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

problem stringlengths 10 1.99k	answer stringlengths 1 74	solution stringclasses 245 values
Calculate the volume of a tetrahedron with vertices at points $A_{1}, A_{2}, A_{3}, A_{4}$, and its height dropped from vertex $A_{4}$ to the face $A_{1} A_{2} A_{3}$. $A_{1}(-2, 0, -4)$ $A_{2}(-1, 7, 1)$ $A_{3}(4, -8, -4)$ $A_{4}(1, -4, 6)$	5\sqrt{2}
How many positive integer multiples of $77$ can be expressed in the form $10^{j} - 10^{i}$, where $i$ and $j$ are integers and $0 \leq i < j \leq 49$?	182
Let $\alpha$ be a nonreal root of $x^4 = 1.$ Compute \[(1 - \alpha + \alpha^2 - \alpha^3)^4 + (1 + \alpha - \alpha^2 + \alpha^3)^4.\]	32
Find the largest prime divisor of $36^2 + 49^2$.	13
In trapezoid \(ABCD\), \(AB\) is parallel to \(DC\) and \(\angle DAF = 90^\circ\). Point \(E\) is on \(DC\) such that \(EB = BC = CE\). Point \(F\) is on \(AB\) such that \(DF\) is parallel to \(EB\). In degrees, what is the measure of \(\angle FDA\)?	30
How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \[\left\{ \begin{aligned} x+3y&=3 \\ \left\| \|x\| - \|y\| \right\| &= 1 \end{aligned}\right.\]	3
In triangle \(ABC\), the side \(BC\) is 19 cm. The perpendicular \(DF\), drawn to side \(AB\) through its midpoint \(D\), intersects side \(BC\) at point \(F\). Find the perimeter of triangle \(AFC\) if side \(AC\) is 10 cm.	29
A doctor told Mikael to take a pill every 75 minutes. He took his first pill at 11:05. At what time did he take his fourth pill?	14:50
Given that (1+ex)<sup>2019</sup>=a<sub>0</sub>+a<sub>1</sub>x+a<sub>2</sub>x<sup>2</sup>+……+a<sub>2019</sub>x<sup>2019</sup>, find the value of: - $$\frac {a_{1}}{e}$$+ $$\frac {a_{2}}{e^{2}}$$\- $$\frac {a_{3}}{e^{3}}$$+ $$\frac {a_{4}}{e^{4}}$$\-……- $$\frac {a_{2019}}{e^{2019}}$$	-1
The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.	294	Let there be $m$ members and $n$ members for the square and $c$ for the number of columns of the other formation. We have $n^2 +5 = c(c+7) \implies n^2+5 = \left(c+\frac{7}{2}\right)^2 -\frac{49}{4} \implies n^2 - \left(c+\frac{7}{2}\right)^2 = -\frac{69}{4} \implies \left(n-c-\frac{7}{2}\right)\left(n + c +\frac{7}{2}\right) \implies (2n-2c-7)(2n+2n+7) = -69.$ To maximize this we let $2n+2c+7 = 68$ and $2n-2c-7 = 1.$ Solving we find $n = 17$ so the desired number of members is $17^2 + 5 = \boxed{294}.$
Phil has 7 green marbles and 3 purple marbles in a bag. He removes a marble at random, records the color, puts it back, and then repeats this process until he has withdrawn 6 marbles. What is the probability that exactly three of the marbles that he removes are green? Express your answer as a decimal rounded to the nearest thousandth.	.185
Given that the equation of line $l_{1}$ is $y=x$, and the equation of line $l_{2}$ is $y=kx-k+1$, find the value of $k$ for which the area of triangle $OAB$ is $2$.	\frac{1}{5}
Given that the midpoint of side $BC$ of triangle $\triangle ABC$ is $D$, point $E$ lies in the plane of $\triangle ABC$, and $\overrightarrow{CD}=3\overrightarrow{CE}-2\overrightarrow{CA}$, if $\overrightarrow{AC}=x\overrightarrow{AB}+y\overrightarrow{BE}$, then determine the value of $x+y$.	11
In a finite sequence of real numbers, the sum of any 7 consecutive terms is negative while the sum of any 11 consecutive terms is positive. What is the maximum number of terms in such a sequence?	16
Quantities $a$ and $b$ vary inversely. When $a$ is $800$, $b$ is $0.5$. If the product of $a$ and $b$ increases by $200$ when $a$ is doubled, what is $b$ when $a$ is $1600$?	0.375
The value of $ 21!$ is $ 51{,}090{,}942{,}171{,}abc{,}440{,}000$ , where $ a$ , $ b$ , and $ c$ are digits. What is the value of $ 100a \plus{} 10b \plus{} c$ ?	709
Let $x$ be the least real number greater than $1$ such that $\sin(x) = \sin(x^2)$, where the arguments are in degrees. What is $x$ rounded up to the closest integer?	13	To solve the problem, we need to find the smallest real number $x > 1$ such that $\sin(x) = \sin(x^2)$. This equation holds true when $x$ and $x^2$ differ by a multiple of $360^\circ$ or when $x^2$ is equivalent to $180^\circ - x + 360^\circ k$ for some integer $k$. We will analyze each choice given in the problem. #### Step 1: Analyze each choice We need to check each choice to see if it satisfies $\sin(x) = \sin(x^2)$. For choice $\textbf{(A)}$, $x = 10$: \[ \sin(10^\circ) \neq \sin(100^\circ) \quad \text{since} \quad \sin(100^\circ) = \sin(80^\circ) \quad \text{and} \quad \sin(10^\circ) \neq \sin(80^\circ). \] For choice $\textbf{(B)}$, $x = 13$: \[ \sin(13^\circ) = \sin(169^\circ) \quad \text{since} \quad \sin(169^\circ) = \sin(11^\circ) \quad \text{and} \quad \sin(13^\circ) \approx \sin(11^\circ). \] This is a potential solution as $\sin(13^\circ) - \sin(11^\circ)$ is very small. For choice $\textbf{(C)}$, $x = 14$: \[ \sin(14^\circ) \neq \sin(196^\circ) \quad \text{since} \quad \sin(196^\circ) = -\sin(16^\circ) \quad \text{and} \quad \sin(14^\circ) \neq -\sin(16^\circ). \] For choice $\textbf{(D)}$, $x = 19$: \[ \sin(19^\circ) \neq \sin(361^\circ) \quad \text{since} \quad \sin(361^\circ) = \sin(1^\circ) \quad \text{and} \quad \sin(19^\circ) \neq \sin(1^\circ). \] For choice $\textbf{(E)}$, $x = 20$: \[ \sin(20^\circ) \neq \sin(400^\circ) \quad \text{since} \quad \sin(400^\circ) = \sin(40^\circ) \quad \text{and} \quad \sin(20^\circ) \neq \sin(40^\circ). \] #### Step 2: Conclusion From the analysis, the only choice where $\sin(x) \approx \sin(x^2)$ is $\textbf{(B)}$ where $x = 13$. The values $\sin(13^\circ)$ and $\sin(11^\circ)$ are very close, making the difference between them nearly zero. Therefore, the answer is $\boxed{\textbf{(B) } 13}$.
A bag contains nine blue marbles, ten ugly marbles, and one special marble. Ryan picks marbles randomly from this bag with replacement until he draws the special marble. He notices that none of the marbles he drew were ugly. Given this information, what is the expected value of the number of total marbles he drew?	\frac{20}{11}	The probability of drawing $k$ marbles is the probability of drawing $k-1$ blue marbles and then the special marble, which is $p_{k}=\left(\frac{9}{20}\right)^{k-1} \times \frac{1}{20}$. The probability of drawing no ugly marbles is therefore $\sum_{k=1}^{\infty} p_{k}=\frac{1}{11}$. Then given that no ugly marbles were drawn, the probability that $k$ marbles were drawn is $11 p_{k}$. The expected number of marbles Ryan drew is $$\sum_{k=1}^{\infty} k\left(11 p_{k}\right)=\frac{11}{20} \sum_{k=1}^{\infty} k\left(\frac{9}{20}\right)^{k-1}=\frac{11}{20} \times \frac{400}{121}=\frac{20}{11}$$ (To compute the sum in the last step, let $S=\sum_{k=1}^{\infty} k\left(\frac{9}{20}\right)^{k-1}$ and note that $\frac{9}{20} S=S-\sum_{k=1}^{\infty}\left(\frac{9}{20}\right)^{k-1}=$ $\left.S-\frac{20}{11}\right)$.
A fair dice is rolled twice, and the scores obtained are denoted as $m$ and $n$ respectively. Let the angle between vector $a=(m,n)$ and vector $b=(1,-1)$ be $\theta$. The probability that $\theta$ is an acute angle is $\_\_\_\_\_\_\_\_\_\_\_\_\_.$	\frac{5}{12}
For how many integers $n$ is $\frac n{20-n}$ the square of an integer?	4	To solve the problem, we need to find the integers $n$ for which $\frac{n}{20-n}$ is a perfect square. 1. Examine the range of $n$: - If $n < 0$ or $n > 20$, the fraction $\frac{n}{20-n}$ is negative, and thus cannot be a perfect square. - If $n = 20$, the fraction is undefined. - If $n \in \{1, 2, \dots, 9\}$, the fraction $\frac{n}{20-n}$ is positive but less than 1, hence cannot be a perfect square. Therefore, we only need to consider $n = 0$ and $n \in \{10, 11, \dots, 19\}$. 2. Check specific values of $n$: - For $n = 0$, $\frac{n}{20-n} = \frac{0}{20} = 0$, which is $0^2$ (a perfect square). - For $n = 10$, $\frac{n}{20-n} = \frac{10}{10} = 1$, which is $1^2$ (a perfect square). 3. Exclude prime values of $n$: - For prime $n$ within the range $\{11, 13, 17, 19\}$, the fraction $\frac{n}{20-n}$ will not be an integer, as the denominator will not contain the prime factor present in the numerator. Thus, these values cannot yield a perfect square. 4. Check remaining composite values of $n$: - For $n = 12$, $\frac{n}{20-n} = \frac{12}{8} = 1.5$, not a perfect square. - For $n = 14$, $\frac{n}{20-n} = \frac{14}{6} = \frac{7}{3}$, not a perfect square. - For $n = 15$, $\frac{n}{20-n} = \frac{15}{5} = 3$, not a perfect square. - For $n = 16$, $\frac{n}{20-n} = \frac{16}{4} = 4$, which is $2^2$ (a perfect square). - For $n = 18$, $\frac{n}{20-n} = \frac{18}{2} = 9$, which is $3^2$ (a perfect square). 5. Conclusion: The integers $n$ that make $\frac{n}{20-n}$ a perfect square are $n = 0, 10, 16, 18$. There are four such integers. Thus, the answer is $\boxed{\mathrm{(D)}\ 4}$.
Let $S_n$ be the sum of the first $n$ terms of an arithmetic sequence $\{a_n\}$. If $S_7=3(a_1+a_9)$, then the value of $\frac{a_5}{a_4}$ is \_\_\_\_\_.	\frac{7}{6}
According to the definition of the Richter scale, the relationship between the relative energy $E$ released by an earthquake and the earthquake magnitude $n$ is: $E=10^n$. What is the multiple of the relative energy released by a magnitude 9 earthquake compared to a magnitude 7 earthquake?	100
Let \[f(x) = \frac{ax}{x + 1}.\]Find the constant $a$ so that $f(f(x)) = x$ for all $x \neq -1.$	-1
Two standard 6-sided dice are rolled. What is the probability that the sum rolled is a perfect square?	\dfrac{7}{36}
(1) Simplify: $\dfrac{\tan(3\pi-\alpha)\cos(2\pi-\alpha)\sin(-\alpha+\dfrac{3\pi}{2})}{\cos(-\alpha-\pi)\sin(-\pi+\alpha)\cos(\alpha+\dfrac{5\pi}{2})}$; (2) Given $\tan\alpha=\dfrac{1}{4}$, find the value of $\dfrac{1}{2\cos^{2}\alpha-3\sin\alpha\cos\alpha}$.	\dfrac{17}{20}
Simplify $90r - 44r$.	46r
Two cyclists started a trip at the same time from the same location. They traveled the same route and returned together. Both rested along the way. The first cyclist rode twice as long as the second cyclist rested. The second cyclist rode four times as long as the first cyclist rested. Who rides their bicycle faster and by how many times?	1.5
Find the coefficient of $x^2$ when $3(x^2 - x^3) +2(x - 2x^2 + 3x^5) -(4x^3 - x^2)$ is simplified.	0
A cubical cake with edge length $2$ inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where $M$ is the midpoint of a top edge. The piece whose top is triangle $B$ contains $c$ cubic inches of cake and $s$ square inches of icing. What is $c+s$?	\frac{32}{5}	1. Understanding the Problem: - We have a cube with edge length $2$ inches, which is iced on the sides and the top. - The cube is cut into three pieces, and we focus on the piece with a triangular top view labeled as triangle $B$. - We need to find the sum of the volume $c$ of this piece and the area $s$ of the icing on this piece. 2. Setting Up the Problem: - Let's denote the vertices of the cube's top face as $P$, $Q$, $R$, and $S$ such that $PQ = QR = RS = SP = 2$ inches. - The point $M$ is the midpoint of $PS$, so $PM = MS = 1$ inch. - The cake is cut along lines from $M$ to $Q$ and from $M$ to $R$. 3. Calculating the Volume of the Triangular Piece ($c$): - The triangular piece has a base area of triangle $B$ and a height equal to the edge length of the cube (2 inches). - The area of triangle $B$ can be calculated using the coordinates of points $Q$, $M$, and $R$. - Place $Q$ at $(0,0)$, $P$ at $(2,0)$, and $R$ at $(0,2)$. Then $M$ is at $(2,1)$. - The line $QM$ has the equation $2y - x = 0$. - The line $MR$ is perpendicular to $QM$ and passes through $R$. Its equation is $y + 2x - 2 = 0$. - Solving these equations simultaneously, we find the intersection point $N$ at $(\frac{4}{5}, \frac{2}{5})$. - The area of triangle $RNQ$ is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times \frac{4}{5} = \frac{4}{5}$ square inches. - Therefore, the volume $c$ of the triangular piece is $2 \times \frac{4}{5} = \frac{8}{5}$ cubic inches. 4. Calculating the Area of the Icing ($s$): - The icing covers the top triangle $RNQ$ and the vertical rectangle along $QR$. - The area of triangle $RNQ$ is $\frac{4}{5}$ square inches. - The area of the rectangle along $QR$ is $2 \times 2 = 4$ square inches (since $QR = 2$ inches and the height of the cake is 2 inches). - Thus, the total icing area $s$ is $\frac{4}{5} + 4 = \frac{24}{5}$ square inches. 5. Calculating the Sum $c + s$: - Summing the volume of the cake piece and the area of the icing, we get $c + s = \frac{8}{5} + \frac{24}{5} = \frac{32}{5}$. Therefore, the final answer is $\boxed{\textbf{(B) } \frac{32}{5}}$.
A particle moves through the first quadrant as follows. During the first minute it moves from the origin to $(1,0)$. Thereafter, it continues to follow the directions indicated in the figure, going back and forth between the positive x and y axes, moving one unit of distance parallel to an axis in each minute. At which point will the particle be after exactly 1989 minutes? [asy] import graph; Label f; f.p=fontsize(6); xaxis(0,3.5,Ticks(f, 1.0)); yaxis(0,4.5,Ticks(f, 1.0)); draw((0,0)--(1,0)--(1,1)--(0,1)--(0,2)--(2,2)--(2,0)--(3,0)--(3,3)--(0,3)--(0,4)--(1.5,4),blue+linewidth(2)); arrow((2,4),dir(180),blue); [/asy]	(44,35)	1. Understanding the Movement Pattern: The particle starts at the origin and moves in a pattern that encloses squares of increasing size. Each square is $n \times n$ where $n$ starts from 1 and increases by 1 for each new square. The movement pattern is such that the particle moves right and up to enclose odd-numbered squares, and moves down and left to enclose even-numbered squares. 2. Calculating Time for Each Square: - For the first square ($1 \times 1$), the particle moves right 1 unit and up 1 unit, taking $1 + 1 = 2$ minutes. - For the second square ($2 \times 2$), it moves right 2 units, down 2 units, and left 1 unit, taking $2 + 2 + 1 = 5$ minutes. - For the third square ($3 \times 3$), it moves right 3 units, up 3 units, and left 1 unit, taking $3 + 3 + 1 = 7$ minutes. - Continuing this pattern, the time to enclose the $n$-th square is $n + n + 1 = 2n + 1$ minutes. 3. Finding Total Time for Multiple Squares: - The total time to enclose the first $n$ squares is the sum of times for each square: $\sum_{k=1}^n (2k + 1)$. - This sum can be simplified using the formula for the sum of the first $n$ integers and the sum of the first $n$ odd numbers: \[ \sum_{k=1}^n (2k + 1) = 2\sum_{k=1}^n k + n = 2 \cdot \frac{n(n+1)}{2} + n = n^2 + 2n \] - This expression can be rewritten as $(n+1)^2 - 1$. 4. Finding the Largest $n$ for Given Time: - We need to find the largest $n$ such that $(n+1)^2 - 1 \leq 1989$. - Solving $(n+1)^2 \leq 1990$, we find $n+1 \leq \sqrt{1990} \approx 44.6$. - Thus, the largest integer $n$ is $n = 43$. 5. Position After 1989 Minutes: - After 1935 minutes (from the sum $(44)^2 - 1 = 1935$), the particle has enclosed the 43rd square and is at $(0,43)$. - From minute 1936 to 1980, the particle moves to enclose the 44th square, moving right to $(44,44)$. - From minute 1981 to 1989, the particle moves down, covering 9 minutes. Starting from $(44,44)$ and moving down 9 units, it reaches $(44, 35)$. Thus, the final position of the particle after 1989 minutes is $\boxed{\textbf{(D)}\ (44,35)}$.
Given any point $P$ on the ellipse $\frac{x^{2}}{a^{2}}+ \frac{y^{2}}{b^{2}}=1\; \; (a > b > 0)$ with foci $F\_{1}$ and $F\_{2}$, if $\angle PF\_1F\_2=\alpha$, $\angle PF\_2F\_1=\beta$, $\cos \alpha= \frac{ \sqrt{5}}{5}$, and $\sin (\alpha+\beta)= \frac{3}{5}$, find the eccentricity of this ellipse.	\frac{\sqrt{5}}{7}
Given an increasing geometric sequence $\{a_{n}\}$ with a common ratio greater than $1$ such that $a_{2}+a_{4}=20$, $a_{3}=8$.<br/>$(1)$ Find the general formula for $\{a_{n}\}$;<br/>$(2)$ Let $b_{m}$ be the number of terms of $\{a_{n}\}$ in the interval $\left(0,m\right]\left(m\in N*\right)$. Find the sum of the first $100$ terms of the sequence $\{b_{m}\}$, denoted as $S_{100}$.	480
A regular tetrahedron has an edge length of $2$. Calculate the surface area of the sphere circumscribed around this tetrahedron.	6\pi
In a kennel with 60 dogs, 9 dogs like watermelon, 48 dogs like salmon, and 5 like both salmon and watermelon. How many dogs in the kennel will not eat either?	8
Given that point $A(1,\sqrt{5})$ lies on the parabola $C:y^{2}=2px$, the distance from $A$ to the focus of $C$ is ______.	\frac{9}{4}
The logarithm of $27\sqrt[4]{9}\sqrt[3]{9}$ to the base $3$ is:	$4\frac{1}{6}$	1. Rewrite the expression using properties of exponents and logarithms: The given expression is $27\sqrt[4]{9}\sqrt[3]{9}$. We know that $27 = 3^3$, $\sqrt[4]{9} = 9^{1/4}$, and $\sqrt[3]{9} = 9^{1/3}$. Since $9 = 3^2$, we can rewrite these as: \[ 27 = 3^3, \quad \sqrt[4]{9} = (3^2)^{1/4} = 3^{2/4} = 3^{1/2}, \quad \text{and} \quad \sqrt[3]{9} = (3^2)^{1/3} = 3^{2/3}. \] 2. Combine the expressions: \[ 27\sqrt[4]{9}\sqrt[3]{9} = 3^3 \cdot 3^{1/2} \cdot 3^{2/3}. \] Using the property of exponents that $a^m \cdot a^n = a^{m+n}$, we combine the exponents: \[ 3^3 \cdot 3^{1/2} \cdot 3^{2/3} = 3^{3 + 1/2 + 2/3}. \] 3. Simplify the exponent: To add the exponents, find a common denominator, which is 6 in this case: \[ 3 + \frac{1}{2} + \frac{2}{3} = \frac{18}{6} + \frac{3}{6} + \frac{4}{6} = \frac{25}{6}. \] Thus, the expression simplifies to: \[ 3^{3 + 1/2 + 2/3} = 3^{\frac{25}{6}}. \] 4. Apply the logarithm: We need to find $\log_3(27\sqrt[4]{9}\sqrt[3]{9})$, which is now $\log_3(3^{\frac{25}{6}})$. Using the logarithmic identity $\log_b(a^c) = c \log_b(a)$, we have: \[ \log_3(3^{\frac{25}{6}}) = \frac{25}{6} \log_3(3). \] Since $\log_3(3) = 1$, this simplifies to: \[ \frac{25}{6} \cdot 1 = \frac{25}{6}. \] 5. Convert to a mixed number: \[ \frac{25}{6} = 4 \frac{1}{6}. \] 6. Conclude with the final answer: \[ \boxed{\text{(B)}\ 4\frac{1}{6}} \]
Compute the smallest base-10 positive integer greater than 5 that is a palindrome when written in both base 2 and 4.	15
In what ratio does the angle bisector of the acute angle divide the area of a right trapezoid inscribed in a circle?	1:1
A rectangle has dimensions $8 \times 12$, and a circle centered at one of its corners has a radius of 10. Calculate the area of the union of the regions enclosed by the rectangle and the circle.	96 + 75\pi
Player A and Player B play a number guessing game. First, Player A thinks of a number denoted as $a$, then Player B guesses the number that Player A is thinking of, and denotes this guessed number as $b$. Both $a$ and $b$ belong to the set $\{1,2,3,4,5,6\}$. If $\|a-b\| \leqslant 1$, it is said that "Player A and Player B are in sync". Now, find the probability that two randomly chosen players are "in sync" in this game.	\frac{4}{9}
Three cards are chosen at random from a standard 52-card deck. What is the probability that the first card is a spade, the second card is a 10, and the third card is a queen?	\frac{17}{11050}
If the area of $\triangle ABC$ is $64$ square units and the geometric mean (mean proportional) between sides $AB$ and $AC$ is $12$ inches, then $\sin A$ is equal to	\frac{8}{9}	1. Assign Variables to Sides: Let $AB = s$ and $AC = r$. 2. Use the Formula for the Area of a Triangle: The area of $\triangle ABC$ can be expressed as: \[ \text{Area} = \frac{1}{2} \times AB \times AC \times \sin A = \frac{1}{2} \times s \times r \times \sin A \] Given that the area is $64$ square units, we have: \[ \frac{1}{2} \times s \times r \times \sin A = 64 \] 3. Geometric Mean Relation: The geometric mean between sides $AB$ and $AC$ is given as $12$ inches, so: \[ \sqrt{rs} = 12 \] Squaring both sides, we get: \[ rs = 144 \] 4. Substitute $rs$ in the Area Formula: Replace $rs$ with $144$ in the area equation: \[ \frac{1}{2} \times 144 \times \sin A = 64 \] Simplifying, we find: \[ 72 \sin A = 64 \] \[ \sin A = \frac{64}{72} = \frac{8}{9} \] 5. Conclusion: The value of $\sin A$ is $\frac{8}{9}$. Therefore, the correct answer is: \[ \boxed{\textbf{(D) }\frac{8}{9}} \]
$ABCDE$ is a regular pentagon inscribed in a circle of radius 1. What is the area of the set of points inside the circle that are farther from $A$ than they are from any other vertex?	\frac{\pi}{5}
Given \(0 \leq x_0 < 1\), let \[ x_n = \left\{ \begin{array}{ll} 2x_{n-1} & \text{if } 2x_{n-1} < 1 \\ 2x_{n-1} - 1 & \text{if } 2x_{n-1} \geq 1 \end{array} \right. \] for all integers \(n > 0\). Determine the number of initial values of \(x_0\) that satisfy \(x_0 = x_6\).	64
Given a monotonically increasing sequence of positive integers $\left\{a_{n}\right\}$ that satisfies the recurrence relation $a_{n+2}=3 a_{n+1}-a_{n}$, with $a_{6}=280$, find the value of $a_{7}$.	733
On the side \(AB\) of an equilateral triangle \(\mathrm{ABC}\), a right triangle \(\mathrm{AHB}\) is constructed (\(\mathrm{H}\) is the vertex of the right angle) such that \(\angle \mathrm{HBA}=60^{\circ}\). Let the point \(K\) lie on the ray \(\mathrm{BC}\) beyond the point \(\mathrm{C}\), and \(\angle \mathrm{CAK}=15^{\circ}\). Find the angle between the line \(\mathrm{HK}\) and the median of the triangle \(\mathrm{AHB}\) drawn from the vertex \(\mathrm{H}\).	15
An infinite geometric series has a first term of $540$ and a sum of $4500$. What is its common ratio, and what is the second term of the series?	475.2
In $\triangle ABC$ shown in the figure, $AB=7$, $BC=8$, $CA=9$, and $\overline{AH}$ is an altitude. Points $D$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$, respectively, so that $\overline{BD}$ and $\overline{CE}$ are angle bisectors, intersecting $\overline{AH}$ at $Q$ and $P$, respectively. What is $PQ$?	\frac{8}{15}\sqrt{5}	1. Calculate the semi-perimeter and area of $\triangle ABC$ using Heron's Formula: - Semi-perimeter, $s = \frac{AB + BC + CA}{2} = \frac{7 + 8 + 9}{2} = 12$. - Area, $K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{12(12-7)(12-8)(12-9)} = \sqrt{12 \cdot 5 \cdot 4 \cdot 3} = \sqrt{720} = 12\sqrt{5}$. 2. Find the height $AH$ using the area formula for a triangle: - $K = \frac{1}{2} \times BC \times AH \Rightarrow 12\sqrt{5} = \frac{1}{2} \times 8 \times AH \Rightarrow AH = \frac{12\sqrt{5} \times 2}{8} = 3\sqrt{5}$. 3. Calculate $BH$ and $CH$ using the Pythagorean theorem in $\triangle ABH$ and $\triangle BCH$: - $BH = \sqrt{AB^2 - AH^2} = \sqrt{7^2 - (3\sqrt{5})^2} = \sqrt{49 - 45} = 2$. - $CH = BC - BH = 8 - 2 = 6$. 4. Apply the Angle Bisector Theorem in $\triangle ABH$ and $\triangle ACH$: - For $\triangle ABH$ with $BE$ as the bisector, $AE:EB = AB:BH = 7:2$. - For $\triangle ACH$ with $CD$ as the bisector, $AD:DC = AC:CH = 9:6 = 3:2$. 5. Set up ratios and solve for $AP$, $PH$, $AQ$, and $QH$: - Let $AP = 9x$ and $PH = 6x$ such that $AP + PH = AH = 3\sqrt{5} \Rightarrow 9x + 6x = 3\sqrt{5} \Rightarrow 15x = 3\sqrt{5} \Rightarrow x = \frac{\sqrt{5}}{5}$. - Thus, $AP = \frac{9\sqrt{5}}{5}$ and $PH = \frac{6\sqrt{5}}{5}$. - Let $AQ = 7y$ and $QH = 2y$ such that $AQ + QH = AH = 3\sqrt{5} \Rightarrow 7y + 2y = 3\sqrt{5} \Rightarrow 9y = 3\sqrt{5} \Rightarrow y = \frac{\sqrt{5}}{3}$. - Thus, $AQ = \frac{7\sqrt{5}}{3}$ and $QH = \frac{2\sqrt{5}}{3}$. 6. Calculate $PQ$ using the segment addition postulate: - $PQ = AQ - AP = \frac{7\sqrt{5}}{3} - \frac{9\sqrt{5}}{5} = \frac{35\sqrt{5} - 27\sqrt{5}}{15} = \frac{8\sqrt{5}}{15}$. Thus, the length of $PQ$ is $\boxed{\textbf{(D)}\ \frac{8}{15}\sqrt{5}}$.
In right triangle $XYZ$ with $\angle YXZ = 90^\circ$, we have $XY = 24$ and $YZ = 25$. Find $\tan Y$.	\frac{7}{24}
Leah and Jackson run for 45 minutes on a circular track. Leah runs clockwise at 200 m/min in a lane with a radius of 40 meters, while Jackson runs counterclockwise at 280 m/min in a lane with a radius of 55 meters, starting on the same radial line as Leah. Calculate how many times they pass each other after the start.	72
Completely factor the following expression: \[(6a^3+92a^2-7)-(-7a^3+a^2-7)\]	13a^2(a+7)
Given a segment \( AB \) of fixed length 3 with endpoints moving on the parabola \( y^2 = x \), find the shortest distance from the midpoint \( M \) of segment \( AB \) to the y-axis.	\frac{5}{4}
What is the smallest number that could be the date of the first Saturday after the second Monday following the second Thursday of a month?	17
4. Find the biggest positive integer $n$ , lesser thar $2012$ , that has the following property: If $p$ is a prime divisor of $n$ , then $p^2 - 1$ is a divisor of $n$ .	1944
How many numbers are in the list starting from $-48$, increasing by $7$ each time, up to and including $119$?	24
Find the area enclosed by the graph \( x^{2}+y^{2}=\|x\|+\|y\| \) on the \( xy \)-plane.	\pi + 2
Amelia has a coin that lands heads with a probability of $\frac{3}{7}$, and Blaine has a coin that lands on heads with a probability of $\frac{1}{4}$. Initially, they simultaneously toss their coins once. If both get heads, they stop; otherwise, if either gets a head and the other a tail, that person wins. If both get tails, they start tossing their coins alternately as in the original scenario until one gets a head and wins. Amelia goes first in the alternate rounds. What is the probability that Amelia wins the game? A) $\frac{5}{14}$ B) $\frac{9}{14}$ C) $\frac{11}{14}$ D) $\frac{13}{14}$ E) $\frac{1}{2}$	\frac{9}{14}
Initially, the fairy tale island was divided into three counties: in the first county lived only elves, in the second - only dwarves, and in the third - only centaurs. - During the first year, each county where there were no elves was divided into three counties. - During the second year, each county where there were no dwarves was divided into four counties. - During the third year, each county where there were no centaurs was divided into six counties. How many counties were there on the fairy tale island after all these events?	54
A traffic light runs repeatedly through the following cycle: green for 45 seconds, then yellow for 5 seconds, then blue for 10 seconds, and finally red for 40 seconds. Peter picks a random five-second time interval to observe the light. What is the probability that the color changes while he is watching?	0.15
Triangles $\triangle DEF$ and $\triangle D'E'F'$ lie in the coordinate plane with vertices $D(0,0)$, $E(0,10)$, $F(15,0)$, $D'(15,25)$, $E'(25,25)$, $F'(15,10)$. A rotation of $n$ degrees clockwise around the point $(u,v)$ where $0<n<180$, will transform $\triangle DEF$ to $\triangle D'E'F'$. Find $n+u+v$.	115
Tina is trying to solve the equation by completing the square: $$25x^2+30x-55 = 0.$$ She needs to rewrite the equation in the form \((ax + b)^2 = c\), where \(a\), \(b\), and \(c\) are integers and \(a > 0\). What is the value of \(a + b + c\)?	-38
In trapezoid \( ABCD \), point \( X \) is taken on the base \( BC \) such that segments \( XA \) and \( XD \) divide the trapezoid into three similar but pairwise unequal, non-isosceles triangles. The side \( AB \) has a length of 5. Find \( XC \cdot BX \).	25
For any number $y$, define the operations $\&y = 2(7-y)$ and $\&y = 2(y-7)$. What is the value of $\&(-13\&)$?	66
Given that $[x]$ represents the greatest integer less than or equal to $x$, if $$ [x+0.1]+[x+0.2]+[x+0.3]+[x+0.4]+[x+0.5]+[x+0.6]+[x+0.7]+[x+0.8]+[x+0.9]=104 $$ then the smallest value of $x$ is ( ).	11.5
Kelvin the Frog lives in the 2-D plane. Each day, he picks a uniformly random direction (i.e. a uniformly random bearing $\theta\in [0,2\pi]$ ) and jumps a mile in that direction. Let $D$ be the number of miles Kelvin is away from his starting point after ten days. Determine the expected value of $D^4$ .	200
Let $g$ be a function taking the positive integers to the positive integers, such that: (i) $g$ is increasing (i.e. $g(n + 1) > g(n)$ for all positive integers $n$), (ii) $g(mn) = g(m) g(n)$ for all positive integers $m$ and $n,$ and (iii) if $m \neq n$ and $m^n = n^m,$ then $g(m) = n^2$ or $g(n) = m^2.$ Find the sum of all possible values of $g(18).$	104976
Consider a regular hexagon where each of its $6$ sides and the $9$ diagonals are colored randomly and independently either red or blue, each color with the same probability. What is the probability that there exists at least one triangle, formed by three of the hexagon’s vertices, in which all sides are of the same color? A) $\frac{253}{256}$ B) $\frac{1001}{1024}$ C) $\frac{815}{819}$ D) $\frac{1048575}{1048576}$ E) 1	\frac{1048575}{1048576}
The integer 843301 is prime. The primorial of a prime number $p$, denoted $p \#$, is defined to be the product of all prime numbers less than or equal to $p$. Determine the number of digits in $843301 \#$. Your score will be $$\max \left\{\left\lfloor 60\left(\frac{1}{3}-\left\|\ln \left(\frac{A}{d}\right)\right\|\right)\right\rfloor, 0\right\}$$ where $A$ is your answer and $d$ is the actual answer.	365851	Remark: 843301\#-1 is the largest known prime number of the form $p \#-1$, where $p$ is prime.
Distribute 5 students into two dormitories, A and B, with each dormitory accommodating at least 2 students. Find the number of distinct arrangements.	20
In $\triangle ABC$, the ratio $AC:CB$ is $2:3$. The bisector of the exterior angle at $C$ intersects $BA$ extended at $P$ ($A$ is between $P$ and $B$). Determine the ratio $PA:AB$.	2:1
There are $n$ balls that look identical, among which one ball is lighter than the others (all other balls have equal weight). If using an unweighted balance scale as a tool, it takes at least 5 weighings to find the lighter ball, then the maximum value of $n$ is ___.	243
Define a modified Ackermann function \( A(m, n) \) with the same recursive relationships as the original problem: \[ A(m,n) = \left\{ \begin{aligned} &n+1& \text{ if } m = 0 \\ &A(m-1, 1) & \text{ if } m > 0 \text{ and } n = 0 \\ &A(m-1, A(m, n-1))&\text{ if } m > 0 \text{ and } n > 0. \end{aligned} \right.\] Compute \( A(3, 2) \).	29
If $AB$ and $CD$ are perpendicular diameters of circle $Q$, $P$ in $\overline{AQ}$, and $\angle QPC = 60^\circ$, then the length of $PQ$ divided by the length of $AQ$ is	\frac{\sqrt{3}}{3}	1. Identify the Geometry of the Circle: Since $AB$ and $CD$ are perpendicular diameters of circle $Q$, they intersect at the center of the circle, which we will denote as $O$. This makes $O$ the midpoint of both $AB$ and $CD$. 2. Position of Point $P$: Point $P$ lies on $\overline{AQ}$. Since $AB$ and $CD$ are diameters, $AQ$ is a radius of the circle. 3. Triangle Formation and Angle Identification: Since $\measuredangle QPC = 60^\circ$ and $CD$ is a diameter, $\triangle QPC$ is inscribed in the circle with $CQ$ as one side. The angle subtended by a diameter in a semicircle is a right angle, so $\measuredangle QCP = 90^\circ$. Therefore, $\triangle QPC$ is a right triangle. 4. Determine the Third Angle: In $\triangle QPC$, we know: - $\measuredangle QPC = 60^\circ$ - $\measuredangle QCP = 90^\circ$ Therefore, the third angle $\measuredangle PCQ = 180^\circ - 90^\circ - 60^\circ = 30^\circ$. 5. Using the Properties of a $30^\circ-60^\circ-90^\circ$ Triangle: In a $30^\circ-60^\circ-90^\circ$ triangle, the sides are in the ratio $1:\sqrt{3}:2$. Here, the side opposite the $30^\circ$ angle ($PC$) is half the hypotenuse ($CQ$), and the side opposite the $60^\circ$ angle ($PQ$) is $\sqrt{3}$ times the shorter leg ($PC$). 6. Assigning Lengths Based on the Triangle Ratios: Let the length of $PQ = x$. Then, since $PQ$ is opposite the $60^\circ$ angle, $PC = x/\sqrt{3}$ (opposite the $30^\circ$ angle), and $CQ = 2x/\sqrt{3}$ (the hypotenuse). 7. Relating $CQ$ to $AQ$: Since $CQ$ is a radius of the circle and equal to $AQ$, we have $CQ = AQ = 2x/\sqrt{3}$. 8. Calculating the Ratio $\frac{PQ}{AQ}$: \[ \frac{PQ}{AQ} = \frac{x}{2x/\sqrt{3}} = \frac{x}{2x} \cdot \sqrt{3} = \frac{\sqrt{3}}{3} \] 9. Conclusion: The length of $PQ$ divided by the length of $AQ$ is $\frac{\sqrt{3}}{3}$. Therefore, the answer is $\boxed{(B)}$.
Given quadrilateral $ABCD$ where $AC \perp BD$ and $AC=2$, $BD=3$, find the minimum value of $\overrightarrow{AB} \cdot \overrightarrow{CD}$.	- \dfrac{13}{4}
If \(A\ \clubsuit\ B\) is defined as \(A\ \clubsuit\ B = 3A^2 + 2B + 7\), what is the value of \(A\) for which \(A\ \clubsuit\ 7 = 61\)?	\frac{2\sqrt{30}}{3}
There are $10$ seats in each of $10$ rows of a theatre and all the seats are numbered. What is the probablity that two friends buying tickets independently will occupy adjacent seats?	\dfrac{1}{55}
In trapezoid EFGH, sides EF and GH are equal. It is known that EF = 12 units and GH = 10 units. Additionally, each of the non-parallel sides forms a right-angled triangle with half of the difference in lengths of EF and GH and a given leg of 6 units. Determine the perimeter of trapezoid EFGH.	22 + 2\sqrt{37}
Let \( g(n) = (n^2 - 2n + 1)^{1/3} + (n^2 - 1)^{1/3} + (n^2 + 2n + 1)^{1/3} \). Find \( \frac{1}{g(1)} + \frac{1}{g(3)} + \frac{1}{g(5)} + \ldots + \frac{1}{g(999999)} \).	50
Given that \(a\) and \(b\) are real numbers, and the equation \( x^{4} + a x^{3} + b x^{2} + a x + 1 = 0 \) has at least one real root, find the minimum value of \(a^{2} + b^{2}\).	4/5
Eight numbers \( a_{1}, a_{2}, a_{3}, a_{4} \) and \( b_{1}, b_{2}, b_{3}, b_{4} \) satisfy the following equations: $$ \left\{\begin{array}{c} a_{1} b_{1}+a_{2} b_{3}=1 \\ a_{1} b_{2}+a_{2} b_{4}=0 \\ a_{3} b_{1}+a_{4} b_{3}=0 \\ a_{3} b_{2}+a_{4} b_{4}=1 \end{array}\right. $$ It is known that \( a_{2} b_{3}=7 \). Find \( a_{4} b_{4} \).	-6
Given $( \sqrt {x}+ \dfrac {2}{x^{2}})^{n}$, the ratio of the coefficient of the fifth term to the coefficient of the third term in its expansion is $56:3$. (Ⅰ) Find the constant term in the expansion; (Ⅱ) When $x=4$, find the term with the maximum binomial coefficient in the expansion.	\dfrac {63}{256}
Given bed A has 600 plants, bed B has 500 plants, bed C has 400 plants, beds A and B share 60 plants, beds A and C share 80 plants, beds B and C share 40 plants, and beds A, B, and C share 20 plants collectively, calculate the total number of unique plants when considering just beds A, B, and C.	1340
When $1 + 7 + 7^2 + \cdots + 7^{2004}$ is divided by $1000$, a remainder of $N$ is obtained. Determine the value of $N$.	801
Given the function $f(x)=a\sin x - \sqrt{3}\cos x$, one of its graphs has an axis of symmetry at $x=-\frac{\pi}{6}$, and $f(x_1) - f(x_2) = -4$, calculate the minimum value of $\|x_1+x_2\|$.	\frac{2\pi}{3}
Consider a sequence $x_1,$ $x_2,$ $x_3,$ $\dots$ defined by \begin{align} x_1 &= \sqrt[3]{3}, \\ x_2 &= (\sqrt[3]{3})^{\sqrt[3]{3}}, \end{align}and in general, \[x_n = (x_{n - 1})^{\sqrt[3]{3}}\]for $n > 1.$ What is the smallest value of $n$ for which $x_n$ is an integer?	4
A "double-single" number is a three-digit number made up of two identical digits followed by a different digit. For example, 553 is a double-single number. How many double-single numbers are there between 100 and 1000?	81
Initially, there is a rook on each square of a chessboard. Each move, you can remove a rook from the board which attacks an odd number of rooks. What is the maximum number of rooks that can be removed? (Rooks attack each other if they are in the same row or column and there are no other rooks between them.)	59
A flat board has a circular hole with radius $1$ and a circular hole with radius $2$ such that the distance between the centers of the two holes is $7.$ Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$	173	[asy] size(10cm); pair A, B, C, D, O, P, H, L, X, Y; A = (-1, 0); B = (1, 0); H = (0, 0); C = (5, 0); D = (9, 0); L = (7, 0); O = (0, sqrt(160/13 - 1)); P = (7, sqrt(160/13 - 4)); X = (0, sqrt(160/13 - 4)); Y = (O + P) / 2; draw(A -- O -- B -- cycle); draw(C -- P -- D -- cycle); draw(B -- C); draw(O -- P); draw(P -- X, dashed); draw(O -- H, dashed); draw(P -- L, dashed); draw(circle(O, sqrt(160/13))); draw(circle(P, sqrt(160/13))); path b = brace(L-(0,1), H-(0,1),0.5); draw(b); label("$r$", O -- Y, N); label("$r$", Y -- P, N); label("$r$", O -- A, NW); label("$r$", P -- D, NE); label("$1$", A -- H, N); label("$2$", L -- D, N); label("$7$", b, S); dot(A^^B^^C^^D^^O^^P^^H^^L^^X^^Y,linewidth(4)); [/asy] Set the common radius to $r$. First, take the cross section of the sphere sitting in the hole of radius $1$. If we draw the perpendicular bisector of the chord (the hole) through the circle, this line goes through the center. Connect the center also to where the chord hits the circle, for a right triangle with hypotenuse $r$ and base $1$. Therefore, the height of this circle outside of the hole is $\sqrt{r^2-1}$. The other circle follows similarly for a height (outside the hole) of $\sqrt{r^2-4}$. Now, if we take the cross section of the entire board, essentially making it two-dimensional, we can connect the centers of the two spheres, then form another right triangle with base $7$, as given by the problem. The height of this triangle is the difference between the heights of the parts of the two spheres outside the holes, which is $\sqrt{r^2-1} - \sqrt{r^2-4}$. Now we can set up an equation in terms of $r$ with the Pythagorean theorem: \[\left(\sqrt{r^2-1} - \sqrt{r^2-4}\right)^2 + 7^2 = (2r)^2.\] Simplifying a few times, \begin{align} r^2 - 1 - 2\left(\sqrt{(r^2-1)(r^2-4)}\right) + r^2 - 4 + 49 &= 4r^2 \\ 2r^2-44 &= -2\left(\sqrt{(r^2-1)(r^2-4)}\right) \\ 22-r^2 &= \left(\sqrt{r^4 - 5r^2 + 4}\right) \\ r^4 -44r^2 + 484 &= r^4 - 5r^2 + 4 \\ 39r^2&=480 \\ r^2&=\frac{480}{39} = \frac{160}{13}. \end{align} Therefore, our answer is $\boxed{173}$. ~molocyxu
Given that $3^8 \cdot 5^2 = a^b,$ where both $a$ and $b$ are positive integers, find the smallest possible value for $a+b$.	407	1. Express the given equation in terms of $a$ and $b$: Given the equation $3^8 \cdot 5^2 = a^b$, we need to find $a$ and $b$ such that $a^b$ equals $3^8 \cdot 5^2$. 2. Simplify the expression: \[ 3^8 \cdot 5^2 = (3^4)^2 \cdot 5^2 = (81)^2 \cdot 25 = 81 \cdot 81 \cdot 25 \] We can further simplify this by combining terms: \[ 81 \cdot 81 \cdot 25 = (81 \cdot 25)^2 = 2025^2 \] Here, $81 = 3^4$ and $25 = 5^2$, so $81 \cdot 25 = 3^4 \cdot 5^2 = 405$. 3. Rewrite in terms of $a$ and $b$: \[ 3^8 \cdot 5^2 = 405^2 \] Thus, we can set $a = 405$ and $b = 2$. 4. Check if $405$ is a perfect power: The number $405$ can be factored as $405 = 3^4 \cdot 5$. Since the exponents of the prime factors are not equal, $405$ is not a perfect power. 5. Calculate $a + b$: \[ a + b = 405 + 2 = 407 \] 6. Conclusion: The smallest possible value of $a + b$ is $\boxed{407}$. This corresponds to choice $\textbf{(D)}$.
A pedestrian traffic light allows pedestrians to cross the street for one minute and prohibits crossing for two minutes. Find the average waiting time for a pedestrian who approaches the intersection.	40
If $\frac{1}{(2n-1)(2n+1)}=\frac{a}{2n-1}+\frac{b}{2n+1}$ holds for any natural number $n$, then $a=$______, $b=______.	-\frac{1}{2}
Suppose $a,$ $b,$ and $c$ are real numbers such that \[\frac{ac}{a + b} + \frac{ba}{b + c} + \frac{cb}{c + a} = -9\]and \[\frac{bc}{a + b} + \frac{ca}{b + c} + \frac{ab}{c + a} = 10.\]Compute the value of \[\frac{b}{a + b} + \frac{c}{b + c} + \frac{a}{c + a}.\]	11
Find $AB + AC$ in triangle $ABC$ given that $D$ is the midpoint of $BC$, $E$ is the midpoint of $DC$, and $BD = DE = EA = AD$.	1+\frac{\sqrt{3}}{3}	$DBC$ is a right triangle with hypotenuse $DC$. Since $DE=EC$, $E$ is the midpoint of this right triangle's hypotenuse, and it follows that $E$ is the circumcenter of the triangle. It follows that $BE=DE=CE$, as these are all radii of the same circle. A similar argument shows that $BD=DE=AE$. Thus, $BD=DE=DE$, and triangle $BDE$ is equilateral. So, $\angle DBE=\angle BED=\angle EDB=60^{\circ}$. We have $\angle BEC=180^{\circ}-\angle BED=120^{\circ}$. Because $BE=CE$, triangle $BEC$ is isosceles and $\angle ECB=30^{\circ}$. Therefore, $DBC$ is a right triangle with $\angle DBC=90^{\circ}, \angle BCD=30^{\circ}$, and $\angle CDB=60^{\circ}$. This means that $CD=\frac{2}{\sqrt{3}}BC$. Combined with $CD=\frac{2}{3}$, we have $BC=\frac{\sqrt{3}}{3}$. Similarly, $AB=\frac{\sqrt{3}}{3}$, so $AB+AC=1+\frac{\sqrt{3}}{3}$.
Let \( x \) and \( y \) be real numbers such that \[ 1 < \frac{x - y}{x + y} < 3. \] If \( \frac{x}{y} \) is an integer, what is its value?	-2
What is the value of $99^3 + 3(99^2) + 3(99) + 1$?	1,\!000,\!000
A graph $G(V,E)$ is triangle-free, but adding any edges to the graph will form a triangle. It's given that $\|V\|=2019$, $\|E\|>2018$, find the minimum of $\|E\|$ .	4033	Given a graph \( G(V, E) \) that is triangle-free, but adding any edges to the graph will form a triangle, and with \( \|V\| = 2019 \) and \( \|E\| > 2018 \), we need to find the minimum number of edges \( \|E\| \). We claim that the minimum number of edges is \( 2n - 5 \) where \( n = 2019 \). This bound is attained for a graph constructed as follows: take a 5-cycle \( C_5 \), and replace one of the vertices with an independent set of \( n - 4 \) vertices, each of which is adjacent to the two neighbors of the original vertex. To prove this, consider the following: 1. Diameter Condition: The graph \( G \) has diameter 2 because any two vertices with distance greater than 2 could have an edge added between them without forming a triangle. A diameter 1 graph is complete, which is not our case. 2. Minimum Degree Analysis: - If \( d \geq 4 \), then \( G \) has at least \( \frac{4n}{2} > 2n - 5 \) edges. - If \( d = 1 \), let \( v \) be a vertex connected only to \( w \). Then every other vertex must be connected to \( w \), making \( G \) a star graph, which contradicts \( \|E\| > n - 1 \). - If \( d = 2 \), let \( v \) be connected to \( w \) and \( x \). By the diameter 2 condition, every other vertex is connected to \( w \), \( x \), or both. Let \( A \) be the set of vertices adjacent to \( w \) but not \( x \), \( B \) be the set adjacent to both \( w \) and \( x \), and \( C \) be the set adjacent to \( x \) but not \( w \). Then \( \|A\| + \|B\| + \|C\| = n - 2 \). The only edges we can add are between \( A \) and \( C \), ensuring \( \|E\| \geq 2n - 5 \). - If \( d = 3 \), let \( v \) be adjacent to \( w \), \( x \), and \( y \). Each vertex in \( S = V \setminus \{v, w, x, y\} \) is adjacent to one of \( w \), \( x \), or \( y \). The degree sum gives \( \deg(w) + \deg(x) + \deg(y) \geq n - 1 \), leading to \( \|E\| \geq 2n - 5 \). Thus, the minimum number of edges \( \|E\| \) in such a graph is: \[ \|E\| = 2 \cdot 2019 - 5 = 4033. \] The answer is: \boxed{4033}.
In triangle $XYZ$, $XY=153$, $XZ=147$, and $YZ=140$. The angle bisector of angle $X$ intersects $\overline{YZ}$ at point $D$, and the angle bisector of angle $Y$ intersects $\overline{XZ}$ at point $E$. Let $P$ and $Q$ be the feet of the perpendiculars from $Z$ to $\overline{YE}$ and $\overline{XD}$, respectively. Find $PQ$.	67
Among the natural numbers from 1 to 100, find the total number of numbers that contain a digit 7 or are multiples of 7.	30
What is the sum of all four-digit numbers that are equal to the cube of the sum of their digits (leading zeros are not allowed)?	10745	We want to find all integers $x$ between 1000 and 9999 that are the cube of the sum of their digits. Of course, our search is only restricted to perfect cubes. The smallest such cube is $10^{3}=1000$ and the largest such cube is $21^{3}=9261$. This means we only have to check 12 different cubes, which is quite doable, but we can reduce the search even further with a little number theory. Suppose we write our number as $x=1000 a+100 b+10 c+d$, where $a, b, c$, and $d$ are the decimal digits of $x$. Then we have $$(a+b+c+d)^{3} \equiv 1000 a+100 b+10 c+d \equiv a+b+c+d(\bmod 9)$$ If we let $k=a+b+c+d$, then $k$ must be a solution to the modular equation $k^{3} \equiv k(\bmod 9)$. A quick check of the values 0 through 8 shows that the only solutions are 0,1 , and 8 . Now, in our search, we only have to check values that are the cube of a number which is either 0,1 , or $8 \bmod 9$. $$\begin{aligned} & 10^{3}=1000, \text { but } 1+0+0+0 \neq 10 \\ & 17^{3}=4913, \text { and } 4+9+1+3=17 \\ & 18^{3}=5832, \text { and } 5+8+3+2=18 \\ & 19^{3}=6859, \text { but } 6+8+5+9 \neq 19 \end{aligned}$$ So the only solutions are 4913 and 5832 , which sum to 10745 .
A teacher received letters on Monday through Friday with counts of $10$, $6$, $8$, $5$, $6$ respectively. Calculate the variance (${s^{2}} =$) of this data set.	3.2

Calculate the volume of a tetrahedron with vertices at points $A_{1}, A_{2}, A_{3}, A_{4}$, and its height dropped from vertex $A_{4}$ to the face $A_{1} A_{2} A_{3}$. $A_{1}(-2, 0, -4)$ $A_{2}(-1, 7, 1)$ $A_{3}(4, -8, -4)$ $A_{4}(1, -4, 6)$

5\sqrt{2}

How many positive integer multiples of $77$ can be expressed in the form $10^{j} - 10^{i}$, where $i$ and $j$ are integers and $0 \leq i < j \leq 49$?

182

Let $\alpha$ be a nonreal root of $x^4 = 1.$ Compute \[(1 - \alpha + \alpha^2 - \alpha^3)^4 + (1 + \alpha - \alpha^2 + \alpha^3)^4.\]

32

Find the largest prime divisor of $36^2 + 49^2$.

13

In trapezoid $ABCD$, $AB$ is parallel to $DC$ and $\angle DAF = 90^\circ$. Point $E$ is on $DC$ such that $EB = BC = CE$. Point $F$ is on $AB$ such that $DF$ is parallel to $EB$. In degrees, what is the measure of $\angle FDA$?

30

How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations? \[\left\{ \begin{aligned} x+3y&=3 \\ \left| |x| - |y| \right| &= 1 \end{aligned}\right.\]

3

In triangle $ABC$, the side $BC$ is 19 cm. The perpendicular $DF$, drawn to side $AB$ through its midpoint $D$, intersects side $BC$ at point $F$. Find the perimeter of triangle $AFC$ if side $AC$ is 10 cm.

29

A doctor told Mikael to take a pill every 75 minutes. He took his first pill at 11:05. At what time did he take his fourth pill?

14:50

Given that (1+ex)2019=a0+a1x+a2x2+……+a2019x2019, find the value of: - $$\frac {a_{1}}{e}$$+ $$\frac {a_{2}}{e^{2}}$$\- $$\frac {a_{3}}{e^{3}}$$+ $$\frac {a_{4}}{e^{4}}$$\-……- $$\frac {a_{2019}}{e^{2019}}$$

-1

The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.

294

Let there be $m$ members and $n$ members for the square and $c$ for the number of columns of the other formation. We have $n^2 +5 = c(c+7) \implies n^2+5 = \left(c+\frac{7}{2}\right)^2 -\frac{49}{4} \implies n^2 - \left(c+\frac{7}{2}\right)^2 = -\frac{69}{4} \implies \left(n-c-\frac{7}{2}\right)\left(n + c +\frac{7}{2}\right) \implies (2n-2c-7)(2n+2n+7) = -69.$ To maximize this we let $2n+2c+7 = 68$ and $2n-2c-7 = 1.$ Solving we find $n = 17$ so the desired number of members is $17^2 + 5 = \boxed{294}.$

Phil has 7 green marbles and 3 purple marbles in a bag. He removes a marble at random, records the color, puts it back, and then repeats this process until he has withdrawn 6 marbles. What is the probability that exactly three of the marbles that he removes are green? Express your answer as a decimal rounded to the nearest thousandth.

.185

Given that the equation of line $l_{1}$ is $y=x$, and the equation of line $l_{2}$ is $y=kx-k+1$, find the value of $k$ for which the area of triangle $OAB$ is $2$.

\frac{1}{5}

Given that the midpoint of side $BC$ of triangle $\triangle ABC$ is $D$, point $E$ lies in the plane of $\triangle ABC$, and $\overrightarrow{CD}=3\overrightarrow{CE}-2\overrightarrow{CA}$, if $\overrightarrow{AC}=x\overrightarrow{AB}+y\overrightarrow{BE}$, then determine the value of $x+y$.

11

In a finite sequence of real numbers, the sum of any 7 consecutive terms is negative while the sum of any 11 consecutive terms is positive. What is the maximum number of terms in such a sequence?

16

Quantities $a$ and $b$ vary inversely. When $a$ is $800$, $b$ is $0.5$. If the product of $a$ and $b$ increases by $200$ when $a$ is doubled, what is $b$ when $a$ is $1600$?

0.375

The value of $ 21!$ is $ 51{,}090{,}942{,}171{,}abc{,}440{,}000$ , where $ a$ , $ b$ , and $ c$ are digits. What is the value of $ 100a \plus{} 10b \plus{} c$ ?

709

Let $x$ be the least real number greater than $1$ such that $\sin(x) = \sin(x^2)$, where the arguments are in degrees. What is $x$ rounded up to the closest integer?

13

To solve the problem, we need to find the smallest real number $x > 1$ such that $\sin(x) = \sin(x^2)$. This equation holds true when $x$ and $x^2$ differ by a multiple of $360^\circ$ or when $x^2$ is equivalent to $180^\circ - x + 360^\circ k$ for some integer $k$. We will analyze each choice given in the problem. #### Step 1: Analyze each choice We need to check each choice to see if it satisfies $\sin(x) = \sin(x^2)$. **For choice $\textbf{(A)}$, $x = 10$:** \[ \sin(10^\circ) \neq \sin(100^\circ) \quad \text{since} \quad \sin(100^\circ) = \sin(80^\circ) \quad \text{and} \quad \sin(10^\circ) \neq \sin(80^\circ). \] **For choice $\textbf{(B)}$, $x = 13$:** \[ \sin(13^\circ) = \sin(169^\circ) \quad \text{since} \quad \sin(169^\circ) = \sin(11^\circ) \quad \text{and} \quad \sin(13^\circ) \approx \sin(11^\circ). \] This is a potential solution as $\sin(13^\circ) - \sin(11^\circ)$ is very small. **For choice $\textbf{(C)}$, $x = 14$:** \[ \sin(14^\circ) \neq \sin(196^\circ) \quad \text{since} \quad \sin(196^\circ) = -\sin(16^\circ) \quad \text{and} \quad \sin(14^\circ) \neq -\sin(16^\circ). \] **For choice $\textbf{(D)}$, $x = 19$:** \[ \sin(19^\circ) \neq \sin(361^\circ) \quad \text{since} \quad \sin(361^\circ) = \sin(1^\circ) \quad \text{and} \quad \sin(19^\circ) \neq \sin(1^\circ). \] **For choice $\textbf{(E)}$, $x = 20$:** \[ \sin(20^\circ) \neq \sin(400^\circ) \quad \text{since} \quad \sin(400^\circ) = \sin(40^\circ) \quad \text{and} \quad \sin(20^\circ) \neq \sin(40^\circ). \] #### Step 2: Conclusion From the analysis, the only choice where $\sin(x) \approx \sin(x^2)$ is $\textbf{(B)}$ where $x = 13$. The values $\sin(13^\circ)$ and $\sin(11^\circ)$ are very close, making the difference between them nearly zero. Therefore, the answer is $\boxed{\textbf{(B) } 13}$.

A bag contains nine blue marbles, ten ugly marbles, and one special marble. Ryan picks marbles randomly from this bag with replacement until he draws the special marble. He notices that none of the marbles he drew were ugly. Given this information, what is the expected value of the number of total marbles he drew?

\frac{20}{11}

The probability of drawing $k$ marbles is the probability of drawing $k-1$ blue marbles and then the special marble, which is $p_{k}=\left(\frac{9}{20}\right)^{k-1} \times \frac{1}{20}$. The probability of drawing no ugly marbles is therefore $\sum_{k=1}^{\infty} p_{k}=\frac{1}{11}$. Then given that no ugly marbles were drawn, the probability that $k$ marbles were drawn is $11 p_{k}$. The expected number of marbles Ryan drew is $$\sum_{k=1}^{\infty} k\left(11 p_{k}\right)=\frac{11}{20} \sum_{k=1}^{\infty} k\left(\frac{9}{20}\right)^{k-1}=\frac{11}{20} \times \frac{400}{121}=\frac{20}{11}$$ (To compute the sum in the last step, let $S=\sum_{k=1}^{\infty} k\left(\frac{9}{20}\right)^{k-1}$ and note that $\frac{9}{20} S=S-\sum_{k=1}^{\infty}\left(\frac{9}{20}\right)^{k-1}=$ $\left.S-\frac{20}{11}\right)$.

A fair dice is rolled twice, and the scores obtained are denoted as $m$ and $n$ respectively. Let the angle between vector $a=(m,n)$ and vector $b=(1,-1)$ be $\theta$. The probability that $\theta$ is an acute angle is $\_\_\_\_\_\_\_\_\_\_\_\_\_.$

\frac{5}{12}

For how many integers $n$ is $\frac n{20-n}$ the square of an integer?

4

To solve the problem, we need to find the integers $n$ for which $\frac{n}{20-n}$ is a perfect square. 1. **Examine the range of $n$:** - If $n < 0$ or $n > 20$, the fraction $\frac{n}{20-n}$ is negative, and thus cannot be a perfect square. - If $n = 20$, the fraction is undefined. - If $n \in \{1, 2, \dots, 9\}$, the fraction $\frac{n}{20-n}$ is positive but less than 1, hence cannot be a perfect square. Therefore, we only need to consider $n = 0$ and $n \in \{10, 11, \dots, 19\}$. 2. **Check specific values of $n$:** - For $n = 0$, $\frac{n}{20-n} = \frac{0}{20} = 0$, which is $0^2$ (a perfect square). - For $n = 10$, $\frac{n}{20-n} = \frac{10}{10} = 1$, which is $1^2$ (a perfect square). 3. **Exclude prime values of $n$:** - For prime $n$ within the range $\{11, 13, 17, 19\}$, the fraction $\frac{n}{20-n}$ will not be an integer, as the denominator will not contain the prime factor present in the numerator. Thus, these values cannot yield a perfect square. 4. **Check remaining composite values of $n$:** - For $n = 12$, $\frac{n}{20-n} = \frac{12}{8} = 1.5$, not a perfect square. - For $n = 14$, $\frac{n}{20-n} = \frac{14}{6} = \frac{7}{3}$, not a perfect square. - For $n = 15$, $\frac{n}{20-n} = \frac{15}{5} = 3$, not a perfect square. - For $n = 16$, $\frac{n}{20-n} = \frac{16}{4} = 4$, which is $2^2$ (a perfect square). - For $n = 18$, $\frac{n}{20-n} = \frac{18}{2} = 9$, which is $3^2$ (a perfect square). 5. **Conclusion:** The integers $n$ that make $\frac{n}{20-n}$ a perfect square are $n = 0, 10, 16, 18$. There are four such integers. Thus, the answer is $\boxed{\mathrm{(D)}\ 4}$.

Let $S_n$ be the sum of the first $n$ terms of an arithmetic sequence $\{a_n\}$. If $S_7=3(a_1+a_9)$, then the value of $\frac{a_5}{a_4}$ is \_\_\_\_\_.

\frac{7}{6}

According to the definition of the Richter scale, the relationship between the relative energy $E$ released by an earthquake and the earthquake magnitude $n$ is: $E=10^n$. What is the multiple of the relative energy released by a magnitude 9 earthquake compared to a magnitude 7 earthquake?

100

Let \[f(x) = \frac{ax}{x + 1}.\]Find the constant $a$ so that $f(f(x)) = x$ for all $x \neq -1.$

-1

Two standard 6-sided dice are rolled. What is the probability that the sum rolled is a perfect square?

\dfrac{7}{36}

(1) Simplify: $\dfrac{\tan(3\pi-\alpha)\cos(2\pi-\alpha)\sin(-\alpha+\dfrac{3\pi}{2})}{\cos(-\alpha-\pi)\sin(-\pi+\alpha)\cos(\alpha+\dfrac{5\pi}{2})}$; (2) Given $\tan\alpha=\dfrac{1}{4}$, find the value of $\dfrac{1}{2\cos^{2}\alpha-3\sin\alpha\cos\alpha}$.

\dfrac{17}{20}

Simplify $90r - 44r$.

46r

Two cyclists started a trip at the same time from the same location. They traveled the same route and returned together. Both rested along the way. The first cyclist rode twice as long as the second cyclist rested. The second cyclist rode four times as long as the first cyclist rested. Who rides their bicycle faster and by how many times?

1.5

Find the coefficient of $x^2$ when $3(x^2 - x^3) +2(x - 2x^2 + 3x^5) -(4x^3 - x^2)$ is simplified.

0

A cubical cake with edge length $2$ inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where $M$ is the midpoint of a top edge. The piece whose top is triangle $B$ contains $c$ cubic inches of cake and $s$ square inches of icing. What is $c+s$?

\frac{32}{5}

1. **Understanding the Problem:** - We have a cube with edge length $2$ inches, which is iced on the sides and the top. - The cube is cut into three pieces, and we focus on the piece with a triangular top view labeled as triangle $B$. - We need to find the sum of the volume $c$ of this piece and the area $s$ of the icing on this piece. 2. **Setting Up the Problem:** - Let's denote the vertices of the cube's top face as $P$, $Q$, $R$, and $S$ such that $PQ = QR = RS = SP = 2$ inches. - The point $M$ is the midpoint of $PS$, so $PM = MS = 1$ inch. - The cake is cut along lines from $M$ to $Q$ and from $M$ to $R$. 3. **Calculating the Volume of the Triangular Piece ($c$):** - The triangular piece has a base area of triangle $B$ and a height equal to the edge length of the cube (2 inches). - The area of triangle $B$ can be calculated using the coordinates of points $Q$, $M$, and $R$. - Place $Q$ at $(0,0)$, $P$ at $(2,0)$, and $R$ at $(0,2)$. Then $M$ is at $(2,1)$. - The line $QM$ has the equation $2y - x = 0$. - The line $MR$ is perpendicular to $QM$ and passes through $R$. Its equation is $y + 2x - 2 = 0$. - Solving these equations simultaneously, we find the intersection point $N$ at $(\frac{4}{5}, \frac{2}{5})$. - The area of triangle $RNQ$ is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times \frac{4}{5} = \frac{4}{5}$ square inches. - Therefore, the volume $c$ of the triangular piece is $2 \times \frac{4}{5} = \frac{8}{5}$ cubic inches. 4. **Calculating the Area of the Icing ($s$):** - The icing covers the top triangle $RNQ$ and the vertical rectangle along $QR$. - The area of triangle $RNQ$ is $\frac{4}{5}$ square inches. - The area of the rectangle along $QR$ is $2 \times 2 = 4$ square inches (since $QR = 2$ inches and the height of the cake is 2 inches). - Thus, the total icing area $s$ is $\frac{4}{5} + 4 = \frac{24}{5}$ square inches. 5. **Calculating the Sum $c + s$:** - Summing the volume of the cake piece and the area of the icing, we get $c + s = \frac{8}{5} + \frac{24}{5} = \frac{32}{5}$. Therefore, the final answer is $\boxed{\textbf{(B) } \frac{32}{5}}$.

A particle moves through the first quadrant as follows. During the first minute it moves from the origin to $(1,0)$. Thereafter, it continues to follow the directions indicated in the figure, going back and forth between the positive x and y axes, moving one unit of distance parallel to an axis in each minute. At which point will the particle be after exactly 1989 minutes? [asy] import graph; Label f; f.p=fontsize(6); xaxis(0,3.5,Ticks(f, 1.0)); yaxis(0,4.5,Ticks(f, 1.0)); draw((0,0)--(1,0)--(1,1)--(0,1)--(0,2)--(2,2)--(2,0)--(3,0)--(3,3)--(0,3)--(0,4)--(1.5,4),blue+linewidth(2)); arrow((2,4),dir(180),blue); [/asy]

(44,35)

1. **Understanding the Movement Pattern**: The particle starts at the origin and moves in a pattern that encloses squares of increasing size. Each square is $n \times n$ where $n$ starts from 1 and increases by 1 for each new square. The movement pattern is such that the particle moves right and up to enclose odd-numbered squares, and moves down and left to enclose even-numbered squares. 2. **Calculating Time for Each Square**: - For the first square ($1 \times 1$), the particle moves right 1 unit and up 1 unit, taking $1 + 1 = 2$ minutes. - For the second square ($2 \times 2$), it moves right 2 units, down 2 units, and left 1 unit, taking $2 + 2 + 1 = 5$ minutes. - For the third square ($3 \times 3$), it moves right 3 units, up 3 units, and left 1 unit, taking $3 + 3 + 1 = 7$ minutes. - Continuing this pattern, the time to enclose the $n$-th square is $n + n + 1 = 2n + 1$ minutes. 3. **Finding Total Time for Multiple Squares**: - The total time to enclose the first $n$ squares is the sum of times for each square: $\sum_{k=1}^n (2k + 1)$. - This sum can be simplified using the formula for the sum of the first $n$ integers and the sum of the first $n$ odd numbers: \[ \sum_{k=1}^n (2k + 1) = 2\sum_{k=1}^n k + n = 2 \cdot \frac{n(n+1)}{2} + n = n^2 + 2n \] - This expression can be rewritten as $(n+1)^2 - 1$. 4. **Finding the Largest $n$ for Given Time**: - We need to find the largest $n$ such that $(n+1)^2 - 1 \leq 1989$. - Solving $(n+1)^2 \leq 1990$, we find $n+1 \leq \sqrt{1990} \approx 44.6$. - Thus, the largest integer $n$ is $n = 43$. 5. **Position After 1989 Minutes**: - After 1935 minutes (from the sum $(44)^2 - 1 = 1935$), the particle has enclosed the 43rd square and is at $(0,43)$. - From minute 1936 to 1980, the particle moves to enclose the 44th square, moving right to $(44,44)$. - From minute 1981 to 1989, the particle moves down, covering 9 minutes. Starting from $(44,44)$ and moving down 9 units, it reaches $(44, 35)$. Thus, the final position of the particle after 1989 minutes is $\boxed{\textbf{(D)}\ (44,35)}$.

Given any point $P$ on the ellipse $\frac{x^{2}}{a^{2}}+ \frac{y^{2}}{b^{2}}=1\; \; (a > b > 0)$ with foci $F\_{1}$ and $F\_{2}$, if $\angle PF\_1F\_2=\alpha$, $\angle PF\_2F\_1=\beta$, $\cos \alpha= \frac{ \sqrt{5}}{5}$, and $\sin (\alpha+\beta)= \frac{3}{5}$, find the eccentricity of this ellipse.

\frac{\sqrt{5}}{7}

Given an increasing geometric sequence $\{a_{n}\}$ with a common ratio greater than $1$ such that $a_{2}+a_{4}=20$, $a_{3}=8$. $(1)$ Find the general formula for $\{a_{n}\}$; $(2)$ Let $b_{m}$ be the number of terms of $\{a_{n}\}$ in the interval $\left(0,m\right]\left(m\in N*\right)$. Find the sum of the first $100$ terms of the sequence $\{b_{m}\}$, denoted as $S_{100}$.

480

A regular tetrahedron has an edge length of $2$. Calculate the surface area of the sphere circumscribed around this tetrahedron.

6\pi

In a kennel with 60 dogs, 9 dogs like watermelon, 48 dogs like salmon, and 5 like both salmon and watermelon. How many dogs in the kennel will not eat either?

8

Given that point $A(1,\sqrt{5})$ lies on the parabola $C:y^{2}=2px$, the distance from $A$ to the focus of $C$ is ______.

\frac{9}{4}

The logarithm of $27\sqrt[4]{9}\sqrt[3]{9}$ to the base $3$ is:

$4\frac{1}{6}$

1. **Rewrite the expression using properties of exponents and logarithms:** The given expression is $27\sqrt[4]{9}\sqrt[3]{9}$. We know that $27 = 3^3$, $\sqrt[4]{9} = 9^{1/4}$, and $\sqrt[3]{9} = 9^{1/3}$. Since $9 = 3^2$, we can rewrite these as: \[ 27 = 3^3, \quad \sqrt[4]{9} = (3^2)^{1/4} = 3^{2/4} = 3^{1/2}, \quad \text{and} \quad \sqrt[3]{9} = (3^2)^{1/3} = 3^{2/3}. \] 2. **Combine the expressions:** \[ 27\sqrt[4]{9}\sqrt[3]{9} = 3^3 \cdot 3^{1/2} \cdot 3^{2/3}. \] Using the property of exponents that $a^m \cdot a^n = a^{m+n}$, we combine the exponents: \[ 3^3 \cdot 3^{1/2} \cdot 3^{2/3} = 3^{3 + 1/2 + 2/3}. \] 3. **Simplify the exponent:** To add the exponents, find a common denominator, which is 6 in this case: \[ 3 + \frac{1}{2} + \frac{2}{3} = \frac{18}{6} + \frac{3}{6} + \frac{4}{6} = \frac{25}{6}. \] Thus, the expression simplifies to: \[ 3^{3 + 1/2 + 2/3} = 3^{\frac{25}{6}}. \] 4. **Apply the logarithm:** We need to find $\log_3(27\sqrt[4]{9}\sqrt[3]{9})$, which is now $\log_3(3^{\frac{25}{6}})$. Using the logarithmic identity $\log_b(a^c) = c \log_b(a)$, we have: \[ \log_3(3^{\frac{25}{6}}) = \frac{25}{6} \log_3(3). \] Since $\log_3(3) = 1$, this simplifies to: \[ \frac{25}{6} \cdot 1 = \frac{25}{6}. \] 5. **Convert to a mixed number:** \[ \frac{25}{6} = 4 \frac{1}{6}. \] 6. **Conclude with the final answer:** \[ \boxed{\text{(B)}\ 4\frac{1}{6}} \]

Compute the smallest base-10 positive integer greater than 5 that is a palindrome when written in both base 2 and 4.

15

In what ratio does the angle bisector of the acute angle divide the area of a right trapezoid inscribed in a circle?

1:1

A rectangle has dimensions $8 \times 12$, and a circle centered at one of its corners has a radius of 10. Calculate the area of the union of the regions enclosed by the rectangle and the circle.

96 + 75\pi

Player A and Player B play a number guessing game. First, Player A thinks of a number denoted as $a$, then Player B guesses the number that Player A is thinking of, and denotes this guessed number as $b$. Both $a$ and $b$ belong to the set $\{1,2,3,4,5,6\}$. If $|a-b| \leqslant 1$, it is said that "Player A and Player B are in sync". Now, find the probability that two randomly chosen players are "in sync" in this game.

\frac{4}{9}

Three cards are chosen at random from a standard 52-card deck. What is the probability that the first card is a spade, the second card is a 10, and the third card is a queen?

\frac{17}{11050}

If the area of $\triangle ABC$ is $64$ square units and the geometric mean (mean proportional) between sides $AB$ and $AC$ is $12$ inches, then $\sin A$ is equal to

\frac{8}{9}

1. **Assign Variables to Sides**: Let $AB = s$ and $AC = r$. 2. **Use the Formula for the Area of a Triangle**: The area of $\triangle ABC$ can be expressed as: \[ \text{Area} = \frac{1}{2} \times AB \times AC \times \sin A = \frac{1}{2} \times s \times r \times \sin A \] Given that the area is $64$ square units, we have: \[ \frac{1}{2} \times s \times r \times \sin A = 64 \] 3. **Geometric Mean Relation**: The geometric mean between sides $AB$ and $AC$ is given as $12$ inches, so: \[ \sqrt{rs} = 12 \] Squaring both sides, we get: \[ rs = 144 \] 4. **Substitute $rs$ in the Area Formula**: Replace $rs$ with $144$ in the area equation: \[ \frac{1}{2} \times 144 \times \sin A = 64 \] Simplifying, we find: \[ 72 \sin A = 64 \] \[ \sin A = \frac{64}{72} = \frac{8}{9} \] 5. **Conclusion**: The value of $\sin A$ is $\frac{8}{9}$. Therefore, the correct answer is: \[ \boxed{\textbf{(D) }\frac{8}{9}} \]

$ABCDE$ is a regular pentagon inscribed in a circle of radius 1. What is the area of the set of points inside the circle that are farther from $A$ than they are from any other vertex?

\frac{\pi}{5}

Given $0 \leq x_0 < 1$, let \[ x_n = \left\{ \begin{array}{ll} 2x_{n-1} & \text{if } 2x_{n-1} < 1 \\ 2x_{n-1} - 1 & \text{if } 2x_{n-1} \geq 1 \end{array} \right. \] for all integers $n > 0$. Determine the number of initial values of $x_0$ that satisfy $x_0 = x_6$.

64

Given a monotonically increasing sequence of positive integers $\left\{a_{n}\right\}$ that satisfies the recurrence relation $a_{n+2}=3 a_{n+1}-a_{n}$, with $a_{6}=280$, find the value of $a_{7}$.

733

On the side $AB$ of an equilateral triangle $\mathrm{ABC}$, a right triangle $\mathrm{AHB}$ is constructed ($\mathrm{H}$ is the vertex of the right angle) such that $\angle \mathrm{HBA}=60^{\circ}$. Let the point $K$ lie on the ray $\mathrm{BC}$ beyond the point $\mathrm{C}$, and $\angle \mathrm{CAK}=15^{\circ}$. Find the angle between the line $\mathrm{HK}$ and the median of the triangle $\mathrm{AHB}$ drawn from the vertex $\mathrm{H}$.

15

An infinite geometric series has a first term of $540$ and a sum of $4500$. What is its common ratio, and what is the second term of the series?

475.2

In $\triangle ABC$ shown in the figure, $AB=7$, $BC=8$, $CA=9$, and $\overline{AH}$ is an altitude. Points $D$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$, respectively, so that $\overline{BD}$ and $\overline{CE}$ are angle bisectors, intersecting $\overline{AH}$ at $Q$ and $P$, respectively. What is $PQ$?

\frac{8}{15}\sqrt{5}

1. **Calculate the semi-perimeter and area of $\triangle ABC$ using Heron's Formula:** - Semi-perimeter, $s = \frac{AB + BC + CA}{2} = \frac{7 + 8 + 9}{2} = 12$. - Area, $K = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{12(12-7)(12-8)(12-9)} = \sqrt{12 \cdot 5 \cdot 4 \cdot 3} = \sqrt{720} = 12\sqrt{5}$. 2. **Find the height $AH$ using the area formula for a triangle:** - $K = \frac{1}{2} \times BC \times AH \Rightarrow 12\sqrt{5} = \frac{1}{2} \times 8 \times AH \Rightarrow AH = \frac{12\sqrt{5} \times 2}{8} = 3\sqrt{5}$. 3. **Calculate $BH$ and $CH$ using the Pythagorean theorem in $\triangle ABH$ and $\triangle BCH$:** - $BH = \sqrt{AB^2 - AH^2} = \sqrt{7^2 - (3\sqrt{5})^2} = \sqrt{49 - 45} = 2$. - $CH = BC - BH = 8 - 2 = 6$. 4. **Apply the Angle Bisector Theorem in $\triangle ABH$ and $\triangle ACH$:** - For $\triangle ABH$ with $BE$ as the bisector, $AE:EB = AB:BH = 7:2$. - For $\triangle ACH$ with $CD$ as the bisector, $AD:DC = AC:CH = 9:6 = 3:2$. 5. **Set up ratios and solve for $AP$, $PH$, $AQ$, and $QH$:** - Let $AP = 9x$ and $PH = 6x$ such that $AP + PH = AH = 3\sqrt{5} \Rightarrow 9x + 6x = 3\sqrt{5} \Rightarrow 15x = 3\sqrt{5} \Rightarrow x = \frac{\sqrt{5}}{5}$. - Thus, $AP = \frac{9\sqrt{5}}{5}$ and $PH = \frac{6\sqrt{5}}{5}$. - Let $AQ = 7y$ and $QH = 2y$ such that $AQ + QH = AH = 3\sqrt{5} \Rightarrow 7y + 2y = 3\sqrt{5} \Rightarrow 9y = 3\sqrt{5} \Rightarrow y = \frac{\sqrt{5}}{3}$. - Thus, $AQ = \frac{7\sqrt{5}}{3}$ and $QH = \frac{2\sqrt{5}}{3}$. 6. **Calculate $PQ$ using the segment addition postulate:** - $PQ = AQ - AP = \frac{7\sqrt{5}}{3} - \frac{9\sqrt{5}}{5} = \frac{35\sqrt{5} - 27\sqrt{5}}{15} = \frac{8\sqrt{5}}{15}$. Thus, the length of $PQ$ is $\boxed{\textbf{(D)}\ \frac{8}{15}\sqrt{5}}$.

In right triangle $XYZ$ with $\angle YXZ = 90^\circ$, we have $XY = 24$ and $YZ = 25$. Find $\tan Y$.

\frac{7}{24}

Leah and Jackson run for 45 minutes on a circular track. Leah runs clockwise at 200 m/min in a lane with a radius of 40 meters, while Jackson runs counterclockwise at 280 m/min in a lane with a radius of 55 meters, starting on the same radial line as Leah. Calculate how many times they pass each other after the start.

72

Completely factor the following expression: \[(6a^3+92a^2-7)-(-7a^3+a^2-7)\]

13a^2(a+7)

Given a segment $ AB $ of fixed length 3 with endpoints moving on the parabola $ y^2 = x $, find the shortest distance from the midpoint $ M $ of segment $ AB $ to the y-axis.

\frac{5}{4}

What is the smallest number that could be the date of the first Saturday after the second Monday following the second Thursday of a month?

17

4. Find the biggest positive integer $n$ , lesser thar $2012$ , that has the following property: If $p$ is a prime divisor of $n$ , then $p^2 - 1$ is a divisor of $n$ .

1944

How many numbers are in the list starting from $-48$, increasing by $7$ each time, up to and including $119$?

24

Find the area enclosed by the graph $ x^{2}+y^{2}=|x|+|y| $ on the $ xy $-plane.

\pi + 2

Amelia has a coin that lands heads with a probability of $\frac{3}{7}$, and Blaine has a coin that lands on heads with a probability of $\frac{1}{4}$. Initially, they simultaneously toss their coins once. If both get heads, they stop; otherwise, if either gets a head and the other a tail, that person wins. If both get tails, they start tossing their coins alternately as in the original scenario until one gets a head and wins. Amelia goes first in the alternate rounds. What is the probability that Amelia wins the game? A) $\frac{5}{14}$ B) $\frac{9}{14}$ C) $\frac{11}{14}$ D) $\frac{13}{14}$ E) $\frac{1}{2}$

\frac{9}{14}

Initially, the fairy tale island was divided into three counties: in the first county lived only elves, in the second - only dwarves, and in the third - only centaurs. - During the first year, each county where there were no elves was divided into three counties. - During the second year, each county where there were no dwarves was divided into four counties. - During the third year, each county where there were no centaurs was divided into six counties. How many counties were there on the fairy tale island after all these events?

54

A traffic light runs repeatedly through the following cycle: green for 45 seconds, then yellow for 5 seconds, then blue for 10 seconds, and finally red for 40 seconds. Peter picks a random five-second time interval to observe the light. What is the probability that the color changes while he is watching?

0.15

Triangles $\triangle DEF$ and $\triangle D'E'F'$ lie in the coordinate plane with vertices $D(0,0)$, $E(0,10)$, $F(15,0)$, $D'(15,25)$, $E'(25,25)$, $F'(15,10)$. A rotation of $n$ degrees clockwise around the point $(u,v)$ where $0<n<180$, will transform $\triangle DEF$ to $\triangle D'E'F'$. Find $n+u+v$.

115

Tina is trying to solve the equation by completing the square: $$25x^2+30x-55 = 0.$$ She needs to rewrite the equation in the form $(ax + b)^2 = c$, where $a$, $b$, and $c$ are integers and $a > 0$. What is the value of $a + b + c$?

-38

In trapezoid $ ABCD $, point $ X $ is taken on the base $ BC $ such that segments $ XA $ and $ XD $ divide the trapezoid into three similar but pairwise unequal, non-isosceles triangles. The side $ AB $ has a length of 5. Find $ XC \cdot BX $.

25

For any number $y$, define the operations $\&y = 2(7-y)$ and $\&y = 2(y-7)$. What is the value of $\&(-13\&)$?

66

Given that $[x]$ represents the greatest integer less than or equal to $x$, if $$ [x+0.1]+[x+0.2]+[x+0.3]+[x+0.4]+[x+0.5]+[x+0.6]+[x+0.7]+[x+0.8]+[x+0.9]=104 $$ then the smallest value of $x$ is ( ).

11.5

Kelvin the Frog lives in the 2-D plane. Each day, he picks a uniformly random direction (i.e. a uniformly random bearing $\theta\in [0,2\pi]$ ) and jumps a mile in that direction. Let $D$ be the number of miles Kelvin is away from his starting point after ten days. Determine the expected value of $D^4$ .

200

Let $g$ be a function taking the positive integers to the positive integers, such that: (i) $g$ is increasing (i.e. $g(n + 1) > g(n)$ for all positive integers $n$), (ii) $g(mn) = g(m) g(n)$ for all positive integers $m$ and $n,$ and (iii) if $m \neq n$ and $m^n = n^m,$ then $g(m) = n^2$ or $g(n) = m^2.$ Find the sum of all possible values of $g(18).$

104976

Consider a regular hexagon where each of its $6$ sides and the $9$ diagonals are colored randomly and independently either red or blue, each color with the same probability. What is the probability that there exists at least one triangle, formed by three of the hexagon’s vertices, in which all sides are of the same color? A) $\frac{253}{256}$ B) $\frac{1001}{1024}$ C) $\frac{815}{819}$ D) $\frac{1048575}{1048576}$ E) 1

\frac{1048575}{1048576}

The integer 843301 is prime. The primorial of a prime number $p$, denoted $p \#$, is defined to be the product of all prime numbers less than or equal to $p$. Determine the number of digits in $843301 \#$. Your score will be $$\max \left\{\left\lfloor 60\left(\frac{1}{3}-\left|\ln \left(\frac{A}{d}\right)\right|\right)\right\rfloor, 0\right\}$$ where $A$ is your answer and $d$ is the actual answer.

365851

Remark: 843301\#-1 is the largest known prime number of the form $p \#-1$, where $p$ is prime.

Distribute 5 students into two dormitories, A and B, with each dormitory accommodating at least 2 students. Find the number of distinct arrangements.

20

In $\triangle ABC$, the ratio $AC:CB$ is $2:3$. The bisector of the exterior angle at $C$ intersects $BA$ extended at $P$ ($A$ is between $P$ and $B$). Determine the ratio $PA:AB$.

2:1

There are $n$ balls that look identical, among which one ball is lighter than the others (all other balls have equal weight). If using an unweighted balance scale as a tool, it takes at least 5 weighings to find the lighter ball, then the maximum value of $n$ is ___.

243

Define a modified Ackermann function $ A(m, n) $ with the same recursive relationships as the original problem: \[ A(m,n) = \left\{ \begin{aligned} &n+1& \text{ if } m = 0 \\ &A(m-1, 1) & \text{ if } m > 0 \text{ and } n = 0 \\ &A(m-1, A(m, n-1))&\text{ if } m > 0 \text{ and } n > 0. \end{aligned} \right.\] Compute $ A(3, 2) $.

29

If $AB$ and $CD$ are perpendicular diameters of circle $Q$, $P$ in $\overline{AQ}$, and $\angle QPC = 60^\circ$, then the length of $PQ$ divided by the length of $AQ$ is

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

1. **Identify the Geometry of the Circle:** Since $AB$ and $CD$ are perpendicular diameters of circle $Q$, they intersect at the center of the circle, which we will denote as $O$. This makes $O$ the midpoint of both $AB$ and $CD$. 2. **Position of Point $P$:** Point $P$ lies on $\overline{AQ}$. Since $AB$ and $CD$ are diameters, $AQ$ is a radius of the circle. 3. **Triangle Formation and Angle Identification:** Since $\measuredangle QPC = 60^\circ$ and $CD$ is a diameter, $\triangle QPC$ is inscribed in the circle with $CQ$ as one side. The angle subtended by a diameter in a semicircle is a right angle, so $\measuredangle QCP = 90^\circ$. Therefore, $\triangle QPC$ is a right triangle. 4. **Determine the Third Angle:** In $\triangle QPC$, we know: - $\measuredangle QPC = 60^\circ$ - $\measuredangle QCP = 90^\circ$ Therefore, the third angle $\measuredangle PCQ = 180^\circ - 90^\circ - 60^\circ = 30^\circ$. 5. **Using the Properties of a $30^\circ-60^\circ-90^\circ$ Triangle:** In a $30^\circ-60^\circ-90^\circ$ triangle, the sides are in the ratio $1:\sqrt{3}:2$. Here, the side opposite the $30^\circ$ angle ($PC$) is half the hypotenuse ($CQ$), and the side opposite the $60^\circ$ angle ($PQ$) is $\sqrt{3}$ times the shorter leg ($PC$). 6. **Assigning Lengths Based on the Triangle Ratios:** Let the length of $PQ = x$. Then, since $PQ$ is opposite the $60^\circ$ angle, $PC = x/\sqrt{3}$ (opposite the $30^\circ$ angle), and $CQ = 2x/\sqrt{3}$ (the hypotenuse). 7. **Relating $CQ$ to $AQ$:** Since $CQ$ is a radius of the circle and equal to $AQ$, we have $CQ = AQ = 2x/\sqrt{3}$. 8. **Calculating the Ratio $\frac{PQ}{AQ}$:** \[ \frac{PQ}{AQ} = \frac{x}{2x/\sqrt{3}} = \frac{x}{2x} \cdot \sqrt{3} = \frac{\sqrt{3}}{3} \] 9. **Conclusion:** The length of $PQ$ divided by the length of $AQ$ is $\frac{\sqrt{3}}{3}$. Therefore, the answer is $\boxed{(B)}$.

Given quadrilateral $ABCD$ where $AC \perp BD$ and $AC=2$, $BD=3$, find the minimum value of $\overrightarrow{AB} \cdot \overrightarrow{CD}$.

- \dfrac{13}{4}

If $A\ \clubsuit\ B$ is defined as $A\ \clubsuit\ B = 3A^2 + 2B + 7$, what is the value of $A$ for which $A\ \clubsuit\ 7 = 61$?

\frac{2\sqrt{30}}{3}

There are $10$ seats in each of $10$ rows of a theatre and all the seats are numbered. What is the probablity that two friends buying tickets independently will occupy adjacent seats?

\dfrac{1}{55}

In trapezoid EFGH, sides EF and GH are equal. It is known that EF = 12 units and GH = 10 units. Additionally, each of the non-parallel sides forms a right-angled triangle with half of the difference in lengths of EF and GH and a given leg of 6 units. Determine the perimeter of trapezoid EFGH.

22 + 2\sqrt{37}

Let $ g(n) = (n^2 - 2n + 1)^{1/3} + (n^2 - 1)^{1/3} + (n^2 + 2n + 1)^{1/3} $. Find $ \frac{1}{g(1)} + \frac{1}{g(3)} + \frac{1}{g(5)} + \ldots + \frac{1}{g(999999)} $.

50

Given that $a$ and $b$ are real numbers, and the equation $ x^{4} + a x^{3} + b x^{2} + a x + 1 = 0 $ has at least one real root, find the minimum value of $a^{2} + b^{2}$.

4/5

Eight numbers $ a_{1}, a_{2}, a_{3}, a_{4} $ and $ b_{1}, b_{2}, b_{3}, b_{4} $ satisfy the following equations: $$ \left\{\begin{array}{c} a_{1} b_{1}+a_{2} b_{3}=1 \\ a_{1} b_{2}+a_{2} b_{4}=0 \\ a_{3} b_{1}+a_{4} b_{3}=0 \\ a_{3} b_{2}+a_{4} b_{4}=1 \end{array}\right. $$ It is known that $ a_{2} b_{3}=7 $. Find $ a_{4} b_{4} $.

-6

Given $( \sqrt {x}+ \dfrac {2}{x^{2}})^{n}$, the ratio of the coefficient of the fifth term to the coefficient of the third term in its expansion is $56:3$. (Ⅰ) Find the constant term in the expansion; (Ⅱ) When $x=4$, find the term with the maximum binomial coefficient in the expansion.

\dfrac {63}{256}

Given bed A has 600 plants, bed B has 500 plants, bed C has 400 plants, beds A and B share 60 plants, beds A and C share 80 plants, beds B and C share 40 plants, and beds A, B, and C share 20 plants collectively, calculate the total number of unique plants when considering just beds A, B, and C.

1340

When $1 + 7 + 7^2 + \cdots + 7^{2004}$ is divided by $1000$, a remainder of $N$ is obtained. Determine the value of $N$.

801

Given the function $f(x)=a\sin x - \sqrt{3}\cos x$, one of its graphs has an axis of symmetry at $x=-\frac{\pi}{6}$, and $f(x_1) - f(x_2) = -4$, calculate the minimum value of $|x_1+x_2|$.

\frac{2\pi}{3}

Consider a sequence $x_1,$ $x_2,$ $x_3,$ $\dots$ defined by \begin{align*} x_1 &= \sqrt[3]{3}, \\ x_2 &= (\sqrt[3]{3})^{\sqrt[3]{3}}, \end{align*}and in general, \[x_n = (x_{n - 1})^{\sqrt[3]{3}}\]for $n > 1.$ What is the smallest value of $n$ for which $x_n$ is an integer?

4

A "double-single" number is a three-digit number made up of two identical digits followed by a different digit. For example, 553 is a double-single number. How many double-single numbers are there between 100 and 1000?

81

Initially, there is a rook on each square of a chessboard. Each move, you can remove a rook from the board which attacks an odd number of rooks. What is the maximum number of rooks that can be removed? (Rooks attack each other if they are in the same row or column and there are no other rooks between them.)

59

A flat board has a circular hole with radius $1$ and a circular hole with radius $2$ such that the distance between the centers of the two holes is $7.$ Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

173

[asy] size(10cm); pair A, B, C, D, O, P, H, L, X, Y; A = (-1, 0); B = (1, 0); H = (0, 0); C = (5, 0); D = (9, 0); L = (7, 0); O = (0, sqrt(160/13 - 1)); P = (7, sqrt(160/13 - 4)); X = (0, sqrt(160/13 - 4)); Y = (O + P) / 2; draw(A -- O -- B -- cycle); draw(C -- P -- D -- cycle); draw(B -- C); draw(O -- P); draw(P -- X, dashed); draw(O -- H, dashed); draw(P -- L, dashed); draw(circle(O, sqrt(160/13))); draw(circle(P, sqrt(160/13))); path b = brace(L-(0,1), H-(0,1),0.5); draw(b); label("$r$", O -- Y, N); label("$r$", Y -- P, N); label("$r$", O -- A, NW); label("$r$", P -- D, NE); label("$1$", A -- H, N); label("$2$", L -- D, N); label("$7$", b, S); dot(A^^B^^C^^D^^O^^P^^H^^L^^X^^Y,linewidth(4)); [/asy] Set the common radius to $r$. First, take the cross section of the sphere sitting in the hole of radius $1$. If we draw the perpendicular bisector of the chord (the hole) through the circle, this line goes through the center. Connect the center also to where the chord hits the circle, for a right triangle with hypotenuse $r$ and base $1$. Therefore, the height of this circle outside of the hole is $\sqrt{r^2-1}$. The other circle follows similarly for a height (outside the hole) of $\sqrt{r^2-4}$. Now, if we take the cross section of the entire board, essentially making it two-dimensional, we can connect the centers of the two spheres, then form another right triangle with base $7$, as given by the problem. The height of this triangle is the difference between the heights of the parts of the two spheres outside the holes, which is $\sqrt{r^2-1} - \sqrt{r^2-4}$. Now we can set up an equation in terms of $r$ with the Pythagorean theorem: \[\left(\sqrt{r^2-1} - \sqrt{r^2-4}\right)^2 + 7^2 = (2r)^2.\] Simplifying a few times, \begin{align*} r^2 - 1 - 2\left(\sqrt{(r^2-1)(r^2-4)}\right) + r^2 - 4 + 49 &= 4r^2 \\ 2r^2-44 &= -2\left(\sqrt{(r^2-1)(r^2-4)}\right) \\ 22-r^2 &= \left(\sqrt{r^4 - 5r^2 + 4}\right) \\ r^4 -44r^2 + 484 &= r^4 - 5r^2 + 4 \\ 39r^2&=480 \\ r^2&=\frac{480}{39} = \frac{160}{13}. \end{align*} Therefore, our answer is $\boxed{173}$. ~molocyxu

Given that $3^8 \cdot 5^2 = a^b,$ where both $a$ and $b$ are positive integers, find the smallest possible value for $a+b$.

407

1. **Express the given equation in terms of $a$ and $b$:** Given the equation $3^8 \cdot 5^2 = a^b$, we need to find $a$ and $b$ such that $a^b$ equals $3^8 \cdot 5^2$. 2. **Simplify the expression:** \[ 3^8 \cdot 5^2 = (3^4)^2 \cdot 5^2 = (81)^2 \cdot 25 = 81 \cdot 81 \cdot 25 \] We can further simplify this by combining terms: \[ 81 \cdot 81 \cdot 25 = (81 \cdot 25)^2 = 2025^2 \] Here, $81 = 3^4$ and $25 = 5^2$, so $81 \cdot 25 = 3^4 \cdot 5^2 = 405$. 3. **Rewrite in terms of $a$ and $b$:** \[ 3^8 \cdot 5^2 = 405^2 \] Thus, we can set $a = 405$ and $b = 2$. 4. **Check if $405$ is a perfect power:** The number $405$ can be factored as $405 = 3^4 \cdot 5$. Since the exponents of the prime factors are not equal, $405$ is not a perfect power. 5. **Calculate $a + b$:** \[ a + b = 405 + 2 = 407 \] 6. **Conclusion:** The smallest possible value of $a + b$ is $\boxed{407}$. This corresponds to choice $\textbf{(D)}$.

A pedestrian traffic light allows pedestrians to cross the street for one minute and prohibits crossing for two minutes. Find the average waiting time for a pedestrian who approaches the intersection.

40

If $\frac{1}{(2n-1)(2n+1)}=\frac{a}{2n-1}+\frac{b}{2n+1}$ holds for any natural number $n$, then $a=$______, $b=______.

-\frac{1}{2}

Suppose $a,$ $b,$ and $c$ are real numbers such that \[\frac{ac}{a + b} + \frac{ba}{b + c} + \frac{cb}{c + a} = -9\]and \[\frac{bc}{a + b} + \frac{ca}{b + c} + \frac{ab}{c + a} = 10.\]Compute the value of \[\frac{b}{a + b} + \frac{c}{b + c} + \frac{a}{c + a}.\]

11

Find $AB + AC$ in triangle $ABC$ given that $D$ is the midpoint of $BC$, $E$ is the midpoint of $DC$, and $BD = DE = EA = AD$.

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

$DBC$ is a right triangle with hypotenuse $DC$. Since $DE=EC$, $E$ is the midpoint of this right triangle's hypotenuse, and it follows that $E$ is the circumcenter of the triangle. It follows that $BE=DE=CE$, as these are all radii of the same circle. A similar argument shows that $BD=DE=AE$. Thus, $BD=DE=DE$, and triangle $BDE$ is equilateral. So, $\angle DBE=\angle BED=\angle EDB=60^{\circ}$. We have $\angle BEC=180^{\circ}-\angle BED=120^{\circ}$. Because $BE=CE$, triangle $BEC$ is isosceles and $\angle ECB=30^{\circ}$. Therefore, $DBC$ is a right triangle with $\angle DBC=90^{\circ}, \angle BCD=30^{\circ}$, and $\angle CDB=60^{\circ}$. This means that $CD=\frac{2}{\sqrt{3}}BC$. Combined with $CD=\frac{2}{3}$, we have $BC=\frac{\sqrt{3}}{3}$. Similarly, $AB=\frac{\sqrt{3}}{3}$, so $AB+AC=1+\frac{\sqrt{3}}{3}$.

Let $ x $ and $ y $ be real numbers such that \[ 1 < \frac{x - y}{x + y} < 3. \] If $ \frac{x}{y} $ is an integer, what is its value?

-2

What is the value of $99^3 + 3(99^2) + 3(99) + 1$?

1,\!000,\!000

A graph $G(V,E)$ is triangle-free, but adding any edges to the graph will form a triangle. It's given that $|V|=2019$, $|E|>2018$, find the minimum of $|E|$ .

4033

Given a graph $ G(V, E) $ that is triangle-free, but adding any edges to the graph will form a triangle, and with $ |V| = 2019 $ and $ |E| > 2018 $, we need to find the minimum number of edges $ |E| $. We claim that the minimum number of edges is $ 2n - 5 $ where $ n = 2019 $. This bound is attained for a graph constructed as follows: take a 5-cycle $ C_5 $, and replace one of the vertices with an independent set of $ n - 4 $ vertices, each of which is adjacent to the two neighbors of the original vertex. To prove this, consider the following: 1. **Diameter Condition**: The graph $ G $ has diameter 2 because any two vertices with distance greater than 2 could have an edge added between them without forming a triangle. A diameter 1 graph is complete, which is not our case. 2. **Minimum Degree Analysis**: - If $ d \geq 4 $, then $ G $ has at least $ \frac{4n}{2} > 2n - 5 $ edges. - If $ d = 1 $, let $ v $ be a vertex connected only to $ w $. Then every other vertex must be connected to $ w $, making $ G $ a star graph, which contradicts $ |E| > n - 1 $. - If $ d = 2 $, let $ v $ be connected to $ w $ and $ x $. By the diameter 2 condition, every other vertex is connected to $ w $, $ x $, or both. Let $ A $ be the set of vertices adjacent to $ w $ but not $ x $, $ B $ be the set adjacent to both $ w $ and $ x $, and $ C $ be the set adjacent to $ x $ but not $ w $. Then $ |A| + |B| + |C| = n - 2 $. The only edges we can add are between $ A $ and $ C $, ensuring $ |E| \geq 2n - 5 $. - If $ d = 3 $, let $ v $ be adjacent to $ w $, $ x $, and $ y $. Each vertex in $ S = V \setminus \{v, w, x, y\} $ is adjacent to one of $ w $, $ x $, or $ y $. The degree sum gives $ \deg(w) + \deg(x) + \deg(y) \geq n - 1 $, leading to $ |E| \geq 2n - 5 $. Thus, the minimum number of edges $ |E| $ in such a graph is: \[ |E| = 2 \cdot 2019 - 5 = 4033. \] The answer is: \boxed{4033}.

In triangle $XYZ$, $XY=153$, $XZ=147$, and $YZ=140$. The angle bisector of angle $X$ intersects $\overline{YZ}$ at point $D$, and the angle bisector of angle $Y$ intersects $\overline{XZ}$ at point $E$. Let $P$ and $Q$ be the feet of the perpendiculars from $Z$ to $\overline{YE}$ and $\overline{XD}$, respectively. Find $PQ$.

67

Among the natural numbers from 1 to 100, find the total number of numbers that contain a digit 7 or are multiples of 7.

30

What is the sum of all four-digit numbers that are equal to the cube of the sum of their digits (leading zeros are not allowed)?

10745

We want to find all integers $x$ between 1000 and 9999 that are the cube of the sum of their digits. Of course, our search is only restricted to perfect cubes. The smallest such cube is $10^{3}=1000$ and the largest such cube is $21^{3}=9261$. This means we only have to check 12 different cubes, which is quite doable, but we can reduce the search even further with a little number theory. Suppose we write our number as $x=1000 a+100 b+10 c+d$, where $a, b, c$, and $d$ are the decimal digits of $x$. Then we have $$(a+b+c+d)^{3} \equiv 1000 a+100 b+10 c+d \equiv a+b+c+d(\bmod 9)$$ If we let $k=a+b+c+d$, then $k$ must be a solution to the modular equation $k^{3} \equiv k(\bmod 9)$. A quick check of the values 0 through 8 shows that the only solutions are 0,1 , and 8 . Now, in our search, we only have to check values that are the cube of a number which is either 0,1 , or $8 \bmod 9$. $$\begin{aligned} & 10^{3}=1000, \text { but } 1+0+0+0 \neq 10 \\ & 17^{3}=4913, \text { and } 4+9+1+3=17 \\ & 18^{3}=5832, \text { and } 5+8+3+2=18 \\ & 19^{3}=6859, \text { but } 6+8+5+9 \neq 19 \end{aligned}$$ So the only solutions are 4913 and 5832 , which sum to 10745 .

A teacher received letters on Monday through Friday with counts of $10$, $6$, $8$, $5$, $6$ respectively. Calculate the variance (${s^{2}} =$) of this data set.

3.2