instruction_seed stringlengths 2 10.7k | solution stringlengths 0 10.8k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 9
values | question_type stringclasses 5
values | problem_is_valid stringclasses 5
values | solution_is_valid stringclasses 4
values | source stringclasses 11
values | synthetic bool 2
classes |
|---|---|---|---|---|---|---|---|---|
9.4. In the castle, there are 9 identical square rooms, forming a $3 \times 3$ square. Nine people, consisting of liars and knights (liars always lie, knights always tell the truth), each occupied one of these rooms. Each of these 9 people said: "At least one of the neighboring rooms to mine is occupied by a liar." Wha... | Answer: 6 knights.
Solution: Note that each knight must have at least one neighbor who is a liar. We will show that there must be at least 3 liars (thus showing that there are no more than 6 knights). Suppose there are no more than 2 liars, then there will be a "vertical row" of rooms where only knights live. But then... | 6 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Given the proposition p: For all $x \in \mathbb{R}$, $\cos x \leq 1$, then ( )
A: $\neg p$: There exists $x \in \mathbb{R}$, $\cos x \geq 1$
B: $\neg p$: There exists $x \in \mathbb{R}$, $\cos x 1$ | The proposition p: For all $x \in \mathbb{R}$, $\cos x \leq 1$, is a universal proposition.
Therefore, $\neg p$: There exists $x \in \mathbb{R}$, $\cos x > 1$,
Hence, the correct choice is $\boxed{\text{D}}$. | \text{D} | Logic and Puzzles | MCQ | Incomplete | Incomplete | cn_k12 | false |
Prove that $46^{2n+1} + 296 \cdot 13^{2n+1}$ is divisible by $1947$. | null | null | Number Theory | proof | Yes | Yes | aops_forum | false |
Nanna and Sofie move in the same direction along two parallel paths, which are $200$ meters apart. Nanna's speed is $3$ meters per second, Sofie's only $1$ meter per second. A tall, cylindrical building with a diameter of $100$ meters is located in the middle between the two paths. Since the building first once the lin... | null | null | Geometry | math-word-problem | Yes | Yes | aops_forum | false |
Benedek scripted a program which calculated the following sum: $1^1+2^2+3^3+. . .+2021^{2021}$. What is the remainder when the sum is divided by $35$? | null | null | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
If a line \( l \) is drawn through vertex \( A \) of an equilateral triangle \( ABC \), what is the probability that \( l \) intersects the line segment \( BC \)? | To determine the probability that a randomly drawn line through point \( A \) intersects \( BC \), we consider all the possible directions that this line could take. Since an equilateral triangle is symmetrical, we can limit our analysis to just the range of angles that line \( l \) can make with respect to one of the ... | P = \frac{1}{3} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
A natural disaster in Tokyo, Japan, caused ¥4 billion in damage. That estimate was in Japanese yen. At the time of the disaster, 100 Japanese yen was worth 0.9 US dollars. Convert the damage into US dollars, and then, given that 1 US dollar is valued at 0.93 euros, convert this amount into euros. | 1. Since 100 Japanese yen is equivalent to 0.9 US dollars, 1 Japanese yen is worth $0.9 / 100 = 0.009$ US dollars.
2. Therefore, the damage in US dollars is $0.009 \times 4,000,000,000 = 36,000,000$ US dollars.
3. Given that 1 US dollar is valued at 0.93 euros, converting $36,000,000 US dollars to euros involve... | \text{€33,480,000} | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
7. In a certain country, the demand function for Moroccan tangerines is $\mathrm{Q}_{\mathrm{d}}(\mathrm{p})=50-\mathrm{p}$, the marginal costs of production and supply of any tangerines are constant and equal to 5, and the tangerine market operates under conditions of perfect competition. An embargo, imposed on tanger... | # Solution:
Before the embargo, there was equilibrium in the competitive market for Moroccan mandarins. The price of the good should be equal to the marginal cost of its production, as the marginal revenue of any supplier is equal to the price of the good. Therefore, Moroccan mandarins were consumed in a volume of $\m... | 1.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $M = \{x | y = \sqrt{x-1}\}$ and $N = \{y | y = \ln x\}$. Find the intersection $M \cap N$. | First, let's determine the domains of the functions that define the sets $M$ and $N$. The function $y = \sqrt{x-1}$ has a domain of $x \geq 1$ because the square root is only defined for non-negative numbers. The function $y = \ln x$ has a domain of $x > 0$ because the natural logarithm is only defined for positive num... | \{(a, \ln a)\} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
A set is called $six\ square$ if it has six pair-wise coprime numbers and for any partition of it into two set with three elements, the sum of the numbers in one of them is perfect square. Prove that there exist infinitely many $six\ square$. | null | null | Number Theory | proof | Yes | Yes | aops_forum | false |
Find the coordinates of the point symmetric to point $(1, -2, 3)$ with respect to the $x$-axis. | To find the coordinates of the point symmetric to a given point with respect to the $x$-axis, we need to change the signs of the $y$ and $z$ coordinates, while keeping the $x$ coordinate unchanged. This is because the reflection across the $x$-axis only affects the $y$ and $z$ coordinates.
Given the point $(1, -2, 3)... | (1, 2, -3) | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
65. Using the sequential compactness lemma and the infinite lemma, provide an independent proof of Proposition 8.5.1. | None
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
Note: The provided instruction is a meta-instruction and not part of the text to be translated. Since the text to be translated is "None", the translation is also "None". ... | notfound | Calculus | proof | Yes | Problem not solved | olympiads | false |
Which of the following functions has a minimum positive period of $π$?
A: $y=\cos^2x$
B: $y=|\sin \frac{x}{2}|$
C: $y=\sin x$
D: $y=\tan \frac{x}{2}$ | For option A, $y=\cos^2x$ can be rewritten using the double-angle identity as $y=\frac{1}{2}\cos 2x + \frac{1}{2}$. The cosine function $\cos 2x$ has a period of $π$, so the minimum positive period of $y=\cos^2x$ is $T=π$.
For option B, considering the function $y=\sin \frac{x}{2}$ has a period of $\frac{2π}{\frac{1}{... | \text{The correct answer is A.} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Given the function $f(x) = ax + \ln{x}$, where $a \in \mathbb{R}$,
(I) Find the extrema of the function $f(x)$.
(II) For any two distinct points $P_1(x_1, y_1), P_2(x_2, y_2)$ on the curve, there exists a point $Q(x_0, y_0)$ on the curve such that $x_1 < x_0 < x_2$ and the tangent line at $Q$, denoted by $l$, is para... | (For simplicity, the proof of uniqueness is omitted here, as the key focus is on showing the existence of the tangent line.)
(I) Starting with the function $f(x) = ax + \ln{x}$, we differentiate it: $f'(x) = a + \frac{1}{x}$, for $x > 0$.
When $a \geq 0$, on the interval $(0, +\infty)$, $f'(x) > 0$, so the function $... | null | Calculus | proof | Yes | Yes | cn_k12 | false |
The force needed to loosen a bolt varies inversely with the length of the handle of the wrench used and directly with a factor depending on the material of the bolt. A steel bolt adds an extra 50 pounds of force regardless of the wrench used due to its toughness. A wrench with a handle length of 12 inches requires 300 ... | The initial setup for the wrench is 12 inches and 300 pounds. When the length changes from 12 inches to 18 inches, we multiply the wrench length by $\frac{18}{12} = 1.5$. Since the force required varries inversely with the wrench length, to keep the product constant, we must multiply the initial force requirement (excl... | 250 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Example 2. Determine all real solutions of the system of equations
$$
\left\{\begin{array}{l}
x+y+z=3, \\
x^{2}+y^{2}+z^{2}=3, \\
x^{5}+y^{5}+z^{5}=3
\end{array}\right.
$$
(2nd United States of America Mathematical Olympiad) | Let the plane $\pi: x+y+z=3$, the sphere $O: x^{2}+y^{2}+z^{2}=3$, and $d=\frac{|-3|}{\sqrt{3}}=\sqrt{3}=R$. By Theorem 2 (2), the plane $\pi$ is tangent to the sphere $O$, and the coordinates of the point of tangency are the solutions to the system of equations $\left\{\begin{array}{l}x+y+z=3, \\ x^{2}+y^{2}+z^{2}=3\e... | x=1, y=1, z=1 | Algebra | math-word-problem | Yes | Yes | cn_contest | false |
In triangle $\triangle ABC$, if $\cos A=\frac{{\sqrt{3}}}{2}$ and $\tan B=\sqrt{3}$, this triangle must be ( )
A: acute triangle
B: right triangle
C: obtuse triangle
D: isosceles triangle | In $\triangle ABC$, given that $\cos A = \frac{\sqrt{3}}{2}$ and $\tan B = \sqrt{3}$, we can determine the angles of the triangle step by step.
First, we find the measure of $\angle A$ using the given cosine value:
\[
\cos A = \frac{\sqrt{3}}{2} \implies \angle A = 30^{\circ}.
\]
This is because the cosine of $30^{\ci... | B | Geometry | MCQ | Yes | Yes | cn_k12 | false |
Let $ABCD$ by a cyclic quadrilateral with circumcenter $O$. Let $P$ be the point of intersection of the diagonals $AC$ and $BD$, and $K, L, M, N$ the circumcenters of triangles $AOP, BOP$, $COP, DOP$, respectively. Prove that $KL = MN$. | 1. **Given**: We have a cyclic quadrilateral \(ABCD\) with circumcenter \(O\). The diagonals \(AC\) and \(BD\) intersect at point \(P\). The circumcenters of triangles \(AOP\), \(BOP\), \(COP\), and \(DOP\) are \(K\), \(L\), \(M\), and \(N\) respectively. We need to prove that \(KL = MN\).
2. **Angles and Radii**:
... | null | Geometry | proof | Yes | Yes | aops_forum | false |
The diagonal lengths of a rhombus are 30 units and 18 units. Calculate the area of the rhombus. | To find the area of a rhombus, we use the same formula provided: \[ \text{Area} = \frac{d_1 \times d_2}{2} \]
Here, the diagonal lengths are $d_1 = 30$ units and $d_2 = 18$ units. Plugging into the formula, we get:
\[ \text{Area} = \frac{30 \times 18}{2} = \frac{540}{2} = \boxed{270} \text{ square units} \]
Conclusion... | 270 | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
If the function $f(x) = kx^2 + (k - 1)x + 3$ is an even function, find the value of $k$. | Since $f(x) = kx^2 + (k - 1)x + 3$ is an even function,
$$\begin{align*}
f(-x) &= f(x) \\
k(-x)^2 + (k - 1)(-x) + 3 &= kx^2 + (k - 1)x + 3 \\
kx^2 - (k - 1)x + 3 &= kx^2 + (k - 1)x + 3 \\
2(k - 1)x &= 0
\end{align*}$$
As $x$ is not always equal to $0$,
$$\begin{align*}
k - 1 &= 0 \\
k &= 1
\end{align*}$$
Therefore, t... | k = 1 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
If the solution set of the inequality $|x-2|+|x+3| \geq a$ is $\mathbb{R}$, then the range of values for $a$ is ____. | The expression $|x-2|+|x+3|$ represents the sum of the distances on the number line from the point corresponding to $x$ to the points corresponding to $2$ and $-3$, respectively. Its minimum value is $5$. Therefore, when $a \leq 5$, the solution set of the inequality $|x-2|+|x+3| \geq a$ is $\mathbb{R}$.
Hence, the an... | (-\infty, 5] | Inequalities | math-word-problem | Yes | Yes | cn_k12 | false |
A grocer makes a pyramid display of cans where the top row has three cans and each successive lower row has four more cans than the row above it. If the display contains 225 cans, how many rows does it contain? | The sequence of cans in each row now starts with 3 and increases by a common difference of 4. The number of cans in each row would be 3, 7, 11, $\dots$, $4n - 1$ for \( n \) rows.
The sum of this sequence can be calculated using the formula for the sum of an arithmetic series:
\[ S_n = \frac{n}{2} \times (\text{first ... | n = 10 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
For the cubic function $f(x)=ax^{3}+bx^{2}+cx+d$ ($a\neq 0$), let's define it; let $f′(x)$ be the derivative of the function $y=f(x)$, and $f″(x)$ be the derivative of the function $f′(x)$. If the equation $f″(x)=0$ has a real solution $x_{0}$, then the point $(x_{0},f(x_{0}))$ is called the "inflection point" of the f... | **Analysis**
This question mainly examines the basic operations of derivatives. Using the given conditions to find the function's center of symmetry is key to solving this problem. The process of summing uses the method of adding in reverse order.
(1) Using the given conditions, taking the second derivative of the ... | 2016 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Given an odd function $f(x)$ defined on $\mathbb{R}$ that satisfies $f(x+2) = -\frac{1}{f(x)}$, and $f(x) = 3^x$ on the interval $(0, 1)$, find the value of $f(\log_{3}{54})$.
A: $\frac{3}{2}$
B: $\frac{2}{3}$
C: $-\frac{3}{2}$
D: $-\frac{2}{3}$ | From the functional equation $f(x+2) = -\frac{1}{f(x)}$, we infer that:
\[ f(x+4) = -\frac{1}{f(x+2)} = f(x). \]
Hence, the function $f(x)$ has a period of 4.
Since $f(x)$ is an odd function defined on $\mathbb{R}$, and $3 < \log_{3}{54} < 4$, and knowing that $f(x) = 3^x$ on the interval $(0, 1)$, we have:
\begin{ali... | -\frac{3}{2} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
The area of a rectangle is 625 square centimeters. If the length of the rectangle is increased by $20\%$ and the width is decreased by $20\%$, what would the new area be? Express your answer as a whole number. | 1. Let the original length \( L \) and width \( W \) of the rectangle be such that \( L \times W = 625 \).
2. The new length and width are \( 1.2L \) and \( 0.8W \) respectively.
3. The new area \( A' \) is calculated as follows:
\[
A' = (1.2L) \times (0.8W) = 0.96LW = 0.96 \times 625 = 600
\]
Thus, the new... | 600\text{ square centimeters} | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
The solution of $ \sqrt{5x\minus{}1}\plus{}\sqrt{x\minus{}1}\equal{}2$ is:
$ \textbf{(A)}\ x\equal{}2,x\equal{}1 \qquad
\textbf{(B)}\ x\equal{}\frac{2}{3} \qquad
\textbf{(C)}\ x\equal{}2 \qquad
\textbf{(D)}\ x\equal{}1 \qquad
\textbf{(E)}\ x\equal{}0$ | null | null | Algebra | MCQ | Yes | Yes | aops_forum | false |
For $c > 0$, when nonzero real numbers $a$ and $b$ satisfy $a^{2}-2ab+2b^{2}-c=0$ and make $a+b$ maximum, the minimum value of $\frac {3}{a}- \frac {4}{b}+ \frac {5}{c}$ is ( ).
A: $- \frac {1}{4}$
B: $\frac {1}{4}$
C: $\frac {1}{2}$
D: $- \frac {1}{3}$ | Given $a^{2}-2ab+2b^{2}-c=0$,
we can rewrite it as:
$$c=(a-b)^{2}+b^{2}.$$
By applying the Cauchy-Schwarz inequality, we have:
$$(a-b)^{2}+b^{2} \geqslant \frac{1}{5}(a+b)^{2}.$$
To maximize $a+b$, we set $a-b = \frac {b}{2}$, which then implies $a= \frac {3}{2}b$.
Substituting this into our expression for $c$, we ge... | A | Algebra | MCQ | Yes | Yes | cn_k12 | false |
## Task 3 - 220713
Two agricultural production cooperatives (LPG) A and B want to clean a drainage ditch that is $2.4 \mathrm{~km}$ long.
LPG A owns $1.5 \mathrm{~km}$ of it, and LPG B owns the remaining $0.9 \mathrm{~km}$.
To ensure that this important work is completed in a short time, LPG C also helps. The three ... | Each LPG cleaned a third of the $2.4 \mathrm{~km}$ long ditch, which is $0.8 \mathrm{~km}$. This length therefore corresponds to the $240 \mathrm{M}$ that LPG C receives. For every $0.1 \mathrm{~km}$, LPG C thus receives $30 \mathrm{M}$. Of the $0.9 \mathrm{~km}$ cleaned by LPG B, $0.8 \mathrm{~km}$ were cleaned with t... | LPG\A\pays\210\M,\LPG\B\pays\30\M | Algebra | math-word-problem | Yes | Yes | olympiads | false |
[ Polygons (inequalities) $]$
a) Convex polygons $A_{1} \ldots A_{\mathrm{n}}$ and $B_{1} \ldots B_{\mathrm{n}}$ are such that all their corresponding sides, except for $A_{1} A_{\mathrm{n}}$ and $B_{1} B_{\mathrm{n}}$, are equal, and $\angle A_{2} \geq \angle B_{2}, \ldots, \angle A_{n-1} \geq \angle B_{\mathrm{n}-1}... | a) Suppose first that $\angle A_{\mathrm{i}} > \angle B_{\mathrm{i}}$, and for all other pairs of angles, equality holds. Arrange the polygons so that the vertices $A_{1}, \ldots, A_{\mathrm{i}}$ coincide with $B_{1}, \ldots, B_{\mathrm{i}}$. In triangles $A_{1} A_{\mathrm{i}} A_{\mathrm{n}}$ and $A_{1} A_{\mathrm{i}} ... | proof | Geometry | proof | Yes | Yes | olympiads | false |
Valentina bought a foot long burger and shared half with his brother. Find the share of each person in inches. | If Valentina bought a foot long burger, that means the burger is 12 inches long (since 1 foot equals 12 inches). If she shared half of the burger with her brother, then each person would get half of the 12 inches.
To find the share of each person, we divide the total length of the burger by 2:
12 inches / 2 = 6 inche... | 6 | Other | math-word-problem | Yes | Yes | orca_math | true |
7 A function $f(x)$ defined on $(0,+\infty)$ satisfies: for any $x, y$,
$$
f(x+y)=f(x y) \text {. }
$$
Prove: $f(x)$ is a constant on $(0,+\infty)$. | 【Analysis】First, we obtain that $T=1$ is a period of $f(x)$. To prove $f(a)=f(b)$, we can consider assigning values to $x$ and $y$ in the given conditions cleverly.
Proof It is easy to see that $T=1$ is a period of $f(x)$.
For any $a, b>0$, since there are infinitely many positive integers, we can always find a positiv... | proof | Algebra | proof | Yes | Yes | cn_contest | false |
42. In the right-angled $\triangle A B C$, find the largest positive real number $k$ such that the inequality $a^{3}+b^{3}+c^{3} \geqslant k(a+$ $b+c)^{3}$ holds. (2006 Iran Mathematical Olympiad) | 42. $a^{3}+b^{3}+c^{3} \geqslant k(a+b+c)^{3} \Leftrightarrow k \leqslant\left(\frac{a}{a+b+c}\right)^{3}+\left(\frac{b}{a+b+c}\right)^{3}+$ $\left(\frac{c}{a+b+c}\right)^{3}$
Let $x=\frac{a}{a+b+c}, y=\frac{b}{a+b+c}, z=\frac{c}{a+b+c}$, then because $a^{2}+b^{2}=c^{2}$, so $x^{2}+y^{2}=z^{2}$
Also, because $x+y+z=1$... | \frac{3 \sqrt{2}-4}{2} | Inequalities | math-word-problem | Yes | Yes | inequalities | false |
7.221. $4^{\lg x+1}-6^{\lg x}-2 \cdot 3^{\lg x^{2}+2}=0$.
7.221. $4^{\lg x+1}-6^{\lg x}-2 \cdot 3^{\lg x^{2}+2}=0$. | ## Solution.
Domain of definition: $x>0$.
From the condition, we have $4 \cdot 2^{2 \lg x}-2^{\lg x} \cdot 3^{\lg x}-18 \cdot 3^{2 \lg x}=0$. Dividing it by $3^{2 \lg x}$, we get $4 \cdot\left(\frac{2}{3}\right)^{2 \lg x}-\left(\frac{2}{3}\right)^{\lg x}-18=0 \Rightarrow\left(\frac{2}{3}\right)^{\lg x}=-2$ (no soluti... | 0.01 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Let $n$ be a four-digit positive integer as the product of three prime factors $x$, $y$, and $10x+y$, where $x < 5$ and $y < 5$. Find the maximum value of $n$ when $x$ and $y$ are distinct primes. | Given $x$ and $y$ are distinct primes below 5, the prime candidates are $2$ and $3$. This leads to two possibilities for primes:
- $10x + y = 10(2) + 3 = 23$
- $10x + y = 10(3) + 2 = 32$
For calculating $n = xy(10x + y)$:
1. With $x = 2$ and $y = 3$, we estimate \(n = 2 \times 3 \times 23 = 138\).
2. With $x = 3$, and... | 138 | Number Theory | math-word-problem | Yes | Problem not solved | synthetic_math | true |
## Task 11/68
If three natural numbers $a, b, c$ form a Pythagorean triple (i.e., $a^{2}+b^{2}=c^{2}$), and one of the two numbers $a$ or $b$ is a prime number, then the other two numbers are consecutive, and the prime number is the smallest of the three numbers.
This statement is to be proven. Furthermore, it is to ... | Assuming that among the three natural numbers $a, b, c$, for which $a^{2}+b^{2}=c^{2}$ holds, $a$ is a prime number: $a=p$. Then it follows that
$$
a^{2}=p^{2}=c^{2}-b^{2}=(c+b)(c-b)
$$
From this, it already follows that $p^{2}c-b$ must, due to the prime number property of $p$, satisfy $p^{2}=c+b$ and $c-b=1$. This i... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
Find the focus of the parabola $y = -2x^2 + 5.$ | The standard form of a parabola with a vertical axis of symmetry is $y = ax^2 + bx + c.$ For the given parabola $y = -2x^2 + 5,$ it can be rewritten to fit the form where $a = -2,$ $b = 0,$ and $c = 5.$
The vertex form of a parabola $y = a(x-h)^2 + k$ is useful because the vertex $(h, k)$ provides information about th... | (0, \frac{39}{8}) | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Find the number of moles of Hydrochrolic acid formed on combining 3 moles of Sodium chloride and 3 moles of Nitric acid | The reaction between sodium chloride (NaCl) and nitric acid (HNO3) does not produce hydrochloric acid (HCl) directly. Instead, the reaction between NaCl and HNO3 typically forms sodium nitrate (NaNO3) and hydrogen chloride (HCl) gas, but this is not a common reaction and usually requires specific conditions to occur.
... | HNO3 | Other | math-word-problem | Incomplete | Yes | orca_math | true |
A rabbit is placed on a $2n\times 2n$ chessboard. Every time the rabbit moves to one of the adjacent squares. (Adjacent means sharing an edge). It is known that the rabbit went through every square and came back to the place where the rabbit started, and the path of the rabbit form a polygon $\mathcal{P}$. Find the max... | 1. **Define the Problem and Variables:**
- We are given a $2n \times 2n$ chessboard.
- A rabbit moves through every square exactly once and returns to the starting point, forming a polygon $\mathcal{P}$.
- We need to find the maximum possible number of vertices of $\mathcal{P}$.
2. **Initial Observation:**
... | null | Combinatorics | math-word-problem | Yes | Yes | aops_forum | false |
Find all primes $p \ge 3$ such that $p- \lfloor p/q \rfloor q$ is a square-free integer for any prime $q<p$. | To solve the problem, we need to find all prime numbers \( p \geq 3 \) such that \( p - \lfloor p/q \rfloor q \) is a square-free integer for any prime \( q < p \).
1. **Check small primes:**
- For \( p = 3 \):
\[
p - \lfloor p/q \rfloor q = 3 - \lfloor 3/q \rfloor q
\]
For \( q = 2 \):
\[
... | \{3, 5, 7, 13\} | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
How many distinct numbers can you get by multiplying two or more different members of the set $\{1, 2, 3, 7, 13, 17\}$ together? | Here, the distinct numbers obtained from multiplying two or more members of the set are considered. Multiplying by $1$ doesn’t change the product and thus will not be considered alone but can combine with other numbers.
We start by counting combinations and respective products ($nCr$ implies combinations of `n` items ... | 57 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
Find the imaginary part of the complex number $\dfrac {i^{2}}{2i-1}$ (where $i$ is the imaginary unit).
A: $\dfrac {1}{5}i$
B: $\dfrac {2}{5}$
C: $- \dfrac {1}{5}i$
D: $- \dfrac {1}{5}$ | First, simplify the given complex number.
$$\dfrac {i^{2}}{2i-1} = \dfrac {-1}{2i-1} = \dfrac {-(-1-2i)}{(-1+2i)(-1-2i)}$$
Multiply the numerator and denominator by the conjugate of the denominator to eliminate the imaginary part of the denominator:
$$\dfrac {-(-1-2i)}{(-1+2i)(-1-2i)} = \dfrac {1+2i}{1+4} = \dfrac {1}{... | \dfrac {2}{5} | Algebra | MCQ | Yes | Yes | cn_k12 | false |
How many moles of MgSO4 are formed when 2 moles of Mg react with 2 moles of H2SO4 along with 2 moles of H2? | To determine the number of moles of MgSO4 formed, we need to look at the balanced chemical equation for the reaction between magnesium (Mg) and sulfuric acid (H2SO4). The balanced equation is:
Mg + H2SO4 → MgSO4 + H2
From the balanced equation, we can see that 1 mole of Mg reacts with 1 mole of H2SO4 to produce 1 mol... | 2 | Other | math-word-problem | Incomplete | Yes | orca_math | true |
50. (10-11 grades) Construct a right triangle given the hypotenuse and the median of one of the legs. | 50. If in a right triangle $ABC$ with a right angle at $B$, the median line $OD$ ($AO=OC$, $BD=DC$) is drawn, then $\angle ODC=90^{\circ}$ and $AB=2OD$. The construction plan is as follows: on the hypotenuse $AC$ and on the segment $OC$, we construct circles, from point $A$ as the center with radius $m_a$, we draw an a... | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false |
Given an arithmetic sequence $\{a_n\}$, the sum of the first $n$ terms is denoted as $S_n$. If $a_1=-11$ and $a_4+a_6=-6$, find the value of $n$ when $S_n$ is minimized. | **Analysis**
This problem mainly examines the properties of arithmetic sequences. Being familiar with the formula for the sum of the first $n$ terms of an arithmetic sequence is key to solving this problem. It is a common type of question in college entrance exams and is considered a basic question.
**Solution**
Giv... | 6 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
In a class of 25 students, the mean number of candies each student takes is 6. If every student takes at least 2 pieces of candy, what is the maximum number of pieces one student could have taken? | 1. Calculate the total number of candies distributed:
\[
\text{Total candies} = 6 \times 25 = 150
\]
2. Calculate the minimum candies taken by the first 24 students, given each takes at least 2 pieces:
\[
\text{Minimum candies for 24 students} = 2 \times 24 = 48
\]
3. Determine the maximum number of... | 102 | Logic and Puzzles | math-word-problem | Yes | Yes | synthetic_math | true |
90 students represent x percent of the boys at jones elementary school . if the boys at jones elementary make up 30 % of the total school population of x students , what is x ? | Let's denote the total number of students at Jones Elementary School as X. According to the information given, boys make up 30% of the total school population. Therefore, the number of boys at the school is 0.30X.
We are also told that 90 students represent some percentage of the boys at the school. This means that 90... | 300 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
10 years ago, the age of Peter was one-third the age of Jacob at that time. The present age of Jacob is 12 years more than the present age of Peter. What is the present age of Peter? | Let's denote Peter's present age as P and Jacob's present age as J.
According to the information given, Jacob is currently 12 years older than Peter. So we can write the following equation:
J = P + 12 (Equation 1)
We are also told that 10 years ago, Peter's age was one-third of Jacob's age at that time. Ten years a... | 16 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Example 1 Draw the approximate graph of the following functions:
(1) $y=\frac{1}{x+1}$;
(2) $y=|x+1|+2|x-3|$;
(3) $y=\left\{\begin{array}{l}x^{2}+x, x \geqslant 0, \\ x^{2}-x, x<0;\end{array}\right.$
(4) $y=2-\left|x-x^{2}\right|$. | Solve Figure 3-4 sequentially as the required function graphs.
$$
y=\frac{1}{x+1} \quad y=|x+1|+2|x-3| \quad y=\left\{\begin{array}{ll}
x^{2}+x, & x \geqslant 0 \\
x^{2}-x, & x<0
\end{array} \quad y=2-\left|x-x^{2}\right|\right.
$$
Figure 3-4
Solution Guidance For (1), from $y=\frac{1}{x+1}$, we get $x \neq -1$ and $y... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false |
What is the degree measure of angle $PTV$ when polygon $PQRSTUVW$ is a regular octagon?
[asy]
draw((1,0)--(0.7071,0.7071)--(0,1)--(-0.7071,0.7071)--(-1,0)--(-0.7071,-0.7071)--(0,-1)--(0.7071,-0.7071)--cycle);
draw((0,-1)--(0.7071,0.7071)--(-0.7071,-0.7071)--cycle);
label("P",(0.7071,-0.7071),SE);
label("Q",(1,0),E);
... | The sum of the interior angles of an $n$-sided polygon is given by the formula $180(n-2)$. For a regular octagon ($n=8$), the interior angles sum to $180(8-2) = 1080^\circ$. Therefore, each interior angle of a regular octagon is
$$ \frac{1080^\circ}{8}=135^\circ. $$
Since $\overline{PV}$ and $\overline{PT}$ are sides... | 22.5^\circ | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
As of November 25, 2021, the total box office revenue of the movie "The Battle at Lake Changjin," which interprets the great spirit of resisting the US and aiding Korea, has exceeded 56.9 billion yuan. Expressing 56.9 billion in scientific notation is ( ).
A: $5.69\times 10^{8}$
B: $5.69\times 10^{9}$
C: $56.9\times... | To express $56.9$ billion in scientific notation, we start by understanding that $1$ billion equals $10^9$. Therefore, $56.9$ billion can be written as $56.9 \times 10^9$ yuan. In scientific notation, we aim to have a single digit before the decimal point, so we adjust $56.9$ to $5.69$ by moving the decimal one place t... | B | Algebra | MCQ | Yes | Yes | cn_k12 | false |
Given that the parabola $y=a\left(x-m\right)\left(x-n\right)$ ($a$, $m$, $n$ are real numbers, $a\neq 0$) intersects with the line $y=kx+b$ at $(1, y_{1})$ and $(6, y_{2})$, which of the following statements is correct?
A: If $m+n \gt 7$, $a \gt 0$, then $k \gt 0$
B: If $m+n \gt 7$, $a \lt 0$, then $k \lt 0$
C: If $... | Given the parabola $y=a(x-m)(x-n)$ intersects with the line $y=kx+b$ at points $(1, y_{1})$ and $(6, y_{2})$, we can set up equations for these intersection points:
1. For the point $(1, y_{1})$, the equation is:
\[a(1-m)(1-n) = k + b\]
2. For the point $(6, y_{2})$, the equation is:
\[a(6-m)(6-n) = 6k + b\]
To find ... | D | Algebra | MCQ | Yes | Yes | cn_k12 | false |
A fair price shopkeeper takes 10% profit on his goods. He lost a certain percentage of goods during theft, and his overall loss percent is 23%. What percentage of goods did he lose during theft? | Let's assume the cost price of the goods is $100.
The shopkeeper takes a 10% profit on his goods, so the selling price would be:
$100 + ($100 * 10/100) = $100 + $10 = $110
Now, let's assume the shopkeeper lost 'x' percent of his goods during theft. This means he could only sell (100 - x)% of his goods.
The value of ... | 30\% | Algebra | math-word-problem | Yes | Yes | orca_math | true |
In the plane figure, what are the values of $x$ and $y$?
[asy]
pair A;
draw(dir(50)--A); draw(dir(250)--A); draw(dir(340)--A);
label("$130^{\circ}$",A,dir(150)); label("$x^{\circ}$",A,dir(250)); label("$y^{\circ}$",A,dir(10));
[/asy] | The key principle remains that the angles around a point sum to $360^\circ$. Let's use that to find $x$ and $y$:
1. The known angle measures $130^\circ$. Formulate the equation with $x$ and $y$:
\[
x^\circ + y^\circ + 130^\circ = 360^\circ
\]
2. Simplify the equation to solve for $x + y$:
\[
x^\circ + ... | 230^\circ | Geometry | math-word-problem | Yes | Yes | synthetic_math | true |
The function is $y=2x+\cos x$, then $y'$ equals to ( )
A: $2+\cos x$
B: $2-\sin x$
C: $2+\sin x$
D: $2x+\sin x$ | Solution: For the function $y=2x+\cos x$, we have $y'=2-\sin x$.
Therefore, the correct option is $\boxed{B}$.
This can be solved using the rules of differentiation.
This question tests the application of differentiation rules and computational skills. | B | Calculus | MCQ | Yes | Yes | cn_k12 | false |
Given $f\left(\frac{1}{2}x-1\right)=2x+3$, if $f\left(t\right)=4$, then $t=\_\_\_\_\_\_$. | To solve for $t$ given $f\left(\frac{1}{2}x-1\right)=2x+3$ and $f\left(t\right)=4$, we follow these steps:
1. Start with the equation $f\left(\frac{1}{2}x-1\right)=2x+3$.
2. Let $\frac{1}{2}x-1=t$. This implies $x=2(t+1) = 2t+2$.
3. Substitute $x$ into $2x+3$ to express $f(t)$ in terms of $t$:
$$f(t) = 2(2t+2)+3 =... | -\frac{3}{4} | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Find $4 \cdot 6 \cdot 8 + 18 \div 3 - 2^3.$ | As per the order of operations, known as PEMDAS/BODMAS (Parentheses/Brackets, Orders (i.e., powers and roots, etc.), Division and Multiplication, Addition and Subtraction), resolve the expression:
1. Perform the multiplications and division:
\[
4 \cdot 6 \cdot 8 = 192, \quad 18 \div 3 = 6
\]
2. Perform the exp... | 190 | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
3.252. $\frac{\cos ^{2}(4 \alpha-3 \pi)-4 \cos ^{2}(2 \alpha-\pi)+3}{\cos ^{2}(4 \alpha+3 \pi)+4 \cos ^{2}(2 \alpha+\pi)-1}$. | Solution.
$$
\begin{aligned}
& \frac{\cos ^{2}(4 \alpha-3 \pi)-4 \cos ^{2}(2 \alpha-\pi)+3}{\cos ^{2}(4 \alpha+3 \pi)+4 \cos ^{2}(2 \alpha+\pi)-1}= \\
& =\frac{(\cos (3 \pi-4 \alpha))^{2}-4(\cos (\pi-2 \alpha))^{2}+3}{(\cos (3 \pi+4 \alpha))^{2}+4(\cos (\pi+2 \alpha))^{2}-1}=\frac{\cos ^{2} 4 \alpha-4 \cos ^{2} 2 \alp... | \operatorname{tg}^{4}2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. (8 points) On a plane, there are 7 points, among which no 3 points are collinear. If 18 line segments are drawn between these 7 points, then these line segments can form at most $\qquad$ triangles. | So the answer is: 23. | 23 | Combinatorics | math-word-problem | Yes | Problem not solved | olympiads | false |
Given a point M with cylindrical coordinates ($\sqrt {2}$, $\frac {5\pi}{4}$, $\sqrt {2}$), find its Cartesian coordinates. | To convert from cylindrical coordinates ($\rho$, $\theta$, $z$) to Cartesian coordinates ($x$, $y$, $z$), we use the following transformation equations:
$$
\begin{cases}
x = \rho \cos \theta \\
y = \rho \sin \theta \\
z = z
\end{cases}
$$
Given that point M has the cylindrical coordinates ($\sqrt {2}$, $\frac {5\pi}... | M(-1, -1, \sqrt {2}) | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Find $\sec(150^\circ)$. | Recall that $\sec(\theta) = \frac{1}{\cos(\theta)}$. We need to calculate $\cos(150^\circ)$ first.
Using the cosine of a sum identity,
\[
\cos(150^\circ) = \cos(180^\circ - 30^\circ) = -\cos(30^\circ),
\]
where we use the fact that cosine is negative in the second quadrant and $\cos(30^\circ) = \frac{\sqrt{3}}{2}$.
T... | -\frac{2\sqrt{3}}{3} | Algebra | math-word-problem | Yes | Yes | synthetic_math | true |
Given: The line $y=x-1$ is shifted $2$ units upward to obtain the line $y=kx+b$. Which of the following statements about the line $y=kx+b$ is correct?
A: Passes through the first, second, and fourth quadrants
B: Intersects the $x$-axis at $\left(1,0\right)$
C: Intersects the $y$-axis at $\left(0,1\right)$
D: $y$ de... | To solve the problem, we start by understanding the effect of shifting the line $y = x - 1$ upwards by $2$ units.
**Step 1: Shifting the Line Upwards**
- The original line is $y = x - 1$.
- Shifting it $2$ units upward, we add $2$ to the $y$-value of the line, resulting in $y = x - 1 + 2$.
**Step 2: Simplifying the ... | C | Algebra | MCQ | Yes | Yes | cn_k12 | false |
4. Let $A, B$ be the vertices of the major axis of the ellipse $\Gamma$, $E, F$ be the two foci of $\Gamma$, $|A B|=4$, $|A F|=2+\sqrt{3}$, and $P$ be a point on $\Gamma$ such that $|P E| \cdot|P F|=2$. Then the area of $\triangle P E F$ is $\qquad$ . | Answer: 1.
Solution: Let's assume the standard equation of $\Gamma$ in the Cartesian coordinate system is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$. According to the conditions, we have $2 a=|A B|=4, a \pm \sqrt{a^{2}-b^{2}}=|A F|=2+\sqrt{3}$, which gives $a=2, b=1$, and $|E F|=2 \sqrt{a^{2}-b^{2}}=2 \sqrt{3}$... | 1 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. (15 points) A and B are playing a game of writing numbers on the blackboard: each time they write a number, it must be a positive integer not exceeding 10, and no repetition is allowed; starting from the third number, each number must also be the difference of two previously written numbers. The player who cannot wr... | 8. Player B has a winning strategy.
Consider the first number written by Player A.
If Player A writes an even number, Player B writes half of this number, then Player A cannot write again.
If Player A writes an odd number:
(1) If Player A writes 1, then Player B writes 2; if Player A writes 3, then Player B writes 6; ... | Player B has a winning strategy. | Logic and Puzzles | math-word-problem | Yes | Yes | cn_contest | false |
Solve the following system:
$$
\left\{\begin{aligned}
x+y+z & =2 \\
x^{2}+y^{2}+z^{2} & =26 \\
x^{3}+y^{3}+z^{3} & =38
\end{aligned}\right.
$$ | We find that the solutions of the system, after calculations, are $(1,-3,4)$ and the associated permutations.
## 3 Polynomials - 2nd part | (1,-3,4) | Algebra | math-word-problem | Yes | Problem not solved | olympiads | false |
[b]p1.[/b] The least prime factor of $a$ is $3$, the least prime factor of $b$ is $7$. Find the least prime factor of $a + b$.
[b]p2.[/b] In a Cartesian coordinate system, the two tangent lines from $P = (39, 52)$ meet the circle defined by $x^2 + y^2 = 625$ at points $Q$ and $R$. Find the length $QR$.
[b]p3.[/b] F... | 1. Since the least prime factor of \(a\) is 3, \(a\) must be divisible by 3 and not by any smaller prime number. Therefore, \(a\) is an odd number.
2. Similarly, since the least prime factor of \(b\) is 7, \(b\) must be divisible by 7 and not by any smaller prime number. Therefore, \(b\) is also an odd number.
3. The s... | 2 | Number Theory | math-word-problem | Yes | Yes | aops_forum | false |
Let $q(x)$ be defined on $3 \le x \le 15$ such that:
$$q(x) = \begin{cases} x + 2 &\quad \lfloor x \rfloor\text{ is an odd prime} \\ q(z) + (x + 2 - \lfloor x \rfloor) &\quad \text{otherwise} \end{cases}$$
where $z$ is the greatest prime factor of $\lfloor x\rfloor.$ Express the range of $q$ in interval notation. | By the definition of $q$, for any odd prime number $x$ such that $3 \le x \le 15$, then $[x+2, x+3) \subset \text{range}\,(q)$. It follows that $[5,6) \cup [7,8) \cup [13,14) \subset \text{range}\,(q)$. For composite numbers, the largest prime factor of a composite number less than or equal to $15$ is $13$. Hence, the ... | [5,10) \cup [13,14) | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
Both \(a\) and \(b\) are positive integers and \(b > 1\). When \(a^b\) is the greatest possible value less than 500, what is the sum of \(a\) and \(b\)? | To solve this problem, we'll examine perfect powers under the revised constraint:
- The largest perfect square less than 500 is \(22^2 = 484\).
- The largest perfect cube less than 500 is \(7^3 = 343\).
- For higher powers, we check the fifth powers. The largest \(a^5\) under 500 is \(3^5 = 243\).
- The seventh power,... | 24 | Number Theory | math-word-problem | Yes | Yes | synthetic_math | true |
59. (LUX 1) (SL\&8-18). | None | notfound | Other | math-word-problem | Incomplete | Problem not solved | olympiads | false |
Given the function $$f(x)= \begin{cases} x^{2}+1 & (x\leq0) \\ -2x & (x>0)\end{cases},$$ then $f(3)=$ ______. | Since $$f(x)= \begin{cases} x^{2}+1 & (x\leq0) \\ -2x & (x>0)\end{cases},$$
then $f(3)=-2\times3=-6.$
Therefore, the answer is $-6.$
To find the value of $f(3),$ we simply apply the function's definition.
This question tests the evaluation of functions, specifically piecewise functions, and is considered a basi... | -6 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of the mixture is X? | Let's assume we have a total of 100 pounds of the mixture of X and Y. We want to find out how many pounds of X and Y we have in this mixture.
Let x be the weight of mixture X in the total mixture, and y be the weight of mixture Y in the total mixture. We have two equations:
1. x + y = 100 (since the total weight of t... | 33.33\% | Algebra | math-word-problem | Yes | Yes | orca_math | true |
A proposition related to positive integers states: If the proposition holds true when $n=k$ ($k \in \mathbb{N}^*$), then it can be deduced that the proposition also holds true when $n=k+1$. It is known that the proposition does not hold true when $n=5$, then we can conclude that
A: The proposition does not hold true w... | The equivalent proposition of "If the proposition holds true when $n=k$, then the proposition also holds true when $n=k+1$" is "If the proposition does not hold true when $n=k+1$, then the proposition also does not hold true when $n=k$." Therefore, the correct answer is $\boxed{\text{A}}$. | \text{A} | Number Theory | MCQ | Yes | Yes | cn_k12 | false |
For example, let $p(x)$ be a polynomial with integer coefficients, and there are six distinct integers $a_{1}, a_{2}, \cdots, a_{5}$ such that $p\left(a_{1}\right)=p\left(a_{2}\right)=\cdots$ $=p\left(a_{6}\right)=-12$. Prove that there does not exist an integer $k$ such that $p(k)=0$. | Prove that if $F(x)=p(x)+12$, then $F\left(a_{1}\right)=F\left(a_{2}\right)=\cdots=F\left(a_{6}\right)=0$. By the Factor Theorem, $x-a_{1}, x-a_{2}, \cdots, x-a_{6}$ are factors of $F(x)$. Therefore, we can set $F(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \cdots\left(x-a_{6}\right) Q_{1}(x)$, where $Q_{1}(x)$ is a po... | proof | Algebra | proof | Yes | Yes | olympiads | false |
Given an ellipse $E$ with eccentricity $\frac{\sqrt{2}}{2}$ and the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a > b > 0$) whose focal distance is 4.
(1) Find the equation of ellipse $E$;
(2) Suppose there is a circle with its center at the origin $O$, and any tangent line to this circle intersects ellipse ... | (1) From the given information, $2c = 4$, $\frac{c}{a} = \frac{\sqrt{2}}{2}$, thus $c = 2$, $a = 2\sqrt{2}$.
Therefore, $b^2 = a^2 - c^2 = 4$, so the equation of ellipse $E$ is $\frac{x^2}{8} + \frac{y^2}{4} = 1$;
(2) Assume there exists a circle centered at the origin such that any tangent line to this circle int... | 2\sqrt{3} | Geometry | math-word-problem | Yes | Yes | cn_k12 | false |
Janaína bought a different deck: there are 180 cards numbered from 1 to 180, with the even cards being red and the odd cards being blue.
a) How many cards are multiples of 7 and red?
b) How many cards are multiples of 7 or red?
c) Janaína randomly chose 80 cards, all, coincidentally, black. What is the minimum numbe... | Solution
a) Since the red cards are even, the red cards that are multiples of 7 must also be multiples of 2, that is, they must be multiples of 14. The first multiple of 14 is $14=1 \cdot 14$, while the last is $168=12 \cdot 14$, which means there are 12 red cards that are multiples of 7.
b) The first odd multiple of... | 12,103,3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Hot dog buns come in packages of 8. For the school picnic, Mr. Gates bought 30 packages of hot dog buns. He has some classes with 30 students in each class. Each of Mr. Gates' students can get 2 hot dog buns. How many classes does Mr. Gates have? | First, let's calculate the total number of hot dog buns Mr. Gates bought:
30 packages * 8 buns per package = 240 buns
Each student gets 2 buns, so we can calculate how many students can be served with 240 buns:
240 buns / 2 buns per student = 120 students
Since each class has 30 students, we can now calculate how m... | 4 | Logic and Puzzles | math-word-problem | Yes | Yes | orca_math | true |
$ 2007^2$ unit squares are arranged forming a $ 2007\times 2007$ table. Arnold and Bernold play the following game: each move by Arnold consists of taking four unit squares that forms a $ 2\times 2$ square; each move by Bernold consists of taking a single unit square. They play anternatively, Arnold being the first. Wh... | 1. **Initial Setup and Definitions**:
- We have a \(2007 \times 2007\) grid of unit squares.
- Arnold's move: taking four unit squares that form a \(2 \times 2\) square.
- Bernold's move: taking a single unit square.
- Arnold starts first, and they alternate turns.
- When Arnold can no longer make a move... | null | Logic and Puzzles | math-word-problem | Yes | Yes | aops_forum | false |
10. (CZE) Find all positive integers $k$ for which the following statement is true:
If $F(x)$ is a polynomial with integer coefficients satisfying the condition
$$
0 \leq F(c) \leq k \quad \text { for each } c \in\{0,1, \ldots, k+1\},
$$
then $F(0)=F(1)=\cdots=F(k+1)$. | 10. Suppose that $k \geq 4$. Consider any polynomial $F(x)$ with integer coefficients such that $0 \leq F(x) \leq k$ for $x=0,1, \ldots, k+1$. Since $F(k+1)-F(0)$ is divisible by $k+\overline{1}$, we must have $F(k+1)=F(0)$. Hence
$$
F(x)-F(0)=x(x-k-1) Q(x)
$$
for some polynomial $Q(x)$ with integer coefficients. In pa... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false |
## Task A-4.5.
Given are 100 points arranged in 10 rows and 10 columns. Each point is either red or blue. A point is in a corner if it is at the end of both a row and a column, and on the edge if it is at the end of a row or a column, but not both. Two points are adjacent if they are in the same row or column, with no... | ## Solution.
Let $P$ be the total number of blue, and $C$ be the total number of red segments.
The total number of points on the boundary is 32, so the number of blue points on the boundary is 16. Exactly two blue points are in the corners, so the total number of blue points in the interior is 30.
Let's count all th... | 37 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
N1) Find all integer values that the expression
$$
\frac{p q+p^{p}+q^{q}}{p+q}
$$
can take, where $p$ and $q$ are prime numbers. | Answer: The only attainable relative integer is 3. Solution: If $p$ and $q$ are odd, the numerator is odd but the denominator is even. Since an even number never divides an odd number, the expression is not a natural integer. Therefore, we can assume that at least one of the two primes is even. Since the expression is ... | 3 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
For every positive integer $ n$ consider
\[ A_n\equal{}\sqrt{49n^2\plus{}0,35n}.
\]
(a) Find the first three digits after decimal point of $ A_1$.
(b) Prove that the first three digits after decimal point of $ A_n$ and $ A_1$ are the same, for every $ n$. | 1. **Finding the first three digits after the decimal point of \( A_1 \):**
Given:
\[
A_1 = \sqrt{49 \cdot 1^2 + 0.35 \cdot 1} = \sqrt{49 + 0.35} = \sqrt{49.35}
\]
To find the first three digits after the decimal point of \( \sqrt{49.35} \), we can approximate it. We know:
\[
\sqrt{49} = 7
\]
... | 024 | Calculus | math-word-problem | Yes | Yes | aops_forum | false |
Each cell of a $100 \times 100$ grid contains an arrow pointing up, down, right, or left. The grid is surrounded by impassable walls, except for an opening at the level of the right side of the cell located at the top-right corner. A chip moves around the grid as follows: when it is on a cell, it moves one cell in the ... | We reason by contradiction, assuming that the chip does not leave the grid. Then it wanders indefinitely on the grid, and there is therefore at least one cell that it visits infinitely many times. Since the arrow turns between consecutive visits, the chip also passes an infinite number of times through all neighboring ... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
If the proposition $(\neg p) \lor (\neg q)$ is a false proposition, then among the following conclusions: ① The proposition $p \land q$ is true;
② The proposition $p \land q$ is false; ③ The proposition $p \lor q$ is true; ④ The proposition $p \lor q$ is false.
The correct ones are: ( )
A: ①③
B: ②④
C: ②③
D:... | Given that the proposition $(\neg p) \lor (\neg q)$ is false, it means that both $\neg p$ and $\neg q$ are false. Therefore, both $p$ and $q$ must be true. If $p$ and $q$ are both true, then the proposition $p \land q$ is true, and the proposition $p \lor q$ is also true. Hence, the correct conclusions are ① and ③.
Th... | \text{A} | Logic and Puzzles | MCQ | Yes | Yes | cn_k12 | false |
Eight distinct points are identified on the circumference of a circle. How many different triangles can be formed if each vertex must be one of these 8 points? | To form a triangle, we must choose 3 of the 8 points to be the vertices. Using the combination formula to select 3 points out of 8, we compute:
\[
\binom{8}{3} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = \frac{336}{6} = 56
\]
Thus, the number of different triangles that can be formed is $\boxed{56}$. | 56 | Combinatorics | math-word-problem | Yes | Yes | synthetic_math | true |
On the minor axis of an ellipse, at a distance $e=O F$ (where $O$ is the center of the ellipse, and $F$ is one of the foci), we take two points and from these, we draw perpendiculars to an arbitrary tangent. Show that the sum of the squares of these perpendiculars is $2 a^{2}$. | (In the October issue, it was stated incorrectly.) We take two points on the minor axis of an ellipse at a distance of $e=O F$ from the center (where $O$ is the center of the ellipse, and $F$ is one of the foci), and from these points, we draw perpendiculars to an arbitrary tangent. Show that the sum of the squares of ... | 2^{2} | Geometry | proof | Yes | Yes | olympiads | false |
4. On all black squares of the first six rows of a chessboard, there are pawns (drawing on the right). In each move, a pawn jumps over one of the pawns that are diagonally adjacent to it and moves two squares diagonally to a free square, and the jumped-over pawn is removed from the board. Can the game be played in such... | Solution. We divide the rows of the chessboard into three classes, so that the first, fourth, and seventh rows are in the first class, the second, fifth, and eighth rows are in the second class, and the third and sixth rows are in the third class (see diagram). At the beginning, each class has eight pawns. Note that af... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
There are some red peaches, 15 yellow peaches, and 8 green peaches in the basket. There are 30 peaches in the basket. How many red peaches are in the basket? | To find out how many red peaches are in the basket, we need to subtract the number of yellow and green peaches from the total number of peaches.
Number of yellow peaches = 15
Number of green peaches = 8
Total number of peaches = 30
Number of red peaches = Total number of peaches - (Number of yellow peaches + Number o... | 7 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Daryl is loading crates at a warehouse and wants to make sure that they are not overloaded. Each crate can weigh up to a certain weight and he has 15 crates he can fill. He has 4 bags of nails to load, each of which weighs 5kg; he has 12 bags of hammers, each of which weighs 5 kg; he also has 10 bags of wooden planks, ... | First, let's calculate the total weight of all the items Daryl has to load into the crates:
4 bags of nails * 5 kg/bag = 20 kg
12 bags of hammers * 5 kg/bag = 60 kg
10 bags of wooden planks * 30 kg/bag = 300 kg
Total weight = 20 kg + 60 kg + 300 kg = 380 kg
Daryl has to leave 80 kg out of the crates, so the weight t... | 20 | Algebra | math-word-problem | Yes | Yes | orca_math | true |
Given that $P\left( \frac{3}{2}, -\frac{3\sqrt{3}}{2}\right)$ is the lowest point on the graph of the function $y=A\sin(\omega x+\varphi)$ ($\omega > 0$), and $M$, $N$ are the two highest points adjacent to $P$. If $\angle MPN=60^{\circ}$, then the smallest positive period of this function is ____.
A: $3$
B: $4$
... | From the given information, $\angle MPN=60^{\circ}$ and $NP=MP$, and the lines $MP$, $NP$ pass through the zeros of the function $y$. Let the intersection points be $E$, $F$. Thus, $\triangle EFP$ is an equilateral triangle.
Since $P\left( \frac{3}{2}, -\frac{3\sqrt{3}}{2}\right)$ is the lowest point on the graph of t... | D | Algebra | MCQ | Yes | Yes | cn_k12 | false |
michael cashed a check for $ 1270 and received only $ 10 and $ 50 bills in return . during the course of a day , he used 15 bills and then lost the rest of the money . if the number of $ 10 bills used was either one more or one less than the number of $ 50 bills used , what is the minimum possible amount of money that ... | Let's denote the number of $10 bills as T and the number of $50 bills as F.
From the problem, we know that Michael used 15 bills in total, so we have:
T + F = 15 (1)
We also know that the number of $10 bills used was either one more or one less than the number of $50 bills used. This gives us two possibilities:
T = ... | \$800 | Number Theory | math-word-problem | Yes | Yes | orca_math | true |
The government plans to introduce a "delayed retirement age policy" to address labor shortages caused by an aging population. To understand people's attitudes towards this policy, a random survey of 100 people aged between 15 and 65 was conducted. Using 45 years old as the dividing point, a 2x2 contingency table was ob... | **Step-by-Step Solution:**
1. **Difference in Support for the "Delayed Retirement Age Policy" Among Different Age Groups:**
To determine if there's a significant difference in support for the policy between the two age groups, we use the Chi-square test statistic formula for a 2x2 contingency table:
\[
K^{2}... | E(X) = 1 \text{ person} | Combinatorics | math-word-problem | Yes | Yes | cn_k12 | false |
Given a sequence $\{a_{n}\}$ where $a_{1}=1$, $a_{2}=4$, and $2{a_n}=\frac{{n-1}}{n}{a_{n-1}}+\frac{{n+1}}{n}{a_{n+1}}$ for $n≥2$, find the maximum value of $\frac{{{a_n}}}{n}$. | To solve for the maximum value of $\frac{a_n}{n}$ given the sequence $\{a_n\}$ where $a_1=1$, $a_2=4$, and $2a_n=\frac{n-1}{n}a_{n-1}+\frac{n+1}{n}a_{n+1}$ for $n \geq 2$, we follow these steps:
1. **Initial Condition Analysis:**
When $n=2$, we have:
\[
2a_2 = \frac{1}{2}a_1 + \frac{3}{2}a_3
\]
Substit... | 2 | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
Example 8 Given $x, y, z \in \mathbf{R}_{+}$. Prove:
$$
\frac{x}{2 x+y+z}+\frac{y}{x+2 y+z}+\frac{z}{x+y+2 z} \leqslant \frac{3}{4} .
$$ | Prove: Let $S=x+y+z$.
Since the function $f(t)=\frac{1}{S+t}$ is decreasing on $(0, S)$, for any $t \in(0, S)$, we always have
$$
\left(t-\frac{S}{3}\right)\left[f(t)-f\left(\frac{S}{3}\right)\right] \leqslant 0 .
$$
From this, it is easy to obtain
$$
\frac{t}{S+t} \leqslant \frac{9 t}{16 S}+\frac{1}{16} \text {. }
$$... | \frac{3}{4} | Inequalities | proof | Yes | Yes | cn_contest | false |
The negation of the proposition "For every $k \in \mathbb{R}$, the line $y=kx+1$ intersects the circle $x^2+y^2=2$" is ( ).
A: For every $k \notin \mathbb{R}$, the line $y=kx+1$ intersects the circle $x^2+y^2=2$
B: For every $k \in \mathbb{R}$, the line $y=kx+1$ does not intersect the circle $x^2+y^2=2$
C: There ex... | Since the negation of a universal proposition is an existential proposition, the negation of the proposition "For every $k \in \mathbb{R}$, the line $y=kx+1$ intersects the circle $x^2+y^2=2$" is: There exists some $k_0 \in \mathbb{R}$ such that the line $y=kx+1$ does not intersect the circle $x^2+y^2=2$.
Therefore, t... | C | Geometry | MCQ | Yes | Yes | cn_k12 | false |
Suppose eight friends are going to a concert. Five of them are only 3/7 sure that they will stay for the entire concert while the other three are certain they'll stay the full time. What is the probability that at least 7 of the eight friends stay for the entire concert? | To solve this, we need to consider two cases:
1) Exactly 7 friends stayed for the entire concert.
2) All 8 friends stayed.
**Case 1: 7 friends stayed the whole time.**
Among the 5 unsure friends, exactly 4 need to stay. The calculation for this is:
\(\binom{5}{4} \times \left(\frac{3}{7}\right)^4 \times \left(\frac{4... | \frac{1539}{16807} | Combinatorics | math-word-problem | Yes | Yes | synthetic_math | true |
There are some employees at company x. 80 are full-time, and 100 have worked at company x for at least a year. There are 20 employees at company x who aren't full-time and haven't worked at company x for at least a year. There are 30 full-time employees of company x who have worked at the company for at least a year. H... | Let's denote the total number of employees at company x as E.
We are given the following information:
- There are 80 full-time employees.
- There are 100 employees who have worked at company x for at least a year.
- There are 20 employees who are neither full-time nor have worked at company x for at least a year.
- Th... | 170 | Combinatorics | math-word-problem | Yes | Yes | orca_math | true |
5. The numerator of each of the 48 fractions is one of the numbers 2, 3, ..., 49, and the denominator is also one of these numbers, with each of these 48 numbers appearing both among the numerators and among the denominators. Prove that either one of these fractions is an integer, or from them, no more than 25 fraction... | Solution. Consider the fraction $a_{1} / 2$ with the denominator 2. If $a_{1}$ is even, we have already obtained an integer. Otherwise, multiply $a_{1} / 2$ by the fraction $a_{2} / a_{1}$, the result - by the fraction $a_{3} / a_{2}$, and so on until some numerator $a_{n}$ becomes even (this will eventually happen bec... | proof | Number Theory | proof | Yes | Yes | olympiads | false |
Given the function $f(x)=\frac{1}{3}x^3-\frac{1}{2}ax^2+(a-1)x+1$, find the range of $a$ such that the function is monotonically decreasing in the interval $(1,4)$ and monotonically increasing in the interval $(6,+\infty)$. | The first derivative of the function is $f'(x)=x^2-ax+a-1$.
Since $f(x)$ is monotonically decreasing in the interval $(1,4)$, we have $f'(x)\leq 0$ in this interval, which implies $a(x-1)\geq x^2-1$. Therefore, $a\geq x+1$.
As $2 7$, we have $f'(x)\geq 0$ in the interval $(6,+\infty)$ when $a\leq 7$.
Thus, the rang... | [5,7] | Calculus | math-word-problem | Yes | Yes | cn_k12 | false |
A certain percentage of x is greater than 1/3rd of x by 110. x is approximately 660. What is the percentage? | Let's call the percentage we are looking for P. Since P is a percentage of x, we can express it as P/100.
The problem states that P/100 of x is greater than 1/3 of x by 110. So we can write the equation:
(P/100) * x = (1/3) * x + 110
We are given that x is approximately 660, so we can substitute x with 660 in the eq... | 50\% | Algebra | math-word-problem | Yes | Yes | orca_math | true |
How many ordered integer pairs $ \left(x,y\right)$ are there such that $ xy \equal{} 4\left(y^{2} \plus{} x\right)$?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 7 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ \text{None}$ | null | null | Algebra | MCQ | Yes | Yes | aops_forum | false |
Given the functions $f(x)=|3x-1|+|3x+k|$ and $g(x)=x+4$,
(Ⅰ) when $k=-3$, find the solution set of the inequality $f(x) \geqslant 4$;
(Ⅱ) let $k > -1$, and for all $x \in \left[ -\frac{k}{3}, \frac{1}{3} \right)$, $f(x) \leqslant g(x)$. Determine the range of values for $k$. | (I) When $k=-3$, the function $f(x)$ can be rewritten as:
$$
f(x)=
\begin{cases}
-6x+4, & \text{if } x 1.
\end{cases}
$$
Thus, the inequality $f(x) \geqslant 4$ is equivalent to solving:
$$
\begin{cases}
6x-4 \geqslant 4, & \text{for } x > 1, \text{ or} \\
2 \geqslant 4, & \text{for } \frac{1}{3} \leqslant x \leqslant... | (-1, \frac{9}{4}] | Algebra | math-word-problem | Yes | Yes | cn_k12 | false |
End of preview. Expand in Data Studio
README.md exists but content is empty.
- Downloads last month
- 4