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

problem stringlengths 10 1.99k	answer stringlengths 1 74	solution stringclasses 245 values
How many three-digit numbers are increased by 99 when their digits are reversed?	80
Find the last two digits (in order) of \( 7^{30105} \).	43
Anton makes custom watches for a jewelry store. Each watch consists of a bracelet, a gemstone, and a clasp. Bracelets are available in silver, gold, and steel. Anton has precious stones: cubic zirconia, emerald, quartz, diamond, and agate - and clasps: classic, butterfly, buckle. Anton is happy only when three watches are displayed in a row from left to right according to the following rules: - There must be a steel watch with a classic clasp and cubic zirconia stone. - Watches with a classic clasp must be flanked by gold and silver watches. - The three watches in a row must have different bracelets, gemstones, and clasps. How many ways are there to make Anton happy?	72
Given a sequence $\{a\_n\}$ that satisfies $a\_1=1$ and $a\_n= \frac{2S\_n^2}{2S\_n-1}$ for $n\geqslant 2$, where $S\_n$ is the sum of the first $n$ terms of the sequence, find the value of $S\_{2016}$.	\frac{1}{4031}
What is the smallest palindrome that is larger than 2015?	2112
The image of the point with coordinates $(-3,-1)$ under the reflection across the line $y=mx+b$ is the point with coordinates $(5,3)$. Find $m+b$.	1
Given that the sum of the first $n$ terms of the arithmetic sequence ${a_n}$ is $S_n$, if $a_2=0$, $S_3+S_4=6$, then the value of $a_5+a_6$ is $\_\_\_\_\_\_$.	21
Given that points $A$ and $B$ lie on the curves $C_{1}: x^{2}-y+1=0$ and $C_{2}: y^{2}-x+1=0$ respectively, what is the minimum value of the distance $\|AB\|$?	\frac{3\sqrt{2}}{4}
Acute-angled $\triangle ABC$ is inscribed in a circle with center at $O$. The measures of arcs are $\stackrel \frown {AB} = 80^\circ$ and $\stackrel \frown {BC} = 100^\circ$. A point $E$ is taken in minor arc $AC$ such that $OE$ is perpendicular to $AC$. Find the ratio of the magnitudes of $\angle OBE$ and $\angle BAC$.	10
Three friends have a total of 6 identical pencils, and each one has at least one pencil. In how many ways can this happen?	10
Given that positive real numbers $x$ and $y$ satisfy $e^{x}=y\ln x+y\ln y$, then the minimum value of $\frac{{e}^{x}}{x}-\ln y$ is ______.	e-1
Find all pairs $(m,n)$ of nonnegative integers for which \[m^2 + 2 \cdot 3^n = m\left(2^{n+1} - 1\right).\] [i]	(9, 3), (6, 3), (9, 5), (54, 5)	We are tasked with finding all pairs \((m, n)\) of nonnegative integers that satisfy the equation: \[ m^2 + 2 \cdot 3^n = m\left(2^{n+1} - 1\right). \] To solve this equation, we rearrange terms to express it in a form that can be factored: \[ m^2 - m(2^{n+1} - 1) + 2 \cdot 3^n = 0. \] This is a quadratic equation in \( m \). To solve for \( m \), we use the quadratic formula: \[ m = \frac{(2^{n+1} - 1) \pm \sqrt{(2^{n+1} - 1)^2 - 8 \cdot 3^n}}{2}. \] For \( m \) to be an integer, the discriminant must be a perfect square: \[ (2^{n+1} - 1)^2 - 8 \cdot 3^n = k^2 \] for some integer \(k\). Let's simplify and check cases for specific values of \( n \): ### Case 1: \( n = 3 \) - Substitute \( n = 3 \) into the equation: \[ (2^{4} - 1)^2 - 8 \cdot 3^3 = (15)^2 - 216 = 225 - 216 = 9 = 3^2. \] Here, the discriminant is a perfect square. Calculate \( m \): \[ m = \frac{15 \pm 3}{2}. \] This gives \[ m = 9 \quad \text{and} \quad m = 6. \] So, the pairs \((m, n)\) are \((9, 3)\) and \((6, 3)\). ### Case 2: \( n = 5 \) - Substitute \( n = 5 \) into the equation: \[ (2^{6} - 1)^2 - 8 \cdot 3^5 = (63)^2 - 1944 = 3969 - 1944 = 2025 = 45^2. \] Again, the discriminant is a perfect square. Calculate \( m \): \[ m = \frac{63 \pm 45}{2}. \] This gives \[ m = 54 \quad \text{and} \quad m = 9. \] So, the pairs \((m, n)\) are \((54, 5)\) and \((9, 5)\). Therefore, the possible pairs \((m, n)\) that satisfy the given equation are: \[ \boxed{(9, 3), (6, 3), (9, 5), (54, 5)}. \]
Let the function \( f(x) \) be defined on \( \mathbb{R} \), and for any \( x \), the condition \( f(x+2) + f(x) = x \) holds. It is also known that \( f(x) = x^3 \) on the interval \( (-2, 0] \). Find \( f(2012) \).	1006
Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?	\frac{145}{147}	1. Identify the vertices and setup the problem: Let $A$, $B$, and $C$ be the vertices of the right triangle field with $AB = 3$ units, $AC = 4$ units, and $\angle BAC = 90^\circ$. Let $S$ be the square in the corner at $A$, and let its vertices be $A$, $M$, $D$, and $N$ with $AM$ and $AN$ along $AB$ and $AC$ respectively. 2. Define additional points and use similarity: Let $X$ be on $BC$ such that $AC \parallel DX$ and $Y$ be on $BC$ such that $AB \parallel DY$. Define $P$ and $Q$ as the feet of the perpendiculars from $X$ to $AC$ and from $Y$ to $AB$ respectively. By similarity of triangles $\triangle ABC \sim \triangle DYX \sim \triangle PXC \sim \triangle QBY$, we have: \[ PC = x \cdot \frac{3}{4}, \quad QB = x \cdot \frac{4}{3} \] where $x$ is the side length of square $S$. 3. Calculate distances using similarity: From the similarity, we also have: \[ DX = 3 - x - \frac{3}{4}x = 3 - \frac{7}{4}x, \quad MQ = 4 - x - \frac{4}{3}x = 4 - \frac{7}{3}x \] and \[ CX = \frac{5}{4}x, \quad BY = \frac{5}{3}x \] leading to \[ XY = 5 - \frac{5}{4}x - \frac{5}{3}x = 5 - \frac{35}{12}x \] 4. Set up an equation using area calculations: The area of $\triangle XDY$ can be calculated in two ways: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (3 - \frac{7}{4}x) \times (4 - \frac{7}{3}x) \] and also as \[ \text{Area} = 2 \times (5 - \frac{35}{12}x) \] Equating these and solving for $x$: \[ 10 - \frac{35}{6}x = \frac{49}{12}x^2 - 14x + 12 \] \[ 49x^2 - 98x + 24 = 0 \] \[ x = \frac{2}{7} \text{ or } \frac{12}{7} \] The value $x = \frac{12}{7}$ is extraneous as it would exceed the dimensions of the triangle. 5. Calculate the area of square $S$ and the planted portion: The area of square $S$ is $x^2 = \left(\frac{2}{7}\right)^2 = \frac{4}{49}$. The total area of the triangle is $\frac{1}{2} \times 3 \times 4 = 6$. The planted area is: \[ 6 - \frac{4}{49} = \frac{294}{49} - \frac{4}{49} = \frac{290}{49} = \frac{145}{24.5} \] The fraction of the field that is planted is: \[ \frac{\frac{290}{49}}{6} = 1 - \frac{\frac{4}{49}}{6} = 1 - \frac{2}{147} = \boxed{\textbf{(D) } \frac{145}{147}} \]
Let \(A B C\) be an acute triangle with circumcenter \(O\) such that \(A B=4, A C=5\), and \(B C=6\). Let \(D\) be the foot of the altitude from \(A\) to \(B C\), and \(E\) be the intersection of \(A O\) with \(B C\). Suppose that \(X\) is on \(B C\) between \(D\) and \(E\) such that there is a point \(Y\) on \(A D\) satisfying \(X Y \parallel A O\) and \(Y O \perp A X\). Determine the length of \(B X\).	\frac{96}{41}	Let \(A X\) intersect the circumcircle of \(\triangle A B C\) again at \(K\). Let \(O Y\) intersect \(A K\) and \(B C\) at \(T\) and \(L\), respectively. We have \(\angle L O A=\angle O Y X=\angle T D X=\angle L A K\), so \(A L\) is tangent to the circumcircle. Furthermore, \(O L \perp A K\), so \(\triangle A L K\) is isosceles with \(A L=A K\), so \(A K\) is also tangent to the circumcircle. Since \(B C\) and the tangents to the circumcircle at \(A\) and \(K\) all intersect at the same point \(L, C L\) is a symmedian of \(\triangle A C K\). Then \(A K\) is a symmedian of \(\triangle A B C\). Then we can use \(\frac{B X}{X C}=\frac{(A B)^{2}}{(A C)^{2}}\) to compute \(B X=\frac{96}{41}\).
Given the sequence $\{a_n\}$ with the sum of its first $n$ terms $S_n = 6n - n^2$, find the sum of the first $20$ terms of the sequence $\left\{ \frac{1}{a_na_{n+1}}\right\}$.	-\frac{4}{35}
In the Cartesian coordinate system, point O is the origin, and the coordinates of three vertices of the parallelogram ABCD are A(2,3), B(-1,-2), and C(-2,-1). (1) Find the lengths of the diagonals AC and BD; (2) If the real number t satisfies $ (\vec{AB}+t\vec{OC})\cdot\vec{OC}=0 $, find the value of t.	-\frac{11}{5}
Solve for $n$, if $9^n\cdot9^n\cdot9^n\cdot9^n=81^4$.	2
$x$, $y$ and $z$ are positive reals such that $x+y+z=xyz$. Find the minimum value of: \[ x^7(yz-1)+y^7(zx-1)+z^7(xy-1) \]	162\sqrt{3}	Given that \( x \), \( y \), and \( z \) are positive reals such that \( x + y + z = xyz \), we aim to find the minimum value of: \[ x^7(yz-1) + y^7(zx-1) + z^7(xy-1). \] First, we use the given condition \( x + y + z = xyz \). By the AM-GM inequality, we have: \[ xyz = x + y + z \geq 3\sqrt[3]{xyz}, \] which implies: \[ xyz \geq 3\sqrt{3}. \] Now, consider the given expression: \[ x^7(yz-1) + y^7(zx-1) + z^7(xy-1). \] Rewriting it, we get: \[ x^7(yz-1) + y^7(zx-1) + z^7(xy-1) = x^6(xyz - x) + y^6(xyz - y) + z^6(xyz - z). \] This simplifies to: \[ x^6(xyz) + y^6(xyz) + z^6(xyz) - (x^7 + y^7 + z^7). \] Factoring out \( xyz \) from the first part, we get: \[ (x^6 + y^6 + z^6)xyz - (x^7 + y^7 + z^7). \] Using the given \( xyz = x + y + z \), we can rewrite it as: \[ (x^6 + y^6 + z^6)(x + y + z) - (x^7 + y^7 + z^7). \] Expanding this, we have: \[ (x^7 + y^7 + z^7) + (x^6y + xy^6 + x^6z + xz^6 + y^6z + yz^6) - (x^7 + y^7 + z^7). \] This simplifies to: \[ x^6y + xy^6 + x^6z + xz^6 + y^6z + yz^6. \] By the AM-GM inequality, we know: \[ x^6y + xy^6 + x^6z + xz^6 + y^6z + yz^6 \geq 6\sqrt[6]{(xyz)^{14}}. \] Given \( xyz \geq 3\sqrt{3} \), we have: \[ 6\sqrt[6]{(3\sqrt{3})^{14}} = 6\sqrt[6]{3^{21}} = 6 \cdot 3^{7/2} = 6 \cdot 27\sqrt{3} = 162\sqrt{3}. \] Therefore, the minimum value is: \[ \boxed{162\sqrt{3}}. \] This minimum is achieved when \( x = y = z = \sqrt{3} \).
A box contains 4 cards, each with one of the following functions defined on \\(R\\): \\(f_{1}(x)={x}^{3}\\), \\(f_{2}(x)=\|x\|\\), \\(f_{3}(x)=\sin x\\), \\(f_{4}(x)=\cos x\\). Now, if we randomly pick 2 cards from the box and multiply the functions on the cards to get a new function, the probability that the resulting function is an odd function is \_\_\_\_\_.	\dfrac{2}{3}
In parallelogram $EFGH$, $EF = 5z + 5$, $FG = 4k^2$, $GH = 40$, and $HE = k + 20$. Determine the values of $z$ and $k$ and find $z \times k$.	\frac{7 + 7\sqrt{321}}{8}
Two people, Person A and Person B, start at the same time from point $A$ to point $B$: Person A is faster than Person B. After reaching point $B$, Person A doubles their speed and immediately returns to point $A$. They meet Person B at a point 240 meters from point $B$. After meeting, Person B also doubles their speed and turns back. When Person A returns to point $A$, Person B is still 120 meters away from point $A$. What is the distance between points $A$ and $B$ in meters?	420
Given that Ben constructs a $4$-step staircase using $26$ toothpicks, determine the number of additional toothpicks needed to extend the staircase to a $6$-step staircase,	22
Let \( f(x) = x^2 + px + q \). It is known that the inequality \( \|f(x)\| > \frac{1}{2} \) has no solutions on the interval \([1, 3]\). Find \( \underbrace{f(f(\ldots f}_{2017}\left(\frac{3+\sqrt{7}}{2}\right)) \ldots) \). If necessary, round your answer to two decimal places.	0.18
Given a geometric progression $\{a_n\}$ with the first term $a_1=2$ and the sum of the first $n$ terms as $S_n$, and the equation $S_5 + 4S_3 = 5S_4$ holds, find the maximum term of the sequence $\left\{ \frac{2\log_{2}a_n + 1}{\log_{2}a_n - 6} \right\}$.	15
In how many ways can I choose 3 captains from a team of 11 people?	165
How many positive integers less that $200$ are relatively prime to either $15$ or $24$ ?	120
The ancient Chinese mathematical classic "The Nine Chapters on the Mathematical Art" contains a problem called "Rice and Grain Separation". During the collection of grain in a granary, 1524 "shi" (a unit of weight) of rice was received, but it was found to contain grains of another type mixed in. A sample of rice was taken and it was found that out of 254 grains, 28 were not rice. Approximately how much of this batch of rice is not rice?	168
Given a triangle $ABC$ with internal angles $A$, $B$, and $C$, and it is known that $$2\sin^{2}(B+C)= \sqrt {3}\sin2A.$$ (Ⅰ) Find the degree measure of $A$; (Ⅱ) If $BC=7$ and $AC=5$, find the area $S$ of $\triangle ABC$.	10 \sqrt {3}
Cindy leaves school at the same time every day. If she cycles at \(20 \ \text{km/h}\), she arrives home at 4:30 in the afternoon. If she cycles at \(10 \ \text{km/h}\), she arrives home at 5:15 in the afternoon. Determine the speed, in \(\text{km/h}\), at which she must cycle to arrive home at 5:00 in the afternoon.	12
Quantities \(r\) and \( s \) vary inversely. When \( r \) is \( 1500 \), \( s \) is \( 0.4 \). Alongside, quantity \( t \) also varies inversely with \( r \) and when \( r \) is \( 1500 \), \( t \) is \( 2.5 \). What is the value of \( s \) and \( t \) when \( r \) is \( 3000 \)? Express your answer as a decimal to the nearest thousandths.	1.25
The volume of the parallelepiped determined by the three-dimensional vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ is 4. Find the volume of the parallelepiped determined by the vectors $\mathbf{a} + \mathbf{b},$ $\mathbf{b} + 3 \mathbf{c},$ and $\mathbf{c} - 7 \mathbf{a}.$	80
Divide 6 volunteers into 4 groups, with two groups having 2 people each and the other two groups having 1 person each, to serve at four different pavilions of the World Expo. How many different allocation schemes are there? (Answer with a number).	1080
Given that $\sin A+\sin B=1$ and $\cos A+\cos B= \frac{3}{2}$, what is the value of $\cos(A-B)$?	\frac{5}{8}
Screws are sold in packs of $10$ and $12$ . Harry and Sam independently go to the hardware store, and by coincidence each of them buys exactly $k$ screws. However, the number of packs of screws Harry buys is different than the number of packs Sam buys. What is the smallest possible value of $k$ ?	60
A function $f$ has the property that $f(3x-1)=x^2+x+1$ for all real numbers $x$. What is $f(5)$?	7
Given a geometric sequence $\{a_n\}$ with the first term $\frac{3}{2}$ and common ratio $-\frac{1}{2}$, and the sum of the first $n$ terms is $S_n$, then when $n\in N^*$, the sum of the maximum and minimum values of $S_n - \frac{1}{S_n}$ is ______.	\frac{1}{4}
Evaluate $\log_\frac{1}{3}9$.	-2
An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Each of the $b$ layers of size $a \times 1 \times c$ parallel to the $(a \times c)$ faces of the box contains exactly $20$ green cubes, exactly $25$ yellow cubes, and some red cubes. Find the smallest possible volume of the box.	180	The total number of green cubes is given by $12a=20b\Longrightarrow a=\frac{5}{3}b$. Let $r$ be the number of red cubes on each one of the $b$ layers then the total number of red cubes is $9a=br$. Substitute $a=\frac{5}{3}b$ gives $r=15$. Repeating the procedure on the number of yellow cubes $y$ on each of the $a$ layers gives $y=15$. Therefore $bc=9+12+15=36$ and $ac=15+20+25=60$. Multiplying yields $abc^2=2160$. Since $abc^2$ is fixed, $abc$ is minimized when $c$ is maximized, which occurs when $a$, $b$ are minimized (since each of $ac$, $bc$ is fixed). Thus $(a,b,c)=(3,5,12)\Longrightarrow abc=\boxed{180}$ ~ Nafer
If an integer of two digits is $k$ times the sum of its digits, the number formed by interchanging the digits is the sum of the digits multiplied by	11-k	1. Define the number and its properties: Let the two-digit integer be represented as $n = 10a + b$, where $a$ and $b$ are the tens and units digits respectively. According to the problem, this number $n$ is $k$ times the sum of its digits. Therefore, we have the equation: \[ 10a + b = k(a + b) \] 2. Expression for the interchanged digits: The number formed by interchanging the digits of $n$ is $10b + a$. We need to find a constant $x$ such that: \[ 10b + a = x(a + b) \] 3. Combine the equations: Adding the equations from step 1 and step 2, we get: \[ (10a + b) + (10b + a) = k(a + b) + x(a + b) \] Simplifying both sides, we have: \[ 11a + 11b = (k + x)(a + b) \] This simplifies further to: \[ 11(a + b) = (k + x)(a + b) \] 4. Solve for $x$: Since $a + b \neq 0$ (as $a$ and $b$ are digits of a two-digit number), we can divide both sides of the equation by $(a + b)$: \[ 11 = k + x \] Solving for $x$, we find: \[ x = 11 - k \] 5. Conclusion: The number formed by interchanging the digits is $11 - k$ times the sum of the digits. Therefore, the correct answer is: \[ \boxed{\textbf{(C) \ } 11-k} \]
The remainder when 111 is divided by 10 is 1. The remainder when 111 is divided by the positive integer $n$ is 6. How many possible values of $n$ are there?	5	Since the remainder when 111 is divided by $n$ is 6, then $111-6=105$ is a multiple of $n$ and $n>6$ (since, by definition, the remainder must be less than the divisor). Since $105=3 \cdot 5 \cdot 7$, the positive divisors of 105 are $1,3,5,7,15,21,35,105$. Therefore, the possible values of $n$ are $7,15,21,35,105$, of which there are 5.
When $x^9-x$ is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is:	5	1. Factor out the greatest common factor: The polynomial $x^9 - x$ can be rewritten by factoring out the common factor of $x$, giving: \[ x^9 - x = x(x^8 - 1) \] 2. Factorize $x^8 - 1$ using difference of squares: The expression $x^8 - 1$ can be factored using the difference of squares formula, $a^2 - b^2 = (a - b)(a + b)$: \[ x^8 - 1 = (x^4 - 1)(x^4 + 1) \] 3. Further factorize $x^4 - 1$: Again using the difference of squares, we factorize $x^4 - 1$: \[ x^4 - 1 = (x^2 - 1)(x^2 + 1) \] Note that $x^2 + 1$ cannot be factored further over the integers. 4. Factorize $x^2 - 1$: Applying the difference of squares one more time: \[ x^2 - 1 = (x - 1)(x + 1) \] 5. Count the total number of factors: We now have the complete factorization of the original polynomial: \[ x^9 - x = x(x^4 + 1)(x^2 + 1)(x - 1)(x + 1) \] This expression consists of 5 factors: $x$, $x^4 + 1$, $x^2 + 1$, $x - 1$, and $x + 1$. 6. Conclusion: Since none of these factors can be factored further into polynomials with integral coefficients, the total number of factors is 5. Thus, the correct answer is $\boxed{\textbf{(B)}\ 5}$.
Find the largest $K$ satisfying the following: Given any closed intervals $A_1,\ldots, A_N$ of length $1$ where $N$ is an arbitrary positive integer. If their union is $[0,2021]$ , then we can always find $K$ intervals from $A_1,\ldots, A_N$ such that the intersection of any two of them is empty.	1011
Working 22 hours in the second week of June, Xenia was able to earn $\$$47.60 more than during the first week of June when she worked 15 hours. If her hourly wage was constant, how many dollars did she earn during the first two weeks of June? Express your answer to the nearest hundredth.	\$ 251.60
In the expansion of $(ax + b)^{2000},$ where $a$ and $b$ are relatively prime positive integers, the coefficients of $x^{2}$ and $x^{3}$ are equal. Find $a + b$.	667	Using the binomial theorem, $\binom{2000}{1998} b^{1998}a^2 = \binom{2000}{1997}b^{1997}a^3 \Longrightarrow b=666a$. Since $a$ and $b$ are positive relatively prime integers, $a=1$ and $b=666$, and $a+b=\boxed{667}$.
There are 17 people at a party, and each has a reputation that is either $1,2,3,4$, or 5. Some of them split into pairs under the condition that within each pair, the two people's reputations differ by at most 1. Compute the largest value of $k$ such that no matter what the reputations of these people are, they are able to form $k$ pairs.	7	First, note that $k=8$ fails when there are $15,0,1,0,1$ people of reputation 1, 2, 3, 4, 5, respectively. This is because the two people with reputation 3 and 5 cannot pair with anyone, and there can only be at maximum $\left\lfloor\frac{15}{2}\right\rfloor=7$ pairs of people with reputation 1. Now, we show that $k=7$ works. Suppose that we keep pairing people until we cannot make a pair anymore. Consider that moment. If there are two people with the same reputation, then these two people can pair up. Thus, there is at most one person for each reputation. Furthermore, if there are at least 4 people, then there must exist two people of consecutive reputations, so they can pair up. Thus, there are at most 3 people left, so we have formed at least $\frac{17-3}{2}=7$ pairs.
Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?	170	1. Understanding the Problem: - We have 6 individuals: Adam, Benin, Chiang, Deshawn, Esther, and Fiona. - Each individual has the same number of internet friends within this group. - We need to determine the number of ways this can happen, considering different possible numbers of friends each person can have. 2. Possible Values for Number of Friends ($n$): - Each person can have between 1 and 4 friends. This is because having 0 friends or all 5 others as friends contradicts the problem's conditions (some, but not all, are friends). 3. Symmetry in Friendships: - The cases for $n=1$ and $n=4$ are symmetric. If a person has 1 friend, then considering the non-friends, they effectively have 4 non-friends, which is equivalent to having 4 friends when considering the inverse situation. - Similarly, the cases for $n=2$ and $n=3$ are symmetric. 4. Counting Configurations for $n=1$: - If each person has exactly one friend, we form 3 disjoint pairs. - Choose a friend for the first person: 5 choices. - The next person (not yet paired) has 3 choices (excluding the first pair). - The last two people must pair with each other. - Total configurations: $5 \times 3 = 15$. 5. Counting Configurations for $n=2$: - Case 1: Triangular Groups - Split the group into two sets of 3, each forming a triangle. - Number of ways to choose 3 people from 6: $\binom{6}{3} = 20$. - Each selection results in exactly one way to form the triangles (since choosing 3 determines the other 3). - However, each triangle configuration is counted twice (once for each triangle), so we divide by 2: $\frac{20}{2} = 10$. - Case 2: Hexagonal Configuration - Each person is a vertex of a hexagon, with edges representing friendships. - Fix one person, and choose 2 friends from the remaining 5: $\binom{5}{2} = 10$ ways. - The remaining 3 people automatically form the other vertices. - Considering rotations and reflections of the hexagon, each configuration is counted multiple times. Correcting for this, we have $\frac{6!}{12} = 60$ configurations. 6. Total Configurations: - For $n=1$ and $n=4$: $15$ configurations each. - For $n=2$ and $n=3$: $10$ (triangles) + $60$ (hexagon) = $70$ configurations each. - Total: $(15 + 70) \times 2 = 170$ configurations. 7. Conclusion: - The total number of different ways the friendships can be configured, given the constraints, is $\boxed{\textbf{(B)}\ 170}$.
A graph has $ 30$ vertices, $ 105$ edges and $ 4822$ unordered edge pairs whose endpoints are disjoint. Find the maximal possible difference of degrees of two vertices in this graph.	22
Find $3^{\frac{1}{3}} \cdot 9^{\frac{1}{9}} \cdot 27^{\frac{1}{27}} \cdot 81^{\frac{1}{81}} \dotsm.$	\sqrt[4]{27}
Lyla and Isabelle run on a circular track both starting at point \( P \). Lyla runs at a constant speed in the clockwise direction. Isabelle also runs in the clockwise direction at a constant speed 25% faster than Lyla. Lyla starts running first and Isabelle starts running when Lyla has completed one third of one lap. When Isabelle passes Lyla for the fifth time, how many times has Lyla returned to point \( P \)?	17
Ponchik was having a snack at a roadside café when a bus passed by. Three pastries after the bus, a motorcycle passed by Ponchik, and three pastries after that, a car passed by. Syrupchik, who was snacking at another café on the same road, saw them in a different order: first the bus, after three pastries the car, and three pastries after that, the motorcycle. It is known that Ponchik and Syrupchik always eat pastries at a constant speed. Find the speed of the bus if the speed of the car is 60 km/h and the speed of the motorcycle is 30 km/h.	40
What is the slope of a line parallel to the line $2x - 4y = 9$? Express your answer as a common fraction.	\frac{1}{2}
In a polar coordinate system, the polar equation of curve C is $\rho=2\cos\theta+2\sin\theta$. Establish a Cartesian coordinate system with the pole as the origin and the positive x-axis as the polar axis. The parametric equation of line l is $\begin{cases} x=1+t \\ y= \sqrt{3}t \end{cases}$ (t is the parameter). Find the length of the chord that curve C cuts off on line l.	\sqrt{7}
Previously, on an old truck, I traveled from village $A$ through $B$ to village $C$. After five minutes, I asked the driver how far we were from $A$. "Half as far as from $B," was the answer. Expressing my concerns about the slow speed of the truck, the driver assured me that while the truck cannot go faster, it maintains its current speed throughout the entire journey. $13$ km after $B$, I inquired again how far we were from $C$. I received exactly the same response as my initial inquiry. A quarter of an hour later, we arrived at our destination. How many kilometers is the journey from $A$ to $C$?	26
If $13^{3n}=\left(\frac{1}{13}\right)^{n-24}$, find $n$.	6
I have two cents and Bill has $n$ cents. Bill wants to buy some pencils, which come in two different packages. One package of pencils costs 6 cents for 7 pencils, and the other package of pencils costs a dime for a dozen pencils (i.e. 10 cents for 12 pencils). Bill notes that he can spend all $n$ of his cents on some combination of pencil packages to get $P$ pencils. However, if I give my two cents to Bill, he then notes that he can instead spend all $n+2$ of his cents on some combination of pencil packages to get fewer than $P$ pencils. What is the smallest value of $n$ for which this is possible?	100	Suppose that Bill buys $a$ packages of 7 and $b$ packages of 12 in the first scenario and $c$ packages of 7 and $d$ packages of 12 in the second scenario. Then we have the following system: $$ \begin{aligned} & 6 a+10 b=n \\ & 6 c+10 d=n+2 \\ & 7 a+12 b>7 c+12 d \end{aligned} $$ Since the packages of 12 give more pencils per cent, we must have $b>d$. Subtract the first two equations and divide by 2 to get $$ 3(c-a)-5(b-d)=1 $$ Note that the last inequality is $12(b-d)>7(c-a)$. The minimal solutions to the equation with $b-d>0$ are $$ (c-a, b-d)=(2,1),(7,4),(12,7),(17,10) $$ $(17,10)$ is the first pair for which $12(b-d)>7(c-a)$. Hence $b \geq 10$ so $n \geq 100$. We can easily verify that $(a, b, c, d, n)=(0,10,17,0,100)$ satisfies the system of equations.
If \( e^{i \theta} = \frac{3 + i \sqrt{2}}{4}, \) then find \( \cos 3\theta. \)	\frac{9}{64}
A jar contains $5$ different colors of gumdrops. $30\%$ are blue, $20\%$ are brown, $15\%$ are red, $10\%$ are yellow, and other $30$ gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?	42	1. Calculate the total number of gumdrops: Given that $30\%$ are blue, $20\%$ are brown, $15\%$ are red, $10\%$ are yellow, and $30$ gumdrops are green. First, we calculate the percentage of gumdrops that are green: \[ 100\% - (30\% + 20\% + 15\% + 10\%) = 100\% - 75\% = 25\% \] Since $25\%$ of the gumdrops are green and there are $30$ green gumdrops, we can find the total number of gumdrops in the jar by setting up the equation: \[ 0.25 \times \text{Total number of gumdrops} = 30 \] Solving for the total number of gumdrops: \[ \text{Total number of gumdrops} = \frac{30}{0.25} = 120 \] 2. Calculate the number of blue and brown gumdrops initially: - Blue gumdrops: $30\%$ of $120$: \[ 0.30 \times 120 = 36 \] - Brown gumdrops: $20\%$ of $120$: \[ 0.20 \times 120 = 24 \] 3. Replace half of the blue gumdrops with brown gumdrops: - Half of the blue gumdrops: \[ \frac{36}{2} = 18 \] - New number of brown gumdrops after replacement: \[ 24 + 18 = 42 \] 4. Conclusion: After replacing half of the blue gumdrops with brown gumdrops, the number of brown gumdrops in the jar becomes $42$. Therefore, the final answer is $\boxed{\textbf{(C)}\ 42}$.
The area of the parallelogram generated by the vectors $\mathbf{a}$ and $\mathbf{b}$ is 8. Find the area of the parallelogram generated by the vectors $2 \mathbf{a} + 3 \mathbf{b}$ and $\mathbf{a} - 5 \mathbf{b}.$	104
Seven cards numbered $1$ through $7$ are to be lined up in a row. Find the number of arrangements of these seven cards where one of the cards can be removed, leaving the remaining six cards in either ascending or descending order.	74
Find the smallest positive integer $N$ with the following property: of the three numbers $N$, $N+1$, and $N+2$, one of them is divisible by $2^2$, one of them is divisible by $3^2$, one is divisible by $5^2$, and one is divisible by $7^2$.	98
In triangle $ABC$, the sides opposite to angles $A$, $B$, and $C$ are denoted as $a$, $b$, and $c$ respectively. It is known that $c=a\cos B+b\sin A$. (I) Find angle $A$. (II) If $a=2$, find the maximum area of $\triangle ABC$.	\sqrt {2}+1
Determine the minimum possible value of the sum \[\frac{a}{3b} + \frac{b}{5c} + \frac{c}{7a},\] where $a,$ $b,$ and $c$ are positive real numbers.	\frac{3}{\sqrt[3]{105}}
A 3x3x3 cube is made of $27$ normal dice. Each die's opposite sides sum to $7$. What is the smallest possible sum of all of the values visible on the $6$ faces of the large cube?	90	1. Understanding the Cube Configuration: - A 3x3x3 cube consists of 27 smaller cubes. - The smaller cubes on the corners have three faces visible. - The smaller cubes on the edges have two faces visible. - The smaller cubes in the center of each face have one face visible. 2. Counting Visible Faces: - There are 8 corner cubes, each with 3 faces visible. - There are 12 edge cubes, each with 2 faces visible. - There are 6 face-center cubes, each with 1 face visible. 3. Calculating Minimum Possible Values on Visible Faces: - For a single die, the opposite faces sum to 7. The pairs are (1,6), (2,5), and (3,4). - The minimum sum for three visible faces on a corner cube is achieved by showing faces 1, 2, and 3. Thus, the sum is $1+2+3=6$. - The minimum sum for two visible faces on an edge cube is achieved by showing faces 1 and 2. Thus, the sum is $1+2=3$. - The minimum sum for one visible face on a face-center cube is achieved by showing face 1. Thus, the sum is $1$. 4. Calculating Total Minimum Sum: - For the 8 corner cubes, the total minimum sum is $8 \times 6 = 48$. - For the 12 edge cubes, the total minimum sum is $12 \times 3 = 36$. - For the 6 face-center cubes, the total minimum sum is $6 \times 1 = 6$. - Adding these sums gives the total minimum sum for all visible faces on the large cube: $48 + 36 + 6 = 90$. 5. Conclusion: - The smallest possible sum of all the values visible on the 6 faces of the large cube is $\boxed{\text{(D)}\ 90}$.
Points $A,B,C,D,E$ and $F$ lie, in that order, on $\overline{AF}$, dividing it into five segments, each of length 1. Point $G$ is not on line $AF$. Point $H$ lies on $\overline{GD}$, and point $J$ lies on $\overline{GF}$. The line segments $\overline{HC}, \overline{JE},$ and $\overline{AG}$ are parallel. Find $HC/JE$.	\frac{5}{3}	1. Identify Key Points and Relationships: - Points $A, B, C, D, E,$ and $F$ are collinear on line $\overline{AF}$, and each segment between consecutive points is of length 1. - Point $G$ is not on line $AF$, and points $H$ and $J$ lie on lines $\overline{GD}$ and $\overline{GF}$ respectively. - Lines $\overline{HC}, \overline{JE},$ and $\overline{AG}$ are parallel. 2. Use of Similar Triangles: - Since $\overline{AG} \parallel \overline{HC}$, triangles $\triangle GAD$ and $\triangle HCD$ are similar by the Basic Proportionality Theorem (or Thales' theorem). - Similarly, since $\overline{AG} \parallel \overline{JE}$, triangles $\triangle GAF$ and $\triangle JEF$ are similar. 3. Calculate Ratios Using Similar Triangles: - For triangles $\triangle GAD$ and $\triangle HCD$, the ratio of corresponding sides is: \[ \frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3} \] Here, $CD = 1$ (distance from $C$ to $D$) and $AD = 3$ (total distance from $A$ to $D$ which is the sum of segments $AB$, $BC$, and $CD$). - For triangles $\triangle GAF$ and $\triangle JEF$, the ratio of corresponding sides is: \[ \frac{JE}{AG} = \frac{EF}{AF} = \frac{1}{5} \] Here, $EF = 1$ (distance from $E$ to $F$) and $AF = 5$ (total distance from $A$ to $F$ which is the sum of all five segments). 4. Find the Desired Ratio $HC/JE$: - Using the ratios from the similar triangles: \[ \frac{HC}{JE} = \frac{\frac{CH}{AG}}{\frac{JE}{AG}} = \frac{\frac{1}{3}}{\frac{1}{5}} = \frac{1}{3} \times \frac{5}{1} = \frac{5}{3} \] 5. Conclusion: - The ratio $\frac{HC}{JE}$ is $\boxed{\frac{5}{3}}$. - The correct answer is $\boxed{(D) \frac{5}{3}}$.
At CMU, markers come in two colors: blue and orange. Zachary fills a hat randomly with three markers such that each color is chosen with equal probability, then Chase shuffles an additional orange marker into the hat. If Zachary chooses one of the markers in the hat at random and it turns out to be orange, the probability that there is a second orange marker in the hat can be expressed as simplified fraction $\tfrac{m}{n}$ . Find $m+n$ .	39
Fill six numbers $1, 3, 5, 7, 9, 11$ into the circles (each circle containing only one number) so that the sum of three numbers on each side of the triangle equals 17. What is the sum of the three numbers in the circles at the vertices of the triangle?	15
Let $\mathcal{S}$ be the set $\lbrace1,2,3,\ldots,10\rbrace$ Let $n$ be the number of sets of two non-empty disjoint subsets of $\mathcal{S}$. (Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when $n$ is divided by $1000$.	501	Let the two disjoint subsets be $A$ and $B$, and let $C = S-(A+B)$. For each $i \in S$, either $i \in A$, $i \in B$, or $i \in C$. So there are $3^{10}$ ways to organize the elements of $S$ into disjoint $A$, $B$, and $C$. However, there are $2^{10}$ ways to organize the elements of $S$ such that $A = \emptyset$ and $S = B+C$, and there are $2^{10}$ ways to organize the elements of $S$ such that $B = \emptyset$ and $S = A+C$. But, the combination such that $A = B = \emptyset$ and $S = C$ is counted twice. Thus, there are $3^{10}-2\cdot2^{10}+1$ ordered pairs of sets $(A,B)$. But since the question asks for the number of unordered sets $\{ A,B \}$, $n = \frac{1}{2}(3^{10}-2\cdot2^{10}+1) = 28501 \equiv \boxed{501} \pmod{1000}$.
A right circular cylinder with radius 3 is inscribed in a hemisphere with radius 7 so that its bases are parallel to the base of the hemisphere. What is the height of this cylinder?	\sqrt{40}
Find all prime numbers whose representation in a base-14 numeral system has the form 101010...101 (alternating ones and zeros).	197
Given $\cos \alpha = \frac{1}{7}$ and $\cos (\alpha-\beta) = \frac{13}{14}$, with $0 < \beta < \alpha < \frac{\pi}{2}$, find $\beta$.	\frac{\pi}{3}
Given the hyperbola $C\_1$: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 (a > b > 0)$ with left and right foci $F\_1$ and $F\_2$, respectively, and hyperbola $C\_2$: $\frac{x^2}{16} - \frac{y^2}{4} = 1$, determine the length of the major axis of hyperbola $C\_1$ given that point $M$ lies on one of the asymptotes of hyperbola $C\_1$, $OM \perp MF\_2$, and the area of $\triangle OMF\_2$ is $16$.	16
Find the area of the region of the \( xy \)-plane defined by the inequality \( \|x\|+\|y\|+\|x+y\| \leq 1 \).	3/4
Given $\|\overrightarrow {a}\|=\sqrt {2}$, $\|\overrightarrow {b}\|=2$, and $(\overrightarrow {a}-\overrightarrow {b})\bot \overrightarrow {a}$, determine the angle between $\overrightarrow {a}$ and $\overrightarrow {b}$.	\frac{\pi}{4}
A triangle with vertices \(A = (4, 3)\), \(B = (6, -2)\), and \(C = (7, 1)\) is reflected about the line \(x = 6\) to create a second triangle. Determine the area of the union of the two triangles.	10
90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 =	945	We are given the sum of consecutive integers from $90$ to $99$. We can calculate this sum using the formula for the sum of an arithmetic series, or by simplifying the expression directly. 1. Identify the series: The series is $90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99$. 2. Calculate the number of terms ($n$): The series starts at $90$ and ends at $99$. The number of terms in the series is $99 - 90 + 1 = 10$. 3. Calculate the sum using the formula for the sum of an arithmetic series: \[ S = \frac{n}{2} \times (\text{first term} + \text{last term}) \] Plugging in the values, we get: \[ S = \frac{10}{2} \times (90 + 99) = 5 \times 189 = 945 \] 4. Verification by expanding the terms: \[ (100-10) + (100-9) + \cdots + (100-1) = 100 \times 10 - (1 + 2 + \cdots + 10) \] The sum of the first 10 positive integers is given by: \[ \frac{10 \times (10 + 1)}{2} = 55 \] Therefore, the sum of the series becomes: \[ 1000 - 55 = 945 \] Both methods confirm that the sum of the integers from $90$ to $99$ is $945$. Thus, the correct answer is $\boxed{\text{(B)}~945}$.
In a bus station in the city, there are 10 waiting seats arranged in a row. Now, if 4 passengers randomly choose some seats to wait, the number of ways to arrange them so that there are exactly 5 consecutive empty seats is $\boxed{480}$.	480
A caterpillar is climbing a 20-meter pole. During the day, it climbs 5 meters, and during the night, it slides down 4 meters. How long will it take for the caterpillar to reach the top of the pole?	16
Four fair coins are tossed once. For every head that appears, two six-sided dice are rolled. What is the probability that the sum of all dice rolled is exactly ten? A) $\frac{1} {48}$ B) $\frac{1} {20}$ C) $\frac{1} {16}$ D) $\frac{1} {30}$	\frac{1} {20}
The graph of the function $f(x)=\frac{x}{x+a}$ is symmetric about the point $(1,1)$, and the function $g(x)=\log_{10}(10^x+1)+bx$ is even. Find the value of $a+b$.	-\frac{3}{2}
Given vectors $\overrightarrow{a}=(\sin \theta, 2)$ and $\overrightarrow{b}=(\cos \theta, 1)$, which are collinear, where $\theta \in (0, \frac{\pi}{2})$. 1. Find the value of $\tan (\theta + \frac{\pi}{4})$. 2. If $5\cos (\theta - \phi)=3 \sqrt{5}\cos \phi, 0 < \phi < \frac{\pi}{2}$, find the value of $\phi$.	\frac{\pi}{4}
Below is a portion of the graph of a function, $y=E(x)$: [asy] import graph; size(8cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.5,xmax=4.5,ymin=-0.99,ymax=6.5; pen cqcqcq=rgb(0.75,0.75,0.75); /grid/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)gx;i<=floor(xmax/gx)gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)gy;i<=floor(ymax/gy)gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis("",xmin,xmax,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis("",ymin,ymax,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); real f1(real x){return sqrt(abs(x+1))+(9/pi)*atan(sqrt(abs(x)));} draw(graph(f1,xmin,xmax),linewidth(1)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); label("$y=E(x)$",(xmax+0.25,f1(xmax)),E); [/asy] The value of $E(3)$ is an integer. What is it?	5
Let $u$ and $v$ be real numbers satisfying the inequalities $2u + 3v \le 10$ and $4u + v \le 9.$ Find the largest possible value of $u + 2v$.	6.1
Solve for $x$: $\sqrt[4]{40x + \sqrt[4]{40x + 24}} = 24.$	8293.8
A positive integer $N$ greater than $1$ is described as special if in its base- $8$ and base- $9$ representations, both the leading and ending digit of $N$ are equal to $1$ . What is the smallest special integer in decimal representation? Proposed by Michael Ren	793
How many ways are there to put 4 balls in 3 boxes if the balls are not distinguishable but the boxes are?	15
If $\Diamond4_7=\Diamond1_{8}$ and $\Diamond$ represents a digit, solve for $\Diamond$.	3
For any real number \( x \), let \( \lfloor x \rfloor \) denote the greatest integer less than or equal to \( x \), and let \( \{ x \} \) denote the fractional part of \( x \). Then $$ \left\{\frac{2014}{2015}\right\}+\left\{\frac{2014^{2}}{2015}\right\}+\cdots+\left\{\frac{2014^{2014}}{2015}\right\} $$ equals:	1007
A car left the city for the village, and simultaneously, a cyclist left the village for the city. When the car and the cyclist met, the car immediately turned around and went back to the city. As a result, the cyclist arrived in the city 35 minutes later than the car. How many minutes did the cyclist spend on the entire trip, given that his speed is 4.5 times less than the speed of the car?	55
A biased coin lands heads with a probability of $\frac{2}{3}$ and tails with $\frac{1}{3}$. A player can choose between Game C and Game D. In Game C, the player tosses the coin five times and wins if either the first three or the last three outcomes are all the same. In Game D, she tosses the coin five times and wins if at least one of the following conditions is met: the first two tosses are the same and the last two tosses are the same, or the middle three tosses are all the same. A) Game C has a higher probability of $\frac{5}{81}$ over Game D. B) Game D has a higher probability of $\frac{5}{81}$ over Game C. C) Game C has a higher probability of $\frac{29}{81}$ over Game D. D) Game D has a higher probability of $\frac{29}{81}$ over Game C.	\frac{29}{81}
In $\triangle ABC$, the median from vertex $A$ is perpendicular to the median from vertex $B$. The lengths of sides $AC$ and $BC$ are 6 and 7 respectively. What is the length of side $AB$?	$\sqrt{17}$
Four complex numbers lie at the vertices of a square in the complex plane. Three of the numbers are $1+2i, -2+i$, and $-1-2i$. The fourth number is	$2-i$	1. Identify the given points in the complex plane: The complex numbers given are $1+2i, -2+i$, and $-1-2i$. We can interpret these as points in the Cartesian plane: $A = (1, 2), B = (-2, 1), C = (-1, -2)$. 2. Determine the slopes of lines AB and BC: - The slope of line $AB$ is calculated using the formula $\frac{y_2 - y_1}{x_2 - x_1}$: \[ \text{slope of } AB = \frac{1 - 2}{-2 - 1} = \frac{-1}{-3} = \frac{1}{3} \] - The slope of line $BC$ is: \[ \text{slope of } BC = \frac{-2 - 1}{-1 + 2} = \frac{-3}{1} = -3 \] 3. Find the equations of the perpendicular bisectors: - The slope of a line perpendicular to another is the negative reciprocal of the original slope. Thus, the slope of the line perpendicular to $AB$ (passing through $A$) is $-3$, and the slope of the line perpendicular to $BC$ (passing through $C$) is $\frac{1}{3}$. - The equation of the line perpendicular to $AB$ through $A(1,2)$: \[ y - 2 = -3(x - 1) \implies y = -3x + 5 \] - The equation of the line perpendicular to $BC$ through $C(-1,-2)$: \[ y + 2 = \frac{1}{3}(x + 1) \implies y = \frac{1}{3}x - \frac{5}{3} \] 4. Find the intersection of these perpendicular bisectors: - Set the equations equal to find $x$: \[ -3x + 5 = \frac{1}{3}x - \frac{5}{3} \] - Solving for $x$: \[ -3x - \frac{1}{3}x = -\frac{5}{3} - 5 \implies -\frac{10}{3}x = -\frac{20}{3} \implies x = 2 \] - Substitute $x = 2$ back into one of the line equations to find $y$: \[ y = -3(2) + 5 = -6 + 5 = -1 \] 5. Convert the Cartesian coordinates back to the complex number: - The Cartesian point $(2, -1)$ corresponds to the complex number $2 - i$. Thus, the fourth vertex of the square in the complex plane is $\boxed{2-i}$.
Let \( a \), \( b \), and \( c \) be the roots of the polynomial \( x^3 - 15x^2 + 25x - 10 = 0 \). Compute \[ (a-b)^2 + (b-c)^2 + (c-a)^2. \]	125
Let $R$ be the rectangle in the Cartesian plane with vertices at $(0,0)$ , $(2,0)$ , $(2,1)$ , and $(0,1)$ . $R$ can be divided into two unit squares, as shown. [asy]size(120); defaultpen(linewidth(0.7)); draw(origin--(2,0)--(2,1)--(0,1)--cycle^^(1,0)--(1,1));[/asy] Pro selects a point $P$ at random in the interior of $R$ . Find the probability that the line through $P$ with slope $\frac{1}{2}$ will pass through both unit squares.	3/4
Let the function \( f(x) = x^3 + a x^2 + b x + c \), where \( a \), \( b \), and \( c \) are non-zero integers. If \( f(a) = a^3 \) and \( f(b) = b^3 \), then the value of \( c \) is:	16
Calculate \(14 \cdot 31\) and \(\left\lfloor\frac{2+\sqrt{2}}{2}\right\rfloor + \left\lfloor\frac{3+\sqrt{3}}{3}\right\rfloor + \left\lfloor\frac{4+\sqrt{4}}{4}\right\rfloor + \cdots + \left\lfloor\frac{1989+\sqrt{1989}}{1989}\right\rfloor + \left\lfloor\frac{1990+\sqrt{1990}}{1990}\right\rfloor\).	1989
Regular hexagon $ABCDEF$ has an area of $n$. Let $m$ be the area of triangle $ACE$. What is $\tfrac{m}{n}?$ A) $\frac{1}{2}$ B) $\frac{2}{3}$ C) $\frac{3}{4}$ D) $\frac{1}{3}$ E) $\frac{3}{2}$	\frac{2}{3}
Billy Bones has two coins — one gold and one silver. One of these coins is fair, while the other is biased. It is unknown which coin is biased but it is known that the biased coin lands heads with a probability of $p=0.6$. Billy Bones tossed the gold coin and it landed heads immediately. Then, he started tossing the silver coin, and it landed heads only on the second toss. Find the probability that the gold coin is the biased one.	5/9
Let $w, x, y$, and $z$ be positive real numbers such that $0 \neq \cos w \cos x \cos y \cos z$, $2 \pi =w+x+y+z$, $3 \tan w =k(1+\sec w)$, $4 \tan x =k(1+\sec x)$, $5 \tan y =k(1+\sec y)$, $6 \tan z =k(1+\sec z)$. Find $k$.	\sqrt{19}	From the identity $\tan \frac{u}{2}=\frac{\sin u}{1+\cos u}$, the conditions work out to $3 \tan \frac{w}{2}=4 \tan \frac{x}{2}=5 \tan \frac{y}{2}=6 \tan \frac{z}{2}=k$. Let $a=\tan \frac{w}{2}, b=\tan \frac{x}{2}, c=\tan \frac{y}{2}$, and $d=\tan \frac{z}{2}$. Using the identity $\tan (M+N)=\frac{\tan M+\tan N}{1-\tan M \tan N}$, we obtain $\tan \left(\frac{w+x}{2}+\frac{y+z}{2}\right) =\frac{\tan \left(\frac{w+x}{2}\right)+\tan \left(\frac{y+z}{2}\right)}{1-\tan \left(\frac{w+x}{2}\right) \tan \left(\frac{y+z}{2}\right)} =\frac{\frac{a+b}{1-a b}+\frac{c+d}{1-c d}}{1-\left(\frac{a+b}{1-a b}\right)\left(\frac{c+d}{1-c d}\right)} =\frac{a+b+c+d-a b c-a b d-b c d-a c d}{1+a b c d-a b-a c-a d-b c-b d-c d}$. Because $x+y+z+w=\pi$, we get that $\tan \left(\frac{x+y+z+w}{2}\right)=0$ and thus $a+b+c+d=a b c+a b d+b c d+a c d$. Substituting $a, b, c, d$ corresponding to the variable $k$, we obtain that $k^{3}-19 k=0$. Therefore, $k$ can be only $0, \sqrt{19},-\sqrt{19}$. However, $k=0$ is impossible as $w, x, y, z$ will all be 0. Also, $k=-\sqrt{19}$ is impossible as $w, x, y, z$ will exceed $\pi$. Therefore, $k=\sqrt{19}$.
If $m$ and $n$ are positive integers with $n > 1$ such that $m^{n} = 2^{25} \times 3^{40}$, what is $m + n$?	209957	Since $m$ and $n$ are positive integers with $n > 1$ and $m^{n} = 2^{25} \times 3^{40}$, then 2 and 3 are prime factors of $m$ (since they are prime factors of $m^{n}$) and must be the only prime factors of $m$ (since if there were other prime factors of $m$, then there would be other prime factors of $m^{n}$). Therefore, $m = 2^{a} \times 3^{b}$ for some positive integers $a$ and $b$ and so $m^{n} = (2^{a} \times 3^{b})^{n} = 2^{an} \times 3^{bn}$. Since $m^{n} = 2^{25} \times 3^{40}$, then we must have $an = 25$ and $bn = 40$. Since $a, b, n$ are positive integers, then $n$ is a common divisor of 25 and 40. Since $n > 1$, then $n = 5$, which means that $a = 5$ and $b = 8$. In this case, $m = 2^{5} \times 3^{8} = 32 \times 6561 = 209952$, which gives $m + n = 209952 + 5 = 209957$.

How many three-digit numbers are increased by 99 when their digits are reversed?

80

Find the last two digits (in order) of $ 7^{30105} $.

43

Anton makes custom watches for a jewelry store. Each watch consists of a bracelet, a gemstone, and a clasp. Bracelets are available in silver, gold, and steel. Anton has precious stones: cubic zirconia, emerald, quartz, diamond, and agate - and clasps: classic, butterfly, buckle. Anton is happy only when three watches are displayed in a row from left to right according to the following rules: - There must be a steel watch with a classic clasp and cubic zirconia stone. - Watches with a classic clasp must be flanked by gold and silver watches. - The three watches in a row must have different bracelets, gemstones, and clasps. How many ways are there to make Anton happy?

72

Given a sequence $\{a\_n\}$ that satisfies $a\_1=1$ and $a\_n= \frac{2S\_n^2}{2S\_n-1}$ for $n\geqslant 2$, where $S\_n$ is the sum of the first $n$ terms of the sequence, find the value of $S\_{2016}$.

\frac{1}{4031}

What is the smallest palindrome that is larger than 2015?

2112

The image of the point with coordinates $(-3,-1)$ under the reflection across the line $y=mx+b$ is the point with coordinates $(5,3)$. Find $m+b$.

1

Given that the sum of the first $n$ terms of the arithmetic sequence ${a_n}$ is $S_n$, if $a_2=0$, $S_3+S_4=6$, then the value of $a_5+a_6$ is $\_\_\_\_\_\_$.

21

Given that points $A$ and $B$ lie on the curves $C_{1}: x^{2}-y+1=0$ and $C_{2}: y^{2}-x+1=0$ respectively, what is the minimum value of the distance $|AB|$?

\frac{3\sqrt{2}}{4}

Acute-angled $\triangle ABC$ is inscribed in a circle with center at $O$. The measures of arcs are $\stackrel \frown {AB} = 80^\circ$ and $\stackrel \frown {BC} = 100^\circ$. A point $E$ is taken in minor arc $AC$ such that $OE$ is perpendicular to $AC$. Find the ratio of the magnitudes of $\angle OBE$ and $\angle BAC$.

10

Three friends have a total of 6 identical pencils, and each one has at least one pencil. In how many ways can this happen?

10

Given that positive real numbers $x$ and $y$ satisfy $e^{x}=y\ln x+y\ln y$, then the minimum value of $\frac{{e}^{x}}{x}-\ln y$ is ______.

e-1

Find all pairs $(m,n)$ of nonnegative integers for which \[m^2 + 2 \cdot 3^n = m\left(2^{n+1} - 1\right).\] [i]

(9, 3), (6, 3), (9, 5), (54, 5)

We are tasked with finding all pairs $(m, n)$ of nonnegative integers that satisfy the equation: \[ m^2 + 2 \cdot 3^n = m\left(2^{n+1} - 1\right). \] To solve this equation, we rearrange terms to express it in a form that can be factored: \[ m^2 - m(2^{n+1} - 1) + 2 \cdot 3^n = 0. \] This is a quadratic equation in $ m $. To solve for $ m $, we use the quadratic formula: \[ m = \frac{(2^{n+1} - 1) \pm \sqrt{(2^{n+1} - 1)^2 - 8 \cdot 3^n}}{2}. \] For $ m $ to be an integer, the discriminant must be a perfect square: \[ (2^{n+1} - 1)^2 - 8 \cdot 3^n = k^2 \] for some integer $k$. Let's simplify and check cases for specific values of $ n $: ### Case 1: $ n = 3 $ - Substitute $ n = 3 $ into the equation: \[ (2^{4} - 1)^2 - 8 \cdot 3^3 = (15)^2 - 216 = 225 - 216 = 9 = 3^2. \] Here, the discriminant is a perfect square. Calculate $ m $: \[ m = \frac{15 \pm 3}{2}. \] This gives \[ m = 9 \quad \text{and} \quad m = 6. \] So, the pairs $(m, n)$ are $(9, 3)$ and $(6, 3)$. ### Case 2: $ n = 5 $ - Substitute $ n = 5 $ into the equation: \[ (2^{6} - 1)^2 - 8 \cdot 3^5 = (63)^2 - 1944 = 3969 - 1944 = 2025 = 45^2. \] Again, the discriminant is a perfect square. Calculate $ m $: \[ m = \frac{63 \pm 45}{2}. \] This gives \[ m = 54 \quad \text{and} \quad m = 9. \] So, the pairs $(m, n)$ are $(54, 5)$ and $(9, 5)$. Therefore, the possible pairs $(m, n)$ that satisfy the given equation are: \[ \boxed{(9, 3), (6, 3), (9, 5), (54, 5)}. \]

Let the function $ f(x) $ be defined on $ \mathbb{R} $, and for any $ x $, the condition $ f(x+2) + f(x) = x $ holds. It is also known that $ f(x) = x^3 $ on the interval $ (-2, 0] $. Find $ f(2012) $.

1006

Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?

\frac{145}{147}

1. **Identify the vertices and setup the problem**: Let $A$, $B$, and $C$ be the vertices of the right triangle field with $AB = 3$ units, $AC = 4$ units, and $\angle BAC = 90^\circ$. Let $S$ be the square in the corner at $A$, and let its vertices be $A$, $M$, $D$, and $N$ with $AM$ and $AN$ along $AB$ and $AC$ respectively. 2. **Define additional points and use similarity**: Let $X$ be on $BC$ such that $AC \parallel DX$ and $Y$ be on $BC$ such that $AB \parallel DY$. Define $P$ and $Q$ as the feet of the perpendiculars from $X$ to $AC$ and from $Y$ to $AB$ respectively. By similarity of triangles $\triangle ABC \sim \triangle DYX \sim \triangle PXC \sim \triangle QBY$, we have: \[ PC = x \cdot \frac{3}{4}, \quad QB = x \cdot \frac{4}{3} \] where $x$ is the side length of square $S$. 3. **Calculate distances using similarity**: From the similarity, we also have: \[ DX = 3 - x - \frac{3}{4}x = 3 - \frac{7}{4}x, \quad MQ = 4 - x - \frac{4}{3}x = 4 - \frac{7}{3}x \] and \[ CX = \frac{5}{4}x, \quad BY = \frac{5}{3}x \] leading to \[ XY = 5 - \frac{5}{4}x - \frac{5}{3}x = 5 - \frac{35}{12}x \] 4. **Set up an equation using area calculations**: The area of $\triangle XDY$ can be calculated in two ways: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (3 - \frac{7}{4}x) \times (4 - \frac{7}{3}x) \] and also as \[ \text{Area} = 2 \times (5 - \frac{35}{12}x) \] Equating these and solving for $x$: \[ 10 - \frac{35}{6}x = \frac{49}{12}x^2 - 14x + 12 \] \[ 49x^2 - 98x + 24 = 0 \] \[ x = \frac{2}{7} \text{ or } \frac{12}{7} \] The value $x = \frac{12}{7}$ is extraneous as it would exceed the dimensions of the triangle. 5. **Calculate the area of square $S$ and the planted portion**: The area of square $S$ is $x^2 = \left(\frac{2}{7}\right)^2 = \frac{4}{49}$. The total area of the triangle is $\frac{1}{2} \times 3 \times 4 = 6$. The planted area is: \[ 6 - \frac{4}{49} = \frac{294}{49} - \frac{4}{49} = \frac{290}{49} = \frac{145}{24.5} \] The fraction of the field that is planted is: \[ \frac{\frac{290}{49}}{6} = 1 - \frac{\frac{4}{49}}{6} = 1 - \frac{2}{147} = \boxed{\textbf{(D) } \frac{145}{147}} \]

Let $A B C$ be an acute triangle with circumcenter $O$ such that $A B=4, A C=5$, and $B C=6$. Let $D$ be the foot of the altitude from $A$ to $B C$, and $E$ be the intersection of $A O$ with $B C$. Suppose that $X$ is on $B C$ between $D$ and $E$ such that there is a point $Y$ on $A D$ satisfying $X Y \parallel A O$ and $Y O \perp A X$. Determine the length of $B X$.

\frac{96}{41}

Let $A X$ intersect the circumcircle of $\triangle A B C$ again at $K$. Let $O Y$ intersect $A K$ and $B C$ at $T$ and $L$, respectively. We have $\angle L O A=\angle O Y X=\angle T D X=\angle L A K$, so $A L$ is tangent to the circumcircle. Furthermore, $O L \perp A K$, so $\triangle A L K$ is isosceles with $A L=A K$, so $A K$ is also tangent to the circumcircle. Since $B C$ and the tangents to the circumcircle at $A$ and $K$ all intersect at the same point $L, C L$ is a symmedian of $\triangle A C K$. Then $A K$ is a symmedian of $\triangle A B C$. Then we can use $\frac{B X}{X C}=\frac{(A B)^{2}}{(A C)^{2}}$ to compute $B X=\frac{96}{41}$.

Given the sequence $\{a_n\}$ with the sum of its first $n$ terms $S_n = 6n - n^2$, find the sum of the first $20$ terms of the sequence $\left\{ \frac{1}{a_na_{n+1}}\right\}$.

-\frac{4}{35}

In the Cartesian coordinate system, point O is the origin, and the coordinates of three vertices of the parallelogram ABCD are A(2,3), B(-1,-2), and C(-2,-1). (1) Find the lengths of the diagonals AC and BD; (2) If the real number t satisfies $ (\vec{AB}+t\vec{OC})\cdot\vec{OC}=0 $, find the value of t.

-\frac{11}{5}

Solve for $n$, if $9^n\cdot9^n\cdot9^n\cdot9^n=81^4$.

2

$x$, $y$ and $z$ are positive reals such that $x+y+z=xyz$. Find the minimum value of: \[ x^7(yz-1)+y^7(zx-1)+z^7(xy-1) \]

162\sqrt{3}

Given that $ x $, $ y $, and $ z $ are positive reals such that $ x + y + z = xyz $, we aim to find the minimum value of: \[ x^7(yz-1) + y^7(zx-1) + z^7(xy-1). \] First, we use the given condition $ x + y + z = xyz $. By the AM-GM inequality, we have: \[ xyz = x + y + z \geq 3\sqrt[3]{xyz}, \] which implies: \[ xyz \geq 3\sqrt{3}. \] Now, consider the given expression: \[ x^7(yz-1) + y^7(zx-1) + z^7(xy-1). \] Rewriting it, we get: \[ x^7(yz-1) + y^7(zx-1) + z^7(xy-1) = x^6(xyz - x) + y^6(xyz - y) + z^6(xyz - z). \] This simplifies to: \[ x^6(xyz) + y^6(xyz) + z^6(xyz) - (x^7 + y^7 + z^7). \] Factoring out $ xyz $ from the first part, we get: \[ (x^6 + y^6 + z^6)xyz - (x^7 + y^7 + z^7). \] Using the given $ xyz = x + y + z $, we can rewrite it as: \[ (x^6 + y^6 + z^6)(x + y + z) - (x^7 + y^7 + z^7). \] Expanding this, we have: \[ (x^7 + y^7 + z^7) + (x^6y + xy^6 + x^6z + xz^6 + y^6z + yz^6) - (x^7 + y^7 + z^7). \] This simplifies to: \[ x^6y + xy^6 + x^6z + xz^6 + y^6z + yz^6. \] By the AM-GM inequality, we know: \[ x^6y + xy^6 + x^6z + xz^6 + y^6z + yz^6 \geq 6\sqrt[6]{(xyz)^{14}}. \] Given $ xyz \geq 3\sqrt{3} $, we have: \[ 6\sqrt[6]{(3\sqrt{3})^{14}} = 6\sqrt[6]{3^{21}} = 6 \cdot 3^{7/2} = 6 \cdot 27\sqrt{3} = 162\sqrt{3}. \] Therefore, the minimum value is: \[ \boxed{162\sqrt{3}}. \] This minimum is achieved when $ x = y = z = \sqrt{3} $.

A box contains 4 cards, each with one of the following functions defined on \$R\$: \$f_{1}(x)={x}^{3}\$, \$f_{2}(x)=|x|\$, \$f_{3}(x)=\sin x\$, \$f_{4}(x)=\cos x\$. Now, if we randomly pick 2 cards from the box and multiply the functions on the cards to get a new function, the probability that the resulting function is an odd function is \_\_\_\_\_.

\dfrac{2}{3}

In parallelogram $EFGH$, $EF = 5z + 5$, $FG = 4k^2$, $GH = 40$, and $HE = k + 20$. Determine the values of $z$ and $k$ and find $z \times k$.

\frac{7 + 7\sqrt{321}}{8}

Two people, Person A and Person B, start at the same time from point $A$ to point $B$: Person A is faster than Person B. After reaching point $B$, Person A doubles their speed and immediately returns to point $A$. They meet Person B at a point 240 meters from point $B$. After meeting, Person B also doubles their speed and turns back. When Person A returns to point $A$, Person B is still 120 meters away from point $A$. What is the distance between points $A$ and $B$ in meters?

420

Given that Ben constructs a $4$-step staircase using $26$ toothpicks, determine the number of additional toothpicks needed to extend the staircase to a $6$-step staircase,

22

Let $ f(x) = x^2 + px + q $. It is known that the inequality $ |f(x)| > \frac{1}{2} $ has no solutions on the interval $[1, 3]$. Find $ \underbrace{f(f(\ldots f}_{2017}\left(\frac{3+\sqrt{7}}{2}\right)) \ldots) $. If necessary, round your answer to two decimal places.

0.18

Given a geometric progression $\{a_n\}$ with the first term $a_1=2$ and the sum of the first $n$ terms as $S_n$, and the equation $S_5 + 4S_3 = 5S_4$ holds, find the maximum term of the sequence $\left\{ \frac{2\log_{2}a_n + 1}{\log_{2}a_n - 6} \right\}$.

15

In how many ways can I choose 3 captains from a team of 11 people?

165

How many positive integers less that $200$ are relatively prime to either $15$ or $24$ ?

120

The ancient Chinese mathematical classic "The Nine Chapters on the Mathematical Art" contains a problem called "Rice and Grain Separation". During the collection of grain in a granary, 1524 "shi" (a unit of weight) of rice was received, but it was found to contain grains of another type mixed in. A sample of rice was taken and it was found that out of 254 grains, 28 were not rice. Approximately how much of this batch of rice is not rice?

168

Given a triangle $ABC$ with internal angles $A$, $B$, and $C$, and it is known that $$2\sin^{2}(B+C)= \sqrt {3}\sin2A.$$ (Ⅰ) Find the degree measure of $A$; (Ⅱ) If $BC=7$ and $AC=5$, find the area $S$ of $\triangle ABC$.

10 \sqrt {3}

Cindy leaves school at the same time every day. If she cycles at $20 \ \text{km/h}$, she arrives home at 4:30 in the afternoon. If she cycles at $10 \ \text{km/h}$, she arrives home at 5:15 in the afternoon. Determine the speed, in $\text{km/h}$, at which she must cycle to arrive home at 5:00 in the afternoon.

12

Quantities $r$ and $ s $ vary inversely. When $ r $ is $ 1500 $, $ s $ is $ 0.4 $. Alongside, quantity $ t $ also varies inversely with $ r $ and when $ r $ is $ 1500 $, $ t $ is $ 2.5 $. What is the value of $ s $ and $ t $ when $ r $ is $ 3000 $? Express your answer as a decimal to the nearest thousandths.

1.25

The volume of the parallelepiped determined by the three-dimensional vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ is 4. Find the volume of the parallelepiped determined by the vectors $\mathbf{a} + \mathbf{b},$ $\mathbf{b} + 3 \mathbf{c},$ and $\mathbf{c} - 7 \mathbf{a}.$

80

Divide 6 volunteers into 4 groups, with two groups having 2 people each and the other two groups having 1 person each, to serve at four different pavilions of the World Expo. How many different allocation schemes are there? (Answer with a number).

1080

Given that $\sin A+\sin B=1$ and $\cos A+\cos B= \frac{3}{2}$, what is the value of $\cos(A-B)$?

\frac{5}{8}

Screws are sold in packs of $10$ and $12$ . Harry and Sam independently go to the hardware store, and by coincidence each of them buys exactly $k$ screws. However, the number of packs of screws Harry buys is different than the number of packs Sam buys. What is the smallest possible value of $k$ ?

60

A function $f$ has the property that $f(3x-1)=x^2+x+1$ for all real numbers $x$. What is $f(5)$?

7

Given a geometric sequence $\{a_n\}$ with the first term $\frac{3}{2}$ and common ratio $-\frac{1}{2}$, and the sum of the first $n$ terms is $S_n$, then when $n\in N^*$, the sum of the maximum and minimum values of $S_n - \frac{1}{S_n}$ is ______.

\frac{1}{4}

Evaluate $\log_\frac{1}{3}9$.

-2

An $a \times b \times c$ rectangular box is built from $a \cdot b \cdot c$ unit cubes. Each unit cube is colored red, green, or yellow. Each of the $a$ layers of size $1 \times b \times c$ parallel to the $(b \times c)$ faces of the box contains exactly $9$ red cubes, exactly $12$ green cubes, and some yellow cubes. Each of the $b$ layers of size $a \times 1 \times c$ parallel to the $(a \times c)$ faces of the box contains exactly $20$ green cubes, exactly $25$ yellow cubes, and some red cubes. Find the smallest possible volume of the box.

180

The total number of green cubes is given by $12a=20b\Longrightarrow a=\frac{5}{3}b$. Let $r$ be the number of red cubes on each one of the $b$ layers then the total number of red cubes is $9a=br$. Substitute $a=\frac{5}{3}b$ gives $r=15$. Repeating the procedure on the number of yellow cubes $y$ on each of the $a$ layers gives $y=15$. Therefore $bc=9+12+15=36$ and $ac=15+20+25=60$. Multiplying yields $abc^2=2160$. Since $abc^2$ is fixed, $abc$ is minimized when $c$ is maximized, which occurs when $a$, $b$ are minimized (since each of $ac$, $bc$ is fixed). Thus $(a,b,c)=(3,5,12)\Longrightarrow abc=\boxed{180}$ ~ Nafer

If an integer of two digits is $k$ times the sum of its digits, the number formed by interchanging the digits is the sum of the digits multiplied by

11-k

1. **Define the number and its properties**: Let the two-digit integer be represented as $n = 10a + b$, where $a$ and $b$ are the tens and units digits respectively. According to the problem, this number $n$ is $k$ times the sum of its digits. Therefore, we have the equation: \[ 10a + b = k(a + b) \] 2. **Expression for the interchanged digits**: The number formed by interchanging the digits of $n$ is $10b + a$. We need to find a constant $x$ such that: \[ 10b + a = x(a + b) \] 3. **Combine the equations**: Adding the equations from step 1 and step 2, we get: \[ (10a + b) + (10b + a) = k(a + b) + x(a + b) \] Simplifying both sides, we have: \[ 11a + 11b = (k + x)(a + b) \] This simplifies further to: \[ 11(a + b) = (k + x)(a + b) \] 4. **Solve for $x$**: Since $a + b \neq 0$ (as $a$ and $b$ are digits of a two-digit number), we can divide both sides of the equation by $(a + b)$: \[ 11 = k + x \] Solving for $x$, we find: \[ x = 11 - k \] 5. **Conclusion**: The number formed by interchanging the digits is $11 - k$ times the sum of the digits. Therefore, the correct answer is: \[ \boxed{\textbf{(C) \ } 11-k} \]

The remainder when 111 is divided by 10 is 1. The remainder when 111 is divided by the positive integer $n$ is 6. How many possible values of $n$ are there?

5

Since the remainder when 111 is divided by $n$ is 6, then $111-6=105$ is a multiple of $n$ and $n>6$ (since, by definition, the remainder must be less than the divisor). Since $105=3 \cdot 5 \cdot 7$, the positive divisors of 105 are $1,3,5,7,15,21,35,105$. Therefore, the possible values of $n$ are $7,15,21,35,105$, of which there are 5.

When $x^9-x$ is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is:

5

1. **Factor out the greatest common factor**: The polynomial $x^9 - x$ can be rewritten by factoring out the common factor of $x$, giving: \[ x^9 - x = x(x^8 - 1) \] 2. **Factorize $x^8 - 1$ using difference of squares**: The expression $x^8 - 1$ can be factored using the difference of squares formula, $a^2 - b^2 = (a - b)(a + b)$: \[ x^8 - 1 = (x^4 - 1)(x^4 + 1) \] 3. **Further factorize $x^4 - 1$**: Again using the difference of squares, we factorize $x^4 - 1$: \[ x^4 - 1 = (x^2 - 1)(x^2 + 1) \] Note that $x^2 + 1$ cannot be factored further over the integers. 4. **Factorize $x^2 - 1$**: Applying the difference of squares one more time: \[ x^2 - 1 = (x - 1)(x + 1) \] 5. **Count the total number of factors**: We now have the complete factorization of the original polynomial: \[ x^9 - x = x(x^4 + 1)(x^2 + 1)(x - 1)(x + 1) \] This expression consists of 5 factors: $x$, $x^4 + 1$, $x^2 + 1$, $x - 1$, and $x + 1$. 6. **Conclusion**: Since none of these factors can be factored further into polynomials with integral coefficients, the total number of factors is 5. Thus, the correct answer is $\boxed{\textbf{(B)}\ 5}$.

Find the largest $K$ satisfying the following: Given any closed intervals $A_1,\ldots, A_N$ of length $1$ where $N$ is an arbitrary positive integer. If their union is $[0,2021]$ , then we can always find $K$ intervals from $A_1,\ldots, A_N$ such that the intersection of any two of them is empty.

1011

Working 22 hours in the second week of June, Xenia was able to earn $\$$47.60 more than during the first week of June when she worked 15 hours. If her hourly wage was constant, how many dollars did she earn during the first two weeks of June? Express your answer to the nearest hundredth.

\$ 251.60

In the expansion of $(ax + b)^{2000},$ where $a$ and $b$ are relatively prime positive integers, the coefficients of $x^{2}$ and $x^{3}$ are equal. Find $a + b$.

667

Using the binomial theorem, $\binom{2000}{1998} b^{1998}a^2 = \binom{2000}{1997}b^{1997}a^3 \Longrightarrow b=666a$. Since $a$ and $b$ are positive relatively prime integers, $a=1$ and $b=666$, and $a+b=\boxed{667}$.

There are 17 people at a party, and each has a reputation that is either $1,2,3,4$, or 5. Some of them split into pairs under the condition that within each pair, the two people's reputations differ by at most 1. Compute the largest value of $k$ such that no matter what the reputations of these people are, they are able to form $k$ pairs.

7

First, note that $k=8$ fails when there are $15,0,1,0,1$ people of reputation 1, 2, 3, 4, 5, respectively. This is because the two people with reputation 3 and 5 cannot pair with anyone, and there can only be at maximum $\left\lfloor\frac{15}{2}\right\rfloor=7$ pairs of people with reputation 1. Now, we show that $k=7$ works. Suppose that we keep pairing people until we cannot make a pair anymore. Consider that moment. If there are two people with the same reputation, then these two people can pair up. Thus, there is at most one person for each reputation. Furthermore, if there are at least 4 people, then there must exist two people of consecutive reputations, so they can pair up. Thus, there are at most 3 people left, so we have formed at least $\frac{17-3}{2}=7$ pairs.

Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?

170

1. **Understanding the Problem:** - We have 6 individuals: Adam, Benin, Chiang, Deshawn, Esther, and Fiona. - Each individual has the same number of internet friends within this group. - We need to determine the number of ways this can happen, considering different possible numbers of friends each person can have. 2. **Possible Values for Number of Friends ($n$):** - Each person can have between 1 and 4 friends. This is because having 0 friends or all 5 others as friends contradicts the problem's conditions (some, but not all, are friends). 3. **Symmetry in Friendships:** - The cases for $n=1$ and $n=4$ are symmetric. If a person has 1 friend, then considering the non-friends, they effectively have 4 non-friends, which is equivalent to having 4 friends when considering the inverse situation. - Similarly, the cases for $n=2$ and $n=3$ are symmetric. 4. **Counting Configurations for $n=1$:** - If each person has exactly one friend, we form 3 disjoint pairs. - Choose a friend for the first person: 5 choices. - The next person (not yet paired) has 3 choices (excluding the first pair). - The last two people must pair with each other. - Total configurations: $5 \times 3 = 15$. 5. **Counting Configurations for $n=2$:** - **Case 1: Triangular Groups** - Split the group into two sets of 3, each forming a triangle. - Number of ways to choose 3 people from 6: $\binom{6}{3} = 20$. - Each selection results in exactly one way to form the triangles (since choosing 3 determines the other 3). - However, each triangle configuration is counted twice (once for each triangle), so we divide by 2: $\frac{20}{2} = 10$. - **Case 2: Hexagonal Configuration** - Each person is a vertex of a hexagon, with edges representing friendships. - Fix one person, and choose 2 friends from the remaining 5: $\binom{5}{2} = 10$ ways. - The remaining 3 people automatically form the other vertices. - Considering rotations and reflections of the hexagon, each configuration is counted multiple times. Correcting for this, we have $\frac{6!}{12} = 60$ configurations. 6. **Total Configurations:** - For $n=1$ and $n=4$: $15$ configurations each. - For $n=2$ and $n=3$: $10$ (triangles) + $60$ (hexagon) = $70$ configurations each. - Total: $(15 + 70) \times 2 = 170$ configurations. 7. **Conclusion:** - The total number of different ways the friendships can be configured, given the constraints, is $\boxed{\textbf{(B)}\ 170}$.

A graph has $ 30$ vertices, $ 105$ edges and $ 4822$ unordered edge pairs whose endpoints are disjoint. Find the maximal possible difference of degrees of two vertices in this graph.

22

Find $3^{\frac{1}{3}} \cdot 9^{\frac{1}{9}} \cdot 27^{\frac{1}{27}} \cdot 81^{\frac{1}{81}} \dotsm.$

\sqrt[4]{27}

Lyla and Isabelle run on a circular track both starting at point $ P $. Lyla runs at a constant speed in the clockwise direction. Isabelle also runs in the clockwise direction at a constant speed 25% faster than Lyla. Lyla starts running first and Isabelle starts running when Lyla has completed one third of one lap. When Isabelle passes Lyla for the fifth time, how many times has Lyla returned to point $ P $?

17

Ponchik was having a snack at a roadside café when a bus passed by. Three pastries after the bus, a motorcycle passed by Ponchik, and three pastries after that, a car passed by. Syrupchik, who was snacking at another café on the same road, saw them in a different order: first the bus, after three pastries the car, and three pastries after that, the motorcycle. It is known that Ponchik and Syrupchik always eat pastries at a constant speed. Find the speed of the bus if the speed of the car is 60 km/h and the speed of the motorcycle is 30 km/h.

40

What is the slope of a line parallel to the line $2x - 4y = 9$? Express your answer as a common fraction.

\frac{1}{2}

In a polar coordinate system, the polar equation of curve C is $\rho=2\cos\theta+2\sin\theta$. Establish a Cartesian coordinate system with the pole as the origin and the positive x-axis as the polar axis. The parametric equation of line l is $\begin{cases} x=1+t \\ y= \sqrt{3}t \end{cases}$ (t is the parameter). Find the length of the chord that curve C cuts off on line l.

\sqrt{7}

Previously, on an old truck, I traveled from village $A$ through $B$ to village $C$. After five minutes, I asked the driver how far we were from $A$. "Half as far as from $B," was the answer. Expressing my concerns about the slow speed of the truck, the driver assured me that while the truck cannot go faster, it maintains its current speed throughout the entire journey. $13$ km after $B$, I inquired again how far we were from $C$. I received exactly the same response as my initial inquiry. A quarter of an hour later, we arrived at our destination. How many kilometers is the journey from $A$ to $C$?

26

If $13^{3n}=\left(\frac{1}{13}\right)^{n-24}$, find $n$.

6

I have two cents and Bill has $n$ cents. Bill wants to buy some pencils, which come in two different packages. One package of pencils costs 6 cents for 7 pencils, and the other package of pencils costs a dime for a dozen pencils (i.e. 10 cents for 12 pencils). Bill notes that he can spend all $n$ of his cents on some combination of pencil packages to get $P$ pencils. However, if I give my two cents to Bill, he then notes that he can instead spend all $n+2$ of his cents on some combination of pencil packages to get fewer than $P$ pencils. What is the smallest value of $n$ for which this is possible?

100

Suppose that Bill buys $a$ packages of 7 and $b$ packages of 12 in the first scenario and $c$ packages of 7 and $d$ packages of 12 in the second scenario. Then we have the following system: $$ \begin{aligned} & 6 a+10 b=n \\ & 6 c+10 d=n+2 \\ & 7 a+12 b>7 c+12 d \end{aligned} $$ Since the packages of 12 give more pencils per cent, we must have $b>d$. Subtract the first two equations and divide by 2 to get $$ 3(c-a)-5(b-d)=1 $$ Note that the last inequality is $12(b-d)>7(c-a)$. The minimal solutions to the equation with $b-d>0$ are $$ (c-a, b-d)=(2,1),(7,4),(12,7),(17,10) $$ $(17,10)$ is the first pair for which $12(b-d)>7(c-a)$. Hence $b \geq 10$ so $n \geq 100$. We can easily verify that $(a, b, c, d, n)=(0,10,17,0,100)$ satisfies the system of equations.

If $ e^{i \theta} = \frac{3 + i \sqrt{2}}{4}, $ then find $ \cos 3\theta. $

\frac{9}{64}

A jar contains $5$ different colors of gumdrops. $30\%$ are blue, $20\%$ are brown, $15\%$ are red, $10\%$ are yellow, and other $30$ gumdrops are green. If half of the blue gumdrops are replaced with brown gumdrops, how many gumdrops will be brown?

42

1. **Calculate the total number of gumdrops**: Given that $30\%$ are blue, $20\%$ are brown, $15\%$ are red, $10\%$ are yellow, and $30$ gumdrops are green. First, we calculate the percentage of gumdrops that are green: \[ 100\% - (30\% + 20\% + 15\% + 10\%) = 100\% - 75\% = 25\% \] Since $25\%$ of the gumdrops are green and there are $30$ green gumdrops, we can find the total number of gumdrops in the jar by setting up the equation: \[ 0.25 \times \text{Total number of gumdrops} = 30 \] Solving for the total number of gumdrops: \[ \text{Total number of gumdrops} = \frac{30}{0.25} = 120 \] 2. **Calculate the number of blue and brown gumdrops initially**: - Blue gumdrops: $30\%$ of $120$: \[ 0.30 \times 120 = 36 \] - Brown gumdrops: $20\%$ of $120$: \[ 0.20 \times 120 = 24 \] 3. **Replace half of the blue gumdrops with brown gumdrops**: - Half of the blue gumdrops: \[ \frac{36}{2} = 18 \] - New number of brown gumdrops after replacement: \[ 24 + 18 = 42 \] 4. **Conclusion**: After replacing half of the blue gumdrops with brown gumdrops, the number of brown gumdrops in the jar becomes $42$. Therefore, the final answer is $\boxed{\textbf{(C)}\ 42}$.

The area of the parallelogram generated by the vectors $\mathbf{a}$ and $\mathbf{b}$ is 8. Find the area of the parallelogram generated by the vectors $2 \mathbf{a} + 3 \mathbf{b}$ and $\mathbf{a} - 5 \mathbf{b}.$

104

Seven cards numbered $1$ through $7$ are to be lined up in a row. Find the number of arrangements of these seven cards where one of the cards can be removed, leaving the remaining six cards in either ascending or descending order.

74

Find the smallest positive integer $N$ with the following property: of the three numbers $N$, $N+1$, and $N+2$, one of them is divisible by $2^2$, one of them is divisible by $3^2$, one is divisible by $5^2$, and one is divisible by $7^2$.

98

In triangle $ABC$, the sides opposite to angles $A$, $B$, and $C$ are denoted as $a$, $b$, and $c$ respectively. It is known that $c=a\cos B+b\sin A$. (I) Find angle $A$. (II) If $a=2$, find the maximum area of $\triangle ABC$.

\sqrt {2}+1

Determine the minimum possible value of the sum \[\frac{a}{3b} + \frac{b}{5c} + \frac{c}{7a},\] where $a,$ $b,$ and $c$ are positive real numbers.

\frac{3}{\sqrt[3]{105}}

A 3x3x3 cube is made of $27$ normal dice. Each die's opposite sides sum to $7$. What is the smallest possible sum of all of the values visible on the $6$ faces of the large cube?

90

1. **Understanding the Cube Configuration**: - A 3x3x3 cube consists of 27 smaller cubes. - The smaller cubes on the corners have three faces visible. - The smaller cubes on the edges have two faces visible. - The smaller cubes in the center of each face have one face visible. 2. **Counting Visible Faces**: - There are 8 corner cubes, each with 3 faces visible. - There are 12 edge cubes, each with 2 faces visible. - There are 6 face-center cubes, each with 1 face visible. 3. **Calculating Minimum Possible Values on Visible Faces**: - For a single die, the opposite faces sum to 7. The pairs are (1,6), (2,5), and (3,4). - The minimum sum for three visible faces on a corner cube is achieved by showing faces 1, 2, and 3. Thus, the sum is $1+2+3=6$. - The minimum sum for two visible faces on an edge cube is achieved by showing faces 1 and 2. Thus, the sum is $1+2=3$. - The minimum sum for one visible face on a face-center cube is achieved by showing face 1. Thus, the sum is $1$. 4. **Calculating Total Minimum Sum**: - For the 8 corner cubes, the total minimum sum is $8 \times 6 = 48$. - For the 12 edge cubes, the total minimum sum is $12 \times 3 = 36$. - For the 6 face-center cubes, the total minimum sum is $6 \times 1 = 6$. - Adding these sums gives the total minimum sum for all visible faces on the large cube: $48 + 36 + 6 = 90$. 5. **Conclusion**: - The smallest possible sum of all the values visible on the 6 faces of the large cube is $\boxed{\text{(D)}\ 90}$.

Points $A,B,C,D,E$ and $F$ lie, in that order, on $\overline{AF}$, dividing it into five segments, each of length 1. Point $G$ is not on line $AF$. Point $H$ lies on $\overline{GD}$, and point $J$ lies on $\overline{GF}$. The line segments $\overline{HC}, \overline{JE},$ and $\overline{AG}$ are parallel. Find $HC/JE$.

\frac{5}{3}

1. **Identify Key Points and Relationships**: - Points $A, B, C, D, E,$ and $F$ are collinear on line $\overline{AF}$, and each segment between consecutive points is of length 1. - Point $G$ is not on line $AF$, and points $H$ and $J$ lie on lines $\overline{GD}$ and $\overline{GF}$ respectively. - Lines $\overline{HC}, \overline{JE},$ and $\overline{AG}$ are parallel. 2. **Use of Similar Triangles**: - Since $\overline{AG} \parallel \overline{HC}$, triangles $\triangle GAD$ and $\triangle HCD$ are similar by the Basic Proportionality Theorem (or Thales' theorem). - Similarly, since $\overline{AG} \parallel \overline{JE}$, triangles $\triangle GAF$ and $\triangle JEF$ are similar. 3. **Calculate Ratios Using Similar Triangles**: - For triangles $\triangle GAD$ and $\triangle HCD$, the ratio of corresponding sides is: \[ \frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3} \] Here, $CD = 1$ (distance from $C$ to $D$) and $AD = 3$ (total distance from $A$ to $D$ which is the sum of segments $AB$, $BC$, and $CD$). - For triangles $\triangle GAF$ and $\triangle JEF$, the ratio of corresponding sides is: \[ \frac{JE}{AG} = \frac{EF}{AF} = \frac{1}{5} \] Here, $EF = 1$ (distance from $E$ to $F$) and $AF = 5$ (total distance from $A$ to $F$ which is the sum of all five segments). 4. **Find the Desired Ratio $HC/JE$**: - Using the ratios from the similar triangles: \[ \frac{HC}{JE} = \frac{\frac{CH}{AG}}{\frac{JE}{AG}} = \frac{\frac{1}{3}}{\frac{1}{5}} = \frac{1}{3} \times \frac{5}{1} = \frac{5}{3} \] 5. **Conclusion**: - The ratio $\frac{HC}{JE}$ is $\boxed{\frac{5}{3}}$. - The correct answer is $\boxed{(D) \frac{5}{3}}$.

At CMU, markers come in two colors: blue and orange. Zachary fills a hat randomly with three markers such that each color is chosen with equal probability, then Chase shuffles an additional orange marker into the hat. If Zachary chooses one of the markers in the hat at random and it turns out to be orange, the probability that there is a second orange marker in the hat can be expressed as simplified fraction $\tfrac{m}{n}$ . Find $m+n$ .

39

Fill six numbers $1, 3, 5, 7, 9, 11$ into the circles (each circle containing only one number) so that the sum of three numbers on each side of the triangle equals 17. What is the sum of the three numbers in the circles at the vertices of the triangle?

15

Let $\mathcal{S}$ be the set $\lbrace1,2,3,\ldots,10\rbrace$ Let $n$ be the number of sets of two non-empty disjoint subsets of $\mathcal{S}$. (Disjoint sets are defined as sets that have no common elements.) Find the remainder obtained when $n$ is divided by $1000$.

501

Let the two disjoint subsets be $A$ and $B$, and let $C = S-(A+B)$. For each $i \in S$, either $i \in A$, $i \in B$, or $i \in C$. So there are $3^{10}$ ways to organize the elements of $S$ into disjoint $A$, $B$, and $C$. However, there are $2^{10}$ ways to organize the elements of $S$ such that $A = \emptyset$ and $S = B+C$, and there are $2^{10}$ ways to organize the elements of $S$ such that $B = \emptyset$ and $S = A+C$. But, the combination such that $A = B = \emptyset$ and $S = C$ is counted twice. Thus, there are $3^{10}-2\cdot2^{10}+1$ ordered pairs of sets $(A,B)$. But since the question asks for the number of unordered sets $\{ A,B \}$, $n = \frac{1}{2}(3^{10}-2\cdot2^{10}+1) = 28501 \equiv \boxed{501} \pmod{1000}$.

A right circular cylinder with radius 3 is inscribed in a hemisphere with radius 7 so that its bases are parallel to the base of the hemisphere. What is the height of this cylinder?

\sqrt{40}

Find all prime numbers whose representation in a base-14 numeral system has the form 101010...101 (alternating ones and zeros).

197

Given $\cos \alpha = \frac{1}{7}$ and $\cos (\alpha-\beta) = \frac{13}{14}$, with $0 < \beta < \alpha < \frac{\pi}{2}$, find $\beta$.

\frac{\pi}{3}

Given the hyperbola $C\_1$: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 (a > b > 0)$ with left and right foci $F\_1$ and $F\_2$, respectively, and hyperbola $C\_2$: $\frac{x^2}{16} - \frac{y^2}{4} = 1$, determine the length of the major axis of hyperbola $C\_1$ given that point $M$ lies on one of the asymptotes of hyperbola $C\_1$, $OM \perp MF\_2$, and the area of $\triangle OMF\_2$ is $16$.

16

Find the area of the region of the $ xy $-plane defined by the inequality $ |x|+|y|+|x+y| \leq 1 $.

3/4

Given $|\overrightarrow {a}|=\sqrt {2}$, $|\overrightarrow {b}|=2$, and $(\overrightarrow {a}-\overrightarrow {b})\bot \overrightarrow {a}$, determine the angle between $\overrightarrow {a}$ and $\overrightarrow {b}$.

\frac{\pi}{4}

A triangle with vertices $A = (4, 3)$, $B = (6, -2)$, and $C = (7, 1)$ is reflected about the line $x = 6$ to create a second triangle. Determine the area of the union of the two triangles.

10

90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 =

945

We are given the sum of consecutive integers from $90$ to $99$. We can calculate this sum using the formula for the sum of an arithmetic series, or by simplifying the expression directly. 1. **Identify the series**: The series is $90 + 91 + 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99$. 2. **Calculate the number of terms ($n$)**: The series starts at $90$ and ends at $99$. The number of terms in the series is $99 - 90 + 1 = 10$. 3. **Calculate the sum using the formula for the sum of an arithmetic series**: \[ S = \frac{n}{2} \times (\text{first term} + \text{last term}) \] Plugging in the values, we get: \[ S = \frac{10}{2} \times (90 + 99) = 5 \times 189 = 945 \] 4. **Verification by expanding the terms**: \[ (100-10) + (100-9) + \cdots + (100-1) = 100 \times 10 - (1 + 2 + \cdots + 10) \] The sum of the first 10 positive integers is given by: \[ \frac{10 \times (10 + 1)}{2} = 55 \] Therefore, the sum of the series becomes: \[ 1000 - 55 = 945 \] Both methods confirm that the sum of the integers from $90$ to $99$ is $945$. Thus, the correct answer is $\boxed{\text{(B)}~945}$.

In a bus station in the city, there are 10 waiting seats arranged in a row. Now, if 4 passengers randomly choose some seats to wait, the number of ways to arrange them so that there are exactly 5 consecutive empty seats is $\boxed{480}$.

480

A caterpillar is climbing a 20-meter pole. During the day, it climbs 5 meters, and during the night, it slides down 4 meters. How long will it take for the caterpillar to reach the top of the pole?

16

Four fair coins are tossed once. For every head that appears, two six-sided dice are rolled. What is the probability that the sum of all dice rolled is exactly ten? A) $\frac{1} {48}$ B) $\frac{1} {20}$ C) $\frac{1} {16}$ D) $\frac{1} {30}$

\frac{1} {20}

The graph of the function $f(x)=\frac{x}{x+a}$ is symmetric about the point $(1,1)$, and the function $g(x)=\log_{10}(10^x+1)+bx$ is even. Find the value of $a+b$.

-\frac{3}{2}

Given vectors $\overrightarrow{a}=(\sin \theta, 2)$ and $\overrightarrow{b}=(\cos \theta, 1)$, which are collinear, where $\theta \in (0, \frac{\pi}{2})$. 1. Find the value of $\tan (\theta + \frac{\pi}{4})$. 2. If $5\cos (\theta - \phi)=3 \sqrt{5}\cos \phi, 0 < \phi < \frac{\pi}{2}$, find the value of $\phi$.

\frac{\pi}{4}

Below is a portion of the graph of a function, $y=E(x)$: [asy] import graph; size(8cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.5,xmax=4.5,ymin=-0.99,ymax=6.5; pen cqcqcq=rgb(0.75,0.75,0.75); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis("",xmin,xmax,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis("",ymin,ymax,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); real f1(real x){return sqrt(abs(x+1))+(9/pi)*atan(sqrt(abs(x)));} draw(graph(f1,xmin,xmax),linewidth(1)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); label("$y=E(x)$",(xmax+0.25,f1(xmax)),E); [/asy] The value of $E(3)$ is an integer. What is it?

5

Let $u$ and $v$ be real numbers satisfying the inequalities $2u + 3v \le 10$ and $4u + v \le 9.$ Find the largest possible value of $u + 2v$.

6.1

Solve for $x$: $\sqrt[4]{40x + \sqrt[4]{40x + 24}} = 24.$

8293.8

A positive integer $N$ greater than $1$ is described as special if in its base- $8$ and base- $9$ representations, both the leading and ending digit of $N$ are equal to $1$ . What is the smallest special integer in decimal representation? *Proposed by Michael Ren*

793

How many ways are there to put 4 balls in 3 boxes if the balls are not distinguishable but the boxes are?

15

If $\Diamond4_7=\Diamond1_{8}$ and $\Diamond$ represents a digit, solve for $\Diamond$.

3

For any real number $ x $, let $ \lfloor x \rfloor $ denote the greatest integer less than or equal to $ x $, and let $ \{ x \} $ denote the fractional part of $ x $. Then $$ \left\{\frac{2014}{2015}\right\}+\left\{\frac{2014^{2}}{2015}\right\}+\cdots+\left\{\frac{2014^{2014}}{2015}\right\} $$ equals:

1007

A car left the city for the village, and simultaneously, a cyclist left the village for the city. When the car and the cyclist met, the car immediately turned around and went back to the city. As a result, the cyclist arrived in the city 35 minutes later than the car. How many minutes did the cyclist spend on the entire trip, given that his speed is 4.5 times less than the speed of the car?

55

A biased coin lands heads with a probability of $\frac{2}{3}$ and tails with $\frac{1}{3}$. A player can choose between Game C and Game D. In Game C, the player tosses the coin five times and wins if either the first three or the last three outcomes are all the same. In Game D, she tosses the coin five times and wins if at least one of the following conditions is met: the first two tosses are the same and the last two tosses are the same, or the middle three tosses are all the same. A) Game C has a higher probability of $\frac{5}{81}$ over Game D. B) Game D has a higher probability of $\frac{5}{81}$ over Game C. C) Game C has a higher probability of $\frac{29}{81}$ over Game D. D) Game D has a higher probability of $\frac{29}{81}$ over Game C.

\frac{29}{81}

In $\triangle ABC$, the median from vertex $A$ is perpendicular to the median from vertex $B$. The lengths of sides $AC$ and $BC$ are 6 and 7 respectively. What is the length of side $AB$?

$\sqrt{17}$

Four complex numbers lie at the vertices of a square in the complex plane. Three of the numbers are $1+2i, -2+i$, and $-1-2i$. The fourth number is

$2-i$

1. **Identify the given points in the complex plane**: The complex numbers given are $1+2i, -2+i$, and $-1-2i$. We can interpret these as points in the Cartesian plane: $A = (1, 2), B = (-2, 1), C = (-1, -2)$. 2. **Determine the slopes of lines AB and BC**: - The slope of line $AB$ is calculated using the formula $\frac{y_2 - y_1}{x_2 - x_1}$: \[ \text{slope of } AB = \frac{1 - 2}{-2 - 1} = \frac{-1}{-3} = \frac{1}{3} \] - The slope of line $BC$ is: \[ \text{slope of } BC = \frac{-2 - 1}{-1 + 2} = \frac{-3}{1} = -3 \] 3. **Find the equations of the perpendicular bisectors**: - The slope of a line perpendicular to another is the negative reciprocal of the original slope. Thus, the slope of the line perpendicular to $AB$ (passing through $A$) is $-3$, and the slope of the line perpendicular to $BC$ (passing through $C$) is $\frac{1}{3}$. - The equation of the line perpendicular to $AB$ through $A(1,2)$: \[ y - 2 = -3(x - 1) \implies y = -3x + 5 \] - The equation of the line perpendicular to $BC$ through $C(-1,-2)$: \[ y + 2 = \frac{1}{3}(x + 1) \implies y = \frac{1}{3}x - \frac{5}{3} \] 4. **Find the intersection of these perpendicular bisectors**: - Set the equations equal to find $x$: \[ -3x + 5 = \frac{1}{3}x - \frac{5}{3} \] - Solving for $x$: \[ -3x - \frac{1}{3}x = -\frac{5}{3} - 5 \implies -\frac{10}{3}x = -\frac{20}{3} \implies x = 2 \] - Substitute $x = 2$ back into one of the line equations to find $y$: \[ y = -3(2) + 5 = -6 + 5 = -1 \] 5. **Convert the Cartesian coordinates back to the complex number**: - The Cartesian point $(2, -1)$ corresponds to the complex number $2 - i$. Thus, the fourth vertex of the square in the complex plane is $\boxed{2-i}$.

Let $ a $, $ b $, and $ c $ be the roots of the polynomial $ x^3 - 15x^2 + 25x - 10 = 0 $. Compute \[ (a-b)^2 + (b-c)^2 + (c-a)^2. \]

125

Let $R$ be the rectangle in the Cartesian plane with vertices at $(0,0)$ , $(2,0)$ , $(2,1)$ , and $(0,1)$ . $R$ can be divided into two unit squares, as shown. [asy]size(120); defaultpen(linewidth(0.7)); draw(origin--(2,0)--(2,1)--(0,1)--cycle^^(1,0)--(1,1));[/asy] Pro selects a point $P$ at random in the interior of $R$ . Find the probability that the line through $P$ with slope $\frac{1}{2}$ will pass through both unit squares.

3/4

Let the function $ f(x) = x^3 + a x^2 + b x + c $, where $ a $, $ b $, and $ c $ are non-zero integers. If $ f(a) = a^3 $ and $ f(b) = b^3 $, then the value of $ c $ is:

16

Calculate $14 \cdot 31$ and $\left\lfloor\frac{2+\sqrt{2}}{2}\right\rfloor + \left\lfloor\frac{3+\sqrt{3}}{3}\right\rfloor + \left\lfloor\frac{4+\sqrt{4}}{4}\right\rfloor + \cdots + \left\lfloor\frac{1989+\sqrt{1989}}{1989}\right\rfloor + \left\lfloor\frac{1990+\sqrt{1990}}{1990}\right\rfloor$.

1989

Regular hexagon $ABCDEF$ has an area of $n$. Let $m$ be the area of triangle $ACE$. What is $\tfrac{m}{n}?$ A) $\frac{1}{2}$ B) $\frac{2}{3}$ C) $\frac{3}{4}$ D) $\frac{1}{3}$ E) $\frac{3}{2}$

\frac{2}{3}

Billy Bones has two coins — one gold and one silver. One of these coins is fair, while the other is biased. It is unknown which coin is biased but it is known that the biased coin lands heads with a probability of $p=0.6$. Billy Bones tossed the gold coin and it landed heads immediately. Then, he started tossing the silver coin, and it landed heads only on the second toss. Find the probability that the gold coin is the biased one.

5/9

Let $w, x, y$, and $z$ be positive real numbers such that $0 \neq \cos w \cos x \cos y \cos z$, $2 \pi =w+x+y+z$, $3 \tan w =k(1+\sec w)$, $4 \tan x =k(1+\sec x)$, $5 \tan y =k(1+\sec y)$, $6 \tan z =k(1+\sec z)$. Find $k$.

\sqrt{19}

From the identity $\tan \frac{u}{2}=\frac{\sin u}{1+\cos u}$, the conditions work out to $3 \tan \frac{w}{2}=4 \tan \frac{x}{2}=5 \tan \frac{y}{2}=6 \tan \frac{z}{2}=k$. Let $a=\tan \frac{w}{2}, b=\tan \frac{x}{2}, c=\tan \frac{y}{2}$, and $d=\tan \frac{z}{2}$. Using the identity $\tan (M+N)=\frac{\tan M+\tan N}{1-\tan M \tan N}$, we obtain $\tan \left(\frac{w+x}{2}+\frac{y+z}{2}\right) =\frac{\tan \left(\frac{w+x}{2}\right)+\tan \left(\frac{y+z}{2}\right)}{1-\tan \left(\frac{w+x}{2}\right) \tan \left(\frac{y+z}{2}\right)} =\frac{\frac{a+b}{1-a b}+\frac{c+d}{1-c d}}{1-\left(\frac{a+b}{1-a b}\right)\left(\frac{c+d}{1-c d}\right)} =\frac{a+b+c+d-a b c-a b d-b c d-a c d}{1+a b c d-a b-a c-a d-b c-b d-c d}$. Because $x+y+z+w=\pi$, we get that $\tan \left(\frac{x+y+z+w}{2}\right)=0$ and thus $a+b+c+d=a b c+a b d+b c d+a c d$. Substituting $a, b, c, d$ corresponding to the variable $k$, we obtain that $k^{3}-19 k=0$. Therefore, $k$ can be only $0, \sqrt{19},-\sqrt{19}$. However, $k=0$ is impossible as $w, x, y, z$ will all be 0. Also, $k=-\sqrt{19}$ is impossible as $w, x, y, z$ will exceed $\pi$. Therefore, $k=\sqrt{19}$.

If $m$ and $n$ are positive integers with $n > 1$ such that $m^{n} = 2^{25} \times 3^{40}$, what is $m + n$?

209957

Since $m$ and $n$ are positive integers with $n > 1$ and $m^{n} = 2^{25} \times 3^{40}$, then 2 and 3 are prime factors of $m$ (since they are prime factors of $m^{n}$) and must be the only prime factors of $m$ (since if there were other prime factors of $m$, then there would be other prime factors of $m^{n}$). Therefore, $m = 2^{a} \times 3^{b}$ for some positive integers $a$ and $b$ and so $m^{n} = (2^{a} \times 3^{b})^{n} = 2^{an} \times 3^{bn}$. Since $m^{n} = 2^{25} \times 3^{40}$, then we must have $an = 25$ and $bn = 40$. Since $a, b, n$ are positive integers, then $n$ is a common divisor of 25 and 40. Since $n > 1$, then $n = 5$, which means that $a = 5$ and $b = 8$. In this case, $m = 2^{5} \times 3^{8} = 32 \times 6561 = 209952$, which gives $m + n = 209952 + 5 = 209957$.