Combinatorics | 5. On the number line, points with integer coordinates are painted red and blue according to the following rules: a) points whose coordinate difference is 7 must be painted the same color; b) points with coordinates 20 and 14 should be painted red, and points with coordinates 71 and 143 - blue. In how many ways can all... | Answer. In eight ways.
Solution. From part a), it follows that the coloring of all points with integer coordinates is uniquely determined by the coloring of the points corresponding to the numbers $0,1,2,3,4,5$, and 6. The point $0=14-2 \cdot 7$ must be colored the same as 14, i.e., red. Similarly, the point $1=71-10 ... | 4f38e3866a17a7f2 | true | 6.67 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
8.5 | AI-MO/NuminaMath-1.5/olympiads | Let $v_D$ be Dima's running speed and $v_V$ be Uncle Vanya's car speed.
Usually, Dima leaves home at $8:10$ AM. Let $T_{usual}$ be the usual travel time from home to school by car. Then the usual arrival time at school is $8:10 \text{ AM} + T_{usual}$.
On Thursday, Dima left home at $7:20$ AM and ran in the opposite d... | Mathematical logic and foundations | 2. Usually, Dima leaves home at $8:10$ AM, gets into Uncle Vanya's car, who delivers him to school by a certain time. But on Thursday, Dima left home at 7:20 and ran in the opposite direction. Uncle Vanya waited for him and at $8:20$ drove after him, caught up with Dima, turned around, and delivered him to school 26 mi... | # Answer: 8.5 times.
## Solution:
The car was on the road for 16 minutes longer than usual, due to spending 8 minutes catching up to Dima and 8 minutes returning home. The car caught up with Dima at 8:28, and during the 68 minutes from 7:20 to 8:28, Dima ran the same distance that the car traveled in 8 minutes, i.e.,... | 9302a67fc7240fd4 | true | 8.5 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
29 | AI-MO/NuminaMath-1.5/olympiads | We want to compute the remainder when $\sum_{k=1}^{30303} k^{k}$ is divided by 101.
We have shown that $\sum_{k=1}^{30303} k^{k} \equiv \sum_{k=1}^{30300} k^{k} + \sum_{k=30301}^{30303} k^{k} \pmod{101}$.
We have $\sum_{k=1}^{30300} k^{k} \equiv 98 \pmod{101}$ and $\sum_{k=30301}^{30303} k^{k} \equiv 32 \pmod{101}$.
Th... | Number theory | 29. [18] Compute the remainder when
$$
\sum_{k=1}^{30303} k^{k}
$$
is divided by 101 . | Answer: 29 The main idea is the following lemma:
Lemma. For any non-negative integer $n$ and prime $p, \sum_{k=n+1}^{n+p^{2}-p} k^{k} \equiv 1(\bmod p)$.
Proof. Note that $a^{b}$ depends only on the value of $a(\bmod p)$ and the value of $b(\bmod p-1)$. Since $p$ and $p-1$ are relatively prime, the Chinese Remainder Th... | d2136761d0326573 | true | 6 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
2 \frac{1}{2} \text{ m} | AI-MO/NuminaMath-1.5/olympiads | Let $x=0$ be the position of the end of the rod where the $20 \mathrm{~kg}$ weight is attached.
The weights are attached at the following positions:
$w_1 = 20 \mathrm{~kg}$ at $x_1 = 0 \mathrm{~m}$
$w_2 = 30 \mathrm{~kg}$ at $x_2 = 1 \mathrm{~m}$
$w_3 = 40 \mathrm{~kg}$ at $x_3 = 2 \mathrm{~m}$
$w_4 = 50 \mathrm{~kg}$ ... | Mechanics of particles and systems | A four-meter-long rod has a $20 \mathrm{~kg}$ weight attached to one end, and $30,40,50 \mathrm{~kg}$ weights are attached at one, two, and three meters from this end, respectively. At the other end of the rod, a $60 \mathrm{~kg}$ weight is attached. Where should the rod be supported so that it remains in balance? The ... | In equilibrium, the algebraic sum of static moments is $0\left(\sum P p=0\right)$. Let $x$ be the distance from the support point to the end of the rod where the $20 \mathrm{~kg}$ weight hangs, then
$$
\begin{gathered}
20 x+(x-1) 30+(x-2) 40+(x-3) 50+(x-4) 60=0 \\
20 x+30 x-30+40 x-80+50 x-150+60 x-240=0 \\
200 x=500 ... | 5c909ddd972b74cc | true | 1.8 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
16a | AI-MO/NuminaMath-1.5/olympiads | To find the length of the cardioid $x=2 a \cos t-a \cos 2 t, y=-2 a \sin t-a \sin 2 t$, we use the formula for the length of a parametric curve:
$L = \int_{0}^{2\pi} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt$
First, we calculate the derivatives of $x$ and $y$ with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(2 a... | Differential geometry | 884. Find the length of the cardioid $x=2 a \cos t-a \cos 2 t, y=$ $-2 a \sin t-a \sin 2 t$. | Solution. We apply formula (1); based on the given parametric equations of the cardioid and the formula for the differential of the arc of a plane curve (Chapter 5, §), we transform the line integral in formula (1) into an ordinary integral with the variable $t$.
$$
\begin{gathered}
\dot{x}=-2 a \sin t+2 a \sin 2 t, \... | 2afb6d56bee6fe8c | true | 3.75 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
74 | AI-MO/NuminaMath-1.5/olympiads | Let $A$ be the set of numbers less than 100 divisible by 2, $B$ be the set of numbers less than 100 divisible by 3, and $C$ be the set of numbers less than 100 divisible by 5. We want to find $|A \cup B \cup C|$.
We use the Principle of Inclusion-Exclusion:
$|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |A \cap C... | Number theory | How many numbers less than 100 are divisible by 2, 3, or 5? | Here there are many possibilities.
- $A_{2}=\left\lfloor\frac{100}{2}\right\rfloor=50$ numbers less than 100 divisible by 2.
- $A_{3}=\left\lfloor\frac{100}{3}\right\rfloor=33$ numbers less than 100 divisible by 3
- $A_{5}=\left\lfloor\frac{100}{5}\right\rfloor=20$ numbers less than 100 divisible by 5
- $A_{2,3}=\... | 5c81816f0547fb9f | true | 1.33 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{2}{3} | AI-MO/NuminaMath-1.5/olympiads | Let $X$ be the number of pairs of adjacent numbers among the 3 selected numbers. We can write $X$ as a sum of indicator random variables. Let $I_i$ be an indicator random variable such that $I_i = 1$ if both $i$ and $i+1$ are selected, and $I_i = 0$ otherwise, for $i = 1, 2, \dots, 8$. Then $X = \sum_{i=1}^8 I_i$.
The ... | Probability theory and stochastic processes | 8. Among the 9 natural numbers $1,2,3, \cdots, 9$, if 3 numbers are randomly selected, let $x$ be the number of pairs of adjacent numbers among these 3 numbers (for example: if the 3 numbers selected are $1,2,3$, then there are two pairs of adjacent numbers, 1,2 and 2,3, at this time the value of $x$ is 2), then the ma... | Solve for $x=0,1,2$, and $P(x=0)=\frac{C_{7}^{3}}{C_{9}^{3}}=\frac{5}{12}, P(x=1)=\frac{6 \times 2+5 \times 6}{C_{9}^{3}}=\frac{1}{2}, P(x=2)=\frac{7}{C_{9}^{3}}=\frac{1}{12}$. Therefore, $E(x)=\frac{1}{2}+\frac{1}{6}=\frac{2}{3}$. | 85590be30782e998 | true | 2 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
16 | AI-MO/NuminaMath-1.5/olympiads | Let $A B C D$ be a square. Points $A, B, G$ are collinear, so $G$ is on the line $A B$. Line segments $A C$ and $D G$ intersect at $E$. Line segments $D G$ and $B C$ intersect at $F$. We are given $D E=15$, $E F=9$, $F G=x$. We want to find $x$.
Since $B C \parallel A D$, we have similar triangles $\triangle E C F \sim... | Geometry | 17. In the diagram below, $A B C D$ is a square. The points $A, B$ and $G$ are collinear. The line segments $A C$ and $D G$ intersect at $E$, and the line segments $D G$ and $B C$ intersect at $F$. Suppose that $D E=15 \mathrm{~cm}, E F=9 \mathrm{~cm}$, and $F G=x \mathrm{~cm}$. Find the value of $x$. | 17. Ans: 16
Let $A B=y \mathrm{~cm}$. Since $\triangle D C F \sim \triangle G B F$, we have
$$
\frac{D C}{D F}=\frac{G B}{G F} \Longrightarrow \frac{y}{24}=\frac{G B}{x} \Longrightarrow G B=\frac{x y}{24} .
$$
Since $\triangle D C E \sim \triangle G A E$, we have
$$
\frac{D C}{D E}=\frac{G A}{G E} \Longrightarrow \fra... | e088e8c13a64aa7e | true | 3 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
105 | AI-MO/NuminaMath-1.5/olympiads | We consider the number of times the digit 2 appears in each position (hundreds, tens, units) for all room numbers.
The room numbers are three-digit numbers. The first digit is the floor number (1 to 5), and the last two digits are the room number on that floor (01 to 35).
First digit (hundreds digit): The first digit i... | Combinatorics | 25. The room numbers of a hotel are all three-digit numbers. The first digit represents the floor and the last two digits represent the room number. The hotel has rooms on five floors, numbered 1 to 5 . It has 35 rooms on each floor, numbered $\mathrm{n} 01$ to $\mathrm{n} 35$ where $\mathrm{n}$ is the number of the fl... | 25. E Each floor has 35 rooms. On every floor except floor 2 , the digit 2 will be used for rooms ' $n 02$ ', ' $n 12$ ', ' $n 20$ ' to ' $n 29$ ' (including ' $n 22$ ') and ' $n 32$ '. Hence the digit 2 will be used 14 times on each floor except floor 2 . On floor 2, the digit 2 will be used an extra 35 times as the f... | 266407231c0c5374 | true | 1.25 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
0, -2, 5 | AI-MO/NuminaMath-1.5/olympiads | We are given the polynomials $A=10 x^{2}-6 x y+3 y^{2}$, $B=a x^{2}+b x y+c y^{2}$, $C=15 y^{4}-36 x y^{3}+62 x^{2} y^{2}-20 x^{3} y$. We are given that $A \times B=C$. We need to find the values of $a, b, c$.
We multiply $A$ and $B$:
$A \times B = (10 x^{2}-6 x y+3 y^{2}) (a x^{2}+b x y+c y^{2})$
Expanding the product... | Commutative algebra | 49. Given the polynomials $A=10 x^{2}-6 x y+3 y^{2}, B=a x^{2}+b x y+c y^{2}, C=15 y^{4}-36 x y^{3}+62 x^{2} y^{2}-$ $20 x^{3} y$. If these three polynomials satisfy $A \times B=C$, then $a=$ $\qquad$ ,$b=$ $\qquad$ , $c=$ . $\qquad$ | Reference answer: $0,-2,5$ | 57efccf7f31e3596 | true | 3.25 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{289}{259} | AI-MO/NuminaMath-1.5/olympiads | Let $r_1, \ldots, r_{2007}$ be the roots of $f(x) = x^{2007} + 17x^{2006} + 1$.
We have $P(y) = c \prod_{j=1}^{2007} (y - (r_j + \frac{1}{r_j}))$.
Then $P(1) = c \prod_{j=1}^{2007} (1 - (r_j + \frac{1}{r_j})) = c \prod_{j=1}^{2007} \frac{r_j - r_j^2 - 1}{r_j} = \frac{(-1)^{2007} c}{\prod_{j=1}^{2007} r_j} \prod_{j=1}^{... | Functions of a complex variable | 10. [8] The polynomial $f(x)=x^{2007}+17 x^{2006}+1$ has distinct zeroes $r_{1}, \ldots, r_{2007}$. A polynomial $P$ of degree 2007 has the property that $P\left(r_{j}+\frac{1}{r_{j}}\right)=0$ for $j=1, \ldots, 2007$. Determine the value of $P(1) / P(-1)$. | Answer: $\frac{289}{259}$. For some constant $k$, we have
$$
P(z)=k \prod_{j=1}^{2007}\left(z-\left(r_{j}+\frac{1}{r_{j}}\right)\right) .
$$
Now writing $\omega^{3}=1$ with $\omega \neq 1$, we have $\omega^{2}+\omega=-1$. Then
$$
\begin{array}{c}
P(1) / P(-1)=\frac{k \prod_{j=1}^{2007}\left(1-\left(r_{j}+\frac{1}{r_{j... | 81e44bd10c578224 | true | 8.8 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
1 | AI-MO/NuminaMath-1.5/amc_aime | We are given that $P(z), Q(z), R(z)$ are polynomials with real coefficients, having degrees $2$, $3$, and $6$, respectively, and constant terms $1$, $2$, and $3$, respectively. We want to find the minimum number of distinct complex numbers $z$ that satisfy the equation $P(z) \cdot Q(z)=R(z)$.
Let $F(z) = P(z) \cdot Q(z... | Field theory and polynomials | Suppose that $P(z), Q(z)$, and $R(z)$ are polynomials with real coefficients, having degrees $2$, $3$, and $6$, respectively, and constant terms $1$, $2$, and $3$, respectively. Let $N$ be the number of distinct complex numbers $z$ that satisfy the equation $P(z) \cdot Q(z)=R(z)$. What is the minimum possible value of ... | The answer cannot be $0,$ as every nonconstant polynomial has at least $1$ distinct complex root (Fundamental Theorem of Algebra). Since $P(z) \cdot Q(z)$ has degree $2 + 3 = 5,$ we conclude that $R(z) - P(z)\cdot Q(z)$ has degree $6$ and is thus nonconstant.
It now suffices to illustrate an example for which $N = 1$: ... | fee6bcf53ae0f073 | true | 1.67 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
V_2 > V_1 | AI-MO/NuminaMath-1.5/olympiads | Final Answer: The cylinder in case b) has larger volume.
In case a), we cut along the segments \(A B\) and \(C D\). The total length of the boundary curves is \(2|A B| + 2|C D| = 2a + 2a = 4a\). If the boundary curves become circular lines of a right cylinder, then the circumference of each circular base is \(C_a = \f... | Geometry | \section*{Problem 3 - 301033}
Let \(F\) be the surface of a regular tetrahedron \(A B C D\). The midpoints of the segments \(A B\) and \(C D\) are \(M\) and \(N\), respectively.
Figure a)... | }
Each of the two resulting cylindrical surfaces has as the circumference of its base and top circle twice the length of one of the curves along which the tetrahedron face \(F\) is cut.
The height of the relevant cylinder is the length of the perpendicular dropped from a point on one of the cutting curves to a - in t... | a5da52365d403b9c | true | 4.25 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
55 | AI-MO/NuminaMath-1.5/olympiads | Final Answer: The final answer is $\boxed{55}$ | Geometry | 8. [6] $A B C D$ is a convex quadrilateral such that $A B<\underline{A D}$. The diagonal $\overline{A C}$ bisects $\angle B A D$, and $m \angle A B D=130^{\circ}$. Let $E$ be a point on the interior of $\overline{A D}$, and $m \angle B A D=40^{\circ}$. Given that $B C=$ $C D=D E$, determine $m \angle A C E$ in degrees. | Answer: $\mathbf{5 5}$. First, we check that $A B C D$ is cyclic. Reflect $B$ over $\overline{A C}$ to $B^{\prime}$ on $\overline{A D}$, and note that $B^{\prime} C=C D$. Therefore, $m \angle A D C=m \angle B^{\prime} D C=m \angle C B^{\prime} D=180^{\circ}-m \angle A B^{\prime} C=180^{\circ}-m \angle C B A$. Now $m \a... | 125499e33fb2d343 | true | 5.8 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$a=6, b=-3$ | AI-MO/NuminaMath-1.5/olympiads | ```text
9. (16 points) Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=a, a_{2}=b, 2 a_{n+2}=a_{n+1}+a_{n}$. If $\lim _{n \rightarrow \infty}\left(a_{1}+a_{2}+\right.$ $\left.\cdots+a_{n}\right)=4$, find the values of $a, b$.
Solution:
The recurrence relation is $2a_{n+2} - a_{n+1} - a_n = 0$. The characteristi... | Sequences, series, summability | 9. (16 points) Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=a, a_{2}=b, 2 a_{n+2}=a_{n+1}+a_{n}$. If $\lim _{n \rightarrow \infty}\left(a_{1}+a_{2}+\right.$ $\left.\cdots+a_{n}\right)=4$, find the values of $a, b$.
保留源文本的换行和格式，直接输出翻译结果。 | 9. From $2 a_{n+2}=a_{n+1}+a_{n}$, we can get
$$
a_{n+2}-a_{n+1}=-\frac{1}{2}\left(a_{n+1}-a_{n}\right).
$$
Thus, $\left\{a_{n+1}-a_{n}\right\}$ is a geometric sequence with the first term $a_{2}-a_{1}=b-a$ and the common ratio $-\frac{1}{2}$, so $a_{n+1}-a_{n}=\left(-\frac{1}{2}\right)^{n-1}(b-a)$.
Taking $n$ as $1,2... | 629ae874651f565e | true | 5 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
-61 | AI-MO/NuminaMath-1.5/olympiads | Let the roots of the equation $x^{4}+a x^{2}+b x+c=0$ be $1, 2, 3, r$.
By Vieta's formulas, the sum of the roots is $1+2+3+r = 0$, since the coefficient of $x^3$ is 0.
Thus, $6+r = 0$, so $r = -6$.
The roots are $1, 2, 3, -6$.
The coefficient $a$ is the sum of the roots taken two at a time.
$a = (1)(2) + (1)(3) + (1)(-... | Field theory and polynomials | $2 \cdot 109$ If $x^{4}+a x^{2}+b x+c=0$ has roots $1, 2, 3$, and another one, then $a+c$ equals
(A) 35.
(B) 24.
(C) -12.
(D) -61.
(E) -63.
(17th American High School Mathematics Examination, 1966) | [Solution] . Since the sum of the four roots of the equation is 0 (the opposite number of the coefficient of $x^{3}$), the fourth root of the equation is -6.
From $(x-1)(x-2)(x-3)(x+6)=x^{4}-25 x^{2}+60 x-36=0$, we know
$$
a+c=-25-36=-61 \text {. }
$$
Therefore, the answer is $(D)$. | 1f3e1f5e5014f235 | true | 4 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
x \equiv 0 \pmod{3}, x \ge 0 | AI-MO/NuminaMath-1.5/olympiads | Final Answer: The values of $x$ for which it is possible for all chameleons in the zoo to be of the same color at some point in time are those non-negative integers $x$ such that $x \not\equiv 2 \pmod{3}$. In other words, $x$ must be of the form $3k$ or $3k+1$ for some non-negative integer $k$.
We can write the answer... | Combinatorics | 11.5. In a zoo, there live chameleons of three colors: $x$ chameleons of yellow color, 2022 chameleons of white color, and 100 chameleons of pink color. Chameleons can meet each other, but only two at a time. If two chameleons of the same color meet, they do not change their color. If two chameleons of different colors... | ## Solution.
Assume that at a certain moment in the zoo, there are: $a$ brown chameleons, $b$ white chameleons, and $c$ pink chameleons, i.e., we have the triplet $(a, b, c)$. After the meeting of any two chameleons, this triplet changes to one of the following three:
a) $(a-1, b-1, c+2)$, or
b) $(a-1, b+2, c-1)$, o... | c52407a0b3874774 | true | 5 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
102 | AI-MO/NuminaMath-1.5/olympiads | We are given the recurrence relation $a_{1}=1$ and $a_{n}=10 a_{n-1}-1$ for $n \geq 2$. We found the closed form for $a_{n}$ to be $a_{n} = \frac{8 \cdot 10^{n-1} + 1}{9}$.
We want to find the minimum $n$ such that $a_{n} > 10^{100}$.
We need to solve the inequality $\frac{8 \cdot 10^{n-1} + 1}{9} > 10^{100}$.
Multiply... | Linear and multilinear algebra; matrix theory | 20. Let $a_{1}, a_{2}, \ldots$ be a sequence satisfying the condition that $a_{1}=1$ and $a_{n}=10 a_{n-1}-1$ for all $n \geq 2$. Find the minimum $n$ such that $a_{n}>10^{100}$. | 20. Ans: 102
Note that from $a_{n}=10 a_{n-1}-1$, we have
$$
a_{n}-\frac{1}{9}=10\left(a_{n-1}-\frac{1}{9}\right)
$$
for all $n \geq 2$. Thus,
$$
a_{n}-\frac{1}{9}=10^{n-1}\left(a_{1}-\frac{1}{9}\right)=10^{n-1} \frac{8}{9}
$$
for all $n \geq 1$. Therefore
$$
a_{n}=\frac{\left(1+8 \times 10^{n-1}\right)}{9} .
$$
Obse... | 7a6175eb3141b0ab | true | 8.75 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
3885 | AI-MO/NuminaMath-1.5/olympiads | We are given the equation $\overline{A B C}+\overline{B C A}+\overline{C A B}+\overline{A B}+\overline{B C}+\overline{C A}+A+B+C=861$.
This can be written as $113A + 123B + 124C = 861$, where A, B, C are digits from 1 to 9.
We tested integer values for A, B, C and found that for A=1, B=1, C=5, the sum is $113(1) + 123(... | Number theory | 7.A, B, C represent digits from 1 to 9, $\overline{A B C}+\overline{B C A}+\overline{C A B}+\overline{A B}+\overline{B C}+\overline{C A}+A+B+C=861$. What is the sum of all three-digit numbers $\overline{A B C}$ that satisfy this condition? $\qquad$ | 【Answer】 3885 | c7d56ab3373d7abd | true | 5.67 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{25+10 \sqrt{10}}{9} | AI-MO/NuminaMath-1.5/olympiads | In triangle $A B C$, we are given $B C=5$, $C K=3, K A=1$, so $A C = C K + K A = 4$. We are given $\cos(\angle A C B) = \frac{4}{5}$.
By the Law of Cosines in triangle $A B C$, $A B^2 = A C^2 + B C^2 - 2 \cdot A C \cdot B C \cos(\angle A C B) = 4^2 + 5^2 - 2 \cdot 4 \cdot 5 \cdot \frac{4}{5} = 16 + 25 - 32 = 9$. Thus $... | Geometry | In triangle $A B C$, side $B C$ is equal to 5. A circle passes through vertices $B$ and $C$ and intersects side $A C$ at point $K$, such that $C K=3, K A=1$. It is known that the cosine of angle $A C B$ is $\frac{4}{5}$. Find the ratio of the radius of the given circle to the radius of the circle inscribed in triangle ... | Apply the Law of Cosines.
## Solution
By the Law of Cosines in triangle $B K C$, we find that
$$
B K=\sqrt{C K^{2}+C B^{2}-2 C K \cdot C B \cos \angle K C B}=\sqrt{9+25-2 \cdot 3 \cdot 5 \cdot \frac{4}{5}}=\sqrt{10}.
$$
If $R$ is the radius of the circumcircle of triangle $B K C$, then
$$
R=\frac{B K}{2 \sin \angl... | 55df515559b412d1 | true | 4 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
4 \text{ g/cm}^3 | AI-MO/NuminaMath-1.5/olympiads | Let $V_1, V_2, V_3$ be the volumes of the three substances, and $\rho_1, \rho_2, \rho_3$ be their densities.
We are given that $V_1 + V_2 + V_3 = 2$ and $m_1 + m_2 + m_3 = \rho_1 V_1 + \rho_2 V_2 + \rho_3 V_3 = 16$.
We are given that $m_2 = 2m_1$, so $\rho_2 V_2 = 2 \rho_1 V_1$.
We are also given that the volume of $4 ... | General algebraic systems | 13.438 By mixing $2 \mathrm{~cm}^{3}$ of three substances, 16 g of the mixture was obtained. It is known that $4 \mathrm{r}$ of the second substance occupies a volume that is $0.5 \mathrm{~cm}^{3}$ larger than $4 \mathrm{r}$ of the third substance. Find the density of the third substance, given that the mass of the sec... | Solution. Let $x, y, z$ be the densities of three substances. Then, according to the condition $\left\{\begin{array}{l}2 x+2 y+2 z=16, \\ \frac{4}{y}-\frac{4}{z}=0.5, \quad \Rightarrow z=4 . \\ y=2 x\end{array}\right.$
Answer: 4 g $/ \mathrm{cm}^{3}$. | ba61f3688041877b | true | 3.67 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
150^{\circ} | AI-MO/NuminaMath-1.5/olympiads | Let $\angle AKB = x$. We assume $\angle AKB = 150^{\circ}$.
Let $\angle BAC = \angle BCA = 45^{\circ}$, $\angle ABC = 90^{\circ}$. Then $\triangle ABC$ is isosceles with $AB=BC$.
Let $AB=BC=CK=1$. Then $AC = \sqrt{AB^2+BC^2} = \sqrt{1^2+1^2} = \sqrt{2}$.
Given $\angle KAC = 30^{\circ}$. Then $\angle BAK = \angle BAC - ... | Linear and multilinear algebra; matrix theory | Bakayev E.v.
Inside an isosceles triangle $\$ \mathrm{ABC} \$$, a point $\$ K \$$ is marked such that $\$ C K=\mathrm{AB}=\mathrm{BC} \$$ and $\angle K A C=30^{\circ}$. Find the angle $\$ A K B \$$.
# | 
line $BC$
$ be the number of 3-element subsets of $\{1, \ldots, 23\}$ whose sum is $k$. We are given that $N(36) = 613$. We want to find the number of 3-element subsets with sum less than 36, which is $\sum_{k=6}^{35} N(k)$.
Consider the transformation that maps a subset $\{x, y, z\}$ to $\{24-x, 24-y, 24-z\}$. If $1 \l... | Combinatorics | 8. For any non-empty finite set $A$ of real numbers, let $s(A)$ be the sum of the elements in $A$. There are exactly 613 -element subsets $A$ of $\{1, \ldots, 23\}$ with $s(A)=36$. Find the number of 3 -element subsets of $\{1, \ldots, 23\}$ with $s(A)<36$. | 8. Ans: 855
The map $\{a, b, c\} \rightarrow\{24-a, 24-b, 24-c\}$ is a bijection from the set of 3-element subsets of $\{1, \ldots, 23\}$ with $s(A)36$. The number of 3 -element subsets of $\{1, \ldots, 23\}$ is $\binom{23}{3}=1771$. Therefore, the number of 3-element subsets of $\{1, \ldots, 23\}$ with $s(A)<36$ is
$... | 8df6ade51e18255b | true | 8 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
303 | AI-MO/NuminaMath-1.5/olympiads | Let $A = \underbrace{11 \cdots 1}_{100}$, $B = \underbrace{22 \cdots 2}_{50}$, $C = \underbrace{44 \cdots 4}_{25}$. We want to find the sum of digits of $S = A+B+C+2010$.
We write $A, B, C$ in terms of powers of 10:
$A = \sum_{i=0}^{99} 1 \cdot 10^i$
$B = \sum_{i=0}^{49} 2 \cdot 10^i$
$C = \sum_{i=0}^{24} 4 \cdot 10^i$... | Geometry | 5. $\underbrace{11 \cdots 1}_{100 \uparrow 1}+\underbrace{22 \cdots 2}_{50 \uparrow 2}+\underbrace{44 \cdots 4}_{25 \uparrow 4}+2010$ The sum of the digits of the calculation result is $\qquad$ | Answer: 303 | 8be935599e270039 | true | 4 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
90 | AI-MO/NuminaMath-1.5/olympiads | Let the interior angles of the isosceles triangle be $A, B, C$.
Since the triangle is isosceles, two of the angles are equal.
We are given that the ratio of the measures of two interior angles is 1:2.
We consider two cases for isosceles triangles:
Case 1: Two angles are equal and smaller than the third angle. Let the e... | Geometry | 9. (6 points) In isosceles $\triangle A B C$, the ratio of the measures of two interior angles is 1:2. Then, the largest angle in $\triangle A B C$ can be $\qquad$ degrees. | 【Solution】Solution: $180^{\circ} \times \frac{2}{1+1+2}$
$$
\begin{array}{l}
=180^{\circ} \times \frac{1}{2} \\
=90^{\circ}
\end{array}
$$
Answer: The maximum angle can be 90 degrees. Therefore, the answer is: 90. | 2f77add8569c5e40 | true | 4 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{\sqrt{2}}{2} | AI-MO/NuminaMath-1.5/olympiads | Let $M$ and $N$ be the midpoints of edges $AB$ and $CD$ respectively.
We find the coordinates of the midpoints $M$ and $N$ using the coordinates of the vertices we found:
$A = (0, 0, 0)$, $B = (1, 0, 0)$, $C = (\frac{1}{2}, \frac{\sqrt{3}}{2}, 0)$, $D = (\frac{1}{2}, \frac{\sqrt{3}}{6}, \frac{\sqrt{6}}{3})$.
Midpoint o... | Geometry | $23 \cdot 6$ In a regular tetrahedron with vertices $A, B, C, D$, each edge has length 1. Point $P$ is on edge $A B$, and point $Q$ is on edge $C D$. Find the shortest possible distance between points $P$ and $Q$.
(A) $\frac{1}{2}$.
(B) $\frac{3}{4}$.
(C) $\frac{\sqrt{2}}{2}$.
(D) $\frac{\sqrt{3}}{2}$.
(E) $\frac{\sqrt... | [Solution] Since the shortest distance from a point to a line is the perpendicular segment from that point to the line, the minimum value of the required distance can be obtained by symmetry.
That is, choose the midpoints of $AB$ and $CD$ as $P$ and $Q$,
Since $PC$ and $PD$ are the altitudes of the equilateral triangle... | edbd7aabd3623155 | true | 3 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
2 | AI-MO/NuminaMath-1.5/olympiads | The two-digit palindromic numbers are 11, 22, 33, 44, 55, 66, 77, 88, 99.
We want to find the smallest three-digit number $N$ that is the sum of three different two-digit palindromic numbers and $N$ is not a palindromic number.
We want to minimize $N$, so we choose the smallest two-digit palindromic numbers.
Let's cons... | Number theory | 13. If a number reads the same from left to right as it does from right to left, it is called a "palindromic number." For example, 12321 is a palindromic number. If $N$ is the smallest three-digit number that satisfies the following conditions: $N$ is equal to the sum of three different two-digit palindromic numbers an... | 13. A.
Notice that all two-digit palindromic numbers $11, 22, \cdots$, 99 are multiples of 11.
Since $N$ is the sum of two-digit palindromic numbers, $N$ is also a multiple of 11.
Among the multiples of 11, the smallest three-digit number is 110, and $110=77+22+11$, so $N=110$.
Therefore, the sum of the digits of $N... | 2cb6a700624bc334 | true | 1 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
84 | AI-MO/NuminaMath-1.5/number_theory | The problem requires us to find the minimum side length of a square base packaging box that can transport four different sizes of chess sets. Each chess set box has a square base with side lengths of 21 cm, 12 cm, 14 cm, and 10.5 cm. The packaging box must be able to completely cover the base of any of these chess set ... | Convex and discrete geometry | 19. Design a packaging box with a square base to transport four different sizes of chess sets, where the base of each chess box is also square, with side lengths of 21 cm, 12 cm, 14 cm, and 10.5 cm, respectively. To ensure that the packaging box can completely cover the base regardless of which size of chess set it is ... | 19. Solution: To ensure that each type of chess box can completely cover the bottom of the packaging box, the side length of the bottom of the packaging box should be a common multiple of the side lengths of the bottom of each chess box. Therefore, the smallest side length of the box bottom is the least common multiple... | 3fd1b7499767d93c | true | 1.5 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
70 | AI-MO/NuminaMath-1.5/olympiads | Final Answer:
Let $N$ be the maximum number of matches, which is 7. The first to win 4 matches is the winner.
We consider the number of matches played in each scenario. The tournament can end in 4, 5, 6, or 7 matches.
Case 1: Tournament ends in 4 matches.
Either Po wins all 4 matches (PPPP) or Golden Monkey wins all 4 ... | Combinatorics | 14. (5 points)
At the martial arts tournament, Panda Po and Golden Monkey are competing, with a maximum of seven matches. The first to win four matches is the winner. How many possible competition scenarios are there? $\qquad$ | $70$ | a7a9491ee6402595 | true | 2.6 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
5 | AI-MO/NuminaMath-1.5/olympiads | Let $M_n$ be the number of mice in month $n$ and $R_n$ be the number of rabbits in month $n$.
We are given that $M_n = 2^{n-1} M_1$ and $R_1 = R_2$, $R_n = R_{n-1} + R_{n-2}$ for $n \ge 3$.
Let $M_1 = m$ and $R_1 = r$. Then $M_n = 2^{n-1} m$.
$R_1 = r$
$R_2 = r$
$R_3 = R_2 + R_1 = 2r$
$R_4 = R_3 + R_2 = 3r$
$R_5 = R_4 ... | Commutative algebra | 10. (12 points) There are now two animals, mice and rabbits, which grow in the following ways: every month, the number of mice doubles from the previous month, and the number of rabbits becomes the sum of the numbers from the previous two months (the second month and the first month have the same number). For example: ... | 【Answer】Solution: According to the problem:
To minimize the number of rabbits, the number of mice is at least 1;
The number of mice doubles in the second month, quadruples in the third month, and so on, until it becomes 64 times in the seventh month, which is 64.
The number of rabbits is 1 in the first month, 1 in the ... | 15dfc67b7b156f62 | true | 3.4 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
3 | AI-MO/NuminaMath-1.5/olympiads | We are given two equations:
1) $x^{2}-t x+36=0$
2) $x^{2}-8 x+t=0$
Subtract equation (2) from equation (1):
$(x^{2}-t x+36) - (x^{2}-8 x+t) = 0 - 0$
$-t x+36 - (-8 x+t) = 0$
$-t x+36 + 8 x-t = 0$
$(8-t)x + (36-t) = 0$
$(8-t)x = t-36$
If $8-t \neq 0$, then $x = \frac{t-36}{8-t}$.
Substitute this expression for $x$ int... | Geometry | Let $t$ be TNYWR.
There is exactly one real number $x$ with the property that both $x^{2}-t x+36=0$ and $x^{2}-8 x+t=0$.
What is the value of $x$ ?
## The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 2022 Canadian Team Mathematics Contest | Suppose $x$ is the real number such that $x^{2}-t x+36=0$ and $x^{2}-8 x+t=0$.
Therefore, we get that $x^{2}-t x+36=x^{2}-8 x+t$ which can be rearranged to get $36-t=$ $t x-8 x=(t-8) x$.
Dividing by $t-8$ gives $x=\frac{36-t}{t-8}$.
Substituting $t=15$ gives $x=\frac{36-15}{15-8}=\frac{21}{7}=3$.
ANSWER: $(4,15,3)$ | 721a32acb591b2d3 | true | 2 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
35 | AI-MO/NuminaMath-1.5/olympiads | We are given that $\overline{\mathrm{ab}}$ is a two-digit number and $\overline{\mathrm{cde}}$ is a three-digit number, and $\overline{\mathrm{ab}}+\overline{\mathrm{cde}}=1079$. We need to find the value of $a+b+c+d+e$.
We can write the numbers in expanded form based on place values:
$\overline{\mathrm{ab}} = 10a + b$... | Geometry | 14. (6 points) $\overline{\mathrm{ab}}, \overline{\mathrm{cde}}$ represent a two-digit number and a three-digit number respectively. If $\overline{\mathrm{ab}}+\overline{\mathrm{cde}}=1079$, then $a+b+c+d+e=$
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
The above text has been translated into English, prese... | 【Solution】Solution: According to the problem, the sum of a two-digit number and a three-digit number is 1079
$$
\begin{array}{l}
\therefore c=9, a+d=17, b+e=9, \\
\therefore a+b+c+d+e=35 .
\end{array}
$$
Therefore, the answer is: 35. | 58869b8f7dac11e9 | true | 2 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
f(x) = x \lg x | AI-MO/NuminaMath-1.5/olympiads | Final Answer: The function that satisfies the conditions is $f(x) = x \lg x$.
We set $g(x) = \frac{f(x)}{x}$. Then the inequality becomes
$$
\lg (x y) \leq g(x) + g(y) \leq g(x y) \text{, for all } x, y>0 \text{. }
$$
Setting $x=y=1$, we get $\lg(1) \leq g(1) + g(1) \leq g(1)$, so $0 \leq 2g(1) \leq g(1)$, which impli... | Difference and functional equations | Problem 3. Determine the functions that satisfy the conditions
$$
x y \lg (x y) \leq y f(x)+x f(y) \leq f(x y) \text{, for all } x, y>0 \text{. }
$$ | Solution. For $x=y=1$, from the statement we get $0 \leq 2 f(1) \leq f(1)$, hence $f(1)=0$.
For $y=1$, the relation from the statement leads to $x \lg x \leq f(x)$, for any $x>0$.
For $y=\frac{1}{x}$, from the statement we obtain $0 \leq \frac{f(x)}{x}+x f\left(\frac{1}{x}\right) \leq 0$, so $f\left(\frac{1}{x}\right... | 237b695d074b2056 | true | 4 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{1}{2}(m-n) | AI-MO/NuminaMath-1.5/olympiads | Given $\lg (1+\sin A)=m$ and $\lg \frac{1}{1-\sin A}=n$.
From the first equation, $1+\sin A = 10^m$.
From the second equation, $\frac{1}{1-\sin A} = 10^n$.
This implies $1-\sin A = \frac{1}{10^n} = 10^{-n}$.
We want to find $\lg \cos A$. We know that $\cos^2 A = 1 - \sin^2 A = (1+\sin A)(1-\sin A)$.
Substituting the ex... | General algebraic systems | 6.44 $A$ is an acute angle, $\lg (1+\sin A)=m, \lg \frac{1}{1-\sin A}=n$, then the value of $\lg \cos A$ is
(A) $m+\frac{1}{n}$.
(B) $m-n$.
(C) $\frac{1}{2}\left(m+\frac{1}{n}\right)$.
(D) $\frac{1}{2}(m-n)$.
(E) $2(m-n)$.
(Beijing Junior High School Mathematics Competition, 1983) | [Solution] Given $\lg (1+\sin A)=m$, and
$$
\begin{array}{rlrl}
& \lg \frac{1}{1-\sin A}= & \lg (1-\sin A)^{-1}=-\lg (1-\sin A), \\
\therefore \quad & \lg (1-\sin A)=-n . \\
\therefore \quad & \lg \cos A & =\frac{1}{2} \lg \cos ^{2} A=\frac{1}{2} \lg \left(1-\sin ^{2} A\right) \\
& =\frac{1}{2} \lg (1+\sin A)(1-\sin A... | 3bdd6d2959d9b172 | true | 1.25 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
66 | AI-MO/NuminaMath-1.5/olympiads | To determine the smallest number of tests required to match twelve keys to twelve padlocks, we consider a strategy where we systematically find the key for each padlock.
For the first padlock, we pick a padlock (say padlock 1) and try keys one by one until we find the key that fits it. In the worst-case scenario, we t... | Combinatorics | Task 2 - 260712 In the material issuance of a company, the keys of twelve padlocks have been mixed up due to a mishap.
Since only one of the twelve keys fits each padlock and only one padlock fits each key, which cannot be distinguished from each other externally, it must be determined which key belongs to which padlo... | From the total of twelve padlocks, exactly one fits each key. If we take one of the twelve keys and try to find out which of the twelve padlocks it fits, in the worst case, it could happen that after eleven attempts, the lock that the key fits has not yet been tried.
Since the key does belong to one of the locks, it m... | 3ffa0720ad8f36a1 | true | 3.67 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
5 | AI-MO/NuminaMath-1.5/olympiads | We are given the four-digit number $(1001)_{z}$ and the two-digit number $(41)_{z}$. We want to find the bases $z$ such that $(1001)_{z}$ is divisible by $(41)_{z}$.
First, convert the numbers to base 10:
$(1001)_{z} = 1 \cdot z^3 + 0 \cdot z^2 + 0 \cdot z^1 + 1 \cdot z^0 = z^3 + 1$
$(41)_{z} = 4 \cdot z^1 + 1 \cdot z^... | Number theory | 1. Find the bases $z$ of all number systems in which the four-digit number $(1001)_{z}$ is divisible by the two-digit number $(41)_{z}$. | 1. Since the digit 4 appears in the notation of a two-digit number, it necessarily follows that $z \geq 5$. From the expanded notations $(1001)_z = z^3 + 1$ and $(41)_z = 4z + 1$, we are looking for exactly those natural numbers $z \geq 5$ for which the number $z^3 + 1$ is a multiple of the number $4z + 1$. Using the E... | 7db003cbfaa7e1d4 | true | 5 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\sqrt[5]{2}-1 | AI-MO/NuminaMath-1.5/olympiads | Final Answer: The final answer is $\boxed{\sqrt[5]{2}-1}$ | General algebraic systems | 123. Rationalize the denominator of the fraction:
$$
\frac{1}{\sqrt[5]{16}+\sqrt[5]{8}+\sqrt[5]{4}+\sqrt[5]{2}+1}
$$ | $\triangle$ Let's use the identity
$$
a^{5}-b^{5}=(a-b)\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right)
$$
(see § 1, item 1.1). Let $a=\sqrt[5]{2}, b=1$; then $a^{5}-b^{5}=2-1=1$. Therefore, the conjugate factor for the denominator of the given fraction is $\sqrt[5]{2}-1$.
Answer: $\sqrt[5]{2}-1$. | ee0b948ddbb1aa74 | true | 3 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
88 | AI-MO/NuminaMath-1.5/olympiads | The numbers that can be expressed in the form $4s + 9r$ for non-negative integers $s$ and $r$ are called Fortune Numbers. We want to find the number of Fortune Numbers among the numbers from $1 \sim 100$.
First, we find the numbers that cannot be expressed in the form $4s + 9r$ for non-negative integers $s$ and $r$. Th... | Number theory | 13. In a restaurant, there are two types of tables: square tables can seat 4 people, and round tables can seat 9 people. If the number of diners can exactly fill a certain number of tables, the restaurant manager calls this number a "Fortune Number." Among the numbers from $1 \sim 100$, there are $\qquad$ "Fortune Numb... | 【Analysis】 88
Classified by the remainder when divided by 9:
$9 a$ type (remainder 0 when divided by 9): 11
$9 a+4$ type (remainder 4 when divided by 9): 11
$9 a+8$ type (remainder 8 when divided by 9): 11
9 - 12 type (remainder 3 when divided by 9): 10
$9 a+16$ type (remainder 7 when divided by 9): 10
$9 a+20$ type (r... | a6a48f32607ca468 | true | 2 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
2 | AI-MO/NuminaMath-1.5/olympiads | Let's assume Luca is a liar, Maria is a liar, Nicola is honest, and Paola is honest.
Luca says: "every girl is always honest". This means Luca claims Maria is honest and Paola is honest. Since Maria is a liar and Paola is honest, Luca's claim "every girl is always honest" is false. This is consistent with Luca being a ... | Mathematical logic and foundations | 4. At a table, there are four people: Luca, Maria, Nicola, and Paola. Each of the four always lies or never lies. Moreover, they do not like to talk about themselves, but rather about their friends; so when asked who among them always lies, their answers are:
Luca: "every girl is always honest"
Maria: "every boy is a... | 4. The answer is $\mathbf{( C )}$. Luca certainly cannot be telling the truth, since the two girls contradict each other, so they cannot both be telling the truth. If Maria were telling the truth, Nicola would be lying; therefore, it is not true that there is a truthful girl, and in particular, Maria is not, which is a... | 4c8adc649bfa936b | true | 3 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
90 | AI-MO/NuminaMath-1.5/olympiads | Final Answer: The number of pairs $(m, n)$ such that the first player has a winning strategy is when the starting position $(1, 1)$ is a winning position. Based on the Nim game analysis, the starting position is a losing position if and only if $(n-1) \oplus (m-1) = 0$, which is equivalent to $n-1 = m-1$, or $n = m$. T... | Combinatorics | 10. (10 points) A rectangular chessboard composed of unit squares of size $m \times n$ (where $m, n$ are positive integers not exceeding 10), has a chess piece placed in the unit square at the bottom-left corner. Two players, A and B, take turns moving the piece. The rules are: move up any number of squares, or move ri... | 【Analysis】When $M=N$, the first player A must lose; when $M$ is not equal to $N$, A can use a unique strategy to return to the starting point of the board, making B go first, and B will lose. There are 10 cases where A will lose, and $100-10=90$ cases where A will win.
【Solution】Assume A goes first, and B goes second,... | 1f98b668b80979de | true | 2.2 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
12 | AI-MO/NuminaMath-1.5/amc_aime | Let the quadrilateral have side lengths $a, b, c, d$. We are given that the perimeter is $26$ and one side has length $4$. Let $a=4$. Then $a+b+c+d = 26$, so $4+b+c+d = 26$, which means $b+c+d = 22$.
For a quadrilateral to exist, the sum of any three sides must be greater than the fourth side. We want to find the great... | Combinatorics | A quadrilateral has all integer sides lengths, a perimeter of $26$, and one side of length $4$. What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$ | Let's use the triangle inequality. We know that for a triangle, the sum of the 2 shorter sides must always be longer than the longest side. This is because if the longest side were to be as long as the sum of the other sides, or longer, we would only have a line.
Similarly, for a convex quadrilateral, the sum of the sh... | 37e8922427d48b69 | true | 3 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
$|k|<|b|$ | AI-MO/NuminaMath-1.5/olympiads | Final Answer: We are given the equation of the line $y=kx+b$. From the graph, we observe that the y-intercept is positive, so $b > 0$. The x-intercept is positive. The slope is negative, so $k < 0$.
The y-intercept is $b$. The x-intercept is found by setting $y=0$, so $kx+b=0$, which gives $x = -b/k$.
From the graph, t... | Real functions | Folklore
The graph of the function $y=kx+b$ is shown in the figure. Compare $|k|$ and $|b|$.
 | First method. The graph intersects the y-axis at the point $(0, b)$. Therefore, $b>0$, which means $|b|=b$. When $x=1$, $y=k+b$, and from the graph, it is clear that $0-k=|k|$.
Second method. Let $(a, 0)$ be the point where the graph intersects the x-axis. From the graph, it is clear that $a>1$. Therefore, $|k|=|b|:|a... | f580a748e520ba19 | true | 1 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
1400 \mathrm{~m} | AI-MO/NuminaMath-1.5/olympiads | The graph shows Spencer's distance from home at different times. To find the total distance he walked, we need to sum the distances he walked in each segment of his journey. We can find the distance walked in each segment by looking at the change in distance from home in each segment.
Segment 1: From time 0 to 5 minut... | Mathematical logic and foundations | Spencer was walking home from school when he realized he had forgotten his homework. He walked back to the school, picked up his homework and then walked home. The graph shows his distance from home at different times. In total, how far did he walk?
(A) $2800 \mathrm{~m}$
(B) $1000 \mathrm{~m}$
(C) $800 \mathrm{~m}$
(D... | Spencer starts $1000 \mathrm{~m}$ from home and walks to a point $800 \mathrm{~m}$ from home, a distance of $200 \mathrm{~m}$. He then walks to a point $1000 \mathrm{~m}$ from home, for a distance of another $200 \mathrm{~m}$. Finally, he walks home, a distance of $1000 \mathrm{~m}$. So Spencer has walked a total of $2... | 4b70ab07a7dbab8e | true | 3.4 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
20486 | AI-MO/NuminaMath-1.5/olympiads | Final Answer:
Polycarp formed the smallest 5-digit number consisting of different even digits. The even digits are 0, 2, 4, 6, 8. To form the smallest 5-digit number, we arrange these digits in ascending order, starting with the smallest non-zero digit in the ten-thousands place. So, Polycarp's number is 20468.
Kolya ... | Mathematical logic and foundations | Excellent student Polycarp and poor student Kolya were forming the smallest 5-digit number consisting of different even digits. Polycarp formed his number correctly, while Kolya made a mistake. However, it turned out that the difference between Kolya's number and the correct answer was less than 100. What numbers did P... | Think about what the first two digits of Polycarp's number and the last two digits of Kolya's number should be.
## Solution
If we proceed as in the previous problem, Polycarp should have formed the number 02468, but the first digit cannot be zero, so Polycarp formed the number 20468. Let's try to find Kolya's number.... | a579c714b9c893a4 | true | 1.5 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{4}{3} | AI-MO/NuminaMath-1.5/olympiads | The equation of circle $C$ is $x^{2}+y^{2}-8 x+15=0$, which can be rewritten as $(x-4)^{2}+y^{2}=1$.
The center of circle $C$ is $O=(4,0)$ and the radius is $R=1$.
Let $P=(x, y)$ be a point on the line $y=k x-2$. We want to find the condition on $k$ such that there exists a point $P$ on the line such that the distance ... | Geometry | 8. Given the equation of circle $C$ as $x^{2}+y^{2}-8 x+15=0$, if there exists at least one point on the line $y=k x-2(k \in \mathbf{R})$ such that a circle with this point as its center and 1 as its radius has a common point with circle $C$, then the maximum value of $k$ is $\qquad$. | The answer is $\frac{4}{3}$.
Analysis: Since the equation of circle $C$ can be transformed into $(x-4)^{2}+y^{2}=1$, the center of circle $C$ is $(4,0)$, and the radius is 1.
If there exists at least one point $A\left(x_{0}, k x_{0}-2\right)$ on the line $y=k x-2$, such that the circle with this point as the center and... | 5d669bea0c65e9ce | true | 2 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
\frac{23}{30} | AI-MO/NuminaMath-1.5/olympiads | To compare the fractions $\frac{2}{3}, \frac{23}{30}, \frac{9}{10}, \frac{11}{15}$, and $\frac{4}{5}$, we find a common denominator. The least common multiple of the denominators 3, 30, 10, 15, and 5 is 30. We rewrite each fraction with a denominator of 30:
$\frac{2}{3} = \frac{2 \times 10}{3 \times 10} = \frac{20}{30}... | Number theory | If $\frac{2}{3}, \frac{23}{30}, \frac{9}{10}, \frac{11}{15}$, and $\frac{4}{5}$ are written from smallest to largest then the middle fraction will be
(A) $\frac{23}{30}$
(B) $\frac{4}{5}$
(C) $\frac{2}{3}$
(D) $\frac{9}{10}$
(E) $\frac{11}{15}$ | If $\frac{2}{3}, \frac{23}{30}, \frac{9}{10}, \frac{11}{15}$, and $\frac{4}{5}$ are written from smallest to largest then the middle fraction will be
(A) $\frac{23}{30}$
(B) $\frac{4}{5}$
(C) $\frac{2}{3}$
(D) $\frac{9}{10}$
(E) $\frac{11}{15}$
## Solution
In order to compare the five fractions, we write them with th... | 9b506cf66cd4bc0b | true | 1 | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
9 | AI-MO/NuminaMath-1.5/olympiads | To find the times when the hour and minute hands of a clock overlap, we need to understand the relative speeds of the hour and minute hands.
* **Minute hand:** Completes a full 360-degree rotation in 60 minutes, or 6 degrees per minute.
* **Hour hand:** Completes a full 360-degree rotation in 12 hours (720 minutes), o... | Ordinary differential equations | 5. A company's working hours are from 8:30 AM to 5:30 PM. During this period, the hour and minute hands of the clock overlap times. | 【Analysis】Method one: At 8:30, the hour hand is between 8 and 9, and the minute hand points to 6. At 9:00, the hour hand points to 9, and the minute hand points to 12, the minute hand has surpassed the hour hand. Therefore, between 8:30 and 9:00, the minute hand and the hour hand overlap once.
Similarly, between 9:00 a... | 20a2d54840c47ac5 | true | 4 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
(1,2] | AI-MO/NuminaMath-1.5/olympiads | Let $\angle AOC = \theta$ and $\angle BOC = \phi$. Since $C$ is on the minor arc $\overparen{AB}$, we have $\theta + \phi = \angle AOB = 120^{\circ}$, and $0 < \theta < 120^{\circ}$ and $0 < \phi < 120^{\circ}$.
We have derived $\lambda + \mu = 2(\cos \theta + \cos \phi)$. Since $\phi = 120^{\circ} - \theta$, we have
$... | Linear and multilinear algebra; matrix theory | 10. Let $A$, $B$, and $C$ be three distinct points on $\odot O$, and $\angle A O B=120^{\circ}$, point $C$ lies on the minor arc $\overparen{A B}$ (point $C$ does not coincide with $A$ or $B$). If $\overrightarrow{O C}=\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}(\lambda, \mu \in \mathbf{R})$, then the range o... | 10. $(1,2]$.
Connect $O C$, intersecting $A B$ at point $D$. Let $\overrightarrow{O D}=m \overrightarrow{O C}$. Then, by the problem, $\overrightarrow{O D}=m \lambda \overrightarrow{O A}+m \mu \overrightarrow{O B}$, and $m \lambda+m \mu=1 \Rightarrow \lambda+\mu=\frac{1}{m}$.
Assume the radius of the circle is 1, and... | 174bd404adee7eee | true | 9.2 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
842 | AI-MO/NuminaMath-1.5/olympiads | Final Answer: The final answer is $\boxed{842}$ | Mathematical logic and foundations | 253. The digits of a three-digit number form a geometric progression with different terms. If this number is decreased by 200, the result is a three-digit number whose digits form an arithmetic progression. Find the original three-digit number. | $\triangle$ The number of three-digit numbers with distinct digits forming a geometric progression is small, and all of them can be easily found by enumeration:
$124,421,139,931,248,842,469,964$.
We will discard the numbers 124 and 139, as they are less than 200. Subtract 200 from the remaining numbers and find the n... | eb7cde275c63c8bf | true | 3.6 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
2 | AI-MO/NuminaMath-1.5/olympiads | We are given that $x, y, z$ are positive numbers and $xyz(x+y+z)=1$. We want to find the minimum value of the expression $(x+y)(x+z)$.
We expand the expression $(x+y)(x+z) = x^2 + xz + xy + yz = x(x+y+z) + yz$.
From the given condition $xyz(x+y+z)=1$, we can write $x(x+y+z) = \frac{1}{yz}$.
Substituting this into the e... | Field theory and polynomials | 4.2. 12 * Given that $x, y, z$ are positive numbers, and satisfy $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(x+z)$.
保留源文本的换行和格式，翻译结果如下：
4.2. 12 * Given that $x, y, z$ are positive numbers, and satisfy $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(x+z)$. | From the arithmetic mean being greater than or equal to the geometric mean, we get
$$
(x+y)(x+z)=y z+x(x+y+z) \geqslant 2 \sqrt{y z \cdot x(x+y+z)}-2,
$$
and when $x=\sqrt{2}-1, y=z=1$, the above inequality holds with equality. Therefore, the minimum value of the expression $(x+y)(x+z)$ is 2. | 4fd451e85a0e049e | true | 4 | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} |
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