solution stringlengths 2 12.9k | answer stringlengths 1 253 | question stringlengths 2 6.18k | math_type stringclasses 9
values | source_type stringclasses 34
values | metadata stringclasses 284
values | domain stringclasses 97
values | thinking_trajectories stringlengths 2 186k | attempt null |
|---|---|---|---|---|---|---|---|---|
Answer: 1.4 times closer to the dog than to the cat.
## Solution:
Let $v$ be the running speed of the dog, $u$ be the eating speed of the cat, and the volume of sausages eaten by each animal be 1.
Then, $2 v$ is the running speed of the cat, and $2 u$ is the eating speed of the dog.
Let the distance from the cat to... | 5x=7y | 4. A kilo of sausages was placed on a straight line between a dog in a kennel and a cat. The animals simultaneously rushed to the sausages. The cat runs twice as fast as the dog, but eats twice as slowly. Upon reaching the sausages, both ate without fighting and ate an equal amount. It is known that the cat could eat a... | Logic and Puzzles | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematics education | Let $v_d$ be the speed of the dog and $v_c$ be the speed of the cat.
Let $e_d$ be the eating rate of the dog and $e_c$ be the eating rate of the cat.
Let $D$ be the position of the dog's kennel, $C$ be the position of the cat, and $S$ be the position of the sausages.
Let $x$ be the distance between the dog and the saus... | null |
【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 ... | 105:44 | 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... | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematical logic and foundations | 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 ... | null |
# Answer: 400
First solution. We will prove that no more than 400 rooks can be placed on the board. In each row or column, there are no more than two rooks; otherwise, the rook that is not at the edge will attack at least two other rooks. Suppose there are $k$ columns with two rooks each. Consider one such pair. They ... | 400 | 5. What is the maximum number of rooks that can be placed on the cells of a $300 \times 300$ board so that each rook attacks no more than one other rook? (A rook attacks all cells it can reach according to chess rules, without passing through other pieces.)
# | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | Final Answer: The final answer is $\boxed{600}$ | null |
$$
\begin{array}{l}
\sqrt{x}-\sqrt{24 \gamma}=4 \sqrt{2} \\
\sqrt{x}=2 \sqrt{6 \gamma}+4 \sqrt{2}
\end{array}
$$
The smallest positive integer $\gamma=3$ | 3 | I2.3 Determine the smallest positive integer $\gamma$ such that the equation $\sqrt{x}-\sqrt{\beta \gamma}=4 \sqrt{2}$ has an integer solution in $x$. | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | We are given the equation $\sqrt{x}-\sqrt{\beta \gamma}=4 \sqrt{2}$, and we want to find the smallest positive integer $\gamma$ such that there exists an integer solution in $x$ for some positive integer $\beta$.
Rearranging the equation, we have $\sqrt{x} = \sqrt{\beta \gamma} + 4 \sqrt{2}$.
Squaring both sides, we ge... | null |
【Answer】D
【Analysis】Considering the four cases respectively, when the gum costs 8 yuan, one person pays 10 yuan (2 five-yuan notes), and another person pays 13 yuan (1 ten-yuan note, 3 one-yuan notes), the first person's 1 five-yuan note can be given as change to the second person, and the second person's 2 one-yuan no... | D | 6. In the Kingdom of Mathematics, the denominations of banknotes in circulation are 1 yuan, 5 yuan, 10 yuan, 20 yuan, 50 yuan, and 100 yuan. One day, two customers each bought a chocolate bar worth 15 yuan at the same grocery store. One of them paid with two 10-yuan banknotes, while the other paid with a 20-yuan and a ... | Logic and Puzzles | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Linear and multilinear algebra; matrix theory | Rationale:
Let's check option (B) 6 yuan.
Price of chewing gum = 6 yuan.
Customer 1 pays 7 yuan (5 yuan + 1 yuan + 1 yuan). Change needed for Customer 1 = 7 - 6 = 1 yuan.
Customer 2 pays 11 yuan (10 yuan + 1 yuan). Change needed for Customer 2 = 11 - 6 = 5 yuan.
Let the banknote given from Customer 1's payment to Custo... | null |
## Solution.
For an index $i$, we say it is good if it satisfies the equality in the problem statement.
Assume there exist two consecutive good indices $i, i+1$. Then we have
$$
\begin{aligned}
a_{i} a_{i+3} & =a_{i} a_{i+1}+a_{i+1} a_{i+2}+a_{i+2} a_{i+3} \\
a_{i+1} a_{i+4} & =a_{i+1} a_{i+2}+a_{i+2} a_{i+3}+a_{i+3... | 999 | ## Task 1.
If $a_{1}, a_{2}, \ldots, a_{2000}$ is a sequence of 2000 positive real numbers, for how many indices $i \in$ $\{1,2, \ldots, 2000\}$ can the equality
$$
a_{i} a_{i+3}=a_{i} a_{i+1}+a_{i+1} a_{i+2}+a_{i+2} a_{i+3} ?
$$
hold? We consider that $a_{j+2000}=a_{j}$ for $j \in\{1,2,3\}$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | 11 | Final Answer: The final answer is $\boxed{1000}$ | null |
From two inclined lines drawn from one point, the one with the larger projection is larger. Therefore, from the inequalities given in the problem, it follows that $O A \geq O B \geq O C \geq O D \geq O A$. Hence, $O A=O B=O C=O D$, which means $O$ is the center of the circle circumscribed around quadrilateral $A B C D$... | 0.5 | Given a quadrilateral $A B C D$ with an area of 1. From its internal point O, perpendiculars OK, OL, OM, and ON are dropped to the sides $A B, B C, C D$, and $D A$ respectively. It is known that $A K \geq K B, B L \geq L C, C M \geq M D$, and $D N \geq N A$. Find the area of the quadrilateral $K L M N$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | 03 | Final Answer: The final answer is $\boxed{1/2}$ | null |
Answer: 1343.
Solution: Note that
$$
\sum_{i=1}^{n} a_{i} \geqslant \sum_{i=1}^{n} i=\frac{n(n+1)}{2} .
$$
A similar inequality is written for the sums $b_{i}$ and $c_{i}$. Adding the three obtained inequalities, we get
$$
\begin{gathered}
3 \cdot \frac{n(n+1)}{2} \leqslant \sum_{i=1}^{n} a_{i}+\sum_{i=1}^{n} b_{i}... | 1343 | 4. Triples of natural numbers $\left(a_{i}, b_{i}, c_{i}\right)$, where $i=1,2, \ldots, n$ satisfy the following conditions:
1) $a_{i}+b_{i}+c_{i}=2017$ for all $i=1,2, \ldots, n$;
2) if $i \neq j$, then $a_{i} \neq a_{j}, b_{i} \neq b_{j}$ and $c_{i} \neq c_{j}$. What is the maximum possible value of $n$? (M. Popov) | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | Let us consider the construction for $n=672$:
$a_i = i$ for $i = 1, 2, \ldots, 672$.
$b_i = 672 + i$ for $i = 1, 2, \ldots, 672$.
$c_i = 2017 - a_i - b_i = 2017 - i - (672 + i) = 1345 - 2i$ for $i = 1, 2, \ldots, 672$.
For $i \neq j$, $a_i = i \neq j = a_j$, $b_i = 672 + i \neq 672 + j = b_j$, $c_i = 1345 - 2i \neq 134... | null |
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... | \frac{7}{8} | 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 $S_{2}$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | 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 $DE$.
$\frac{S_{ACE}}{S_{CDE}} = \frac{AE}{DE} = \frac{AE}{2AE} = \frac{1}{2}$.... | null |
The number 22 does not appear on any round. On the other hand, the numbers 1 through 21 appear as follows. \begin{tabular}{c|c|c} Number & Round & Problem \\ \hline 1 & Guts & 21 \\ 2 & Guts & 13 \\ 3 & Guts & 17 \\ 4 & Guts & 13 \\ 5 & Guts & 14 \\ 6 & Guts & 2 \\ 7 & Guts & 10 \\ 8 & Guts & 13 \\ 9 & Guts & 28 \\ 10 ... | 22 | Compute the smallest positive integer that does not appear in any problem statement on any round at HMMT November 2023. | KbsdJames/Omni-MATH/HMMT_11 | {'domain': ['Mathematics -> Discrete Mathematics -> Combinatorics'], 'difficulty': 3.0} | Number theory | We go through each problem statement in HMMT November 2023 and list all positive integers that appear in the problem statements.
Combinatorics Round:
1. 3, 2, 5, 4
2. 1, 100, 101
3. 3, 2, 4
4. 5, 8, 7, 3
5. 1, 500, 3, 5, 7
6. 2
7. 10, 5, 7
8. 1, 2, 3, 4, 5, 6, 2
9. 1, 2, 10
10. 52, 2, 3, 4, 5, 6, 7, 8, 9, 10
Geometry... | null | |
$(x-2)^2$ = 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
$4+5+6+0+(-1)+(-2)=\boxed{\textbf{(B)} 12}$ | 12 | What is the sum of all integer solutions to $1<(x-2)^2<25$?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 25$ | Inequalities | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | 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... | null |
$$
a^{2}+4 b^{2}+9 c^{2}=4 b+12 c-2 \Rightarrow a^{2}+(2 b-1)^{2}+(3 c-2)^{2}=3,
$$
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(1+1+1)\left[a^{2}+(2 b-1)^{2}+(3 c-2)^{2}\right] \geqslant(a+2 b+3 c-3)^{2} \\
\Rightarrow(a+2 b+3 c-3)^{2} \leqslant 9 \Rightarrow 0<a+2 b+3 c \leqslant 6 .
\end{array}
$$... | 6 | 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 $\frac{1}{a}+\frac{2}{b}+\frac{3}{c}$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Real functions | 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... | null |
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. | 66 | 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 $\{A_1,A_2,\dots,A_9\}$?
$\text{(A) }30 \qquad \text{(B) }36 \qquad \text{(C) }63 \qquad \text{(D) }66 \qquad \text{(E) }72$ | Combinatorics | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | 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_... | null |
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... | 36 | [ [ 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 $B$.
# | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | 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... | null |
Geometry, cleverly finding area, cutting and supplementing.
(Method 1)
Take the midpoint $E$ of $B C$, connect $D E$; connect $B D$;
$S_{\triangle A B D}=A B \times B E \div 2=28 \times 14 \div 2=196$ square centimeters;
$S_{\text {sector } B E D}=\frac{1}{4} \times \pi \times B E^{2}=\frac{1}{4} \times \pi \times 14^{... | 252 | As shown in the figure, $\triangle A B C$ is an isosceles right triangle, $A B=28 \mathrm{~cm}$. A semicircle is drawn with $B C$ as the diameter, and point $D$ is the midpoint of the semicircle arc. Try to find the area of the shaded part. (Take $\pi=\frac{22}{7}$.) | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | 05 | Given that $\triangle ABC$ is an isosceles right triangle and $AB=28 \mathrm{~cm}$. From the figure, it is clear that $\angle BAC = 90^\circ$. Thus, $AB = AC = 28 \mathrm{~cm}$.
By Pythagorean theorem, $BC^2 = AB^2 + AC^2 = 28^2 + 28^2 = 2 \times 28^2$. So $BC = 28\sqrt{2} \mathrm{~cm}$.
A semicircle is drawn with $BC$... | null |
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... | \frac{5}{7} | 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 ... | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | 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... | null |
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(\... | (2,4,4),(2,3,6),(3,3,3) | 3-ча 1. The sum of the reciprocals of three positive integers is equal to 1. What are these numbers? Find all solutions. | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | 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... | null |
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
$\left(\frac{228}{x}-0.9\right)(2 x+5)=408$,
... | 85 | 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... | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | General algebraic systems | 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... | null |
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 ... | 0.6 | 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. | Other | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Probability theory and stochastic processes | 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... | null |
Answer: 20.
Solution. Players No. 5 and No. 9 together scored $72+84=156=12 \cdot 13$ points. In one spin, they can together score no more than 12 points. Therefore, in each of the 13 spins, they together scored 12 points. Note that the 12 points they score can be one of the sums $8+4, 7+5$, $6+6, 5+7$ or $4+8$, when ... | 20 | 13.1. [7-8.7 (20 points), 9.8 (15 points), 10.8 (20 points)] There is a rotating round table with 16 sectors, on which numbers $0,1,2, \ldots, 7,8,7,6, \ldots, 2,1$ are written in a circle. 16 players are sitting around the table, numbered in order. After each rotation of the table, each player receives as many points ... | Logic and Puzzles | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | Let $v(s)$ be the score for sector $s$. The values are $v(0)=0, v(1)=1, 2, 3, 4, 5, 6, 7, v(8)=8, v(9)=7, 6, 5, 4, 3, 2, v(15)=1$.
So $v(s) = \begin{cases} s & 0 \le s \le 8 \\ 16-s & 9 \le s \le 15 \end{cases}$.
For player $p$, in rotation $k$, the sector is $(p-1+k) \pmod{16}$.
Total score for player $p$ after 13 rot... | null |
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). | 400 | 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) $500 \%$ | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematics education | 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... | null |
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=... | f(x)=x | Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying:
$$
f(x(1+y))=f(x)(1+f(y))
$$ | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Difference and functional equations | Final Answer: The solutions are $\boxed{f(x) = 0, f(x) = x}$ | null |
【Analysis】By applying the commutative and associative laws of multiplication to integrate $11 \times 101 \times 1001 \times 10001 \times 1000001 \times 111$, then organize it into $11111111 \times 111111111111$, we can derive that the last 8 digits of $11 \times 101 \times 1001 \times 10001 \times 1000001 \times 111$ a... | 87654321 | 7. (10 points) The last 8 digits of $11 \times 101 \times 1001 \times 10001 \times 1000001 \times 111$ are | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | We want to find the last 8 digits of $11 \times 101 \times 1001 \times 10001 \times 1000001 \times 111$.
We compute the product modulo $10^8$.
$11 \times 101 = 1111 \pmod{10^8}$
$1111 \times 1001 = 1111000 + 1111 = 1112111 \pmod{10^8}$
$1112111 \times 10001 = 1112111 + 11121110000 = 11122222111 \equiv 22222111 \pmod{10... | null |
【Answer】Solution: As shown in the figure,
To see the maximum number from the outside, it is necessary to make the rectangles seen from the outside as "deeply" inside the square as possible, the result is as follows: a total of $6 \times 3+3 \times 4+3 \times 1+1=31$ (pieces). Therefore, the answer is: 31. | 31 | 14. (12 points) Use 36 solid rectangular prisms of size $3 \times 2 \times 1$ to form a large cube of size $6 \times 6 \times 6$. Among all possible arrangements, the maximum number of small rectangular prisms that can be seen from a point outside the large cube is $\qquad$. | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Group theory and generalizations | Final Answer: We want to maximize the number of visible prisms from outside the large cube. This is equivalent to maximizing the number of prisms that are on the surface of the large cube. To maximize the number of prisms on the surface, we need to minimize the number of prisms in the interior of the large cube. The in... | null |
【Analysis】This problem seems difficult to approach, but it is related to the sum of digits, and generally, we would try modulo 3 first. It can be observed that $m$ modulo 3 should leave a remainder of 1, and the construction method is the same as Example 2.
Solution Assume $a b c$ is a multiple of 3. Then $m$ is a mul... | m | Example 3 Find all positive integers $m$ such that there exist pairwise coprime positive integers $a, b, c$ satisfying
$$
S(ab)=S(ac)=S(bc)=m .
$$ | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | Final Answer: The final answer is $\boxed{1}$ | null |
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 $\boxed{\textbf{(E)}\ \dfrac{5}{2}}$ | \textbf{(E)}\\frac{5}{2} | $650$ 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}... | Other | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematics education | 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... | null |
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... | 3323 | 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? | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Numerical analysis | 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/... | null |
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... | (1,2),(9,22),(8,9) | 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. | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | 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... | null |
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... | 999 | 5. Let $\Sigma(n)$ denote the sum of the digits of the number $n$. Find the smallest three-digit $n$ such that $\Sigma(n)=\Sigma(2 n)=\Sigma(3 n)=\ldots=\Sigma\left(n^{2}\right)$ | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | Final Answer: The final answer is $\boxed{999}$ | null |
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 $\sin (2 x)=0$,
$$
\sin (2.5 x) \cos (1.5 x)=\cos (2 x) \sin (3 x... | 320 | 20. [10] There exist several solutions to the equation
$$
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. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | General algebraic systems | 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}... | null |
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}$. | f(x)=\frac{(x-2)^{2}}{2x+2} | 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. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Real functions | 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 >... | null |
Isosceles triangles $E O F, G O H$ and $E F G$ are similar.
## Otвет
$3: 1$.
## Answer
$3: 1$. | 3:1 | 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$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | 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... | null |
The answer is: $2 \leq k \leq n \leq 3$ or $3 \leq k \leq n$.
Obviously, $k \leq n$ according to the problem statement, and $k \geq 2$, because for $k=1$ all points would have the same color, while the number $n$ of colors should be $\geq 2$. We number the circles and the colors from 1 to n and denote by $F(i, j)$ the ... | 2 \leq k \leq n \leq 3 \text{ or } 3 \leq k \leq n | Given two positive integers $n$ and $k$. In the plane, there are $n$ circles ($n \geq 2$) such that each circle intersects every other circle at two points, and all these intersection points are pairwise distinct.
Each intersection point is colored with one of $n$ colors such that each color is used at least once and o... | Combinatorics | AI-MO/NuminaMath-1.5/olympiads_ref | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | Final Answer: The possible values of $n$ and $k$ are:
\begin{itemize}
\item If $n=2$, then $k=2$.
\item If $n \geq 3$, then $k=n-1$ or $k=n$.
\end{itemize}
In summary, for $n=2$, $k=2$. For $n \geq 3$, $k \in \{n-1, n\}$.
We have shown that for $n=2$, $k=2$ is possible. For $n=3$, $k=2$ and $k=3$ are possible.... | null |
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... | \frac{1}{5} | 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... | Combinatorics | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | 51 | 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 ... | null |
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,... | \frac{\sqrt{6}}{4} | 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... | Geometry | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | 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)... | null |
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,... | \frac{2+\sqrt{2}}{4} | 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 | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Linear and multilinear algebra; matrix theory | 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 ... | null |
【Answer】Solution: The number of squares with a side length of 1 unit is 12;
The number of squares with a side length of 2 units is 6;
The number of squares with a side length of 3 units is 2;
The maximum side length is 3 units, any larger and it would not form a square;
In total, there are squares: $12+6+2=20$ (squares... | 20 | 11. (3 points) There are 20 points below, with each adjacent pair of points being equidistant. By connecting four points with straight lines, you can form a square. Using this method, you can form $\qquad$ squares.
The text above has been translated into English, preserving the original text's line breaks and format... | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | Final Answer: The final answer is $\boxed{28}$ | null |
## Solution.
For the solutions of the equation $x^{2}-\left(2^{m-1}-5\right) x+1=0$, we have
$$
x_{1}+x_{2}=2^{m-1}-5 \quad \text{and} \quad x_{1} x_{2}=1
$$
Therefore,
$$
\begin{aligned}
\left(\frac{x_{1}}{x_{2}}\right)^{2}+\left(\frac{x_{2}}{x_{1}}\right)^{2} & =\frac{x_{1}^{4}+x_{2}^{4}}{x_{1}^{2} x_{2}^{2}}=\fr... | \in(2,4) | ## Task B-2.1.
For which values of the real parameter $m$ does the inequality $\left(\frac{x_{1}}{x_{2}}\right)^{2}+\left(\frac{x_{2}}{x_{1}}\right)^{2}<47$ hold if $x_{1}$ and $x_{2}$ are the solutions of the equation $x^{2}-\left(2^{m-1}-5\right) x+1=0$? | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | General algebraic systems | Let $y = 2^{m-1}-5$. Then $x_{1}+x_{2} = y$ and $x_{1}x_{2} = 1$.
The inequality is $\left(\frac{x_{1}}{x_{2}}\right)^{2}+\left(\frac{x_{2}}{x_{1}}\right)^{2} < 47$, which is equivalent to $x_{1}^{4}+x_{2}^{4} < 47$.
We have $x_{1}^{4}+x_{2}^{4} = ((x_{1}+x_{2})^{2}-2x_{1}x_{2})^{2} - 2(x_{1}x_{2})^{2} = (y^{2}-2)^{2} ... | null |
# 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... | 13 | 1. Find the maximum value of the expression $x+y$, where $x, y-$ are integer solutions of the equation $3 x^{2}+5 y^{2}=345$ | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | 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} + ... | null |
The elements of the sequence $\left(x_{n}\right)$ are positive real numbers, and for every positive integer $n$,
$$
2\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{4}=\left(x_{1}^{5}+x_{2}^{5}+\ldots+x_{n}^{5}\right)+\left(x_{1}^{7}+x_{2}^{7}+\ldots+x_{n}^{7}\right)
$$
Determine the elements of the sequence.
Solution. We wi... | x_{n}=n | The elements of the sequence $\mathrm{Az}\left(x_{n}\right)$ are positive real numbers, and for every positive integer $n$,
$$
2\left(x_{1}+x_{2}+\ldots+x_{n}\right)^{4}=\left(x_{1}^{5}+x_{2}^{5}+\ldots+x_{n}^{5}\right)+\left(x_{1}^{7}+x_{2}^{7}+\ldots+x_{n}^{7}\right)
$$
Determine the elements of the sequence. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Sequences, series, summability | Final Answer: The elements of the sequence are given by $x_{n}=n$ for every positive integer $n$. We verify this solution.
Let $x_{k}=k$ for $k=1, 2, \ldots, n$. Then $x_{1}+x_{2}+\ldots+x_{n} = \sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.
Also, $x_{k}^{5}=k^{5}$ and $x_{k}^{7}=k^{7}$.
We need to check if $2\left(\sum_{k=1}^{... | null |
Consider the number of beans Alice has in base 5. Note that $2008=31013_{5}, 42=132_{5}$, and $100=400_{5}$. Now, suppose Alice has $d_{k} \cdots d_{2} d_{1}$ beans when she wins; the conditions for winning mean that these digits must satisfy $d_{2} d_{1}=32, d_{k} \cdots d_{3} \geq 310$, and $d_{k} \cdots d_{3}=4i+1$ ... | 35 | Alice and the Cheshire Cat play a game. At each step, Alice either (1) gives the cat a penny, which causes the cat to change the number of (magic) beans that Alice has from $n$ to $5n$ or (2) gives the cat a nickel, which causes the cat to give Alice another bean. Alice wins (and the cat disappears) as soon as the numb... | KbsdJames/Omni-MATH/HMMT_2 | {'domain': ['Mathematics -> Number Theory -> Other', 'Mathematics -> Algebra -> Prealgebra -> Integers', 'Mathematics -> Algebra -> Algebra -> Equations and Inequalities'], 'difficulty': 4.5} | Number theory | We want to find the minimum cost to reach a number of beans greater than 2008 and ending in 42. We use dynamic programming to find the minimum cost $C(n)$ to reach $n$ beans.
We have the recurrence relation:
$C(0) = 0$
$C(n) = C(n-1) + 5$ if $n$ is not divisible by 5, for $n > 0$
$C(n) = \min(C(n-1) + 5, C(n/5) + 1)$ i... | null | |
18. Since $2002=2 \times 7 \times 11 \times 13$, we can choose certain vertices among the 2002 vertices to form regular 7-, 11-, 13-, $\cdots$ polygons (where the number of sides runs through divisors of 2002 greater than 2 and less than 1001). Since $2002=7 \times 286$, so there are at least 286 different positive int... | 287287 | 18. (2 marks) Let $A_{1} A_{2} \cdots A_{2002}$ be a regular 2002-sided polygon. Each vertex $A_{i}$ is associated with a positive integer $a_{i}$ such that the following condition is satisfied: If $j_{1}, j_{2}, \cdots, j_{k}$ are positive integers such that $k<500$ and $A_{j_{1}} A_{j_{2}} \cdots A_{j_{k}}$ is a regu... | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Commutative algebra | Let $M=499$. Define $a_i = (i-1) \pmod{499} + 1$ for $i=1, 2, \cdots, 2002$.
We need to check that for any regular $k$-sided polygon $A_{j_1} A_{j_2} \cdots A_{j_k}$ with $k < 500$, the values $a_{j_1}, a_{j_2}, \cdots, a_{j_k}$ are distinct.
The vertices of a regular $k$-sided polygon are of the form $A_j, A_{j+m}, \c... | null |
Answer. For $n=6$.
Solution. For $n=6$, we can set $a_{1}=a_{2}=a_{3}=a_{4}=1$ and $a_{5}=a_{6}=-1$; then the quadratic trinomial from the condition becomes $x^{2}-8 x+7$ and has two integer roots: 1 and 7. It remains to show that this is the smallest possible value of $n$.
Suppose the numbers $a_{1}, a_{2}, \ldots, ... | 6 | 9.2. For what least natural $n$ do there exist integers $a_{1}, a_{2}, \ldots, a_{n}$ such that the quadratic trinomial
$$
x^{2}-2\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{2} x+\left(a_{1}^{4}+a_{2}^{4}+\ldots+a_{n}^{4}+1\right)
$$
has at least one integer root?
(P. Kozlov) | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | Final Answer: The final answer is $\boxed{3}$ | null |
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,... | 7or14 | 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. | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | 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, $8+nx = \frac{(n+2)(n+1)}{2}$.
$16+2nx = (n+2)(n+1) = n^2+3n+2$.
$2nx = n^2+3n-14$.
$... | null |
(1) $\left[0, \frac{\sqrt{6}}{6}\right]$ | [0,\frac{\sqrt{6}}{6}] | (1) The range of the function $f(x)=\sqrt{\frac{x+1}{x^{2}+4 x+7}}$ is $\qquad$ | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Real functions | 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... | null |
2. (10 points) Five pieces of paper are written with $1, 2, 3, 4, 5$ respectively, facing upwards from smallest to largest, stacked into one pile. Now, the $1, 3,$ and $5$ are flipped to their backs and placed back in their original positions. If the entire stack is split at any one piece of paper into two stacks, and ... | 5 | 2. (10 points) Five pieces of paper are written with $1$, $2$, $3$, $4$, and $5$, facing up from smallest to largest, stacked in a pile. Now, the 1, 3, and 5 are flipped to their backs, and still placed in their original positions. If the entire stack of paper is split at any one piece of paper into two stacks, and the... | Logic and Puzzles | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | Final Answer: The final answer is $\boxed{4}$ | null |
[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
$$
\begin{aligned}
& \left(2 k_{1}+1\right)^{2}+\left(2 k_{2}+1\right)^{2}+\cdots+\left(2 k_{6}+1\right)^{2} \\
= & 4\left(k_{1}^{2}+k_{2}^{2}+\cdots+k... | C | $30 \cdot 35$ Six odd numbers, their sum is 42, their sum of squares can only be
(A) 280.
(B) 368.
(C) 382.
(D) 423.
(China Jiangxi Nanchang Junior High School Mathematics Competition, 1990) | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | Let the six odd numbers be $x_1, x_2, x_3, x_4, x_5, x_6$. We are given that $\sum_{i=1}^6 x_i = 42$. Since the sum of six odd numbers is even, this is possible.
Let $x_i = 7 + \delta_i$, where $\sum_{i=1}^6 \delta_i = 0$. Since $x_i$ are odd, $\delta_i = x_i - 7$ must be even integers. Let $\delta_i = 2k_i$, where $k_... | null |
Answer: 8.
Solution. Let the number at the $i$-th position be $a_{i}(i=1, \ldots, 15$.) Fix 5 consecutive numbers. The numbers to the left and right of this quintet must match. Therefore, $a_{i}=a_{i+6}$. Let's go in a circle, marking the same numbers:
$$
a_{1}=a_{7}=a_{13}=a_{4}=a_{10}=a_{1} .
$$
Now it is clear th... | 8 | 2. Fifteen numbers are arranged in a circle. The sum of any six consecutive numbers is 50. Petya covered one of the numbers with a card. The two numbers adjacent to the card are 7 and 10. What number is under the card? | Logic and Puzzles | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | Let the fifteen numbers be $a_1, a_2, \dots, a_{15}$ in a circle. We are given that the sum of any six consecutive numbers is 50, which implies that $a_{i+6} = a_i$ for all $i$. Let the covered number be $x$. Let the adjacent numbers be 7 and 10. So we have ..., 7, x, 10, ... or ..., 10, x, 7, ... in clockwise order.
L... | null |
Let $S$ be the area of the base of the pyramid, $S_1$ and $S_2$ be the areas of the largest and smallest sections, respectively. The largest and smallest sections are polygons similar to the polygon of the base of the pyramid with coefficients $\frac{99}{100}$ and $\frac{1}{100}$, respectively. Therefore,
$$
S_1 = \le... | 9801 | [ [ 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.
# | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | 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... | null |
First, let us list the statement "If a quadrilateral is a square, then it is a rectangle" as a statement of the form "If $p$, then $q$". In this case, $p$ is "a quadrilateral is a square", and $q$ is "it is a rectangle".
The converse is then: "If $q$, then $p$". Plugging in, we get "If a quadrilateral is a rectangle, t... | D | Given the true statement: If a quadrilateral is a square, then it is a rectangle.
It follows that, of the converse and the inverse of this true statement is:
$\textbf{(A)}\ \text{only the converse is true} \qquad \\ \textbf{(B)}\ \text{only the inverse is true }\qquad \\ \textbf{(C)}\ \text{both are true} \qquad \\ \... | Logic and Puzzles | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematical logic and foundations | Let the given statement be "If a quadrilateral is a square, then it is a rectangle."
Let P be the statement "A quadrilateral is a square."
Let Q be the statement "It is a rectangle."
The given statement is in the form "If P, then Q", which is $P \implies Q$. We are given that this statement is true.
The converse of th... | null |
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
$\sqrt{6}$. | \sqrt{6} | [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. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | 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... | null |
$1347$ | 1347 | 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]+[... | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Real functions | 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... | null |
3. When $n=3$, the conclusion is 3.
When $n>4$, construct a regular $n$-sided polygon and slightly adjust its shape to meet the conditions described in the problem. In this $n$-sided polygon, it can be directly calculated that for any vertex, the lines connecting it to its two adjacent vertices are the closest.
When $... | 3 | 3. Given $n(n>2)$ points on a plane, satisfying that no three points are collinear. Draw a line through any two points, and mark the point among the remaining points that is closest to this line (it is known that in any case, the point satisfying this condition is unique). Question: For each given $n$, what is the maxi... | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | Final Answer: For each given $n$, we want to find the maximum number of points that can be marked.
We consider small values of $n$.
For $n=3$, let the points be $A, B, C$.
For line $AB$, the closest point is $C$.
For line $AC$, the closest point is $B$.
For line $BC$, the closest point is $A$.
All 3 points $A, B, C$ ar... | null |
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... | 30\mathrm{~}^{2} | 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) | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | 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... | null |
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... | 59 | $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... | Algebra | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematical logic and foundations | 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... | null |
According to the inequality between the quadratic mean and the arithmetic mean
$$
\begin{aligned}
& \sqrt{2}\left(\sqrt{x_{1}^{2}+\left(1-x_{2}\right)^{2}}+\sqrt{x_{2}^{2}+\left(1-x_{3}\right)^{2}}+\ldots+\sqrt{x_{n}^{2}+\left(1-x_{1}\right)^{2}}\right) \geq \\
& \geq\left|x_{1}\right|+\left|1-x_{2}\right|+\left|x_{2}... | \frac{n}{\sqrt{2}} | [ $\underline{\text { Classical Inequalities (Miscellaneous) })]}$
$$
\sqrt{x_{1}^{2}+\left(1-x_{2}\right)^{2}}+\sqrt{x_{2}^{2}+\left(1-x_{3}\right)^{2}}+\ldots+\sqrt{x_{2 n}^{2}+\left(1-x_{1}\right)^{2}}
$$ | Inequalities | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Real functions | We want to find the minimum value of the expression
$$ S = \sqrt{x_{1}^{2}+\left(1-x_{2}\right)^{2}}+\sqrt{x_{2}^{2}+\left(1-x_{3}\right)^{2}}+\ldots+\sqrt{x_{2 n}^{2}+\left(1-x_{1}\right)^{2}} $$
We can use Minkowski's inequality, which states that for real numbers $a_i, b_i$,
$$ \sum_{i=1}^{n} \sqrt{a_i^2 + b_i^2} \g... | null |
Reference: 1995 HG5 . . $37^{100} \ldots$ 157-digit number, $37^{15} \ldots n$-digit ....
$$
\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... | 23 | 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... | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Field theory and polynomials | 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}... | null |
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... | 58 | 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 $\angle BAG$. | Geometry | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | 08 | 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... | null |
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 $223 + 1 + 1 = \boxed{\textbf{(D)}\ 225}.$ | 225 | 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?
$\textbf{(A)}\ 202\qquad\textbf{(B)}\ 223\qquad\textbf{(C)}\ 224\qquad\textbf{(D)}\ 225\qquad\textbf{(E)}\ 234$ | Combinatorics | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | 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... | null |
Answer: 1) two points
$$
\text { 2) } x=2+\frac{\pi}{3}, y=2-\frac{\pi}{3}
$$ | 2)2+\frac{\pi}{3},2-\frac{\pi}{3} | 2. The coordinates $(x ; y)$ of points in the square $\{(x ; y):-\pi \leq x \leq \pi, 0 \leq y \leq 2 \pi\}$ satisfy the system of equations $\left\{\begin{array}{c}\sin x+\sin y=\sin 2 \\ \cos x+\cos y=\cos 2\end{array}\right.$. How many such points are there in the square? Find the coordinates $(x ; y)$ of the point ... | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Real functions | Final Answer: The number of such points is $\boxed{1}$. The coordinates of the point with the smallest ordinate is $\boxed{\left(2-\frac{\pi}{3} ; 2+\frac{\pi}{3}\right)}$ | null |
Answer: 4 links. According to the solution of problem 5 for grades $7-8$, for a chain consisting of $n$ links, where $64 \leq n \leq 159$, it is sufficient to unfasten 4 links. | 4 | 10,11
There is a piece of chain consisting of 150 links, each weighing 1 g. What is the smallest number of links that need to be broken so that from the resulting parts, all weights of 1 g, 2 g, 3 g, ..., 150 g can be formed (a broken link also weighs 1 g)? | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | To form all weights from 1 g to 150 g, we need to find a set of weights such that any integer weight from 1 to 150 can be obtained by summing a subset of these weights. To minimize the number of broken links, we want to minimize the number of parts of the chain.
We can use the weights that are powers of 2: 1, 2, 4, 8,... | null |
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 ... | \frac{1}{58} | 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 $20 ?$ | KbsdJames/Omni-MATH/HMMT_2 | {'domain': ['Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations'], 'difficulty': 4.0} | Combinatorics | 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... | null | |
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... | 2 | 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... | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | 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... | null |
12.23 The function is defined for $x>0$ and $x \neq 1$. We have
$$
y^{\prime}=\frac{\ln x - x \cdot \frac{1}{x}}{\ln ^{2} x}=\frac{\ln x - 1}{\ln ^{2} x}
$$
Thus, $y^{\prime}=0$ when $\ln x - 1 = 0$, i.e., when $x=e$. Let's construct a table:
| Interval | $(0,1)$ | $(1, e)$ | $e$ | $(e, \infty)$ |
| :---: | :---: | ... | (e) | $12.23 y=\frac{x}{\ln x}$.
The above text is translated into English, please retain the original text's line breaks and format, and output the translation result directly.
However, since the provided text is already in a mathematical format which is universal and does not require translation, the translation is as f... | Calculus | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Real functions | $12.23 y=\frac{x}{\ln x}$. | null |
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 $E D=F C-F E-D C=(30 \mathrm{~cm})-(5 \mathrm{~cm})-(x \mathrm{~cm})=(25-x) \mathrm{cm}$... | 17\mathrm{~} | 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 ... | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | 91 | 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... | null |
- Omвem: for $n=5$. Let's number the elephants in ascending order of their size, then it is enough to verify that $C_{1}+C_{4}=C_{2}+C_{3}, C_{1}+C_{5}=C_{2}+C_{4}$ and $C_{2}+$ $C_{5}=C_{3}+C_{4}$. These equalities are equivalent to $C_{4}-C_{3}=C_{2}-C_{1}$, $C_{2}-C_{1}=C_{5}-C_{4}$ and $C_{5}-C_{4}=C_{3}-C_{2}$, th... | 5 | 6. A factory produces sets of $n>2$ elephants of different sizes. According to the standard, the difference in mass between adjacent elephants within each set should be the same. The inspector checks the sets one by one using a balance scale without weights. For what smallest $n$ is this possible? | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | Final Answer: The final answer is $\boxed{4}$ | null |
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... | \frac{1}{3} | 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 $\$ 50$... | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Probability theory and stochastic processes | 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 $... | null |
Rong Ge 42.
Analysis Note $x y \leq \frac{1}{4} x^{2}+y^{2}, 8 x \leq x^{2}+16, y \leq \frac{1}{4} y^{2}+1$, adding these three inequalities yields
$$
x y+8 x+y \leq \frac{5}{4}\left(x^{2}+y^{2}\right)+17=42
$$
and when $x=4, y=2$, the equality can be achieved, so the maximum value of $x y+8 x+y$ is 42.
Alternatively,... | 42 | 10. Real numbers $x, y$ satisfy $x^{2}+y^{2}=20$, then the maximum value of $x y+8 x+y$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Real functions | 10. Real numbers $x, y$ satisfy $x^{2}+y^{2}=20$, then the maximum value of $x y+8 x+y$ is | null |
6.11 Let's use formula (6.10), valid under the condition $|x|<1$:
a) $2 x+1+\frac{x^{2}}{1+x}=\frac{13}{6} \Rightarrow x_{1}=\frac{1}{2}, x_{2}=-\frac{7}{9}$;
b) $\frac{1}{x}+\frac{x}{1-x}=\frac{7}{2} \Rightarrow x_{1}=\frac{1}{3}, x_{2}=\frac{2}{3}$.
Answer: a) $x_{1}=\frac{1}{2}, x_{2}=-\frac{7}{9} ;$ b) $x_{1}=\f... | )x_{1}=\frac{1}{2},x_{2}=-\frac{7}{9};b)x_{1}=\frac{1}{3},x_{2}=\frac{2}{3} | 6.11 Solve the equations:
a) $2 x+1+x^{2}-x^{3}+x^{4}-x^{5}+\ldots=\frac{13}{6}$, where $|x|<1$;
b) $\frac{1}{x}+x+x^{2}+\ldots+x^{n}+\ldots=\frac{7}{2}$, where $|x|<1$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Sequences, series, summability | 6.11 Solve the equations:
a) $2 x+1+x^{2}-x^{3}+x^{4}-x^{5}+\ldots=\frac{13}{6}$, where $|x|<1$;
We can rearrange the terms as:
$1 + (2x + x^{2}-x^{3}+x^{4}-x^{5}+\ldots)=\frac{13}{6}$
Consider the series $S = 2x + x^{2}-x^{3}+x^{4}-x^{5}+\ldots = 2x + (x^{2}-x^{3}+x^{4}-x^{5}+\ldots)$.
The series in the parenthesis ... | null |
Solution.
Rewrite the equation as $3^{2 x}+2^{x} \cdot 3^{x}-2 \cdot 2^{2 x}=0$ and divide it by $2^{2 x} \neq 0$. Then $\left(\frac{3}{2}\right)^{2 x}+\left(\frac{3}{2}\right)^{x}-2=0 \Rightarrow\left(\left(\frac{3}{2}\right)^{x}\right)=-2$ (no solutions) or $\left(\left(\frac{3}{2}\right)^{x}\right)_{2}=1 \Rightarro... | 0 | 7.214. $9^{x}+6^{x}=2^{2 x+1}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
7.214. $9^{x}+6^{x}=2^{2 x+1}$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | General algebraic systems | 7.214. $9^{x}+6^{x}=2^{2 x+1}$. | null |
The sum is always 300. Pairs are (43, 257), (100, 200), (98, 202), (125, 175) | 300 | ## Task 4
Pair the following numbers so that they always add up to the same sum.
$\begin{array}{llllllll}43 & 202 & 100 & 175 & 98 & 257 & 125 & 200\end{array}$ | Logic and Puzzles | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematical logic and foundations | 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... | null |
Solution
a) 4. We need to choose 3 points out of a total of 4. In situations like this, when we choose any triangle, for example, triangle $A B C$, all permutations of points $A, B$, and $C$ still form triangle $A B C$. Therefore, when using the fundamental principle of counting, we must divide by the number of permut... | 4,0,12 | Juquinha marks points on a circumference and draws triangles by connecting 3 of these points. The lengths of the arcs between 2 consecutive points are equal.
a) By marking 4 points on the circumference, how many triangles can he draw?
b) By marking 5 points on the circumference, how many equilateral triangles can he ... | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | Final Answer:
a) By marking 4 points on the circumference, the number of triangles he can draw is the number of ways to choose 3 points out of 4, which is $\binom{4}{3} = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4$.
b) By marking 5 points on the circumference, we want to find the number of equilateral triangl... | null |
With a fair die that has faces numbered from 1 to 6, the probability of rolling each of 1 to 6 is $\frac{1}{6}$. We calculate the probability for each of the five choices. There are 4 values of $x$ that satisfy $x>2$, so the probability is $\frac{4}{6}=\frac{2}{3}$. There are 2 values of $x$ that satisfy $x=4$ or $x=5$... | x > 2 | After a fair die with faces numbered 1 to 6 is rolled, the number on the top face is $x$. What is the most likely outcome? | KbsdJames/Omni-MATH/cayley | {'domain': ['Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations'], 'difficulty': 1.0} | Probability theory and stochastic processes | A fair die with faces numbered 1 to 6 is rolled.
The possible outcomes are {1, 2, 3, 4, 5, 6}.
Since the die is fair, each outcome has an equal probability of occurring.
The probability of each outcome is $1/6$.
Let $P(x=n)$ be the probability that the number on the top face is $n$, where $n \in \{1, 2, 3, 4, 5, 6\}$.
... | null | |
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. $\angle CAP = \angle PCG$,... | 90 | 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 $AB$? | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | 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 ... | null |
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... | =1,n=4 | 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... | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | 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... | null |
$13$ | 13 | 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 $\qquad$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | 51 | 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... | null |
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
$$
2 p(0)+p(2022)=p(674)=0 \Rightarrow p(2022)=-2 p(0)=5392 \text {. }
$$ | 5392 | 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. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Commutative algebra | 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)... | null |
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... | 7 | 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。 | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Algebraic geometry | 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... | null |
25. Let's schematically represent the windows and measure them:
a) Lighting all four windows gives one signal;
b) Lighting one of the windows is perceived as one signal, as in the dark, th... | 10 | 25. A scout is in a house with four windows arranged in a rectangular shape. He needs to signal to the sea at night by lighting a window or several windows. How many different signals can he send? | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | Let $n$ be the number of windows, which is 4 in this case.
For each window, there are two possibilities: either it is lit or it is not lit.
If we consider all combinations of lighting or not lighting each window, we have $2^n$ possible combinations.
In this case, with 4 windows, there are $2^4 = 16$ possible combinatio... | null |
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
$$
\begin{aligned}
\mathrm{P}(D \mid C)=\frac{\mathrm{P}(D C)}{\mathrm{P}(C)}=\frac{\mathrm{P}(A)}... | \frac{2}{3} | 14. A ball was added to an urn containing one white ball - either white or black (with equal probabilities of selection). After this, one ball was randomly drawn from the urn. It turned out to be white. What is the conditional probability that the remaining ball in the urn is also white? | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Probability theory and stochastic processes | Let $W$ be the event that a white ball is added, and $B$ be the event that a black ball is added. We are given $P(W) = P(B) = 1/2$.
Let $D$ be the event that a white ball is drawn from the urn.
We want to find the conditional probability that the remaining ball in the urn is also white, given that a white ball was draw... | null |
(1) For any filling method that satisfies the conditions, the number filled at point $a_{i}$ is still denoted as $a_{i}, i=1,2, \cdots, 12$. If the sum of the four numbers on each line is $s$, then from $6 s=2(1+2+\cdots+12)$, we get $s=26$;
In $\triangle a_{1} a_{3} a_{5}$, on the three sides, we have
$$
\begin{array}... | 24 | 25 Two equilateral triangles overlap to form a six-pointed star (as shown in the figure). The first 12 positive integers 1, $2, \cdots, 12$ are to be placed at the 12 nodes of the figure, such that the sum of the four numbers on each straight line is equal.
(1) Find the minimum sum of the numbers placed at the six vert... | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | Final Answer:
(1) The minimum sum of the numbers placed at the six vertices $a_{1}$, $a_{2}, \cdots, a_{6}$ of the six-pointed star is $\boxed{21}$
(2) Proof:
Let $X$ be the set of all valid filling schemes. We define a transformation $f: X \to X$ as follows: for any filling scheme $\pi \in X$, we define $f(\pi) = \pi... | null |
【Analysis】For a proper fraction, when the difference between the numerator and the denominator is constant, the larger the denominator, the larger the fraction.
【Solution】Solution: $\frac{2007}{2009}<\frac{2008}{2010}, \frac{2008}{2010}<\frac{2009}{2011}, \frac{2009}{2011}<\frac{2010}{2012}$,
Therefore: $\frac{2007}{20... | C | 5. (3 points) $a=\frac{2007 \times 2008}{2009 \times 2010}, b=\frac{2008 \times 2009}{2010 \times 2011}, c=\frac{2009 \times 2010}{2011 \times 2012}$, then we have ( )
A. $a>b>c$
B. $a>c>b$
C. $a<b<c$
D. $a<c<b$ | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | 05 | We are given $a=\frac{2007 \times 2008}{2009 \times 2010}, b=\frac{2008 \times 2009}{2010 \times 2011}, c=\frac{2009 \times 2010}{2011 \times 2012}$.
Let $n=2007$. Then $a=\frac{n(n+1)}{(n+2)(n+3)}, b=\frac{(n+1)(n+2)}{(n+3)(n+4)}, c=\frac{(n+2)(n+3)}{(n+4)(n+5)}$.
To compare $a$ and $b$, we consider the ratio $\frac{a... | null |
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. | 10 | 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 | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | 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... | null |
Answer: $2015^{2}-1=4060224$. Solution. Note that the common divisor will also divide $(x+2015 y)-2015(y+2015 x)=\left(1-2015^{2}\right) x$. Similarly, it divides $\left(1-2015^{2}\right) y$, and since $(x, y)=1$, it divides $\left(1-2015^{2}\right)$. On the other hand, if we take $x=1, y=2015^{2}-2016$, then we get $\... | 4060224 | 6. Find the greatest possible value of $\gcd(x+2015 y, y+2015 x)$, given that $x$ and $y$ are coprime numbers. | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | Let $d = \gcd(x+2015 y, y+2015 x)$.
We have $d \mid (x+2015 y) - 2015(y+2015 x) = x+2015 y - 2015 y - 2015^2 x = x - 2015^2 x = (1-2015^2) x = -(2015^2-1) x = -2014 \cdot 2016 x$.
So $d \mid 2014 \cdot 2016 x$.
Also $d \mid (y+2015 x) - 2015(x+2015 y) = y+2015 x - 2015 x - 2015^2 y = y - 2015^2 y = (1-2015^2) y = -(201... | null |
Whenever Ann farms a patch $P$, she also farms all the patches due west of $P$ and due south of $P$. So, the only way she can put a scarecrow on $P$ is if Keith farms the patch immediately north of $P$ and the patch immediately east of $P$, in which case Ann cannot farm any of the patches due north of $P$ or due east o... | 7 | The Dingoberry Farm is a 10 mile by 10 mile square, broken up into 1 mile by 1 mile patches. Each patch is farmed either by Farmer Keith or by Farmer Ann. Whenever Ann farms a patch, she also farms all the patches due west of it and all the patches due south of it. Ann puts up a scarecrow on each of her patches that is... | KbsdJames/Omni-MATH/HMMT_2 | {'domain': ['Mathematics -> Applied Mathematics -> Math Word Problems'], 'difficulty': 5.0} | Combinatorics | Final Answer: The final answer is $\boxed{20}$ | null | |
7. $\sqrt{2007}$ Hint: $f(x)=x^{4}\left(x^{2}-2 \sqrt{2006} x-1\right)+x\left(x^{2}-2 \sqrt{2007} x+1\right)+x-\sqrt{2006}-\sqrt{2007}$ $+\sqrt{2007}, f(\sqrt{2006}+\sqrt{2007})=\sqrt{2007}$. | \sqrt{2007} | $$
\begin{array}{l}
\text { 7. If } f(x)=x^{6}-2 \sqrt{2006} x^{5}-x^{4}+x^{3}-2 \sqrt{2007} x^{2}+2 x-\sqrt{2006} \text {, then } f(\sqrt{2006}+ \\
\sqrt{2007})= \\
\end{array}
$$ | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Field theory and polynomials | 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... | null |
(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... | 3\sqrt{4+\sqrt{3}} | 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... | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | 39 | Final Answer: The final answer is $\boxed{(A)}$ | null |
Solve $2 \sqrt{2}-1$.
Analysis $\frac{1}{|a|}+\frac{2|a|}{b}=\frac{a+b}{|a|}+\frac{2|a|}{b}=\frac{a}{|a|}+\left(\frac{b}{|a|}+\frac{2|a|}{b}\right) \geq \frac{a}{|a|}+2 \sqrt{\frac{b}{|a|} \cdot \frac{2|a|}{b}}=2 \sqrt{2}+\frac{a}{|a|}$, where the equality holds when $\frac{b}{|a|}=\frac{2|a|}{b}$, i.e., $b^{2}=2 a^{2}... | 2\sqrt{2}-1 | 9. Let $a+b=1, b>0, a \neq 0$, then the minimum value of $\frac{1}{|a|}+\frac{2|a|}{b}$ is
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Real functions | 9. Let $a+b=1, b>0, a \neq 0$, then the minimum value of $\frac{1}{|a|}+\frac{2|a|}{b}$ is | null |
3. Let the desired value of the number $l$ be denoted by $l_{n}$. Trivially, $l_{1}=1$ and $l_{2}=3$. The examples in the figure show that $l_{3} \leqslant 4$ and, in general, $l_{n} \leqslant 2 n-2$ for $n \geqslant 3$.
Assume that the broken line $\mathcal{C}$ of length $l$ covers all $n^{2}$ points. Let there be $a... | 2n-2 | 3. Given $n^{2}$ points arranged in the plane in the form of a square grid $n \times n$. A broken line of length $l$ is the union of closed segments $A_{0} A_{1}$, $A_{1} A_{2}$, $A_{2} A_{3}, \ldots, A_{l-1} A_{l}$ in that plane. Determine the smallest natural number $l$ for which there exists a broken line of length ... | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | Final Answer: The smallest natural number $l$ is given by
$$ l = \begin{cases} 1 & \text{if } n=1 \\ n^2 - 1 & \text{if } n \ge 2 \end{cases} $$
We can also write this as $l = \max(1, n^2 - 1)$.
Final Answer: The final answer is $\boxed{n^2-1}$ | null |
$$
V_{K L M D}=\frac{D K}{D A} \cdot \frac{D L}{D B} \cdot \frac{D M}{D C} \cdot V_{A B C D}=\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{2}{5} \cdot 1=\frac{4}{45}
$$
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
$\frac{41}{45}$ | \frac{41}{45} | 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 $ABCKLM$.
# | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | 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 ... | null |
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 ... | 8 | 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... | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | 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 \... | null |
15. (3 points) There are three different sizes of cubic wooden blocks, A, B, and C, where the edge length of A is $\frac{1}{2}$ of the edge length of B, and the edge length of B is $\frac{2}{3}$ of the edge length of C. If A, B, and C blocks are used to form a large cube with the smallest possible volume (using at leas... | 50 | 15. (3 points) There are three different sizes of cubic wooden blocks, A, B, and C, where the edge length of A is $\frac{1}{2}$ of the edge length of B, and the edge length of B is $\frac{2}{3}$ of the edge length of C. If A, B, and C blocks are used to form a large cube with the smallest possible volume (using at leas... | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Combinatorics | Final Answer: The final answer is $\boxed{13}$ | null |
## Solution.
Domain: $x \in R$.
$\frac{2 x^{4}+9 x^{2}}{\left(x^{2}+4\right)\left(x^{2}+5\right)}=0 \Leftrightarrow 2 x^{4}+9 x^{2}=0 \Leftrightarrow x^{2}\left(2 x^{2}+9\right)=0$,
$x^{2}=0, x_{1}=0$ or $2 x^{2}+9=0, x_{2,3} \in \varnothing$.
Answer: $x=0$. | 0 | 6.014. $\frac{4}{x^{2}+4}+\frac{5}{x^{2}+5}=2$. | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Field theory and polynomials | To solve the equation $\frac{4}{x^{2}+4}+\frac{5}{x^{2}+5}=2$, we first combine the fractions on the left side of the equation.
The common denominator is $(x^{2}+4)(x^{2}+5)$.
$$ \frac{4(x^{2}+5)}{(x^{2}+4)(x^{2}+5)} + \frac{5(x^{2}+4)}{(x^{2}+4)(x^{2}+5)} = 2 $$
$$ \frac{4(x^{2}+5) + 5(x^{2}+4)}{(x^{2}+4)(x^{2}+5)} = ... | null |
# 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.,... | 8.5 | 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... | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mathematical logic and foundations | 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... | null |
2. First, for any odd number $n$ not less than 3, take $M$ as the $n$ distinct complex roots of the equation $z^{n}-1=0$. Clearly, the set $M$ satisfies the requirements.
Next, consider even numbers $n$ not less than 6.
Let $\frac{n}{2}=k$, first consider the decomposition of $\frac{1}{2}$.
From $\frac{1}{n}=\frac{1}{n... | {n\in{N}\midn\geqslant3,n\neq4} | 2. Find all positive integers $n$ $(n \geqslant 3)$ such that there exists a set $M$ with $n$ elements, where the elements are distinct non-zero vectors of equal length, and the following conditions are satisfied: $\sum_{u \in M} u=0$, and for any $v, w \in M$, $\boldsymbol{v}+\boldsymbol{w} \neq \mathbf{0}$. | Combinatorics | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Linear and multilinear algebra; matrix theory | Final Answer: The set of all possible values of $n$ is the set of all integers greater than or equal to 3. We have shown that for any integer $n \geqslant 3$, we can construct a set $M$ of $n$ distinct non-zero vectors of equal length such that $\sum_{u \in M} u = 0$ and for any $v, w \in M$, $v+w \neq 0$. We can take ... | null |
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, \... | 16a | 884. Find the length of the cardioid $x=2 a \cos t-a \cos 2 t, y=$ $-2 a \sin t-a \sin 2 t$. | Calculus | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Differential geometry | 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... | null |
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... | 29 | 29. [18] Compute the remainder when
$$
\sum_{k=1}^{30303} k^{k}
$$
is divided by 101 . | Number Theory | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Number theory | 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... | null |
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
$$
\begin{gathered}
20 x+(x-1) 30+(x-2) 40+(x-3) 50+(x-4) 60=0 \\
20 x+30 x-30+40 x-80+50 x-150+60 x-240=0 \\
200 x=500 ... | 2\frac{1}{2} | A four-meter-long rod has a $20 \mathrm{~kg}$ weight attached to one end, and $30,40,50 \mathrm{~kg}$ weights are attached at one, two, and three meters from this end, respectively. At the other end of the rod, a $60 \mathrm{~kg}$ weight is attached. Where should the rod be supported so that it remains in balance? The ... | Algebra | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Mechanics of particles and systems | Let $x=0$ be the position of the end of the rod where the $20 \mathrm{~kg}$ weight is attached.
The weights are attached at the following positions:
$w_1 = 20 \mathrm{~kg}$ at $x_1 = 0 \mathrm{~m}$
$w_2 = 30 \mathrm{~kg}$ at $x_2 = 1 \mathrm{~m}$
$w_3 = 40 \mathrm{~kg}$ at $x_3 = 2 \mathrm{~m}$
$w_4 = 50 \mathrm{~kg}$ ... | null |
17. Ans: 16
Let $A B=y \mathrm{~cm}$. Since $\triangle D C F \sim \triangle G B F$, we have
$$
\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... | 16 | 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 $x$. | Geometry | AI-MO/NuminaMath-1.5/olympiads | {'question_type': 'math-word-problem', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Geometry | 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... | null |
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 $N = 1$: ... | 1 | 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 ... | Algebra | AI-MO/NuminaMath-1.5/amc_aime | {'question_type': 'MCQ', 'problem_is_valid': 'Yes', 'solution_is_valid': 'Yes', 'synthetic': False} | Field theory and polynomials | 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... | null |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.