minqi/DeepScaleR-1k-subset · Datasets at Fast360

problem stringlengths 10 1.99k	answer stringlengths 1 74	solution stringclasses 245 values
Find the sum of the reciprocals of all the (positive) divisors of 144.	403/144	As $d$ ranges over the divisors of 144 , so does $144 / d$, so the sum of $1 / d$ is $1 / 144$ times the sum of the divisors of 144 . Using the formula for the sum of the divisors of a number (or just counting them out by hand), we get that this sum is 403, so the answer is 403/144.
Line $m$ in the coordinate plane has the equation $2x - 3y + 30 = 0$. This line is rotated $30^{\circ}$ counterclockwise about the point $(10, 10)$ to form line $n$. Find the $x$-coordinate of the $x$-intercept of line $n$.	\frac{20\sqrt{3} + 20}{2\sqrt{3} + 3}
Given $F(x) = \int_{0}^{x} (t^{2} + 2t - 8) \, dt$, where $x > 0$. 1. Determine the intervals of monotonicity for $F(x)$. 2. Find the maximum and minimum values of the function $F(x)$ on the interval $[1, 3]$.	-\frac{28}{3}
Each of the variables $a, b, c, d$, and $e$ represents a positive integer with the properties that $b+d>a+d$, $c+e>b+e$, $b+d=c$, $a+c=b+e$. Which of the variables has the greatest value?	c	Since $b+d>a+d$, then $b>a$. This means that $a$ does not have the greatest value. Since $c+e>b+e$, then $c>b$. This means that $b$ does not have the greatest value. Since $b+d=c$ and each of $b, c, d$ is positive, then $d<c$, which means that $d$ does not have the greatest value. Consider the last equation $a+c=b+e$ along with the fact that $a<b<c$. From this, we see that $e=c+(a-b)$. Since $a<b$, then $a-b$ is negative and so $e<c$. This means that $c$ has the greatest value.
Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m + n ?$	471	1. Understanding the Problem: - Ang, Ben, and Jasmin each have 5 blocks of different colors and there are 5 boxes. - They independently place one block in each box. - We need to find the probability that at least one box ends up with 3 blocks of the same color. 2. Setting Up the Problem: - Each person has 5! ways to place their blocks into the boxes. - We need to calculate the probability that at least one box has 3 blocks of the same color. 3. Using Combinatorics and Derangements: - Let $!n$ denote the number of derangements of $n$ elements. - Recall the recursive formula for derangements: $!0 = 1, !1 = 0, !n = (n-1)(!(n-1) + !(n-2))$. 4. Calculating the Probability: - After Ang places blocks, each box has a fixed color. - Ben has $\binom{5}{k} \cdot !k$ ways to disqualify $k$ candidates from being uniform-color, where $0 \leq k \leq 5$. - Jasmin's placement is calculated similarly, but we need to consider the remaining candidates after Ben's placement. 5. Recursive Calculation for Jasmin's Placement: - Define $D(n, k)$ as the number of ways Jasmin can place blocks such that no box remains a candidate, with $n$ boxes and $k$ non-candidates. - We have $D(n, 0) = !n$ and the recursive relation $D(n, k) = D(n-1, k-1) + D(n, k-1)$. 6. Constructing the Table for $D(n, k)$: - The table is constructed based on the recursive relation and initial conditions. - The values of $D(5, k)$ are extracted from the table for $k = 0$ to $5$. 7. Calculating the Desired Probability: - The probability that at least one box is of uniform color is calculated using: \[ \frac{\sum_{k=0}^{5}{\left(\binom{5}{k}\cdot{!}k\right)(5!-D(5,k))}}{(5!)^2} \] - Simplifying the expression, we find: \[ \frac{76 + 0 + 560 + 840 + 1080 + 0}{(5!)^2} = \frac{2556}{14400} = \frac{639}{3600} = \frac{71}{400} \] 8. Final Answer: - The probability that at least one box receives 3 blocks all of the same color is $\frac{71}{400}$. - The problem asks for $m+n$ where $\frac{m}{n} = \frac{71}{400}$. Thus, $m+n = 71 + 400 = \boxed{471}$.
An isosceles triangle and a rectangle have the same area. The base of the triangle is equal to the width of the rectangle, and this dimension is 10 units. The length of the rectangle is twice its width. What is the height of the triangle, $h$, in terms of the dimensions of the rectangle?	40
The lock opens only if a specific three-digit number is entered. An attempt consists of randomly selecting three digits from a given set of five. The code was guessed correctly only on the last of all attempts. How many attempts preceded the successful one?	124
Let \( A = \{1, 2, \cdots, 2004\} \) and \( f: A \rightarrow A \) be a bijection satisfying \( f^{[2004]}(x) = f(x) \), where \( f^{[2004]}(x) \) denotes applying \( f \) 2004 times to \( x \). How many such functions \( f \) are there?	1 + 2004!
Given the parabola $y^{2}=4x$, and the line $l$: $y=- \frac {1}{2}x+b$ intersects the parabola at points $A$ and $B$. (I) If the $x$-axis is tangent to the circle with $AB$ as its diameter, find the equation of the circle; (II) If the line $l$ intersects the negative semi-axis of $y$, find the maximum area of $\triangle AOB$.	\frac {32 \sqrt {3}}{9}
In how many ways can 8 people be seated in a row of chairs if three of the people, John, Wilma, and Paul, refuse to sit in three consecutive seats?	36000
There are 8 white balls and 2 red balls in a bag. Each time a ball is randomly taken out, a white ball is put back in. What is the probability that all red balls are taken out exactly by the 4th draw?	353/5000
The "Tiao Ri Method", invented by mathematician He Chengtian during the Southern and Northern Dynasties of China, is an algorithm for finding a more accurate fraction to represent a numerical value. Its theoretical basis is as follows: If the deficient approximate value and the excess approximate value of a real number $x$ are $\frac{b}{a}$ and $\frac{d}{c}$ ($a, b, c, d \in \mathbb{N}^*$) respectively, then $\frac{b+d}{a+c}$ is a more accurate deficient approximate value or excess approximate value of $x$. We know that $\pi = 3.14159...$, and if we let $\frac{31}{10} < \pi < \frac{49}{15}$, then after using the "Tiao Ri Method" once, we get $\frac{16}{5}$ as a more accurate excess approximate value of $\pi$, i.e., $\frac{31}{10} < \pi < \frac{16}{5}$. If we always choose the simplest fraction, then what approximate fraction can we get for $\pi$ after using the "Tiao Ri Method" four times?	\frac{22}{7}
Let $a \bowtie b = a + \sqrt{b + \sqrt{b + \sqrt{b + \ldots}}}$. If $5 \bowtie x = 12$, find the value of $x$.	42
In the sequence \( \left\{a_{n}\right\} \), if \( a_{k}+a_{k+1}=2k+1 \) (where \( k \in \mathbf{N}^{*} \)), then \( a_{1}+a_{100} \) equals?	101
Attempt to obtain one billion (1,000,000,000) by multiplying two integers, each of which contains no zeros.	512 * 1953125
Determine the number of zeros in the quotient $Q = R_{30}/R_6$, where $R_k$ is a number consisting of $k$ repeated digits of 1 in base-ten.	25
In $\triangle ABC$, the sides opposite to angles $A$, $B$, $C$ are $a$, $b$, $c$ respectively. If $\sin (A-B)+ \sin C= \sqrt {2}\sin A$. (I) Find the value of angle $B$; (II) If $b=2$, find the maximum value of $a^{2}+c^{2}$, and find the values of angles $A$ and $C$ when the maximum value is obtained.	8+4 \sqrt {2}
Two fair, six-sided dice are rolled. What is the probability that the sum of the two numbers showing is less than or equal to 10 and at least one die shows a number greater than 3?	\frac{2}{3}
Circles with centers $A$, $B$, and $C$ each have radius $r$, where $1 < r < 2$. The distance between each pair of centers is $2$. If $B'$ is the point of intersection of circle $A$ and circle $C$ which is outside circle $B$, and if $C'$ is the point of intersection of circle $A$ and circle $B$ which is outside circle $C$, then length $B'C'$ equals	1+\sqrt{3(r^2-1)}	1. Setting up the problem: We are given three circles with centers at $A$, $B$, and $C$, each having radius $r$ where $1 < r < 2$. The distance between each pair of centers is $2$. We need to find the length of the line segment $B'C'$, where $B'$ and $C'$ are points of intersection of the circles centered at $A$ with circles centered at $C$ and $B$ respectively, and each is outside the other circle. 2. Positioning the circles: We can position the centers of the circles in the Cartesian plane as follows: - $A$ at $(0, \sqrt{3})$ - $B$ at $(1, 0)$ - $C$ at $(-1, 0)$ Each circle has radius $r$, so their equations are: - Circle $A$: \(x^2 + (y - \sqrt{3})^2 = r^2\) - Circle $B$: \((x - 1)^2 + y^2 = r^2\) - Circle $C$: \((x + 1)^2 + y^2 = r^2\) 3. Finding intersection points: - Solve the equations of Circle $A$ and Circle $B$: \[ x^2 + (y - \sqrt{3})^2 = r^2 \quad \text{and} \quad (x - 1)^2 + y^2 = r^2 \] Expanding and simplifying, we get: \[ x^2 + y^2 - 2\sqrt{3}y + 3 = r^2 \quad \text{and} \quad x^2 - 2x + 1 + y^2 = r^2 \] Subtracting these equations: \[ 2x - 2\sqrt{3}y + 2 = 0 \quad \Rightarrow \quad x = \sqrt{3}y - 1 \] Substituting $x = \sqrt{3}y - 1$ back into the equation of Circle $A$: \[ (\sqrt{3}y - 1)^2 + (y - \sqrt{3})^2 = r^2 \] Solving this equation gives: \[ y = \frac{\sqrt{3} \pm \sqrt{r^2 - 1}}{2} \] We select the larger root $y = \frac{\sqrt{3} + \sqrt{r^2 - 1}}{2}$ for $B'$, and the corresponding $x$ value is $x = -\frac{1 + \sqrt{3(r^2 - 1)}}{2}$. 4. Calculating $B'C'$: - By symmetry, the $x$-coordinates of $B'$ and $C'$ are negatives of each other, and they have the same $y$-coordinate. Thus, the length of $B'C'$ is twice the absolute value of the $x$-coordinate of $B'$: \[ B'C' = 2 \left\| -\frac{1 + \sqrt{3(r^2 - 1)}}{2} \right\| = 1 + \sqrt{3(r^2 - 1)} \] 5. Conclusion: - The length of $B'C'$ is $1 + \sqrt{3(r^2 - 1)}$, which corresponds to choice $\textbf{(D)}$. Thus, the final answer is $\boxed{D}$.
Let $f : [0, 1] \rightarrow \mathbb{R}$ be a monotonically increasing function such that $$ f\left(\frac{x}{3}\right) = \frac{f(x)}{2} $$ $$ f(1 0 x) = 2018 - f(x). $$ If $f(1) = 2018$ , find $f\left(\dfrac{12}{13}\right)$ .	2018
(1) Given $0 < x < \frac{1}{2}$, find the maximum value of $y= \frac{1}{2}x(1-2x)$; (2) Given $x > 0$, find the maximum value of $y=2-x- \frac{4}{x}$; (3) Given $x$, $y\in\mathbb{R}_{+}$, and $x+y=4$, find the minimum value of $\frac{1}{x}+ \frac{3}{y}$.	1+ \frac{ \sqrt{3}}{2}
The square quilt block shown is made from sixteen unit squares, where eight of these squares have been divided in half diagonally to form triangles. Each triangle is shaded. What fraction of the square quilt is shaded? Express your answer as a common fraction.	\frac{1}{4}
One material particle entered the opening of a pipe, and after 6.8 minutes, a second particle entered the same opening. Upon entering the pipe, each particle immediately began linear motion along the pipe: the first particle moved uniformly at a speed of 5 meters per minute, while the second particle covered 3 meters in the first minute and in each subsequent minute covered 0.5 meters more than in the previous minute. How many minutes will it take for the second particle to catch up with the first?	17
Among 6 courses, if person A and person B each choose 3 courses, the number of ways in which exactly 1 course is chosen by both A and B is \_\_\_\_\_\_.	180
The store owner bought 2000 pens at $0.15 each and plans to sell them at $0.30 each, calculate the number of pens he needs to sell to make a profit of exactly $150.	1000
Given that the point $(9,7)$ is on the graph of $y=f(x)$, there is one point that must be on the graph of $2y=\frac{f(2x)}2+2$. What is the sum of coordinates of that point?	\frac{29}4
You start with a single piece of chalk of length 1. Every second, you choose a piece of chalk that you have uniformly at random and break it in half. You continue this until you have 8 pieces of chalk. What is the probability that they all have length $\frac{1}{8}$ ?	\frac{1}{63}	There are 7! total ways to break the chalks. How many of these result in all having length $\frac{1}{8}$ ? The first move gives you no choice. Then, among the remaining 6 moves, you must apply 3 breaks on the left side and 3 breaks on the right side, so there are $\binom{6}{3}=20$ ways to order those. On each side, you can either break the left side or the right side first. So the final answer is $$\frac{20 \cdot 2^{2}}{7!}=\frac{1}{63}$$
Given that \(\theta\) is an angle in the third quadrant, and \(\sin^{4}\theta + \cos^{4}\theta = \frac{5}{9}\), determine the value of \(\sin 2\theta\).	-\frac{2\sqrt{2}}{3}
Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?	70	1. Define Variables: Let $L$ be the length of the ship, $E$ be the length of Emily's step, and $S$ be the length of the ship's step (the distance the ship travels while Emily takes one step). 2. Set Up Equations: - When Emily walks from the back of the ship to the front, she takes $210$ steps, covering a distance of $210E$. During this time, the front of the ship also moves forward by $210S$. The total distance covered by Emily relative to the ship is the length of the ship plus the distance the ship has moved, which gives us: \[ 210E = L + 210S \quad \text{(Equation 1)} \] - When Emily walks from the front of the ship to the back, she takes $42$ steps, covering a distance of $42E$. During this time, the back of the ship moves forward by $42S$. The total distance covered by Emily relative to the ship is the length of the ship minus the distance the ship has moved, which gives us: \[ 42E = L - 42S \quad \text{(Equation 2)} \] 3. Manipulate Equations: - First, rearrange Equation 1: \[ 210E - 210S = L \] - Rearrange Equation 2: \[ 42E + 42S = L \] - Multiply Equation 2 by $5$ to align the coefficients of $E$ with those in Equation 1: \[ 5(42E + 42S) = 5L \implies 210E + 210S = 5L \quad \text{(Equation 3)} \] - Subtract Equation 1 from Equation 3: \[ (210E + 210S) - (210E - 210S) = 5L - L \implies 420S = 4L \] - Simplify to find $L$ in terms of $S$: \[ L = 105S \] 4. Solve for $\frac{L}{E}$: - From Equation 1, substitute $L = 105S$: \[ 210E = 105S + 210S \implies 210E = 315S \] - Simplify to find the ratio of $E$ to $S$: \[ E = 1.5S \] - Substitute $E = 1.5S$ into the expression for $L$: \[ L = 105S = 70 \times 1.5S = 70E \] 5. Conclusion: - The length of the ship in terms of Emily's steps is $\boxed{70}$.
The number \( C \) is defined as the sum of all the positive integers \( n \) such that \( n-6 \) is the second largest factor of \( n \). What is the value of \( 11C \)?	308
A semicircle is inscribed in an isosceles triangle with base 16 and height 15 so that the diameter of the semicircle is contained in the base of the triangle. What is the radius of the semicircle?	\frac{120}{17}	1. Identify the Triangle and its Properties: Let's denote the isosceles triangle as $\triangle ABD$ where $AB = AD$ and $BD = 16$. The height from $A$ to $BD$ is given as $15$. Let $M$ be the midpoint of $BD$. 2. Divide the Triangle: Draw a line from $A$ to $M$. This line divides $\triangle ABD$ into two right triangles, $\triangle ABM$ and $\triangle ADM$. Since $M$ is the midpoint of $BD$, $BM = MD = \frac{16}{2} = 8$. 3. Use the Pythagorean Theorem: In $\triangle ABM$, we know $BM = 8$ and $AM = 15$. We can find $AB$ using the Pythagorean theorem: \[ AB^2 = AM^2 + BM^2 = 15^2 + 8^2 = 225 + 64 = 289 \implies AB = \sqrt{289} = 17. \] Thus, $AB = AD = 17$. 4. Reflect the Triangle: Reflect $\triangle ABD$ over its base $BD$ to form a rhombus $ABCD$ with $AB = BC = CD = DA = 17$. 5. Calculate the Semiperimeter and Area: The semiperimeter $s$ of rhombus $ABCD$ is: \[ s = \frac{AB + BC + CD + DA}{2} = \frac{4 \times 17}{2} = 34. \] The area $[ABCD]$ of the rhombus is twice the area of $\triangle ABD$: \[ [ABCD] = 2 \times \left(\frac{1}{2} \times BD \times AM\right) = 2 \times (16 \times 15 \div 2) = 240. \] 6. Apply the Formula for the Incircle of a Quadrilateral: The formula relating the area, semiperimeter, and inradius $r$ of a quadrilateral is $s \cdot r = [ABCD]$. Substituting the known values: \[ 34r = 240 \implies r = \frac{240}{34} = \frac{120}{17}. \] 7. Conclusion: The radius of the semicircle inscribed in the isosceles triangle $\triangle ABD$ is $\boxed{\textbf{(B) }\frac{120}{17}}$.
Suppose $d$ is a digit. For how many values of $d$ is $2.00d5 > 2.005$?	5	To solve this problem, we need to compare the number $2.00d5$ with $2.005$ for each possible digit $d$ (where $d$ can be any digit from $0$ to $9$). 1. Expression of $2.00d5$: The number $2.00d5$ can be expressed as $2.00d5 = 2 + 0.00d + 0.0005$. Here, $0.00d$ represents the contribution of the digit $d$ in the thousandths place. 2. Comparison with $2.005$: We need to find when $2.00d5 > 2.005$. \[ 2 + 0.00d + 0.0005 > 2.005 \] Simplifying the inequality: \[ 0.00d + 0.0005 > 0.005 \] \[ 0.00d > 0.005 - 0.0005 \] \[ 0.00d > 0.0045 \] 3. Value of $0.00d$: Since $d$ is a digit, $0.00d$ can take values $0.000, 0.001, 0.002, \ldots, 0.009$. We need to find for which values of $d$, $0.00d > 0.0045$. 4. Finding the values of $d$: - If $d = 0$, then $0.00d = 0.000$ (which is not greater than $0.0045$). - If $d = 1$, then $0.00d = 0.001$ (which is not greater than $0.0045$). - If $d = 2$, then $0.00d = 0.002$ (which is not greater than $0.0045$). - If $d = 3$, then $0.00d = 0.003$ (which is not greater than $0.0045$). - If $d = 4$, then $0.00d = 0.004$ (which is not greater than $0.0045$). - If $d = 5$, then $0.00d = 0.005$ (which is greater than $0.0045$). - If $d = 6$, then $0.00d = 0.006$ (which is greater than $0.0045$). - If $d = 7$, then $0.00d = 0.007$ (which is greater than $0.0045$). - If $d = 8$, then $0.00d = 0.008$ (which is greater than $0.0045$). - If $d = 9$, then $0.00d = 0.009$ (which is greater than $0.0045$). 5. Counting the valid values of $d$: The values of $d$ that satisfy $0.00d > 0.0045$ are $5, 6, 7, 8, 9$. There are $5$ such values. Thus, there are $\boxed{\textbf{(C)}\ 5}$ values of $d$ for which $2.00d5 > 2.005$.
(1) Let the function $f(x)=\|x+2\|+\|x-a\|$. If the inequality $f(x) \geqslant 3$ always holds for $x$ in $\mathbb{R}$, find the range of the real number $a$. (2) Given positive numbers $x$, $y$, $z$ satisfying $x+2y+3z=1$, find the minimum value of $\frac{3}{x}+\frac{2}{y}+\frac{1}{z}$.	16+8\sqrt{3}
In the complex plane, the graph of \( \|z - 5\| = 3\|z + 5\| \) intersects the graph of \( \|z\| = k \) in exactly one point. Find all possible values of \( k \).	12.5
Nine people are practicing the triangle dance, which is a dance that requires a group of three people. During each round of practice, the nine people split off into three groups of three people each, and each group practices independently. Two rounds of practice are different if there exists some person who does not dance with the same pair in both rounds. How many different rounds of practice can take place?	280
Given the function $f\left(x+ \frac {1}{2}\right)= \frac {2x^{4}+x^{2}\sin x+4}{x^{4}+2}$, calculate the value of $f\left( \frac {1}{2017}\right)+f\left( \frac {2}{2017}\right)+\ldots+f\left( \frac {2016}{2017}\right)$.	4032
Vasya has three cans of paint of different colors. In how many different ways can he paint a fence consisting of 10 planks so that any two adjacent planks are different colors and he uses all three colors? Provide a justification for your answer.	1530
Calculate:<br/>$(1)(1\frac{3}{4}-\frac{3}{8}+\frac{5}{6})÷(-\frac{1}{24})$;<br/>$(2)-2^2+(-4)÷2×\frac{1}{2}+\|-3\|$.	-2
For a bijective function $g : R \to R$ , we say that a function $f : R \to R$ is its superinverse if it satisfies the following identity $(f \circ g)(x) = g^{-1}(x)$ , where $g^{-1}$ is the inverse of $g$ . Given $g(x) = x^3 + 9x^2 + 27x + 81$ and $f$ is its superinverse, find $\|f(-289)\|$ .	10
A factory implements a time-based wage system, where each worker is paid 6 yuan for every hour worked, for a total of 8 hours per day. However, the clock used for timing is inaccurate: it takes 69 minutes for the minute hand to coincide with the hour hand once. Calculate the amount of wages underpaid to each worker per day.	2.60
For how many non-negative real values of $x$ is $\sqrt{169-\sqrt[4]{x}}$ an integer?	14
When three standard dice are tossed, the numbers $a, b, c$ are obtained. Find the probability that $abc = 216$.	\frac{1}{216}
The numbers $1,2, \ldots, 10$ are randomly arranged in a circle. Let $p$ be the probability that for every positive integer $k<10$, there exists an integer $k^{\prime}>k$ such that there is at most one number between $k$ and $k^{\prime}$ in the circle. If $p$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$, compute $100 a+b$.	1390	Let $n=10$ and call two numbers close if there is at most one number between them and an circular permutation focused if only $n$ is greater than all numbers close to it. Let $A_{n}$ be the number of focused circular permutations of $\{1,2, \ldots, n\}$. If $n \geq 5$, then there are 2 cases: $n-1$ is either one or two positions from $n$. If $n-1$ is one position from $n$, it is either on its left or right. In this case, one can check a permutation is focused if and only if removing $n$ yields a focused permutation, so there are $2 A_{n-1}$ permutations in this case. If $n-1$ is two positions from $n$, there are $n-2$ choices for $k$, the element that lies between $n$ and $n-1$. One can show that this permutation is focused if and only if removing both $n$ and $k$ and relabeling the numbers yields a focused permutation, so there are $2(n-2) A_{n-2}$ permutations in this case. Thus, we have $A_{n}=2 A_{n-1}+2(n-2) A_{n-2}$. If we let $p_{n}=A_{n} /(n-1)$ ! the probability that a random circular permutation is focused, then this becomes $$p_{n}=\frac{2 p_{n-1}+2 p_{n-2}}{n-1}$$ Since $p_{3}=p_{4}=1$, we may now use this recursion to calculate $$p_{5}=1, p_{6}=\frac{4}{5}, p_{7}=\frac{3}{5}, p_{8}=\frac{2}{5}, p_{9}=\frac{1}{4}, p_{10}=\frac{13}{90}$$
Given $A(-1,\cos \theta)$, $B(\sin \theta,1)$, if $\|\overrightarrow{OA}+\overrightarrow{OB}\|=\|\overrightarrow{OA}-\overrightarrow{OB}\|$, then find the value of the acute angle $\theta$.	\frac{\pi}{4}
Given an infinite geometric sequence $\{a_n\}$, the product of its first $n$ terms is $T_n$, and $a_1 > 1$, $a_{2008}a_{2009} > 1$, $(a_{2008} - 1)(a_{2009} - 1) < 0$, determine the maximum positive integer $n$ for which $T_n > 1$.	4016
For all non-zero real numbers $x$ and $y$ such that $x-y=xy$, $\frac{1}{x}-\frac{1}{y}$ equals	-1	1. Start with the given equation and manipulate it: Given that \(x - y = xy\), we can rearrange this equation to: \[ x - y - xy = 0 \] Adding 1 to both sides, we get: \[ x - y - xy + 1 = 1 \] Factoring the left-hand side, we have: \[ (x-1)(y+1) = 1 \] 2. Analyze the expression \(\frac{1}{x} - \frac{1}{y}\): We rewrite the expression using a common denominator: \[ \frac{1}{x} - \frac{1}{y} = \frac{y-x}{xy} \] Since \(x - y = xy\), substituting \(xy\) for \(x - y\) in the numerator, we get: \[ \frac{1}{x} - \frac{1}{y} = \frac{-xy}{xy} = -1 \] 3. Conclusion: The expression \(\frac{1}{x} - \frac{1}{y}\) simplifies to \(-1\). Therefore, the correct answer is: \[ \boxed{\textbf{(D) } -1} \]
Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted questions worth a total of 500 points. Alpha scored 160 points out of 300 points attempted on the first day, and scored 140 points out of 200 points attempted on the second day. Beta who did not attempt 300 points on the first day, had a positive integer score on each of the two days, and Beta's daily success rate (points scored divided by points attempted) on each day was less than Alpha's on that day. Alpha's two-day success ratio was 300/500 = 3/5. The largest possible two-day success ratio that Beta could achieve is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m+n$?	849	Let $q$ be the number of questions Beta takes on day 1 and $a$ be the number he gets right. Let $b$ be the number he gets right on day 2. These inequalities follow: \[\frac{a}{q} < \frac{160}{300} = \frac{8}{15}\] \[\frac{b}{500-q} < \frac{140}{200} = \frac{7}{10}\] Solving for a and b and adding the two inequalities: \[a + b < \frac{8}{15}q + (350 - \frac{7}{10}q)\] \[a + b < 350 - \frac{1}{6}q\] From here, we see the largest possible value of $a+b$ is $349$. Checking our conditions, we know that $a$ must be positive so therefore $q$ must be positive. A quick check shows that for $2\le q \le 5$, $q$ follows all the conditions and results in $a+b=349$. This makes Beta's success ratio $\frac{349}{500}$. Thus, the answer is $m+n = 349 + 500 = \boxed{849}$.
Read the following material and complete the corresponding tasks. 平衡多项式定义:对于一组多项式$x+a$,$x+b$,$x+c$,$x+d(a$,$b$,$c$,$d$是常数）,当其中两个多项式的乘积与另外两个多项式乘积的差是一个常数$p$时,称这样的四个多项式是一组平衡多项式,$p$的绝对值是这组平衡多项式的平衡因子. 例如:对于多项式$x+1$,$x+2$,$x+5$,$x+6$,因为$(x+1)(x+6)-(x+2)(x+5)=(x^{2}+7x+6)-(x^{2}+7x+10)=-4$,所以多项式$x+1$,$x+2$,$x+5$,$x+6$是一组平衡多项式,其平衡因子为$\|-4\|=4$. 任务: $(1)$小明发现多项式$x+3$,$x+4$,$x+6$,$x+7$是一组平衡多项式,在求其平衡因子时,列式如下:$(x+3)(x+7)-(x+4)(x+6)$,根据他的思路求该组平衡多项式的平衡因子； $A$.判断多项式$x-1$,$x-2$,$x-4$,$x-5$是否为一组平衡多项式,若是,求出其平衡因子；若不是,说明理由. $B$.若多项式$x+2$,$x-4$,$x+1$,$x+m(m$是常数）是一组平衡多项式,求$m$的值.	-5
Given the quadratic function $f(x) = mx^2 - 2x - 3$, if the solution set of the inequality $f(x) < 0$ is $(-1, n)$. (1) Solve the inequality about $x$: $2x^2 - 4x + n > (m + 1)x - 1$; (2) Determine whether there exists a real number $a \in (0, 1)$, such that the minimum value of the function $y = f(a^x) - 4a^{x+1}$ ($x \in [1, 2]$) is $-4$. If it exists, find the value of $a$; if not, explain why.	\frac{1}{3}
Given an arithmetic sequence, the sum of the first four terms is 26, the sum of the last four terms is 110, and the sum of all terms in the sequence is 187. Determine the total number of terms in the sequence.	11
Our school's basketball team has won the national middle school basketball championship multiple times! In one competition, including our school's basketball team, 7 basketball teams need to be randomly divided into two groups (one group with 3 teams and the other with 4 teams) for the group preliminaries. The probability that our school's basketball team and the strongest team among the other 6 teams end up in the same group is ______.	\frac{3}{7}
Twelve chess players played a round-robin tournament. Each player then wrote 12 lists. In the first list, only the player himself was included, and in the $(k+1)$-th list, the players included those who were in the $k$-th list as well as those whom they defeated. It turned out that each player's 12th list differed from the 11th list. How many draws were there?	54
Given a triangle \(ABC\) where \(AB = AC\) and \(\angle A = 80^\circ\). Inside triangle \(ABC\) is a point \(M\) such that \(\angle MBC = 30^\circ\) and \(\angle MCB = 10^\circ\). Find \(\angle AMC\).	70
What is the smallest positive integer $n$ such that $\frac{n}{n+53}$ is equal to a terminating decimal?	11
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(f(x)+y)+xf(y)=f(xy+y)+f(x)$$ for reals $x, y$.	f(x) = x \text{ or } f(x) = 0	Let \( f: \mathbb{R} \rightarrow \mathbb{R} \) be a function satisfying the functional equation: \[ f(f(x) + y) + x f(y) = f(xy + y) + f(x) \] for all real numbers \( x \) and \( y \). ### Step 1: Initial Substitution Start by substituting \( y = 0 \) into the equation: \[ f(f(x)) + x f(0) = f(x) + f(x) \] Simplifying gives: \[ f(f(x)) + x f(0) = 2f(x) \] ### Step 2: Exploring Constant Solutions Suppose \( f(x) = c \) for all \( x \), where \( c \) is a constant. Then: \[ f(f(x) + y) = f(c + y) = c \quad \text{and} \quad f(xy + y) = f(y(x + 1)) = c \] Hence, substituting back into the original equation: \[ c + x \cdot c = c + c \] This implies \( xc = c \). If \( c \neq 0 \), this equation has no solution for all \( x \). Thus, \( c = 0 \) is the only constant function solution. Therefore, one solution is: \[ f(x) = 0 \quad \text{for all } x \in \mathbb{R} \] ### Step 3: Exploring Non-Constant Solutions Assume \( f \) is non-constant. Substitute \( f(x) = x \) in the original equation: \[ f(f(x) + y) = f(x + y) \] \[ x f(y) = x \cdot y \] \[ f(xy + y) = xy + y \] \[ f(x) = x \] Substituting \( f(x) = x \) into the original equation: \[ f(f(x) + y) + x f(y) = f(xy + y) + f(x) \] Results in: \[ f(x + y) + xy = xy + y + x \] Thus both sides are equal, confirming that: \[ f(x) = x \quad \text{for all } x \in \mathbb{R} \] ### Conclusion The solutions to the functional equation are: \[ f(x) = x \quad \text{or} \quad f(x) = 0 \quad \text{for all } x \in \mathbb{R} \] Thus, the complete set of solutions is: \[ \boxed{f(x) = x \text{ or } f(x) = 0} \] Both solutions satisfy the initial functional equation over all reals.
If the height of an external tangent cone of a sphere is three times the radius of the sphere, determine the ratio of the lateral surface area of the cone to the surface area of the sphere.	\frac{3}{2}
Calculate: $1.23 \times 67 + 8.2 \times 12.3 - 90 \times 0.123$	172.2
A certain organization consists of five leaders and some number of regular members. Every year, the current leaders are kicked out of the organization. Next, each regular member must find two new people to join as regular members. Finally, five new people are elected from outside the organization to become leaders. In the beginning, there are fifteen people in the organization total. How many people total will be in the organization five years from now?	2435
James wrote a different integer from 1 to 9 in each cell of a table. He then calculated the sum of the integers in each of the rows and in each of the columns of the table. Five of his answers were 12, 13, 15, 16, and 17, in some order. What was his sixth answer?	17
Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0,0)$, $B(0,12)$, $C(16,0)$, $A'(24,18)$, $B'(36,18)$, $C'(24,2)$. A rotation of $m$ degrees clockwise around the point $(x,y)$ where $0<m<180$, will transform $\triangle ABC$ to $\triangle A'B'C'$. Find $m+x+y$.	108	After sketching, it is clear a $90^{\circ}$ rotation is done about $(x,y)$. Looking between $A$ and $A'$, $x+y=18$ and $x-y=24$. Solving gives $(x,y)\implies(21,-3)$. Thus $90+21-3=\boxed{108}$. ~mn28407
If the surface area of a cone is $3\pi$, and its lateral surface unfolds into a semicircle, then the diameter of the base of the cone is ___.	\sqrt{6}
Let $C(k)$ denotes the sum of all different prime divisors of a positive integer $k$. For example, $C(1)=0$, $C(2)=2, C(45)=8$. Find all positive integers $n$ such that $C(2^{n}+1)=C(n)$	n=3	Let $P(t)$ be the largest prime divisor of a positive integer $t>1$. Let $m$ be the largest odd divisor of $n: n=2^{k} m$. Then $2^{n}+1=2^{2^{k} m}+1=a^{m}+1$, where $a=2^{2^{k}}$. If $k>0$, that is, $n$ is even, then $C(n)=C(m)+2$ and $C(2^{n}+1)=C(a^{m}+1)$. We need the following two lemmas. Lemma 1. For every prime $p>2$ we have $P(\frac{a^{p}+1}{a+1})=p$ or $P(\frac{a^{p}+1}{a+1}) \geqslant 2 p+1$. Proof. Let $P(\frac{a^{p}+1}{a+1})=q$. It follows from Fermat's little theorem that $q$ divides $2^{q-1}-1$ and therefore $(a^{2 p}-1, a^{q-1}-1)=a^{(2 p, q-1)}-1$. The greatest common divisor $(2 p, q-1)$ is even and must equal $2 p$ or 2. In the first case $2 p$ divides $q-1$, whence $q \geqslant 2 p+1$. In the second case $q$ divides $a^{2}-1$ but not $a-1$ (because $a^{p}+1$ is divisible by $q$), that is, $a \equiv-1 \pmod{q}$. Then $\frac{a^{p}+1}{a+1}=a^{p-1}-\ldots+1 \equiv p \pmod{q}$ and $p=q$. Lemma 2. If $p_{1}$ and $p_{2}$ are different odd primes then $P(\frac{a^{p_{1}}+1}{a+1}) \neq P(\frac{a^{p_{2}}+1}{a+1})$. Proof. If $P(\frac{a^{p_{1}}+1}{a+1})=P(\frac{a^{p_{2}}+1}{a+1})=q$ then $q$ divides $a^{2 p_{1}}-1$ and $a^{2 p_{2}}-1$, therefore $(a^{2 p_{1}}-1, a^{2 p_{2}}-1)=a^{(2 p_{1}, 2 p_{2})}-1=a^{2}-1$ and hence $a+1$, but then $p_{1}=q$ and $p_{2}=q$, a contradiction. We are ready now to solve the problem. Let $p_{1}, \ldots, p_{s}$ be all the prime divisors of $n$. It follows from Lemma 2 that $$C(2^{n}+1) \geqslant P(\frac{a^{p_{1}}+1}{a+1})+\ldots+P(\frac{a^{p_{s}}+1}{a+1})$$ If $C(2^{n}+1)>P(\frac{a^{p_{1}}+1}{a+1})+\ldots+P(\frac{a^{p_{s}}+1}{a+1})$, then $2^{n}+1$ has at least one prime divisor not summed in the L.H.S., that is, $$C(2^{n}+1) \geqslant P(\frac{a^{p_{1}}+1}{a+1})+\ldots+P(\frac{a^{p_{s}}+1}{a+1})+3 \geqslant p_{1}+\ldots+p_{s}+3>C(n)$$ Therefore we can assume the equality: $$C(2^{n}+1)=P(\frac{a^{p_{1}}+1}{a+1})+\ldots+P(\frac{a^{p_{s}}+1}{a+1})$$ If in this case there is an $i$ such that $P(\frac{a^{p_{i}}+1}{a+1}) \neq p_{i}$, then $C(2^{n}+1) \geqslant p_{1}+\ldots+p_{s}+p_{i}+1>C(n)$. It remains to consider the case when $P(\frac{a^{p_{i}}+1}{a+1})=p_{i}$ for all $i$. In this case we have $C(n)=C(2^{n}+1)=p_{1}+\ldots+p_{s}$, so $n$ must be odd and $a=2$. But $2^{p} \equiv 2 \pmod{p}$ for all odd prime $p$, therefore $p>3$ cannot divide $2^{p}+1$. Thus $s=1, p=3, n=3^{r}$ with some positive integral $r$. The number $2^{n}+1=2^{3^{r}}+1$ must be a power of 3. However 19 divides this number for $r=2$ and consequently for all $r \geqslant 2$. Thus the only remaining case is $n=3$, which obviously satisfies the condition.
Jacqueline has 40% less sugar than Liliane, and Bob has 30% less sugar than Liliane. Express the relationship between the amounts of sugar that Jacqueline and Bob have as a percentage.	14.29\%
Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$. Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot \| a_{n-2}-a_{n-3} \|$ for all $n\ge 4$. Find the number of such sequences for which $a_n=0$ for some $n$.	494	Let $a_1=x, a_2=y, a_3=z$. First note that if any absolute value equals 0, then $a_n=0$. Also note that if at any position, $a_n=a_{n-1}$, then $a_{n+2}=0$. Then, if any absolute value equals 1, then $a_n=0$. Therefore, if either $\|y-x\|$ or $\|z-y\|$ is less than or equal to 1, then that ordered triple meets the criteria. Assume that to be the only way the criteria is met. To prove, let $\|y-x\|>1$, and $\|z-y\|>1$. Then, $a_4 \ge 2z$, $a_5 \ge 4z$, and $a_6 \ge 4z$. However, since the minimum values of $a_5$ and $a_6$ are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be $z=1$, $\|y-x\|=2$. Again assume that any other scenario will not meet criteria. To prove, divide the other scenarios into two cases: $z>1$, $\|y-x\|>1$, and $\|z-y\|>1$; and $z=1$, $\|y-x\|>2$, and $\|z-y\|>1$. For the first one, $a_4 \ge 2z$, $a_5 \ge 4z$, $a_6 \ge 8z$, and $a_7 \ge 16z$, by which point we see that this function diverges. For the second one, $a_4 \ge 3$, $a_5 \ge 6$, $a_6 \ge 18$, and $a_7 \ge 54$, by which point we see that this function diverges. Therefore, the only scenarios where $a_n=0$ is when any of the following are met: $\|y-x\|<2$ (280 options) $\|z-y\|<2$ (280 options, 80 of which coincide with option 1) $z=1$, $\|y-x\|=2$. (16 options, 2 of which coincide with either option 1 or option 2) Adding the total number of such ordered triples yields $280+280-80+16-2=\boxed{494}$. Note to author: Because $a_4 \ge 2z$, $a_5 \ge 4z$, $a_6 \ge 8z$, and $a_7 \ge 16z$ doesn't mean the function diverges. What if $z = 7$, $a_4 = 60$, and $a_5 = 30$ too? Note to the Note to author: This isn't possible because the difference between is either 0, 1, or some number greater than 1. If $a_4 = 63$ (since it must be a multiple of $z$), then a_5 is either 0, 63, or some number greater than 63.
Compute $\begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}^3.$	\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}.
A club has between 300 and 400 members. The members gather every weekend and are divided into eight distinct groups. If two members are absent, the groups can all have the same number of members. What is the sum of all possible numbers of members in the club?	4200
Given that point $P$ moves on the ellipse $\frac{x^{2}}{4}+y^{2}=1$, find the minimum distance from point $P$ to line $l$: $x+y-2\sqrt{5}=0$.	\frac{\sqrt{10}}{2}
For a natural number $n \ge 3$ , we draw $n - 3$ internal diagonals in a non self-intersecting, but not necessarily convex, n-gon, cutting the $n$ -gon into $n - 2$ triangles. It is known that the value (in degrees) of any angle in any of these triangles is a natural number and no two of these angle values are equal. What is the largest possible value of $n$ ?	41
Matt's four cousins are coming to visit. There are four identical rooms that they can stay in. If any number of the cousins can stay in one room, how many different ways are there to put the cousins in the rooms?	15
A circle has center $(-10, -4)$ and has radius $13$. Another circle has center $(3, 9)$ and radius $\sqrt{65}$. The line passing through the two points of intersection of the two circles has equation $x+y=c$. What is $c$?	3	1. Identify the equations of the circles: The first circle has center $(-10, -4)$ and radius $13$. Its equation is: \[ (x+10)^2 + (y+4)^2 = 169. \] The second circle has center $(3, 9)$ and radius $\sqrt{65}$. Its equation is: \[ (x-3)^2 + (y-9)^2 = 65. \] 2. Set up the equation for the intersection of the circles: The points of intersection satisfy both circle equations. Setting the left-hand sides of the equations equal gives: \[ (x+10)^2 + (y+4)^2 = (x-3)^2 + (y-9)^2. \] 3. Simplify the equation: Expand both sides: \[ x^2 + 20x + 100 + y^2 + 8y + 16 = x^2 - 6x + 9 + y^2 - 18y + 81. \] Simplify by canceling $x^2$ and $y^2$ from both sides: \[ 20x + 8y + 116 = -6x - 18y + 90. \] Combine like terms: \[ 26x + 26y + 26 = 104. \] Divide through by 26: \[ x + y + 1 = 4 \implies x + y = 3. \] 4. Conclusion: The equation of the line passing through the intersection points of the circles is $x + y = 3$. Therefore, the value of $c$ is $\boxed{\textbf{(A)}\ 3}$.
There are four distinct codes $A, B, C, D$ used by an intelligence station, with one code being used each week. Each week, a code is chosen randomly with equal probability from the three codes that were not used the previous week. Given that code $A$ is used in the first week, what is the probability that code $A$ is also used in the seventh week? (Express your answer as a simplified fraction.)	61/243
After Natasha ate a third of the peaches from the jar, the level of the compote lowered by one quarter. By how much (relative to the new level) will the level of the compote lower if all the remaining peaches are eaten?	\frac{2}{9}
In a mathematics class, the probability of earning an A is 0.6 times the probability of earning a B, and the probability of earning a C is 1.2 times the probability of earning a B. Assuming that all grades are A, B, or C, how many B's will there be in a mathematics class of 40 students?	14
Given \(\tan \theta = \frac{5}{12}\), where \(180^{\circ} \leq \theta \leq 270^{\circ}\). If \(A = \cos \theta + \sin \theta\), find the value of \(A\).	-\frac{17}{13}
Let \[f(x) = x^3 + 6x^2 + 16x + 28.\]The graphs of $y = f(x)$ and $y = f^{-1}(x)$ intersect at exactly one point $(a,b).$ Enter the ordered pair $(a,b).$	(-4,-4)
Triangle $DEF$ is isosceles with angle $E$ congruent to angle $F$. The measure of angle $F$ is three times the measure of angle $D$. What is the number of degrees in the measure of angle $E$?	\frac{540}{7}
Let $O$ be the origin. There exists a scalar $m$ such that for any points $E,$ $F,$ $G,$ and $H$ satisfying the vector equation: \[4 \overrightarrow{OE} - 3 \overrightarrow{OF} + 2 \overrightarrow{OG} + m \overrightarrow{OH} = \mathbf{0},\] the four points $E,$ $F,$ $G,$ and $H$ are coplanar. Find the value of $m.$	-3
For how many integer values of $n$ between 1 and 1000 inclusive does the decimal representation of $\frac{n}{1260}$ terminate?	47
In a trapezoid, the lengths of the diagonals are known to be 6 and 8, and the length of the midline is 5. Find the height of the trapezoid.	4.8
Given an equilateral triangle \( \triangle ABC \) with a side length of 1, \[ \overrightarrow{AP} = \frac{1}{3}(\overrightarrow{AB} + \overrightarrow{AC}), \quad \overrightarrow{AQ} = \overrightarrow{AP} + \frac{1}{2}\overrightarrow{BC}. \] Find the area of \( \triangle APQ \).	\frac{\sqrt{3}}{12}

Find the sum of the reciprocals of all the (positive) divisors of 144.

403/144

As $d$ ranges over the divisors of 144 , so does $144 / d$, so the sum of $1 / d$ is $1 / 144$ times the sum of the divisors of 144 . Using the formula for the sum of the divisors of a number (or just counting them out by hand), we get that this sum is 403, so the answer is 403/144.

Line $m$ in the coordinate plane has the equation $2x - 3y + 30 = 0$. This line is rotated $30^{\circ}$ counterclockwise about the point $(10, 10)$ to form line $n$. Find the $x$-coordinate of the $x$-intercept of line $n$.

\frac{20\sqrt{3} + 20}{2\sqrt{3} + 3}

Given $F(x) = \int_{0}^{x} (t^{2} + 2t - 8) \, dt$, where $x > 0$. 1. Determine the intervals of monotonicity for $F(x)$. 2. Find the maximum and minimum values of the function $F(x)$ on the interval $[1, 3]$.

-\frac{28}{3}

Each of the variables $a, b, c, d$, and $e$ represents a positive integer with the properties that $b+d>a+d$, $c+e>b+e$, $b+d=c$, $a+c=b+e$. Which of the variables has the greatest value?

c

Since $b+d>a+d$, then $b>a$. This means that $a$ does not have the greatest value. Since $c+e>b+e$, then $c>b$. This means that $b$ does not have the greatest value. Since $b+d=c$ and each of $b, c, d$ is positive, then $d<c$, which means that $d$ does not have the greatest value. Consider the last equation $a+c=b+e$ along with the fact that $a<b<c$. From this, we see that $e=c+(a-b)$. Since $a<b$, then $a-b$ is negative and so $e<c$. This means that $c$ has the greatest value.

Ang, Ben, and Jasmin each have $5$ blocks, colored red, blue, yellow, white, and green; and there are $5$ empty boxes. Each of the people randomly and independently of the other two people places one of their blocks into each box. The probability that at least one box receives $3$ blocks all of the same color is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m + n ?$

471

1. **Understanding the Problem:** - Ang, Ben, and Jasmin each have 5 blocks of different colors and there are 5 boxes. - They independently place one block in each box. - We need to find the probability that at least one box ends up with 3 blocks of the same color. 2. **Setting Up the Problem:** - Each person has 5! ways to place their blocks into the boxes. - We need to calculate the probability that at least one box has 3 blocks of the same color. 3. **Using Combinatorics and Derangements:** - Let $!n$ denote the number of derangements of $n$ elements. - Recall the recursive formula for derangements: $!0 = 1, !1 = 0, !n = (n-1)(!(n-1) + !(n-2))$. 4. **Calculating the Probability:** - After Ang places blocks, each box has a fixed color. - Ben has $\binom{5}{k} \cdot !k$ ways to disqualify $k$ candidates from being uniform-color, where $0 \leq k \leq 5$. - Jasmin's placement is calculated similarly, but we need to consider the remaining candidates after Ben's placement. 5. **Recursive Calculation for Jasmin's Placement:** - Define $D(n, k)$ as the number of ways Jasmin can place blocks such that no box remains a candidate, with $n$ boxes and $k$ non-candidates. - We have $D(n, 0) = !n$ and the recursive relation $D(n, k) = D(n-1, k-1) + D(n, k-1)$. 6. **Constructing the Table for $D(n, k)$:** - The table is constructed based on the recursive relation and initial conditions. - The values of $D(5, k)$ are extracted from the table for $k = 0$ to $5$. 7. **Calculating the Desired Probability:** - The probability that at least one box is of uniform color is calculated using: \[ \frac{\sum_{k=0}^{5}{\left(\binom{5}{k}\cdot{!}k\right)(5!-D(5,k))}}{(5!)^2} \] - Simplifying the expression, we find: \[ \frac{76 + 0 + 560 + 840 + 1080 + 0}{(5!)^2} = \frac{2556}{14400} = \frac{639}{3600} = \frac{71}{400} \] 8. **Final Answer:** - The probability that at least one box receives 3 blocks all of the same color is $\frac{71}{400}$. - The problem asks for $m+n$ where $\frac{m}{n} = \frac{71}{400}$. Thus, $m+n = 71 + 400 = \boxed{471}$.

An isosceles triangle and a rectangle have the same area. The base of the triangle is equal to the width of the rectangle, and this dimension is 10 units. The length of the rectangle is twice its width. What is the height of the triangle, $h$, in terms of the dimensions of the rectangle?

40

The lock opens only if a specific three-digit number is entered. An attempt consists of randomly selecting three digits from a given set of five. The code was guessed correctly only on the last of all attempts. How many attempts preceded the successful one?

124

Let $ A = \{1, 2, \cdots, 2004\} $ and $ f: A \rightarrow A $ be a bijection satisfying $ f^{[2004]}(x) = f(x) $, where $ f^{[2004]}(x) $ denotes applying $ f $ 2004 times to $ x $. How many such functions $ f $ are there?

1 + 2004!

Given the parabola $y^{2}=4x$, and the line $l$: $y=- \frac {1}{2}x+b$ intersects the parabola at points $A$ and $B$. (I) If the $x$-axis is tangent to the circle with $AB$ as its diameter, find the equation of the circle; (II) If the line $l$ intersects the negative semi-axis of $y$, find the maximum area of $\triangle AOB$.

\frac {32 \sqrt {3}}{9}

In how many ways can 8 people be seated in a row of chairs if three of the people, John, Wilma, and Paul, refuse to sit in three consecutive seats?

36000

There are 8 white balls and 2 red balls in a bag. Each time a ball is randomly taken out, a white ball is put back in. What is the probability that all red balls are taken out exactly by the 4th draw?

353/5000

The "Tiao Ri Method", invented by mathematician He Chengtian during the Southern and Northern Dynasties of China, is an algorithm for finding a more accurate fraction to represent a numerical value. Its theoretical basis is as follows: If the deficient approximate value and the excess approximate value of a real number $x$ are $\frac{b}{a}$ and $\frac{d}{c}$ ($a, b, c, d \in \mathbb{N}^*$) respectively, then $\frac{b+d}{a+c}$ is a more accurate deficient approximate value or excess approximate value of $x$. We know that $\pi = 3.14159...$, and if we let $\frac{31}{10} < \pi < \frac{49}{15}$, then after using the "Tiao Ri Method" once, we get $\frac{16}{5}$ as a more accurate excess approximate value of $\pi$, i.e., $\frac{31}{10} < \pi < \frac{16}{5}$. If we always choose the simplest fraction, then what approximate fraction can we get for $\pi$ after using the "Tiao Ri Method" four times?

\frac{22}{7}

Let $a \bowtie b = a + \sqrt{b + \sqrt{b + \sqrt{b + \ldots}}}$. If $5 \bowtie x = 12$, find the value of $x$.

42

In the sequence $ \left\{a_{n}\right\} $, if $ a_{k}+a_{k+1}=2k+1 $ (where $ k \in \mathbf{N}^{*} $), then $ a_{1}+a_{100} $ equals?

101

Attempt to obtain one billion (1,000,000,000) by multiplying two integers, each of which contains no zeros.

512 * 1953125

Determine the number of zeros in the quotient $Q = R_{30}/R_6$, where $R_k$ is a number consisting of $k$ repeated digits of 1 in base-ten.

25

In $\triangle ABC$, the sides opposite to angles $A$, $B$, $C$ are $a$, $b$, $c$ respectively. If $\sin (A-B)+ \sin C= \sqrt {2}\sin A$. (I) Find the value of angle $B$; (II) If $b=2$, find the maximum value of $a^{2}+c^{2}$, and find the values of angles $A$ and $C$ when the maximum value is obtained.

8+4 \sqrt {2}

Two fair, six-sided dice are rolled. What is the probability that the sum of the two numbers showing is less than or equal to 10 and at least one die shows a number greater than 3?

\frac{2}{3}

Circles with centers $A$, $B$, and $C$ each have radius $r$, where $1 < r < 2$. The distance between each pair of centers is $2$. If $B'$ is the point of intersection of circle $A$ and circle $C$ which is outside circle $B$, and if $C'$ is the point of intersection of circle $A$ and circle $B$ which is outside circle $C$, then length $B'C'$ equals

1+\sqrt{3(r^2-1)}

1. **Setting up the problem**: We are given three circles with centers at $A$, $B$, and $C$, each having radius $r$ where $1 < r < 2$. The distance between each pair of centers is $2$. We need to find the length of the line segment $B'C'$, where $B'$ and $C'$ are points of intersection of the circles centered at $A$ with circles centered at $C$ and $B$ respectively, and each is outside the other circle. 2. **Positioning the circles**: We can position the centers of the circles in the Cartesian plane as follows: - $A$ at $(0, \sqrt{3})$ - $B$ at $(1, 0)$ - $C$ at $(-1, 0)$ Each circle has radius $r$, so their equations are: - Circle $A$: $x^2 + (y - \sqrt{3})^2 = r^2$ - Circle $B$: $(x - 1)^2 + y^2 = r^2$ - Circle $C$: $(x + 1)^2 + y^2 = r^2$ 3. **Finding intersection points**: - Solve the equations of Circle $A$ and Circle $B$: \[ x^2 + (y - \sqrt{3})^2 = r^2 \quad \text{and} \quad (x - 1)^2 + y^2 = r^2 \] Expanding and simplifying, we get: \[ x^2 + y^2 - 2\sqrt{3}y + 3 = r^2 \quad \text{and} \quad x^2 - 2x + 1 + y^2 = r^2 \] Subtracting these equations: \[ 2x - 2\sqrt{3}y + 2 = 0 \quad \Rightarrow \quad x = \sqrt{3}y - 1 \] Substituting $x = \sqrt{3}y - 1$ back into the equation of Circle $A$: \[ (\sqrt{3}y - 1)^2 + (y - \sqrt{3})^2 = r^2 \] Solving this equation gives: \[ y = \frac{\sqrt{3} \pm \sqrt{r^2 - 1}}{2} \] We select the larger root $y = \frac{\sqrt{3} + \sqrt{r^2 - 1}}{2}$ for $B'$, and the corresponding $x$ value is $x = -\frac{1 + \sqrt{3(r^2 - 1)}}{2}$. 4. **Calculating $B'C'$**: - By symmetry, the $x$-coordinates of $B'$ and $C'$ are negatives of each other, and they have the same $y$-coordinate. Thus, the length of $B'C'$ is twice the absolute value of the $x$-coordinate of $B'$: \[ B'C' = 2 \left| -\frac{1 + \sqrt{3(r^2 - 1)}}{2} \right| = 1 + \sqrt{3(r^2 - 1)} \] 5. **Conclusion**: - The length of $B'C'$ is $1 + \sqrt{3(r^2 - 1)}$, which corresponds to choice $\textbf{(D)}$. Thus, the final answer is $\boxed{D}$.

Let $f : [0, 1] \rightarrow \mathbb{R}$ be a monotonically increasing function such that $$ f\left(\frac{x}{3}\right) = \frac{f(x)}{2} $$ $$ f(1 0 x) = 2018 - f(x). $$ If $f(1) = 2018$ , find $f\left(\dfrac{12}{13}\right)$ .

2018

(1) Given $0 < x < \frac{1}{2}$, find the maximum value of $y= \frac{1}{2}x(1-2x)$; (2) Given $x > 0$, find the maximum value of $y=2-x- \frac{4}{x}$; (3) Given $x$, $y\in\mathbb{R}_{+}$, and $x+y=4$, find the minimum value of $\frac{1}{x}+ \frac{3}{y}$.

1+ \frac{ \sqrt{3}}{2}

The square quilt block shown is made from sixteen unit squares, where eight of these squares have been divided in half diagonally to form triangles. Each triangle is shaded. What fraction of the square quilt is shaded? Express your answer as a common fraction.

\frac{1}{4}

One material particle entered the opening of a pipe, and after 6.8 minutes, a second particle entered the same opening. Upon entering the pipe, each particle immediately began linear motion along the pipe: the first particle moved uniformly at a speed of 5 meters per minute, while the second particle covered 3 meters in the first minute and in each subsequent minute covered 0.5 meters more than in the previous minute. How many minutes will it take for the second particle to catch up with the first?

17

Among 6 courses, if person A and person B each choose 3 courses, the number of ways in which exactly 1 course is chosen by both A and B is \_\_\_\_\_\_.

180

The store owner bought 2000 pens at $0.15 each and plans to sell them at $0.30 each, calculate the number of pens he needs to sell to make a profit of exactly $150.

1000

Given that the point $(9,7)$ is on the graph of $y=f(x)$, there is one point that must be on the graph of $2y=\frac{f(2x)}2+2$. What is the sum of coordinates of that point?

\frac{29}4

You start with a single piece of chalk of length 1. Every second, you choose a piece of chalk that you have uniformly at random and break it in half. You continue this until you have 8 pieces of chalk. What is the probability that they all have length $\frac{1}{8}$ ?

\frac{1}{63}

There are 7! total ways to break the chalks. How many of these result in all having length $\frac{1}{8}$ ? The first move gives you no choice. Then, among the remaining 6 moves, you must apply 3 breaks on the left side and 3 breaks on the right side, so there are $\binom{6}{3}=20$ ways to order those. On each side, you can either break the left side or the right side first. So the final answer is $$\frac{20 \cdot 2^{2}}{7!}=\frac{1}{63}$$

Given that $\theta$ is an angle in the third quadrant, and $\sin^{4}\theta + \cos^{4}\theta = \frac{5}{9}$, determine the value of $\sin 2\theta$.

-\frac{2\sqrt{2}}{3}

Emily sees a ship traveling at a constant speed along a straight section of a river. She walks parallel to the riverbank at a uniform rate faster than the ship. She counts $210$ equal steps walking from the back of the ship to the front. Walking in the opposite direction, she counts $42$ steps of the same size from the front of the ship to the back. In terms of Emily's equal steps, what is the length of the ship?

70

1. **Define Variables:** Let $L$ be the length of the ship, $E$ be the length of Emily's step, and $S$ be the length of the ship's step (the distance the ship travels while Emily takes one step). 2. **Set Up Equations:** - When Emily walks from the back of the ship to the front, she takes $210$ steps, covering a distance of $210E$. During this time, the front of the ship also moves forward by $210S$. The total distance covered by Emily relative to the ship is the length of the ship plus the distance the ship has moved, which gives us: \[ 210E = L + 210S \quad \text{(Equation 1)} \] - When Emily walks from the front of the ship to the back, she takes $42$ steps, covering a distance of $42E$. During this time, the back of the ship moves forward by $42S$. The total distance covered by Emily relative to the ship is the length of the ship minus the distance the ship has moved, which gives us: \[ 42E = L - 42S \quad \text{(Equation 2)} \] 3. **Manipulate Equations:** - First, rearrange Equation 1: \[ 210E - 210S = L \] - Rearrange Equation 2: \[ 42E + 42S = L \] - Multiply Equation 2 by $5$ to align the coefficients of $E$ with those in Equation 1: \[ 5(42E + 42S) = 5L \implies 210E + 210S = 5L \quad \text{(Equation 3)} \] - Subtract Equation 1 from Equation 3: \[ (210E + 210S) - (210E - 210S) = 5L - L \implies 420S = 4L \] - Simplify to find $L$ in terms of $S$: \[ L = 105S \] 4. **Solve for $\frac{L}{E}$:** - From Equation 1, substitute $L = 105S$: \[ 210E = 105S + 210S \implies 210E = 315S \] - Simplify to find the ratio of $E$ to $S$: \[ E = 1.5S \] - Substitute $E = 1.5S$ into the expression for $L$: \[ L = 105S = 70 \times 1.5S = 70E \] 5. **Conclusion:** - The length of the ship in terms of Emily's steps is $\boxed{70}$.

The number $ C $ is defined as the sum of all the positive integers $ n $ such that $ n-6 $ is the second largest factor of $ n $. What is the value of $ 11C $?

308

A semicircle is inscribed in an isosceles triangle with base 16 and height 15 so that the diameter of the semicircle is contained in the base of the triangle. What is the radius of the semicircle?

\frac{120}{17}

1. **Identify the Triangle and its Properties**: Let's denote the isosceles triangle as $\triangle ABD$ where $AB = AD$ and $BD = 16$. The height from $A$ to $BD$ is given as $15$. Let $M$ be the midpoint of $BD$. 2. **Divide the Triangle**: Draw a line from $A$ to $M$. This line divides $\triangle ABD$ into two right triangles, $\triangle ABM$ and $\triangle ADM$. Since $M$ is the midpoint of $BD$, $BM = MD = \frac{16}{2} = 8$. 3. **Use the Pythagorean Theorem**: In $\triangle ABM$, we know $BM = 8$ and $AM = 15$. We can find $AB$ using the Pythagorean theorem: \[ AB^2 = AM^2 + BM^2 = 15^2 + 8^2 = 225 + 64 = 289 \implies AB = \sqrt{289} = 17. \] Thus, $AB = AD = 17$. 4. **Reflect the Triangle**: Reflect $\triangle ABD$ over its base $BD$ to form a rhombus $ABCD$ with $AB = BC = CD = DA = 17$. 5. **Calculate the Semiperimeter and Area**: The semiperimeter $s$ of rhombus $ABCD$ is: \[ s = \frac{AB + BC + CD + DA}{2} = \frac{4 \times 17}{2} = 34. \] The area $[ABCD]$ of the rhombus is twice the area of $\triangle ABD$: \[ [ABCD] = 2 \times \left(\frac{1}{2} \times BD \times AM\right) = 2 \times (16 \times 15 \div 2) = 240. \] 6. **Apply the Formula for the Incircle of a Quadrilateral**: The formula relating the area, semiperimeter, and inradius $r$ of a quadrilateral is $s \cdot r = [ABCD]$. Substituting the known values: \[ 34r = 240 \implies r = \frac{240}{34} = \frac{120}{17}. \] 7. **Conclusion**: The radius of the semicircle inscribed in the isosceles triangle $\triangle ABD$ is $\boxed{\textbf{(B) }\frac{120}{17}}$.

Suppose $d$ is a digit. For how many values of $d$ is $2.00d5 > 2.005$?

5

To solve this problem, we need to compare the number $2.00d5$ with $2.005$ for each possible digit $d$ (where $d$ can be any digit from $0$ to $9$). 1. **Expression of $2.00d5$:** The number $2.00d5$ can be expressed as $2.00d5 = 2 + 0.00d + 0.0005$. Here, $0.00d$ represents the contribution of the digit $d$ in the thousandths place. 2. **Comparison with $2.005$:** We need to find when $2.00d5 > 2.005$. \[ 2 + 0.00d + 0.0005 > 2.005 \] Simplifying the inequality: \[ 0.00d + 0.0005 > 0.005 \] \[ 0.00d > 0.005 - 0.0005 \] \[ 0.00d > 0.0045 \] 3. **Value of $0.00d$:** Since $d$ is a digit, $0.00d$ can take values $0.000, 0.001, 0.002, \ldots, 0.009$. We need to find for which values of $d$, $0.00d > 0.0045$. 4. **Finding the values of $d$:** - If $d = 0$, then $0.00d = 0.000$ (which is not greater than $0.0045$). - If $d = 1$, then $0.00d = 0.001$ (which is not greater than $0.0045$). - If $d = 2$, then $0.00d = 0.002$ (which is not greater than $0.0045$). - If $d = 3$, then $0.00d = 0.003$ (which is not greater than $0.0045$). - If $d = 4$, then $0.00d = 0.004$ (which is not greater than $0.0045$). - If $d = 5$, then $0.00d = 0.005$ (which is greater than $0.0045$). - If $d = 6$, then $0.00d = 0.006$ (which is greater than $0.0045$). - If $d = 7$, then $0.00d = 0.007$ (which is greater than $0.0045$). - If $d = 8$, then $0.00d = 0.008$ (which is greater than $0.0045$). - If $d = 9$, then $0.00d = 0.009$ (which is greater than $0.0045$). 5. **Counting the valid values of $d$:** The values of $d$ that satisfy $0.00d > 0.0045$ are $5, 6, 7, 8, 9$. There are $5$ such values. Thus, there are $\boxed{\textbf{(C)}\ 5}$ values of $d$ for which $2.00d5 > 2.005$.

(1) Let the function $f(x)=|x+2|+|x-a|$. If the inequality $f(x) \geqslant 3$ always holds for $x$ in $\mathbb{R}$, find the range of the real number $a$. (2) Given positive numbers $x$, $y$, $z$ satisfying $x+2y+3z=1$, find the minimum value of $\frac{3}{x}+\frac{2}{y}+\frac{1}{z}$.

16+8\sqrt{3}

In the complex plane, the graph of $ |z - 5| = 3|z + 5| $ intersects the graph of $ |z| = k $ in exactly one point. Find all possible values of $ k $.

12.5

Nine people are practicing the triangle dance, which is a dance that requires a group of three people. During each round of practice, the nine people split off into three groups of three people each, and each group practices independently. Two rounds of practice are different if there exists some person who does not dance with the same pair in both rounds. How many different rounds of practice can take place?

280

Given the function $f\left(x+ \frac {1}{2}\right)= \frac {2x^{4}+x^{2}\sin x+4}{x^{4}+2}$, calculate the value of $f\left( \frac {1}{2017}\right)+f\left( \frac {2}{2017}\right)+\ldots+f\left( \frac {2016}{2017}\right)$.

4032

Vasya has three cans of paint of different colors. In how many different ways can he paint a fence consisting of 10 planks so that any two adjacent planks are different colors and he uses all three colors? Provide a justification for your answer.

1530

Calculate:<br/>$(1)(1\frac{3}{4}-\frac{3}{8}+\frac{5}{6})÷(-\frac{1}{24})$;<br/>$(2)-2^2+(-4)÷2×\frac{1}{2}+|-3|$.

-2

For a bijective function $g : R \to R$ , we say that a function $f : R \to R$ is its superinverse if it satisfies the following identity $(f \circ g)(x) = g^{-1}(x)$ , where $g^{-1}$ is the inverse of $g$ . Given $g(x) = x^3 + 9x^2 + 27x + 81$ and $f$ is its superinverse, find $|f(-289)|$ .

10

A factory implements a time-based wage system, where each worker is paid 6 yuan for every hour worked, for a total of 8 hours per day. However, the clock used for timing is inaccurate: it takes 69 minutes for the minute hand to coincide with the hour hand once. Calculate the amount of wages underpaid to each worker per day.

2.60

For how many non-negative real values of $x$ is $\sqrt{169-\sqrt[4]{x}}$ an integer?

14

When three standard dice are tossed, the numbers $a, b, c$ are obtained. Find the probability that $abc = 216$.

\frac{1}{216}

The numbers $1,2, \ldots, 10$ are randomly arranged in a circle. Let $p$ be the probability that for every positive integer $k<10$, there exists an integer $k^{\prime}>k$ such that there is at most one number between $k$ and $k^{\prime}$ in the circle. If $p$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$, compute $100 a+b$.

1390

Let $n=10$ and call two numbers close if there is at most one number between them and an circular permutation focused if only $n$ is greater than all numbers close to it. Let $A_{n}$ be the number of focused circular permutations of $\{1,2, \ldots, n\}$. If $n \geq 5$, then there are 2 cases: $n-1$ is either one or two positions from $n$. If $n-1$ is one position from $n$, it is either on its left or right. In this case, one can check a permutation is focused if and only if removing $n$ yields a focused permutation, so there are $2 A_{n-1}$ permutations in this case. If $n-1$ is two positions from $n$, there are $n-2$ choices for $k$, the element that lies between $n$ and $n-1$. One can show that this permutation is focused if and only if removing both $n$ and $k$ and relabeling the numbers yields a focused permutation, so there are $2(n-2) A_{n-2}$ permutations in this case. Thus, we have $A_{n}=2 A_{n-1}+2(n-2) A_{n-2}$. If we let $p_{n}=A_{n} /(n-1)$ ! the probability that a random circular permutation is focused, then this becomes $$p_{n}=\frac{2 p_{n-1}+2 p_{n-2}}{n-1}$$ Since $p_{3}=p_{4}=1$, we may now use this recursion to calculate $$p_{5}=1, p_{6}=\frac{4}{5}, p_{7}=\frac{3}{5}, p_{8}=\frac{2}{5}, p_{9}=\frac{1}{4}, p_{10}=\frac{13}{90}$$

Given $A(-1,\cos \theta)$, $B(\sin \theta,1)$, if $|\overrightarrow{OA}+\overrightarrow{OB}|=|\overrightarrow{OA}-\overrightarrow{OB}|$, then find the value of the acute angle $\theta$.

\frac{\pi}{4}

Given an infinite geometric sequence $\{a_n\}$, the product of its first $n$ terms is $T_n$, and $a_1 > 1$, $a_{2008}a_{2009} > 1$, $(a_{2008} - 1)(a_{2009} - 1) < 0$, determine the maximum positive integer $n$ for which $T_n > 1$.

4016

For all non-zero real numbers $x$ and $y$ such that $x-y=xy$, $\frac{1}{x}-\frac{1}{y}$ equals

-1

1. **Start with the given equation and manipulate it:** Given that $x - y = xy$, we can rearrange this equation to: \[ x - y - xy = 0 \] Adding 1 to both sides, we get: \[ x - y - xy + 1 = 1 \] Factoring the left-hand side, we have: \[ (x-1)(y+1) = 1 \] 2. **Analyze the expression $\frac{1}{x} - \frac{1}{y}$:** We rewrite the expression using a common denominator: \[ \frac{1}{x} - \frac{1}{y} = \frac{y-x}{xy} \] Since $x - y = xy$, substituting $xy$ for $x - y$ in the numerator, we get: \[ \frac{1}{x} - \frac{1}{y} = \frac{-xy}{xy} = -1 \] 3. **Conclusion:** The expression $\frac{1}{x} - \frac{1}{y}$ simplifies to $-1$. Therefore, the correct answer is: \[ \boxed{\textbf{(D) } -1} \]

Alpha and Beta both took part in a two-day problem-solving competition. At the end of the second day, each had attempted questions worth a total of 500 points. Alpha scored 160 points out of 300 points attempted on the first day, and scored 140 points out of 200 points attempted on the second day. Beta who did not attempt 300 points on the first day, had a positive integer score on each of the two days, and Beta's daily success rate (points scored divided by points attempted) on each day was less than Alpha's on that day. Alpha's two-day success ratio was 300/500 = 3/5. The largest possible two-day success ratio that Beta could achieve is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

849

Let $q$ be the number of questions Beta takes on day 1 and $a$ be the number he gets right. Let $b$ be the number he gets right on day 2. These inequalities follow: \[\frac{a}{q} < \frac{160}{300} = \frac{8}{15}\] \[\frac{b}{500-q} < \frac{140}{200} = \frac{7}{10}\] Solving for a and b and adding the two inequalities: \[a + b < \frac{8}{15}q + (350 - \frac{7}{10}q)\] \[a + b < 350 - \frac{1}{6}q\] From here, we see the largest possible value of $a+b$ is $349$. Checking our conditions, we know that $a$ must be positive so therefore $q$ must be positive. A quick check shows that for $2\le q \le 5$, $q$ follows all the conditions and results in $a+b=349$. This makes Beta's success ratio $\frac{349}{500}$. Thus, the answer is $m+n = 349 + 500 = \boxed{849}$.

Read the following material and complete the corresponding tasks. 平衡多项式定义:对于一组多项式$x+a$,$x+b$,$x+c$,$x+d(a$,$b$,$c$,$d$是常数）,当其中两个多项式的乘积与另外两个多项式乘积的差是一个常数$p$时,称这样的四个多项式是一组平衡多项式,$p$的绝对值是这组平衡多项式的平衡因子. 例如:对于多项式$x+1$,$x+2$,$x+5$,$x+6$,因为$(x+1)(x+6)-(x+2)(x+5)=(x^{2}+7x+6)-(x^{2}+7x+10)=-4$,所以多项式$x+1$,$x+2$,$x+5$,$x+6$是一组平衡多项式,其平衡因子为$|-4|=4$. 任务: $(1)$小明发现多项式$x+3$,$x+4$,$x+6$,$x+7$是一组平衡多项式,在求其平衡因子时,列式如下:$(x+3)(x+7)-(x+4)(x+6)$,根据他的思路求该组平衡多项式的平衡因子； $A$.判断多项式$x-1$,$x-2$,$x-4$,$x-5$是否为一组平衡多项式,若是,求出其平衡因子；若不是,说明理由. $B$.若多项式$x+2$,$x-4$,$x+1$,$x+m(m$是常数）是一组平衡多项式,求$m$的值.

-5

Given the quadratic function $f(x) = mx^2 - 2x - 3$, if the solution set of the inequality $f(x) < 0$ is $(-1, n)$. (1) Solve the inequality about $x$: $2x^2 - 4x + n > (m + 1)x - 1$; (2) Determine whether there exists a real number $a \in (0, 1)$, such that the minimum value of the function $y = f(a^x) - 4a^{x+1}$ ($x \in [1, 2]$) is $-4$. If it exists, find the value of $a$; if not, explain why.

\frac{1}{3}

Given an arithmetic sequence, the sum of the first four terms is 26, the sum of the last four terms is 110, and the sum of all terms in the sequence is 187. Determine the total number of terms in the sequence.

11

Our school's basketball team has won the national middle school basketball championship multiple times! In one competition, including our school's basketball team, 7 basketball teams need to be randomly divided into two groups (one group with 3 teams and the other with 4 teams) for the group preliminaries. The probability that our school's basketball team and the strongest team among the other 6 teams end up in the same group is ______.

\frac{3}{7}

Twelve chess players played a round-robin tournament. Each player then wrote 12 lists. In the first list, only the player himself was included, and in the $(k+1)$-th list, the players included those who were in the $k$-th list as well as those whom they defeated. It turned out that each player's 12th list differed from the 11th list. How many draws were there?

54

Given a triangle $ABC$ where $AB = AC$ and $\angle A = 80^\circ$. Inside triangle $ABC$ is a point $M$ such that $\angle MBC = 30^\circ$ and $\angle MCB = 10^\circ$. Find $\angle AMC$.

70

What is the smallest positive integer $n$ such that $\frac{n}{n+53}$ is equal to a terminating decimal?

11

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(f(x)+y)+xf(y)=f(xy+y)+f(x)$$ for reals $x, y$.

f(x) = x \text{ or } f(x) = 0

Let $ f: \mathbb{R} \rightarrow \mathbb{R} $ be a function satisfying the functional equation: \[ f(f(x) + y) + x f(y) = f(xy + y) + f(x) \] for all real numbers $ x $ and $ y $. ### Step 1: Initial Substitution Start by substituting $ y = 0 $ into the equation: \[ f(f(x)) + x f(0) = f(x) + f(x) \] Simplifying gives: \[ f(f(x)) + x f(0) = 2f(x) \] ### Step 2: Exploring Constant Solutions Suppose $ f(x) = c $ for all $ x $, where $ c $ is a constant. Then: \[ f(f(x) + y) = f(c + y) = c \quad \text{and} \quad f(xy + y) = f(y(x + 1)) = c \] Hence, substituting back into the original equation: \[ c + x \cdot c = c + c \] This implies $ xc = c $. If $ c \neq 0 $, this equation has no solution for all $ x $. Thus, $ c = 0 $ is the only constant function solution. Therefore, one solution is: \[ f(x) = 0 \quad \text{for all } x \in \mathbb{R} \] ### Step 3: Exploring Non-Constant Solutions Assume $ f $ is non-constant. Substitute $ f(x) = x $ in the original equation: \[ f(f(x) + y) = f(x + y) \] \[ x f(y) = x \cdot y \] \[ f(xy + y) = xy + y \] \[ f(x) = x \] Substituting $ f(x) = x $ into the original equation: \[ f(f(x) + y) + x f(y) = f(xy + y) + f(x) \] Results in: \[ f(x + y) + xy = xy + y + x \] Thus both sides are equal, confirming that: \[ f(x) = x \quad \text{for all } x \in \mathbb{R} \] ### Conclusion The solutions to the functional equation are: \[ f(x) = x \quad \text{or} \quad f(x) = 0 \quad \text{for all } x \in \mathbb{R} \] Thus, the complete set of solutions is: \[ \boxed{f(x) = x \text{ or } f(x) = 0} \] Both solutions satisfy the initial functional equation over all reals.

If the height of an external tangent cone of a sphere is three times the radius of the sphere, determine the ratio of the lateral surface area of the cone to the surface area of the sphere.

\frac{3}{2}

Calculate: $1.23 \times 67 + 8.2 \times 12.3 - 90 \times 0.123$

172.2

A certain organization consists of five leaders and some number of regular members. Every year, the current leaders are kicked out of the organization. Next, each regular member must find two new people to join as regular members. Finally, five new people are elected from outside the organization to become leaders. In the beginning, there are fifteen people in the organization total. How many people total will be in the organization five years from now?

2435

James wrote a different integer from 1 to 9 in each cell of a table. He then calculated the sum of the integers in each of the rows and in each of the columns of the table. Five of his answers were 12, 13, 15, 16, and 17, in some order. What was his sixth answer?

17

Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0,0)$, $B(0,12)$, $C(16,0)$, $A'(24,18)$, $B'(36,18)$, $C'(24,2)$. A rotation of $m$ degrees clockwise around the point $(x,y)$ where $0<m<180$, will transform $\triangle ABC$ to $\triangle A'B'C'$. Find $m+x+y$.

108

After sketching, it is clear a $90^{\circ}$ rotation is done about $(x,y)$. Looking between $A$ and $A'$, $x+y=18$ and $x-y=24$. Solving gives $(x,y)\implies(21,-3)$. Thus $90+21-3=\boxed{108}$. ~mn28407

If the surface area of a cone is $3\pi$, and its lateral surface unfolds into a semicircle, then the diameter of the base of the cone is ___.

\sqrt{6}

Let $C(k)$ denotes the sum of all different prime divisors of a positive integer $k$. For example, $C(1)=0$, $C(2)=2, C(45)=8$. Find all positive integers $n$ such that $C(2^{n}+1)=C(n)$

n=3

Let $P(t)$ be the largest prime divisor of a positive integer $t>1$. Let $m$ be the largest odd divisor of $n: n=2^{k} m$. Then $2^{n}+1=2^{2^{k} m}+1=a^{m}+1$, where $a=2^{2^{k}}$. If $k>0$, that is, $n$ is even, then $C(n)=C(m)+2$ and $C(2^{n}+1)=C(a^{m}+1)$. We need the following two lemmas. Lemma 1. For every prime $p>2$ we have $P(\frac{a^{p}+1}{a+1})=p$ or $P(\frac{a^{p}+1}{a+1}) \geqslant 2 p+1$. Proof. Let $P(\frac{a^{p}+1}{a+1})=q$. It follows from Fermat's little theorem that $q$ divides $2^{q-1}-1$ and therefore $(a^{2 p}-1, a^{q-1}-1)=a^{(2 p, q-1)}-1$. The greatest common divisor $(2 p, q-1)$ is even and must equal $2 p$ or 2. In the first case $2 p$ divides $q-1$, whence $q \geqslant 2 p+1$. In the second case $q$ divides $a^{2}-1$ but not $a-1$ (because $a^{p}+1$ is divisible by $q$), that is, $a \equiv-1 \pmod{q}$. Then $\frac{a^{p}+1}{a+1}=a^{p-1}-\ldots+1 \equiv p \pmod{q}$ and $p=q$. Lemma 2. If $p_{1}$ and $p_{2}$ are different odd primes then $P(\frac{a^{p_{1}}+1}{a+1}) \neq P(\frac{a^{p_{2}}+1}{a+1})$. Proof. If $P(\frac{a^{p_{1}}+1}{a+1})=P(\frac{a^{p_{2}}+1}{a+1})=q$ then $q$ divides $a^{2 p_{1}}-1$ and $a^{2 p_{2}}-1$, therefore $(a^{2 p_{1}}-1, a^{2 p_{2}}-1)=a^{(2 p_{1}, 2 p_{2})}-1=a^{2}-1$ and hence $a+1$, but then $p_{1}=q$ and $p_{2}=q$, a contradiction. We are ready now to solve the problem. Let $p_{1}, \ldots, p_{s}$ be all the prime divisors of $n$. It follows from Lemma 2 that $$C(2^{n}+1) \geqslant P(\frac{a^{p_{1}}+1}{a+1})+\ldots+P(\frac{a^{p_{s}}+1}{a+1})$$ If $C(2^{n}+1)>P(\frac{a^{p_{1}}+1}{a+1})+\ldots+P(\frac{a^{p_{s}}+1}{a+1})$, then $2^{n}+1$ has at least one prime divisor not summed in the L.H.S., that is, $$C(2^{n}+1) \geqslant P(\frac{a^{p_{1}}+1}{a+1})+\ldots+P(\frac{a^{p_{s}}+1}{a+1})+3 \geqslant p_{1}+\ldots+p_{s}+3>C(n)$$ Therefore we can assume the equality: $$C(2^{n}+1)=P(\frac{a^{p_{1}}+1}{a+1})+\ldots+P(\frac{a^{p_{s}}+1}{a+1})$$ If in this case there is an $i$ such that $P(\frac{a^{p_{i}}+1}{a+1}) \neq p_{i}$, then $C(2^{n}+1) \geqslant p_{1}+\ldots+p_{s}+p_{i}+1>C(n)$. It remains to consider the case when $P(\frac{a^{p_{i}}+1}{a+1})=p_{i}$ for all $i$. In this case we have $C(n)=C(2^{n}+1)=p_{1}+\ldots+p_{s}$, so $n$ must be odd and $a=2$. But $2^{p} \equiv 2 \pmod{p}$ for all odd prime $p$, therefore $p>3$ cannot divide $2^{p}+1$. Thus $s=1, p=3, n=3^{r}$ with some positive integral $r$. The number $2^{n}+1=2^{3^{r}}+1$ must be a power of 3. However 19 divides this number for $r=2$ and consequently for all $r \geqslant 2$. Thus the only remaining case is $n=3$, which obviously satisfies the condition.

Jacqueline has 40% less sugar than Liliane, and Bob has 30% less sugar than Liliane. Express the relationship between the amounts of sugar that Jacqueline and Bob have as a percentage.

14.29\%

Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$. Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$. Find the number of such sequences for which $a_n=0$ for some $n$.

494

Let $a_1=x, a_2=y, a_3=z$. First note that if any absolute value equals 0, then $a_n=0$. Also note that if at any position, $a_n=a_{n-1}$, then $a_{n+2}=0$. Then, if any absolute value equals 1, then $a_n=0$. Therefore, if either $|y-x|$ or $|z-y|$ is less than or equal to 1, then that ordered triple meets the criteria. Assume that to be the only way the criteria is met. To prove, let $|y-x|>1$, and $|z-y|>1$. Then, $a_4 \ge 2z$, $a_5 \ge 4z$, and $a_6 \ge 4z$. However, since the minimum values of $a_5$ and $a_6$ are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be $z=1$, $|y-x|=2$. Again assume that any other scenario will not meet criteria. To prove, divide the other scenarios into two cases: $z>1$, $|y-x|>1$, and $|z-y|>1$; and $z=1$, $|y-x|>2$, and $|z-y|>1$. For the first one, $a_4 \ge 2z$, $a_5 \ge 4z$, $a_6 \ge 8z$, and $a_7 \ge 16z$, by which point we see that this function diverges. For the second one, $a_4 \ge 3$, $a_5 \ge 6$, $a_6 \ge 18$, and $a_7 \ge 54$, by which point we see that this function diverges. Therefore, the only scenarios where $a_n=0$ is when any of the following are met: $|y-x|<2$ (280 options) $|z-y|<2$ (280 options, 80 of which coincide with option 1) $z=1$, $|y-x|=2$. (16 options, 2 of which coincide with either option 1 or option 2) Adding the total number of such ordered triples yields $280+280-80+16-2=\boxed{494}$. Note to author: Because $a_4 \ge 2z$, $a_5 \ge 4z$, $a_6 \ge 8z$, and $a_7 \ge 16z$ doesn't mean the function diverges. What if $z = 7$, $a_4 = 60$, and $a_5 = 30$ too? Note to the Note to author: This isn't possible because the difference between is either 0, 1, or some number greater than 1. If $a_4 = 63$ (since it must be a multiple of $z$), then a_5 is either 0, 63, or some number greater than 63.

Compute $\begin{pmatrix} 1 & -1 \\ 1 & 0 \end{pmatrix}^3.$

\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}.

A club has between 300 and 400 members. The members gather every weekend and are divided into eight distinct groups. If two members are absent, the groups can all have the same number of members. What is the sum of all possible numbers of members in the club?

4200

Given that point $P$ moves on the ellipse $\frac{x^{2}}{4}+y^{2}=1$, find the minimum distance from point $P$ to line $l$: $x+y-2\sqrt{5}=0$.

\frac{\sqrt{10}}{2}

For a natural number $n \ge 3$ , we draw $n - 3$ internal diagonals in a non self-intersecting, but not necessarily convex, n-gon, cutting the $n$ -gon into $n - 2$ triangles. It is known that the value (in degrees) of any angle in any of these triangles is a natural number and no two of these angle values are equal. What is the largest possible value of $n$ ?

41

Matt's four cousins are coming to visit. There are four identical rooms that they can stay in. If any number of the cousins can stay in one room, how many different ways are there to put the cousins in the rooms?

15

A circle has center $(-10, -4)$ and has radius $13$. Another circle has center $(3, 9)$ and radius $\sqrt{65}$. The line passing through the two points of intersection of the two circles has equation $x+y=c$. What is $c$?

3

1. **Identify the equations of the circles**: The first circle has center $(-10, -4)$ and radius $13$. Its equation is: \[ (x+10)^2 + (y+4)^2 = 169. \] The second circle has center $(3, 9)$ and radius $\sqrt{65}$. Its equation is: \[ (x-3)^2 + (y-9)^2 = 65. \] 2. **Set up the equation for the intersection of the circles**: The points of intersection satisfy both circle equations. Setting the left-hand sides of the equations equal gives: \[ (x+10)^2 + (y+4)^2 = (x-3)^2 + (y-9)^2. \] 3. **Simplify the equation**: Expand both sides: \[ x^2 + 20x + 100 + y^2 + 8y + 16 = x^2 - 6x + 9 + y^2 - 18y + 81. \] Simplify by canceling $x^2$ and $y^2$ from both sides: \[ 20x + 8y + 116 = -6x - 18y + 90. \] Combine like terms: \[ 26x + 26y + 26 = 104. \] Divide through by 26: \[ x + y + 1 = 4 \implies x + y = 3. \] 4. **Conclusion**: The equation of the line passing through the intersection points of the circles is $x + y = 3$. Therefore, the value of $c$ is $\boxed{\textbf{(A)}\ 3}$.

There are four distinct codes $A, B, C, D$ used by an intelligence station, with one code being used each week. Each week, a code is chosen randomly with equal probability from the three codes that were not used the previous week. Given that code $A$ is used in the first week, what is the probability that code $A$ is also used in the seventh week? (Express your answer as a simplified fraction.)

61/243

After Natasha ate a third of the peaches from the jar, the level of the compote lowered by one quarter. By how much (relative to the new level) will the level of the compote lower if all the remaining peaches are eaten?

\frac{2}{9}

In a mathematics class, the probability of earning an A is 0.6 times the probability of earning a B, and the probability of earning a C is 1.2 times the probability of earning a B. Assuming that all grades are A, B, or C, how many B's will there be in a mathematics class of 40 students?

14

Given $\tan \theta = \frac{5}{12}$, where $180^{\circ} \leq \theta \leq 270^{\circ}$. If $A = \cos \theta + \sin \theta$, find the value of $A$.

-\frac{17}{13}

Let \[f(x) = x^3 + 6x^2 + 16x + 28.\]The graphs of $y = f(x)$ and $y = f^{-1}(x)$ intersect at exactly one point $(a,b).$ Enter the ordered pair $(a,b).$

(-4,-4)

Triangle $DEF$ is isosceles with angle $E$ congruent to angle $F$. The measure of angle $F$ is three times the measure of angle $D$. What is the number of degrees in the measure of angle $E$?

\frac{540}{7}

Let $O$ be the origin. There exists a scalar $m$ such that for any points $E,$ $F,$ $G,$ and $H$ satisfying the vector equation: \[4 \overrightarrow{OE} - 3 \overrightarrow{OF} + 2 \overrightarrow{OG} + m \overrightarrow{OH} = \mathbf{0},\] the four points $E,$ $F,$ $G,$ and $H$ are coplanar. Find the value of $m.$

-3

For how many integer values of $n$ between 1 and 1000 inclusive does the decimal representation of $\frac{n}{1260}$ terminate?

47

In a trapezoid, the lengths of the diagonals are known to be 6 and 8, and the length of the midline is 5. Find the height of the trapezoid.

4.8

Given an equilateral triangle $ \triangle ABC $ with a side length of 1, \[ \overrightarrow{AP} = \frac{1}{3}(\overrightarrow{AB} + \overrightarrow{AC}), \quad \overrightarrow{AQ} = \overrightarrow{AP} + \frac{1}{2}\overrightarrow{BC}. \] Find the area of $ \triangle APQ $.

\frac{\sqrt{3}}{12}