FlyLee/deepscaler-randomsampled-10k · Datasets at Hugging Face

problem stringlengths 11 4.03k	answer stringlengths 0 157	solution stringlengths 0 11.1k
Given triangle ABC, where sides $a$, $b$, and $c$ correspond to angles A, B, and C respectively, and $a=4$, $\cos{B}=\frac{4}{5}$. (1) If $b=6$, find the value of $\sin{A}$; (2) If the area of triangle ABC, $S=12$, find the values of $b$ and $c$.	2\sqrt{13}
A ball inside a rectangular container of width 7 and height 12 is launched from the lower-left vertex of the container. It first strikes the right side of the container after traveling a distance of $\sqrt{53}$ (and strikes no other sides between its launch and its impact with the right side). How many times does the ball bounce before it returns to a vertex? (The final contact with a vertex does not count as a bounce.)	5	Every segment the ball traverses between bounces takes it 7 units horizontally and 2 units up. Thus, after 5 bounces it has traveled up 10 units, and the final segment traversed takes it directly to the upper right vertex of the rectangle.
Ilya Muromets encounters the three-headed Dragon, Gorynych. Each minute, Ilya chops off one head of the dragon. Let $x$ be the dragon's resilience ($x > 0$). The probability $p_{s}$ that $s$ new heads will grow in place of a chopped-off one ($s=0,1,2$) is given by $\frac{x^{s}}{1+x+x^{2}}$. During the first 10 minutes of the battle, Ilya recorded the number of heads that grew back for each chopped-off one. The vector obtained is: $K=(1,2,2,1,0,2,1,0,1,2)$. Find the value of the dragon's resilience $x$ that maximizes the probability of vector $K$.	\frac{1 + \sqrt{97}}{8}
(Full score for this problem is 12 points) Given $f(x) = e^x - ax - 1$. (1) Find the intervals where $f(x)$ is monotonically increasing. (2) If $f(x)$ is monotonically increasing on the domain $\mathbb{R}$, find the range of possible values for $a$. (3) Does there exist a value of $a$ such that $f(x)$ is monotonically decreasing on $(-\infty, 0]$ and monotonically increasing on $[0, +\infty)$? If so, find the value of $a$; if not, explain why.	a = 1
If the sum of the lengths of the six edges of a trirectangular tetrahedron $PABC$ (i.e., $\angle APB=\angle BPC=\angle CPA=90^o$ ) is $S$ , determine its maximum volume.	\[ \frac{S^3(\sqrt{2}-1)^3}{162} \]	Let the side lengths of $AP$ , $BP$ , and $CP$ be $a$ , $b$ , and $c$ , respectively. Therefore $S=a+b+c+\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}$ . Let the volume of the tetrahedron be $V$ . Therefore $V=\frac{abc}{6}$ . Note that $(a-b)^2\geq 0$ implies $\frac{a^2-2ab+b^2}{2}\geq 0$ , which means $\frac{a^2+b^2}{2}\geq ab$ , which implies $a^2+b^2\geq ab+\frac{a^2+b^2}{2}$ , which means $a^2+b^2\geq \frac{(a+b)^2}{2}$ , which implies $\sqrt{a^2+b^2}\geq \frac{1}{\sqrt{2}} \cdot (a+b)$ . Equality holds only when $a=b$ . Therefore $S\geq a+b+c+\frac{1}{\sqrt{2}} \cdot (a+b)+\frac{1}{\sqrt{2}} \cdot (c+b)+\frac{1}{\sqrt{2}} \cdot (a+c)$ $=(a+b+c)(1+\sqrt{2})$ . $\frac{a+b+c}{3}\geq \sqrt[3]{abc}$ is true from AM-GM, with equality only when $a=b=c$ . So $S\geq (a+b+c)(1+\sqrt{2})\geq 3(1+\sqrt{2})\sqrt[3]{abc}=3(1+\sqrt{2})\sqrt[3]{6V}$ . This means that $\frac{S}{3(1+\sqrt{2})}=\frac{S(\sqrt{2}-1)}{3}\geq \sqrt[3]{6V}$ , or $6V\leq \frac{S^3(\sqrt{2}-1)^3}{27}$ , or $V\leq \frac{S^3(\sqrt{2}-1)^3}{162}$ , with equality only when $a=b=c$ . Therefore the maximum volume is $\frac{S^3(\sqrt{2}-1)^3}{162}$ .
For some positive integers $p$, there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$, right angles at $B$ and $C$, $AB=2$, and $CD=AD$. How many different values of $p<2015$ are possible? $\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$	31
It is known that the only solution to the equation $$ \pi / 4 = \operatorname{arcctg} 2 + \operatorname{arcctg} 5 + \operatorname{arcctg} 13 + \operatorname{arcctg} 34 + \operatorname{arcctg} 89 + \operatorname{arcctg}(x / 14) $$ is a natural number. Find it.	2016
Find the largest positive integer solution of the equation $\left\lfloor\frac{N}{3}\right\rfloor=\left\lfloor\frac{N}{5}\right\rfloor+\left\lfloor\frac{N}{7}\right\rfloor-\left\lfloor\frac{N}{35}\right\rfloor$.	65	For $N$ to be a solution, it is necessary that $\frac{N-2}{3}+\frac{N-34}{35} \leq \frac{N}{5}+\frac{N}{7}$, which simplifies to $N \leq 86$. However, if $N \geq 70$, then $N \leq 59$, contradicting $N \geq 70$. It follows that $N$ must be at most 69. Checking for $N \leq 69$, we find that when $N=65$, the equation holds. Thus the answer is $N=65$.
Let $(a_1, a_2, a_3, \ldots, a_{15})$ be a permutation of $(1, 2, 3, \ldots, 15)$ for which $a_1 > a_2 > a_3 > a_4 > a_5 > a_6 > a_7 \mathrm{\ and \ } a_7 < a_8 < a_9 < a_{10} < a_{11} < a_{12} < a_{13} < a_{14} < a_{15}.$ Find the number of such permutations.	3003
What integer $n$ satisfies $0\le n<{101}$ and $$100n\equiv 72\pmod {101}~?$$	29
The area of triangle \(ABC\) is 1. Points \(B'\), \(C'\), and \(A'\) are placed respectively on the rays \(AB\), \(BC\), and \(CA\) such that: \[ BB' = 2 AB, \quad CC' = 3 BC, \quad AA' = 4 CA. \] Calculate the area of triangle \(A'B'C'\).	39
A line is parameterized by \[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -3 \end{pmatrix}.\]A second line is parameterized by \[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ -9 \end{pmatrix} + u \begin{pmatrix} 4 \\ 2 \end{pmatrix}.\]Find the point where the lines intersect.	\begin{pmatrix} 7 \\ -8 \end{pmatrix}
Solve \[\sqrt{1 + \sqrt{2 + \sqrt{x}}} = \sqrt[3]{1 + \sqrt{x}}.\]	49
Let $b_n$ be the integer obtained by writing the integers from $5$ to $n+4$ from left to right. For example, $b_2 = 567$, and $b_{10} = 567891011121314$. Compute the remainder when $b_{25}$ is divided by $55$ (which is the product of $5$ and $11$ for the application of the Chinese Remainder Theorem).	39
Three numbers, $a_1\,$, $a_2\,$, $a_3\,$, are drawn randomly and without replacement from the set $\{1, 2, 3, \dots, 1000\}\,$. Three other numbers, $b_1\,$, $b_2\,$, $b_3\,$, are then drawn randomly and without replacement from the remaining set of 997 numbers. Let $p\,$ be the probability that, after a suitable rotation, a brick of dimensions $a_1 \times a_2 \times a_3\,$ can be enclosed in a box of dimensions $b_1 \times b_2 \times b_3\,$, with the sides of the brick parallel to the sides of the box. If $p\,$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?	5
Express the quotient $1021_3 \div 11_3$ in base $3$.	22_3
Danielle Bellatrix Robinson is organizing a poker tournament with 9 people. The tournament will have 4 rounds, and in each round the 9 players are split into 3 groups of 3. During the tournament, each player plays every other player exactly once. How many different ways can Danielle divide the 9 people into three groups in each round to satisfy these requirements?	20160	We first split the 9 people up arbitrarily into groups of 3. There are $\frac{\binom{9}{3}\binom{6}{3}\binom{3}{3}}{3!}=280$ ways of doing this. Without loss of generality, label the people 1 through 9 so that the first round groups are $\{1,2,3\},\{4,5,6\}$, and $\{7,8,9\}$. We will use this numbering to count the number of ways DBR can divide the 9 players in rounds 2,3, and 4. In round 2, because players 1,2, and 3 are together in the first round, they must be in separate groups, and likewise for $\{4,5,6\}$ and $\{7,8,9\}$. Disregarding ordering of the three groups in a single round, we will first place 1,2, and 3 into their groups, then count the number of ways to place $\{4,5,6\}$ and $\{7,8,9\}$ in the groups with them. We do this by placing one member from each of $\{4,5,6\}$ and $\{7,8,9\}$ into each group. There are $(3!)^{2}$ ways to do this. Now, because of symmetry, we can use the round 2 grouping $\{1,4,7\},\{2,5,8\},\{3,6,9\}$ to list out the remaining possibilities for round 3 and 4 groupings. Casework shows that there are 2 ways to group the players in the remaining two rounds. We multiply $280 \cdot(3!)^{2} \cdot 2$ to get 20160
Two adjacent faces of a tetrahedron, which are equilateral triangles with side length 1, form a dihedral angle of 60 degrees. The tetrahedron rotates around the common edge of these faces. Find the maximum area of the projection of the rotating tetrahedron on the plane containing the given edge. (12 points)	\frac{\sqrt{3}}{4}
The integer 119 is a multiple of which number?	7	The ones digit of 119 is not even, so 119 is not a multiple of 2. The ones digit of 119 is not 0 or 5, so 119 is not a multiple of 5. Since $120=3 \times 40$, then 119 is 1 less than a multiple of 3 so is not itself a multiple of 3. Since $110=11 \times 10$ and $121=11 \times 11$, then 119 is between two consecutive multiples of 11, so is not itself a multiple of 11. Finally, $119 \div 7=17$, so 119 is a multiple of 7.
A traffic light runs repeatedly through the following cycle: green for 45 seconds, then yellow for 5 seconds, then blue for 10 seconds, and finally red for 40 seconds. Peter picks a random five-second time interval to observe the light. What is the probability that the color changes while he is watching?	0.15
A bug crawls along a number line, starting at $-2$. It crawls to $-6$, then turns around and crawls to $5$. How many units does the bug crawl altogether?	15	1. Calculate the distance from $-2$ to $-6$: The distance on a number line is the absolute difference between the two points. Thus, the distance from $-2$ to $-6$ is: \[ \|-6 - (-2)\| = \|-6 + 2\| = \|-4\| = 4 \text{ units} \] 2. Calculate the distance from $-6$ to $5$: Similarly, the distance from $-6$ to $5$ is: \[ \|5 - (-6)\| = \|5 + 6\| = \|11\| = 11 \text{ units} \] 3. Add the distances to find the total distance crawled: The total distance the bug crawls is the sum of the distances calculated in steps 1 and 2: \[ 4 \text{ units} + 11 \text{ units} = 15 \text{ units} \] Thus, the total distance the bug crawls is $\boxed{\textbf{(E)}\ 15}$.
Find $\sec (-300^\circ).$	2
Compute the following expressions: (1) $2 \sqrt{12} -6 \sqrt{ \frac{1}{3}} + \sqrt{48}$ (2) $(\sqrt{3}-\pi)^{0}-\frac{\sqrt{20}-\sqrt{15}}{\sqrt{5}}+(-1)^{2017}$	\sqrt{3} - 2
A triangle has vertices $(0,0)$, $(1,1)$, and $(6m,0)$. The line $y = mx$ divides the triangle into two triangles of equal area. What is the sum of all possible values of $m$?	- \frac {1}{6}	1. Identify the vertices and the line dividing the triangle: Let's label the vertices of the triangle as $A = (0,0)$, $B = (1,1)$, and $C = (6m,0)$. The line $y = mx$ is supposed to divide the triangle into two triangles of equal area. 2. Determine the intersection point: The line $y = mx$ will intersect the line segment $BC$. Let $D$ be the point of intersection. The coordinates of $D$ can be found by setting the $y$-coordinates of the line $y = mx$ and the line segment $BC$ equal. The line segment $BC$ can be represented by the equation $y = -\frac{1}{6m-1}(x-1)$ (using point-slope form and the points $B$ and $C$). Setting $mx = -\frac{1}{6m-1}(x-1)$ and solving for $x$, we find the $x$-coordinate of $D$. 3. Condition for equal areas: The triangles $ABD$ and $ACD$ will have equal areas if $D$ is the midpoint of $BC$. The midpoint of $BC$ is given by $\left(\frac{6m+1}{2}, \frac{1}{2}\right)$. For $D$ to be on the line $y = mx$, the coordinates must satisfy $y = mx$. Substituting the midpoint coordinates into the line equation, we get: \[ \frac{1}{2} = m \cdot \frac{6m+1}{2} \] Simplifying this equation: \[ 1 = m(6m+1) \implies 6m^2 + m - 1 = 0 \] 4. Solve the quadratic equation: The quadratic equation $6m^2 + m - 1 = 0$ can be solved using the quadratic formula: \[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \quad \text{where } a = 6, b = 1, c = -1 \] \[ m = \frac{-1 \pm \sqrt{1 + 24}}{12} = \frac{-1 \pm 5}{12} \] \[ m = \frac{4}{12} = \frac{1}{3}, \quad m = \frac{-6}{12} = -\frac{1}{2} \] 5. Sum of all possible values of $m$: Using Vieta's formulas, the sum of the roots of the equation $6m^2 + m - 1 = 0$ is given by $-\frac{b}{a} = -\frac{1}{6}$. Thus, the sum of all possible values of $m$ is $\boxed{\textbf{(B)} - \!\frac {1}{6}}$.
Huahua is writing letters to Yuanyuan with a pen. When she finishes the 3rd pen refill, she is working on the 4th letter; when she finishes the 5th letter, the 4th pen refill is not yet used up. If Huahua uses the same amount of ink for each letter, how many pen refills does she need to write 16 letters?	13
Students guess that Norb's age is $24, 28, 30, 32, 36, 38, 41, 44, 47$, and $49$. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?	37	1. Identify the conditions given by Norb: - At least half of the guesses are too low. - Two guesses are off by one. - Norb's age is a prime number. 2. List of guesses: $24, 28, 30, 32, 36, 38, 41, 44, 47, 49$. 3. Determine possible values for Norb's age ($x$) based on guesses off by one: - If two guesses are off by one, then $x-1$ and $x+1$ must both be in the list. - Possible pairs $(x-1, x+1)$ from the list are: - $(28, 30)$, hence $x = 29$ - $(30, 32)$, hence $x = 31$ - $(36, 38)$, hence $x = 37$ - $(47, 49)$, hence $x = 48$ 4. Check which of these values are prime numbers: - $29$ is prime. - $31$ is prime. - $37$ is prime. - $48$ is not prime. So, possible values for $x$ are $29$, $31$, and $37$. 5. Check the condition that at least half of the guesses are too low: - If $x = 29$, the guesses too low are $24, 28$. This is not half. - If $x = 31$, the guesses too low are $24, 28, 30$. This is not half. - If $x = 37$, the guesses too low are $24, 28, 30, 32, 36$. This is exactly half of the total guesses (5 out of 10). 6. Conclusion: - The only value that satisfies all conditions (prime number, two guesses off by one, and at least half of the guesses are too low) is $37$. Thus, Norb's age is $\boxed{\textbf{(C)}\ 37}$.
Tim's quiz scores were 85, 87, 92, 94, 78, and 96. Calculate his mean score and find the range of his scores.	18
Find the only positive real number $x$ for which $\displaystyle \frac{x-4}{9} = \frac{4}{x-9}$.	13
Evaluate the product \[\left(\sqrt{5}+\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{6}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).\]	104	We group the first and last factors as well as the two middle factors, then apply the difference of squares repeatedly: \begin{align} \left(\left(\sqrt{6} + \sqrt{7}\right)^2 - \sqrt{5}^2\right)\left(\sqrt{5}^2 - \left(\sqrt{6} - \sqrt{7}\right)^2\right) &= \left(13 + 2\sqrt{42} - 5\right)\left(5 - \left(13 - 2\sqrt{42}\right)\right) \\ &= \left(2\sqrt{42} + 8\right)\left(2\sqrt{42} - 8\right) \\ &= \left(2\sqrt{42}\right)^2 - 8^2 \\ &= \boxed{104}. \end{align} ~Azjps (Solution) ~MRENTHUSIASM (Revision)
Find the smallest positive integer $n$ such that the polynomial $(x+1)^{n}-1$ is "divisible by $x^{2}+1$ modulo 3", or more precisely, either of the following equivalent conditions holds: there exist polynomials $P, Q$ with integer coefficients such that $(x+1)^{n}-1=\left(x^{2}+1\right) P(x)+3 Q(x)$; or more conceptually, the remainder when (the polynomial) $(x+1)^{n}-1$ is divided by (the polynomial) $x^{2}+1$ is a polynomial with (integer) coefficients all divisible by 3.	8	We have $(x+1)^{2}=x^{2}+2 x+1 \equiv 2 x,(x+1)^{4} \equiv(2 x)^{2} \equiv-4 \equiv-1$, and $(x+1)^{8} \equiv(-1)^{2}=1$. So the order $n$ divides 8, as $x+1$ and $x^{2}+1$ are relatively prime polynomials modulo 3 (or more conceptually, in $\mathbb{F}_{3}[x]$ ), but cannot be smaller by our computations of the 2 nd and 4 th powers.
The distances between the points on a line are given as $2, 4, 5, 7, 8, k, 13, 15, 17, 19$. Determine the value of $k$.	12
The ship decided to determine the depth of the ocean at its location. The signal sent by the echo sounder was received on the ship 8 seconds later. The speed of sound in water is 1.5 km/s. Determine the depth of the ocean.	6000
Given the numbers 1, 3, 5 and 2, 4, 6, calculate the total number of different three-digit numbers that can be formed when arranging these numbers on three cards.	48
In triangle $\triangle ABC$, $A=60^{\circ}$, $a=\sqrt{6}$, $b=2$. $(1)$ Find $\angle B$; $(2)$ Find the area of $\triangle ABC$.	\frac{3 + \sqrt{3}}{2}
Quadrilateral $PQRS$ is a square. A circle with center $S$ has arc $PXC$. A circle with center $R$ has arc $PYC$. If $PQ = 3$ cm, what is the total number of square centimeters in the football-shaped area of regions II and III combined? Express your answer as a decimal to the nearest tenth.	5.1
The equation \( \sin^2 x + \sin^2 3x + \sin^2 5x + \sin^2 6x = \frac{81}{32} \) can be reduced to the equivalent equation \[ \cos ax \cos bx \cos cx = 0, \] for some positive integers \( a, \) \( b, \) and \( c. \) Find \( a + b + c. \)	14
Suppose $X$ and $Y$ are digits in base $d > 8$ such that $\overline{XY}_d + \overline{XX}_d = 234_d$. Find $X_d - Y_d$ in base $d$.	-2
Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio $3: 2: 1,$what is the least possible total for the number of bananas?	408	Let $A,B,C$ be the fraction of bananas taken by the first, second, and third monkeys respectively. Then we have the system of equations \[\frac{3}{4}A+\frac{3}{8}B+\frac{11}{24}C=\frac{1}{2}\] \[\frac{1}{8}A+\frac{1}{4}B+\frac{11}{24}C=\frac{1}{3}\] \[\frac{1}{8}A+\frac{3}{8}B+\frac{2}{24}C=\frac{1}{6}.\] Solve this your favorite way to get that \[(A,B,C)=\left( \frac{11}{51}, \frac{13}{51}, \frac{9}{17} \right).\] We need the amount taken by the first and second monkeys to be divisible by 8 and the third by 24 (but for the third, we already have divisibility by 9). Thus our minimum is $8 \cdot 51 = \boxed{408}.$ ~Dhillonr25
Denote by $\mathbb{Z}^2$ the set of all points $(x,y)$ in the plane with integer coordinates. For each integer $n \geq 0$, let $P_n$ be the subset of $\mathbb{Z}^2$ consisting of the point $(0,0)$ together with all points $(x,y)$ such that $x^2 + y^2 = 2^k$ for some integer $k \leq n$. Determine, as a function of $n$, the number of four-point subsets of $P_n$ whose elements are the vertices of a square.	5n+1	The answer is $5n+1$. We first determine the set $P_n$. Let $Q_n$ be the set of points in $\mathbb{Z}^2$ of the form $(0, \pm 2^k)$ or $(\pm 2^k, 0)$ for some $k \leq n$. Let $R_n$ be the set of points in $\mathbb{Z}^2$ of the form $(\pm 2^k, \pm 2^k)$ for some $k \leq n$ (the two signs being chosen independently). We prove by induction on $n$ that \[ P_n = \{(0,0)\} \cup Q_{\lfloor n/2 \rfloor} \cup R_{\lfloor (n-1)/2 \rfloor}. \] We take as base cases the straightforward computations \begin{align} P_0 &= \{(0,0), (\pm 1, 0), (0, \pm 1)\} \\ P_1 &= P_0 \cup \{(\pm 1, \pm 1)\}. \end{align} For $n \geq 2$, it is clear that $\{(0,0)\} \cup Q_{\lfloor n/2 \rfloor} \cup R_{\lfloor (n-1)/2 \rfloor} \subseteq P_n$, so it remains to prove the reverse inclusion. For $(x,y) \in P_n$, note that $x^2 + y^2 \equiv 0 \pmod{4}$; since every perfect square is congruent to either 0 or 1 modulo 4, $x$ and $y$ must both be even. Consequently, $(x/2, y/2) \in P_{n-2}$, so we may appeal to the induction hypothesis to conclude. We next identify all of the squares with vertices in $P_n$. In the following discussion, let $(a,b)$ and $(c,d)$ be two opposite vertices of a square, so that the other two vertices are \[ \left( \frac{a-b+c+d}{2}, \frac{a+b-c+d}{2} \right) \] and \[ \left( \frac{a+b+c-d}{2}, \frac{-a+b+c+d}{2} \right). \] \begin{itemize} \item Suppose that $(a,b) = (0,0)$. Then $(c,d)$ may be any element of $P_n$ not contained in $P_0$. The number of such squares is $4n$. \item Suppose that $(a,b), (c,d) \in Q_k$ for some $k$. There is one such square with vertices \[ \{(0, 2^k), (0, 2^{-k}), (2^k, 0), (2^{-k}, 0)\} \] for $k = 0,\dots,\lfloor \frac{n}{2} \rfloor$, for a total of $\lfloor \frac{n}{2} \rfloor + 1$. To show that there are no others, by symmetry it suffices to rule out the existence of a square with opposite vertices $(a,0)$ and $(c,0)$ where $a > \left\| c \right\|$. The other two vertices of this square would be $((a+c)/2, (a-c)/2)$ and $((a+c)/2, (-a+c)/2)$. These cannot belong to any $Q_k$, or be equal to $(0,0)$, because $\|a+c\|, \|a-c\| \geq a - \|c\| > 0$ by the triangle inequality. These also cannot belong to any $R_k$ because $(a + \|c\|)/2 > (a - \|c\|)/2$. (One can also phrase this argument in geometric terms.) \item Suppose that $(a,b), (c,d) \in R_k$ for some $k$. There is one such square with vertices \[ \{(2^k, 2^k), (2^k, -2^k), (-2^k, 2^k), (-2^k, -2^k)\} \] for $k=0,\dots, \lfloor \frac{n-1}{2} \rfloor$, for a total of $\lfloor \frac{n+1}{2} \rfloor$. To show that there are no others, we may reduce to the previous case: rotating by an angle of $\frac{\pi}{4}$ and then rescaling by a factor of $\sqrt{2}$ would yield a square with two opposite vertices in some $Q_k$ not centered at $(0,0)$, which we have already ruled out. \item It remains to show that we cannot have $(a,b) \in Q_k$ and $(c,d) \in R_k$ for some $k$. By symmetry, we may reduce to the case where $(a,b) = (0, 2^k)$ and $(c,d) = (2^\ell, \pm 2^\ell)$. If $d>0$, then the third vertex $(2^{k-1}, 2^{k-1} + 2^\ell)$ is impossible. If $d<0$, then the third vertex $(-2^{k-1}, 2^{k-1} - 2^\ell)$ is impossible. \end{itemize} Summing up, we obtain \[ 4n + \left\lfloor \frac{n}{2} \right\rfloor + 1 + \left\lfloor \frac{n+1}{2} \right\rfloor = 5n+1 \] squares, proving the claim.
Seven students count from 1 to 1000 as follows: Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000. Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers. Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers. Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers. Finally, George says the only number that no one else says. What number does George say?	365	1. Alice's Numbers: Alice says all numbers except those that are the middle number in each consecutive group of three numbers. This means Alice skips numbers of the form $3n - 1$ for $n = 1, 2, 3, \ldots, 333$. The numbers Alice skips are: \[ 2, 5, 8, \ldots, 998 \] which are calculated as: \[ 3 \times 1 - 1, 3 \times 2 - 1, 3 \times 3 - 1, \ldots, 3 \times 333 - 1 \] 2. Barbara's Numbers: Barbara says the numbers that Alice doesn't say, but she also skips the middle number in each consecutive group of three numbers among these. The numbers Barbara skips are those of the form $3(3n - 1) - 1$ for $n = 1, 2, 3, \ldots$. Calculating a few terms: \[ 3(3 \times 1 - 1) - 1 = 5, \quad 3(3 \times 2 - 1) - 1 = 14, \quad 3(3 \times 3 - 1) - 1 = 23, \ldots \] 3. Candice's Numbers: Candice says the numbers that neither Alice nor Barbara says, but also skips the middle number in each consecutive group of three numbers. The numbers Candice skips are those of the form $3(3(3n - 1) - 1) - 1$ for $n = 1, 2, 3, \ldots$. Calculating a few terms: \[ 3(3(3 \times 1 - 1) - 1) - 1 = 14, \quad 3(3(3 \times 2 - 1) - 1) - 1 = 41, \ldots \] 4. Continuing the Pattern: Following the same pattern for Debbie, Eliza, and Fatima, each skips numbers of the form $3(\text{previous pattern}) - 1$. 5. George's Number: George says the only number that no one else says. Following the pattern to George, the first number he says is calculated as: \[ 3(3(3(3(3(3 \times 1 - 1) - 1) - 1) - 1) - 1) - 1 = 365 \] Checking the next number George would say: \[ 3(3(3(3(3(3 \times 2 - 1) - 1) - 1) - 1) - 1) - 1 > 1000 \] Thus, the only number George says within the range 1 to 1000 is 365. $\boxed{\textbf{(C)}\ 365}$
Let \( a_{n} = 1 + 2 + \cdots + n \), where \( n \in \mathbf{N}_{+} \), and \( S_{m} = a_{1} + a_{2} + \cdots + a_{m} \), \( m = 1, 2, \cdots, m \). Find the number of values among \( S_{1}, S_{2}, \cdots, S_{2017} \) that are divisible by 2 but not by 4.	252
If \(\frac{1}{4} + 4\left(\frac{1}{2013} + \frac{1}{x}\right) = \frac{7}{4}\), find the value of \(1872 + 48 \times \left(\frac{2013 x}{x + 2013}\right)\).	2000
The natural number \(a\) is divisible by 35 and has 75 different divisors, including 1 and \(a\). Find the smallest such \(a\).	490000
How many three-digit positive integers $x$ satisfy $3874x+481\equiv 1205 \pmod{23}$?	40
(In the preliminaries of optimal method and experimental design) When using the 0.618 method to find the optimal amount to add in an experiment, if the current range of excellence is $[628, 774]$ and the good point is 718, then the value of the addition point for the current experiment is ________.	684
Remove all perfect squares from the sequence of positive integers \(1, 2, 3, \cdots\) to get a new sequence. What is the 2003rd term of this new sequence?	2048
Simplify $\frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{4}}{\sqrt{5}} \cdot \frac{\sqrt{6}}{\sqrt{7}}$ and rationalize the denominator of the resulting fraction.	\frac{4\sqrt{35}}{35}
Given $\triangle ABC$, where the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively, and it satisfies $a\cos 2C+2c\cos A\cos C+a+b=0$. $(1)$ Find the size of angle $C$; $(2)$ If $b=4\sin B$, find the maximum value of the area $S$ of $\triangle ABC$.	\sqrt{3}
If a die is rolled, event \( A = \{1, 2, 3\} \) consists of rolling one of the faces 1, 2, or 3. Similarly, event \( B = \{1, 2, 4\} \) consists of rolling one of the faces 1, 2, or 4. The die is rolled 10 times. It is known that event \( A \) occurred exactly 6 times. a) Find the probability that under this condition, event \( B \) did not occur at all. b) Find the expected value of the random variable \( X \), which represents the number of occurrences of event \( B \).	\frac{16}{3}
The graph of the function $y=\sin(\omega x+ \frac {5\pi}{6})$ where $0<\omega<\pi$ intersects with the coordinate axes at points closest to the origin, which are $(0, \frac {1}{2})$ and $( \frac {1}{2}, 0)$. Determine the axis of symmetry of this graph closest to the y-axis.	-1
Let \( p, q, r, s \) be distinct real numbers such that the roots of \( x^2 - 12px - 13q = 0 \) are \( r \) and \( s \), and the roots of \( x^2 - 12rx - 13s = 0 \) are \( p \) and \( q \). Find the value of \( p + q + r + s \).	-13
Given two lines $l_{1}$: $(a-1)x+2y+1=0$, $l_{2}$: $x+ay+1=0$, find the value of $a$ that satisfies the following conditions: $(1) l_{1} \parallel l_{2}$ $(2) l_{1} \perp l_{2}$	\frac{1}{3}
How many three-digit whole numbers have at least one 7 or at least one 9 as digits?	452
The Bank of Springfield's Super High Yield savings account compounds annually at a rate of one percent. If Lisa invests 1000 dollars in one of these accounts, then how much interest will she earn after five years? (Give your answer to the nearest dollar.)	51
In square $ABCD$, points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA}$, respectively, so that $AE=AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD}$, respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$. See the figure below. Triangle $AEH$, quadrilateral $BFIE$, quadrilateral $DHJG$, and pentagon $FCGJI$ each has area $1.$ What is $FI^2$? [asy] real x=2sqrt(2); real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); real z=2sqrt(8-4sqrt(2)); pair A, B, C, D, E, F, G, H, I, J; A = (0,0); B = (4,0); C = (4,4); D = (0,4); E = (x,0); F = (4,y); G = (y,4); H = (0,x); I = F + z * dir(225); J = G + z * dir(225); draw(A--B--C--D--A); draw(H--E); draw(J--G^^F--I); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$E$", E, S); dot("$F$", F, dir(0)); dot("$G$", G, N); dot("$H$", H, W); dot("$I$", I, SW); dot("$J$", J, SW); [/asy] $\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2$	8-4\sqrt{2}
There are 100 positive integers written on a board. At each step, Alex composes 50 fractions using each number written on the board exactly once, brings these fractions to their irreducible form, and then replaces the 100 numbers on the board with the new numerators and denominators to create 100 new numbers. Find the smallest positive integer $n{}$ such that regardless of the values of the initial 100 numbers, after $n{}$ steps Alex can arrange to have on the board only pairwise coprime numbers.	99	To solve this problem, we aim to find the smallest positive integer \( n \) such that after \( n \) steps, the 100 numbers on the board are all pairwise coprime regardless of their initial values. ### Key Observations 1. Irreducible Fractions: At each step, Alex forms 50 fractions out of the 100 numbers. Each fraction \(\frac{a}{b}\) is reduced to its irreducible form \(\frac{p}{q}\), where \(\gcd(p, q) = 1\). 2. Numerators and Denominators: The new set of numbers on the board after each step are the numerators and denominators of these 50 irreducible fractions. 3. Pairwise Coprimeness: For numbers to be pairwise coprime, each pair of numbers has a greatest common divisor of 1. ### Strategy - Step Progression: As we progress with the steps, fractions are reduced to irreducible form, potentially introducing many coprime pairs. However, we need them to all become pairwise coprime eventually. - Minimizing the Steps: To get the numbers pairwise coprime, consider the worst-case scenario: Starting with 100 numbers where no two numbers are coprime. ### Execution Analyzing each pair of numbers: - Each step incorporates forming pairs that guarantee at least one pair becomes coprime. By the nature of reduction to irreducible form, this iteration slowly increases the number of coprime pairs among the set. - After the first step, observe that some coprime pairs necessarily occur due to the fraction reduction process. - Due to the properties of the Euclidean algorithm, composed during the fraction process, this coprime nature spreads as the steps progress. - After 99 steps, according to the pigeonhole principle and the iterative application of number theory principles, all numbers can be arranged to be pairwise coprime. Thus, regardless of initial values, the minimum number of steps required to achieve pairwise coprimeness among the 100 numbers is: \[ \boxed{99} \]
The figure may be folded along the lines shown to form a number cube. Three number faces come together at each corner of the cube. What is the largest sum of three numbers whose faces come together at a corner?	14	To solve this problem, we need to understand how a cube is formed and how the numbers on the faces are arranged. In a standard die, opposite faces sum up to 7. This means: - If one face shows 1, the opposite face shows 6. - If one face shows 2, the opposite face shows 5. - If one face shows 3, the opposite face shows 4. Given this arrangement, no two of the highest numbers (6, 5, 4) can be adjacent because they are on opposite faces. Therefore, we need to find the highest possible sum of three numbers that can meet at a vertex. 1. Check the possibility of the sum of 6, 5, and 4: - These numbers are on opposite faces, so they cannot meet at a vertex. 2. Check the possibility of the sum of 6, 5, and 3: - Since 6 and 5 are on opposite faces, they cannot be adjacent, but 6 and 3 can be adjacent, and 5 and 3 can also be adjacent. - The numbers 6, 5, and 3 can meet at a vertex because they are not on directly opposite faces. 3. Calculate the sum of 6, 5, and 3: \[ 6 + 5 + 3 = 14 \] 4. Verify if there is a higher possible sum: - The next highest possible combinations would involve numbers less than 3 since 4 is opposite to 3 and cannot be at the same vertex with 6 or 5. - Any combination involving numbers less than 3 will result in a sum less than 14. Since no other combination of three numbers that can meet at a vertex exceeds the sum of 14, the largest sum of three numbers whose faces come together at a corner of the cube is 14. Thus, the answer is $\boxed{\text{D}}$.
If $n$ is $1$ less than a multiple of $50$, then what is the remainder when $n^2+2n+3$ is divided by $50$?	2
A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are $6.1$ cm, $8.2$ cm and $9.7$ cm. What is the area of the square in square centimeters?	36
Given a number $\overline{abcd}$ , where $a$ , $b$ , $c$ , and $d$ , represent the digits of $\overline{abcd}$ , find the minimum value of \[\frac{\overline{abcd}}{a+b+c+d}\] where $a$ , $b$ , $c$ , and $d$ are distinct <details><summary>Answer</summary>$\overline{abcd}=1089$ , minimum value of $\dfrac{\overline{abcd}}{a+b+c+d}=60.5$</details>	60.5
If $x+y=\frac{7}{13}$ and $x-y=\frac{1}{91}$, what is the value of $x^2-y^2$? Express your answer as a common fraction.	\frac{1}{169}
One of the following four-digit numbers is not divisible by 4: 3544, 3554, 3564, 3572, 3576. What is the product of the units digit and the tens digit of that number?	20
The average of $x+6$, $6x+2$, and $2x+7$ is $4x-7$. What is $x$?	12
The nine horizontal and nine vertical lines on an $8\times8$ checkerboard form $r$ rectangles, of which $s$ are squares. The number $s/r$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$	125	To determine the two horizontal sides of a rectangle, we have to pick two of the horizontal lines of the checkerboard, or ${9\choose 2} = 36$. Similarly, there are ${9\choose 2}$ ways to pick the vertical sides, giving us $r = 1296$ rectangles. For $s$, there are $8^2$ unit squares, $7^2$ of the $2\times2$ squares, and so on until $1^2$ of the $8\times 8$ squares. Using the sum of squares formula, that gives us $s=1^2+2^2+\cdots+8^2=\dfrac{(8)(8+1)(2\cdot8+1)}{6}=12*17=204$. Thus $\frac sr = \dfrac{204}{1296}=\dfrac{17}{108}$, and $m+n=\boxed{125}$.
Yan is somewhere between his office and a concert hall. To get to the concert hall, he can either walk directly there, or walk to his office and then take a scooter to the concert hall. He rides 5 times as fast as he walks, and both choices take the same amount of time. What is the ratio of Yan's distance from his office to his distance from the concert hall?	\frac{2}{3}
Ana and Luíza train every day for the Big Race that will take place at the end of the year at school, each running at the same speed. The training starts at point $A$ and ends at point $B$, which are $3000 \mathrm{~m}$ apart. They start at the same time, but when Luíza finishes the race, Ana still has $120 \mathrm{~m}$ to reach point $B$. Yesterday, Luíza gave Ana a chance: "We will start at the same time, but I will start some meters before point $A$ so that we arrive together." How many meters before point $A$ should Luíza start?	125
In triangle $ABC$, $AB = 3$, $BC = 4$, $AC = 5$, and $BD$ is the angle bisector from vertex $B$. If $BD = k \sqrt{2}$, then find $k$.	\frac{12}{7}
Given $2^{3x} = 128$, calculate the value of $2^{-x}$.	\frac{1}{2^{\frac{7}{3}}}
Triangle $\triangle ABC$ in the figure has area $10$. Points $D, E$ and $F$, all distinct from $A, B$ and $C$, are on sides $AB, BC$ and $CA$ respectively, and $AD = 2, DB = 3$. If triangle $\triangle ABE$ and quadrilateral $DBEF$ have equal areas, then that area is	6	1. Identify Key Relationships: Given that $AD = 2$, $DB = 3$, and $[ABE] = [DBEF]$, where $[X]$ denotes the area of figure $X$. 2. Area of $\triangle ABC$: The area of $\triangle ABC$ is given as $10$. 3. Ratio of Segments on $AB$: Since $AD = 2$ and $DB = 3$, the total length of $AB = AD + DB = 2 + 3 = 5$. Thus, $AD:DB = 2:3$. 4. Area of $\triangle ABE$ Using Similarity: Since $D$ and $E$ are points on $AB$ and $BC$ respectively, and $[ABE] = [DBEF]$, we consider the implications of the areas being equal. Let $G$ be the intersection of $AE$ and $DF$. We have: \[ [DBEF] = [DBEG] + [EFG] \quad \text{and} \quad [ABE] = [ABEG] + [ADG]. \] Since $[DBEF] = [ABE]$, it follows that $[DBEG] + [EFG] = [ABEG] + [ADG]$. Given $[DBEG] = [ABEG]$, we deduce $[EFG] = [ADG]$. 5. Implication of Equal Areas: The equality $[EFG] = [ADG]$ implies that $[ADF] = [AFE]$ because $[ADF] = [ADG] + [AGF]$ and $[AFE] = [EFG] + [AGF]$. This equality suggests that $AF \parallel DE$. 6. Similarity of Triangles: Since $AF \parallel DE$, $\triangle DBE \sim \triangle ABC$. The ratio of similarity is the same as the ratio of the corresponding sides on $AB$, which is $\frac{BD}{BA} = \frac{3}{5}$. 7. Area of $\triangle ABE$: Since $\triangle ABE$ is similar to $\triangle ABC$ with a ratio of $\frac{3}{5}$, the area of $\triangle ABE$ is: \[ [ABE] = \left(\frac{3}{5}\right)^2 \times [ABC] = \frac{9}{25} \times 10 = 3.6. \] However, this calculation seems incorrect as it contradicts the given areas are equal and the choice options. We need to re-evaluate the area calculation considering the linear ratio for height, not the square of the ratio (since area scales with the square of the linear dimensions, but here we consider the height directly proportional, not the square). Correcting the area calculation: \[ [ABE] = \frac{3}{5} \times [ABC] = \frac{3}{5} \times 10 = 6. \] 8. Conclusion: The area of $\triangle ABE$ (and hence $[DBEF]$) is $\boxed{6}$. $\blacksquare$
In the final stage of a professional bowling tournament, the competition between the top five players is conducted as follows: the fifth and fourth place players compete first, the loser gets the 5th place prize; the winner competes with the third place player, the loser gets the 4th place prize; the winner competes with the second place player, the loser gets the 3rd place prize; the winner competes with the first place player, the loser gets the 2nd place prize, and the winner gets the 1st place prize. How many different possible outcomes of the prize distribution are there?	16
The $\textit{arithmetic derivative}$ $D(n)$ of a positive integer $n$ is defined via the following rules: - $D(1) = 0$ ; - $D(p)=1$ for all primes $p$ ; - $D(ab)=D(a)b+aD(b)$ for all positive integers $a$ and $b$ . Find the sum of all positive integers $n$ below $1000$ satisfying $D(n)=n$ .	31
Let point $P$ be a point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 (a > b > 0)$. Let $F_1$ and $F_2$ respectively be the left and right foci of the ellipse, and let $I$ be the incenter of $\triangle PF_1F_2$. If $S_{\triangle IPF_1} + S_{\triangle IPF_2} = 2S_{\triangle IF_1F_2}$, then the eccentricity of the ellipse is ______.	\frac{1}{2}
Given $f(x) = x^2$ and $g(x) = \|x - 1\|$, let $f_1(x) = g(f(x))$, $f_{n+1}(x) = g(f_n(x))$, calculate the number of solutions to the equation $f_{2015}(x) = 1$.	2017
Pirate Bob shares his treasure with Pirate Sam in a peculiar manner. Bob first declares, ``One for me, one for you,'' keeping one coin for himself and starting Sam's pile with one coin. Then Bob says, ``Two for me, and two for you,'' adding two more coins to his pile but updating Sam's total to two coins. This continues until Bob says, ``$x$ for me, $x$ for you,'' at which he takes $x$ more coins and makes Sam's total $x$ coins in total. After all coins are distributed, Pirate Bob has exactly three times as many coins as Pirate Sam. Find out how many gold coins they have between them?	20
There are six clearly distinguishable frogs sitting in a row. Two are green, three are red, and one is blue. Green frogs refuse to sit next to the red frogs, for they are highly poisonous. In how many ways can the frogs be arranged?	24
Let $x,$ $y,$ and $z$ be positive real numbers such that $x + y + z = 1.$ Find the maximum value of $x^3 y^2 z.$	\frac{1}{432}
Compute: $9 \cdot \frac{1}{13} \cdot 26.$	18
Given that a square $S_1$ has an area of $25$, the area of the square $S_3$ constructed by bisecting the sides of $S_2$ is formed by the points of bisection of $S_2$.	6.25
On the lateral side \( CD \) of the trapezoid \( ABCD (AD \parallel BC) \), point \( M \) is marked. From vertex \( A \), a perpendicular \( AH \) is dropped onto segment \( BM \). It turns out that \( AD = HD \). Find the length of segment \( AD \) if it is known that \( BC = 16 \), \( CM = 8 \), and \( MD = 9 \).	18
Consider the case when all numbers are equal. $\frac{5}{4} n + \frac{5}{4} = n$. If the first number is -5, then all numbers will be equal to -5. The same applies to all cases where the first number is equal to $-5 + 1024n$, $n \in \mathbb{Z}$.	-5
Big Al, the ape, ate 100 bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many bananas did Big Al eat on May 5?	32	1. Identify the sequence type: Big Al's banana consumption forms an arithmetic sequence because he eats six more bananas each day than the previous day. 2. Define the terms of the sequence: Let $a$ be the number of bananas Big Al ate on May 1. Then, the number of bananas he ate on subsequent days can be expressed as: - May 2: $a + 6$ - May 3: $a + 12$ - May 4: $a + 18$ - May 5: $a + 24$ 3. Set up the equation for the sum of the sequence: The sum of an arithmetic sequence can be calculated using the formula: \[ S = \frac{n}{2} \times (\text{first term} + \text{last term}) \] where $n$ is the number of terms. Here, $n = 5$, the first term is $a$, and the last term is $a + 24$. The total number of bananas eaten over the five days is given as 100. Therefore, we have: \[ \frac{5}{2} \times (a + (a + 24)) = 100 \] 4. Simplify and solve for $a$: \[ \frac{5}{2} \times (2a + 24) = 100 \] \[ 5a + 60 = 100 \] \[ 5a = 40 \] \[ a = 8 \] 5. Calculate the number of bananas eaten on May 5: Since $a = 8$, the number of bananas eaten on May 5 is: \[ a + 24 = 8 + 24 = 32 \] 6. Conclusion: Big Al ate $\boxed{32}$ bananas on May 5, which corresponds to choice $\textbf{(D)}\ 32$.
Given the coefficient of determination R^2 for four different regression models, where the R^2 values are 0.98, 0.67, 0.85, and 0.36, determine which model has the best fitting effect.	0.98
Find all values of $x$ that satisfy $x=1-x+x^{2}-x^{3}+x^{4}-x^{5}+\cdots$ (be careful; this is tricky).	x=\frac{-1+\sqrt{5}}{2}	Multiplying both sides by $1+x$ gives $(1+x) x=1$, or $x=\frac{-1 \pm \sqrt{5}}{2}$. However, the series only converges for $\|x\|<1$, so only the answer $x=\frac{-1+\sqrt{5}}{2}$ makes sense.
What is the sum of the proper divisors of 256?	255
Find all functions $f : \mathbb{N}\rightarrow \mathbb{N}$ satisfying following condition: \[f(n+1)>f(f(n)), \quad \forall n \in \mathbb{N}.\]	{f(n)=n}	We are tasked with finding all functions \( f: \mathbb{N} \rightarrow \mathbb{N} \) that satisfy the condition: \[ f(n+1) > f(f(n)), \quad \forall n \in \mathbb{N}. \] To solve this problem, let us first analyze the condition given: \[ f(n+1) > f(f(n)). \] This inequality implies that the function \( f \) must order the values of its output in a certain way: the value of \( f(n+1) \) must always be greater than the value of \( f(f(n)) \). ### Testing Simple Functions #### Assume \( f(n) = n \) Substituting \( f(n) = n \) into the inequality gives: \[ f(n+1) = n+1 \quad \text{and} \quad f(f(n)) = f(n) = n. \] The inequality becomes: \[ n+1 > n, \] which is obviously true. Thus, \( f(n) = n \) satisfies the condition. ### Considering Other Types of Functions Next, let us consider whether other non-linear or non-identity functions might satisfy this inequality. Suppose \( f \) is a strictly increasing function with \( f(1) = 1 \): - For \( n = 1 \), we have: \[ f(2) > f(f(1)) = f(1) = 1. \] - Also, for higher \( n \), the condition consistently requires the output to be greater than the iterated value. Since any deviation from \( f(n) = n \) leads to failure under \( f(n+1) > f(f(n)) \) if \( f \) does not grow at the rate of identity (for instance, significant jumps or non-monotonic behaviors), it must preserve the identity nature to maintain all properties aligned. Thus, any form of function where \( f(n) \neq n \) leads to contradictions or failure in maintaining strict inequality for all \( n \). Given these arguments, the only function that meets the criteria is: \[ f(n) = n. \] Therefore, the solution to the functional equation under the given condition is: \[ \boxed{f(n) = n}. \]
A certain university needs $40L$ of helium gas to make balloon decorations for its centennial celebration. The chemistry club voluntarily took on this task. The club's equipment can produce a maximum of $8L$ of helium gas per day. According to the plan, the club must complete the production within 30 days. Upon receiving the task, the club members immediately started producing helium gas at a rate of $xL$ per day. It is known that the cost of raw materials for producing $1L$ of helium gas is $100$ yuan. If the daily production of helium gas is less than $4L$, the additional cost per day is $W_1=4x^2+16$ yuan. If the daily production of helium gas is greater than or equal to $4L$, the additional cost per day is $W_2=17x+\frac{9}{x}-3$ yuan. The production cost consists of raw material cost and additional cost. $(1)$ Write the relationship between the total cost $W$ (in yuan) and the daily production $x$ (in $L$). $(2)$ When the club produces how many liters of helium gas per day, the total cost is minimized? What is the minimum cost?	4640
Given that $x$ is a root of the equation $x^{2}+x-6=0$, simplify $\frac{x-1}{\frac{2}{{x-1}}-1}$ and find its value.	\frac{8}{3}
For how many integer values of $a$ does the equation $$x^2 + ax + 12a = 0$$ have integer solutions for $x$?	16
Given a circle $x^2 + (y-1)^2 = 1$ with its tangent line $l$, which intersects the positive x-axis at point A and the positive y-axis at point B. Determine the y-intercept of the tangent line $l$ when the distance AB is minimized.	\frac{3+\sqrt{5}}{2}
For each even positive integer $x$, let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square.	899	First note that $g(k)=1$ if $k$ is odd and $2g(k/2)$ if $k$ is even. so $S_n=\sum_{k=1}^{2^{n-1}}g(2k). = \sum_{k=1}^{2^{n-1}}2g(k) = 2\sum_{k=1}^{2^{n-1}}g(k) = 2\sum_{k=1}^{2^{n-2}} g(2k-1)+g(2k).$ $2k-1$ must be odd so this reduces to $2\sum_{k=1}^{2^{n-2}}1+g(2k) = 2(2^{n-2}+\sum_{k=1}^{2^n-2}g(2k)).$ Thus $S_n=2(2^{n-2}+S_{n-1})=2^{n-1}+2S_{n-1}.$ Further noting that $S_0=1$ we can see that $S_n=2^{n-1}\cdot (n-1)+2^n\cdot S_0=2^{n-1}\cdot (n-1)+2^{n-1}\cdot2=2^{n-1}\cdot (n+1).$ which is the same as above. To simplify the process of finding the largest square $S_n$ we can note that if $n-1$ is odd then $n+1$ must be exactly divisible by an odd power of $2$. However, this means $n+1$ is even but it cannot be. Thus $n-1$ is even and $n+1$ is a large even square. The largest even square $< 1000$ is $900$ so $n+1= 900 => n= \boxed{899}$
Charlie folds an $\frac{17}{2}$-inch by 11-inch piece of paper in half twice, each time along a straight line parallel to one of the paper's edges. What is the smallest possible perimeter of the piece after two such folds?	$\frac{39}{2}$	Note that a piece of paper is folded in half, one pair of opposite sides is preserved and the other pair is halved. Hence, the net effect on the perimeter is to decrease it by one of the side lengths. Hence, the original perimeter is $2\left(\frac{17}{2}\right)+2 \cdot 11=39$ and by considering the cases of folding twice along one edge or folding once along each edge, one can see that this perimeter can be decreased by at most $11+\frac{17}{2}=\frac{39}{2}$. Hence, the minimal perimeter is $\frac{39}{2}$.
Given the natural numbers $1,2,3,\ldots,10,11,12$, divide them into two groups such that the quotient of the product of all numbers in the first group by the product of all numbers in the second group is an integer and takes on the smallest possible value. What is this quotient?	231
What is the smallest positive integer with exactly 10 positive integer divisors?	48
A, B, C, D, and E are five students who obtained the top 5 positions (no ties) in a math competition. When taking a photo, they stood in a line and each of them made the following statement: A said: The two students next to me have rankings lower than mine; B said: The two students next to me have rankings adjacent to mine; C said: All students to my right (at least one) have higher rankings than mine; D said: All students to my left (at least one) have lower rankings than mine; E said: I am standing in second from the right. Given that all their statements are true, determine the five-digit number $\overline{\mathrm{ABCDE}}$.	23514
The area of the closed region formed by the line $y = nx$ and the curve $y = x^2$ is \_\_\_\_\_\_ when the binomial coefficients of the third and fourth terms in the expansion of $(x - \frac{2}{x})^n$ are equal.	\frac{125}{6}
The inverse of $f(x) = \frac{2x-1}{x+5}$ may be written in the form $f^{-1}(x)=\frac{ax+b}{cx+d}$, where $a$, $b$, $c$, and $d$ are real numbers. Find $a/c$.	-5
A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode (most frequent value). Let $D$ be the difference between the mode and the arithmetic mean of the sample. What is the largest possible value of $\lfloor D\rfloor$? (For real $x$, $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.)	947	With the same reasoning as Solution 1, in order to get largest possible value of D, we can construct that our set of numbers as $\underbrace{1,1,1...1,}_\text{n times}\underbrace{2,2,2...2,}_\text{n times}\underbrace{3,3,3...3,}_\text{n times}........\underbrace{1000,1000,1000....}_\text{n+1 times}$ And, we need to find the value of n that makes the sum as low as possible. And, we can create a formula to make it easier. It isn't hard to find the sum. The numbers which are not 1000, average to $\frac{120}{2n}$ or $\frac{60}{n}$, and there are $120-n$ of them. So, they sum to $\frac{60}{n}(120-n)$. And, the sum of the numbers that are 1000 is $1000(n+1)$ so, our total sum gets us $1000(n+1)+120/2n(120-n)$ We want to minimize it, since the mode will always be 1000. And, testing the values n = 1, n = 2, n = 3, n = 4, we get these results. $n = 1: 2000+60119 = 9140$ $n = 2: 3000+30118 = 6540$ $n = 3: 4000+20117 = 6340$ $n = 4: 5000+15116 = 6740$ And, as n grows larger and larger from 4, the values will start increasing. Thus, the lowest possible sum is 6340. Dividing by 121, the lowest possible mean is 52.396...., and thus, the highest possible value of $D$ is 947.604, and the floor of that is $\boxed{947}$ - AlexLikeMath Notes ^ In fact, when $n = 2,3,4,5$ (which some simple testing shows that the maximum will occur around), it turns out that $\frac{121-n}{n-1}$ is an integer anyway, so indeed $k = \left\lfloor \frac{121-n}{n-1} \right\rfloor = \frac{121-n}{n-1}$.
On a sphere, there are four points A, B, C, and D satisfying $AB=1$, $BC=\sqrt{3}$, $AC=2$. If the maximum volume of tetrahedron D-ABC is $\frac{\sqrt{3}}{2}$, then the surface area of this sphere is _______.	\frac{100\pi}{9}
James is six years older than Louise. Eight years from now, James will be four times as old as Louise was four years before now. What is the sum of their current ages?	26
Suppose \( a, b \), and \( c \) are real numbers with \( a < b < 0 < c \). Let \( f(x) \) be the quadratic function \( f(x) = (x-a)(x-c) \) and \( g(x) \) be the cubic function \( g(x) = (x-a)(x-b)(x-c) \). Both \( f(x) \) and \( g(x) \) have the same \( y \)-intercept of -8 and \( g(x) \) passes through the point \( (-a, 8) \). Determine the value of \( c \).	\frac{8}{3}

Given triangle ABC, where sides $a$, $b$, and $c$ correspond to angles A, B, and C respectively, and $a=4$, $\cos{B}=\frac{4}{5}$. (1) If $b=6$, find the value of $\sin{A}$; (2) If the area of triangle ABC, $S=12$, find the values of $b$ and $c$.

2\sqrt{13}

A ball inside a rectangular container of width 7 and height 12 is launched from the lower-left vertex of the container. It first strikes the right side of the container after traveling a distance of $\sqrt{53}$ (and strikes no other sides between its launch and its impact with the right side). How many times does the ball bounce before it returns to a vertex? (The final contact with a vertex does not count as a bounce.)

5

Every segment the ball traverses between bounces takes it 7 units horizontally and 2 units up. Thus, after 5 bounces it has traveled up 10 units, and the final segment traversed takes it directly to the upper right vertex of the rectangle.

Ilya Muromets encounters the three-headed Dragon, Gorynych. Each minute, Ilya chops off one head of the dragon. Let $x$ be the dragon's resilience ($x > 0$). The probability $p_{s}$ that $s$ new heads will grow in place of a chopped-off one ($s=0,1,2$) is given by $\frac{x^{s}}{1+x+x^{2}}$. During the first 10 minutes of the battle, Ilya recorded the number of heads that grew back for each chopped-off one. The vector obtained is: $K=(1,2,2,1,0,2,1,0,1,2)$. Find the value of the dragon's resilience $x$ that maximizes the probability of vector $K$.

\frac{1 + \sqrt{97}}{8}

(Full score for this problem is 12 points) Given $f(x) = e^x - ax - 1$. (1) Find the intervals where $f(x)$ is monotonically increasing. (2) If $f(x)$ is monotonically increasing on the domain $\mathbb{R}$, find the range of possible values for $a$. (3) Does there exist a value of $a$ such that $f(x)$ is monotonically decreasing on $(-\infty, 0]$ and monotonically increasing on $[0, +\infty)$? If so, find the value of $a$; if not, explain why.

a = 1

If the sum of the lengths of the six edges of a trirectangular tetrahedron $PABC$ (i.e., $\angle APB=\angle BPC=\angle CPA=90^o$ ) is $S$ , determine its maximum volume.

\[ \frac{S^3(\sqrt{2}-1)^3}{162} \]

Let the side lengths of $AP$ , $BP$ , and $CP$ be $a$ , $b$ , and $c$ , respectively. Therefore $S=a+b+c+\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}$ . Let the volume of the tetrahedron be $V$ . Therefore $V=\frac{abc}{6}$ . Note that $(a-b)^2\geq 0$ implies $\frac{a^2-2ab+b^2}{2}\geq 0$ , which means $\frac{a^2+b^2}{2}\geq ab$ , which implies $a^2+b^2\geq ab+\frac{a^2+b^2}{2}$ , which means $a^2+b^2\geq \frac{(a+b)^2}{2}$ , which implies $\sqrt{a^2+b^2}\geq \frac{1}{\sqrt{2}} \cdot (a+b)$ . Equality holds only when $a=b$ . Therefore $S\geq a+b+c+\frac{1}{\sqrt{2}} \cdot (a+b)+\frac{1}{\sqrt{2}} \cdot (c+b)+\frac{1}{\sqrt{2}} \cdot (a+c)$ $=(a+b+c)(1+\sqrt{2})$ . $\frac{a+b+c}{3}\geq \sqrt[3]{abc}$ is true from AM-GM, with equality only when $a=b=c$ . So $S\geq (a+b+c)(1+\sqrt{2})\geq 3(1+\sqrt{2})\sqrt[3]{abc}=3(1+\sqrt{2})\sqrt[3]{6V}$ . This means that $\frac{S}{3(1+\sqrt{2})}=\frac{S(\sqrt{2}-1)}{3}\geq \sqrt[3]{6V}$ , or $6V\leq \frac{S^3(\sqrt{2}-1)^3}{27}$ , or $V\leq \frac{S^3(\sqrt{2}-1)^3}{162}$ , with equality only when $a=b=c$ . Therefore the maximum volume is $\frac{S^3(\sqrt{2}-1)^3}{162}$ .

For some positive integers $p$, there is a quadrilateral $ABCD$ with positive integer side lengths, perimeter $p$, right angles at $B$ and $C$, $AB=2$, and $CD=AD$. How many different values of $p<2015$ are possible? $\textbf{(A) }30\qquad\textbf{(B) }31\qquad\textbf{(C) }61\qquad\textbf{(D) }62\qquad\textbf{(E) }63$

31

It is known that the only solution to the equation $$ \pi / 4 = \operatorname{arcctg} 2 + \operatorname{arcctg} 5 + \operatorname{arcctg} 13 + \operatorname{arcctg} 34 + \operatorname{arcctg} 89 + \operatorname{arcctg}(x / 14) $$ is a natural number. Find it.

2016

Find the largest positive integer solution of the equation $\left\lfloor\frac{N}{3}\right\rfloor=\left\lfloor\frac{N}{5}\right\rfloor+\left\lfloor\frac{N}{7}\right\rfloor-\left\lfloor\frac{N}{35}\right\rfloor$.

65

For $N$ to be a solution, it is necessary that $\frac{N-2}{3}+\frac{N-34}{35} \leq \frac{N}{5}+\frac{N}{7}$, which simplifies to $N \leq 86$. However, if $N \geq 70$, then $N \leq 59$, contradicting $N \geq 70$. It follows that $N$ must be at most 69. Checking for $N \leq 69$, we find that when $N=65$, the equation holds. Thus the answer is $N=65$.

Let $(a_1, a_2, a_3, \ldots, a_{15})$ be a permutation of $(1, 2, 3, \ldots, 15)$ for which $a_1 > a_2 > a_3 > a_4 > a_5 > a_6 > a_7 \mathrm{\ and \ } a_7 < a_8 < a_9 < a_{10} < a_{11} < a_{12} < a_{13} < a_{14} < a_{15}.$ Find the number of such permutations.

3003

What integer $n$ satisfies $0\le n<{101}$ and $$100n\equiv 72\pmod {101}~?$$

29

The area of triangle $ABC$ is 1. Points $B'$, $C'$, and $A'$ are placed respectively on the rays $AB$, $BC$, and $CA$ such that: \[ BB' = 2 AB, \quad CC' = 3 BC, \quad AA' = 4 CA. \] Calculate the area of triangle $A'B'C'$.

39

A line is parameterized by \[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -3 \end{pmatrix}.\]A second line is parameterized by \[\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ -9 \end{pmatrix} + u \begin{pmatrix} 4 \\ 2 \end{pmatrix}.\]Find the point where the lines intersect.

\begin{pmatrix} 7 \\ -8 \end{pmatrix}

Solve \[\sqrt{1 + \sqrt{2 + \sqrt{x}}} = \sqrt[3]{1 + \sqrt{x}}.\]

49

Let $b_n$ be the integer obtained by writing the integers from $5$ to $n+4$ from left to right. For example, $b_2 = 567$, and $b_{10} = 567891011121314$. Compute the remainder when $b_{25}$ is divided by $55$ (which is the product of $5$ and $11$ for the application of the Chinese Remainder Theorem).

39

Three numbers, $a_1\,$, $a_2\,$, $a_3\,$, are drawn randomly and without replacement from the set $\{1, 2, 3, \dots, 1000\}\,$. Three other numbers, $b_1\,$, $b_2\,$, $b_3\,$, are then drawn randomly and without replacement from the remaining set of 997 numbers. Let $p\,$ be the probability that, after a suitable rotation, a brick of dimensions $a_1 \times a_2 \times a_3\,$ can be enclosed in a box of dimensions $b_1 \times b_2 \times b_3\,$, with the sides of the brick parallel to the sides of the box. If $p\,$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

5

Express the quotient $1021_3 \div 11_3$ in base $3$.

22_3

Danielle Bellatrix Robinson is organizing a poker tournament with 9 people. The tournament will have 4 rounds, and in each round the 9 players are split into 3 groups of 3. During the tournament, each player plays every other player exactly once. How many different ways can Danielle divide the 9 people into three groups in each round to satisfy these requirements?

20160

We first split the 9 people up arbitrarily into groups of 3. There are $\frac{\binom{9}{3}\binom{6}{3}\binom{3}{3}}{3!}=280$ ways of doing this. Without loss of generality, label the people 1 through 9 so that the first round groups are $\{1,2,3\},\{4,5,6\}$, and $\{7,8,9\}$. We will use this numbering to count the number of ways DBR can divide the 9 players in rounds 2,3, and 4. In round 2, because players 1,2, and 3 are together in the first round, they must be in separate groups, and likewise for $\{4,5,6\}$ and $\{7,8,9\}$. Disregarding ordering of the three groups in a single round, we will first place 1,2, and 3 into their groups, then count the number of ways to place $\{4,5,6\}$ and $\{7,8,9\}$ in the groups with them. We do this by placing one member from each of $\{4,5,6\}$ and $\{7,8,9\}$ into each group. There are $(3!)^{2}$ ways to do this. Now, because of symmetry, we can use the round 2 grouping $\{1,4,7\},\{2,5,8\},\{3,6,9\}$ to list out the remaining possibilities for round 3 and 4 groupings. Casework shows that there are 2 ways to group the players in the remaining two rounds. We multiply $280 \cdot(3!)^{2} \cdot 2$ to get 20160

Two adjacent faces of a tetrahedron, which are equilateral triangles with side length 1, form a dihedral angle of 60 degrees. The tetrahedron rotates around the common edge of these faces. Find the maximum area of the projection of the rotating tetrahedron on the plane containing the given edge. (12 points)

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

The integer 119 is a multiple of which number?

7

The ones digit of 119 is not even, so 119 is not a multiple of 2. The ones digit of 119 is not 0 or 5, so 119 is not a multiple of 5. Since $120=3 \times 40$, then 119 is 1 less than a multiple of 3 so is not itself a multiple of 3. Since $110=11 \times 10$ and $121=11 \times 11$, then 119 is between two consecutive multiples of 11, so is not itself a multiple of 11. Finally, $119 \div 7=17$, so 119 is a multiple of 7.

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

0.15

A bug crawls along a number line, starting at $-2$. It crawls to $-6$, then turns around and crawls to $5$. How many units does the bug crawl altogether?

15

1. **Calculate the distance from $-2$ to $-6$:** The distance on a number line is the absolute difference between the two points. Thus, the distance from $-2$ to $-6$ is: \[ |-6 - (-2)| = |-6 + 2| = |-4| = 4 \text{ units} \] 2. **Calculate the distance from $-6$ to $5$:** Similarly, the distance from $-6$ to $5$ is: \[ |5 - (-6)| = |5 + 6| = |11| = 11 \text{ units} \] 3. **Add the distances to find the total distance crawled:** The total distance the bug crawls is the sum of the distances calculated in steps 1 and 2: \[ 4 \text{ units} + 11 \text{ units} = 15 \text{ units} \] Thus, the total distance the bug crawls is $\boxed{\textbf{(E)}\ 15}$.

Find $\sec (-300^\circ).$

2

Compute the following expressions: (1) $2 \sqrt{12} -6 \sqrt{ \frac{1}{3}} + \sqrt{48}$ (2) $(\sqrt{3}-\pi)^{0}-\frac{\sqrt{20}-\sqrt{15}}{\sqrt{5}}+(-1)^{2017}$

\sqrt{3} - 2

A triangle has vertices $(0,0)$, $(1,1)$, and $(6m,0)$. The line $y = mx$ divides the triangle into two triangles of equal area. What is the sum of all possible values of $m$?

- \frac {1}{6}

1. **Identify the vertices and the line dividing the triangle**: Let's label the vertices of the triangle as $A = (0,0)$, $B = (1,1)$, and $C = (6m,0)$. The line $y = mx$ is supposed to divide the triangle into two triangles of equal area. 2. **Determine the intersection point**: The line $y = mx$ will intersect the line segment $BC$. Let $D$ be the point of intersection. The coordinates of $D$ can be found by setting the $y$-coordinates of the line $y = mx$ and the line segment $BC$ equal. The line segment $BC$ can be represented by the equation $y = -\frac{1}{6m-1}(x-1)$ (using point-slope form and the points $B$ and $C$). Setting $mx = -\frac{1}{6m-1}(x-1)$ and solving for $x$, we find the $x$-coordinate of $D$. 3. **Condition for equal areas**: The triangles $ABD$ and $ACD$ will have equal areas if $D$ is the midpoint of $BC$. The midpoint of $BC$ is given by $\left(\frac{6m+1}{2}, \frac{1}{2}\right)$. For $D$ to be on the line $y = mx$, the coordinates must satisfy $y = mx$. Substituting the midpoint coordinates into the line equation, we get: \[ \frac{1}{2} = m \cdot \frac{6m+1}{2} \] Simplifying this equation: \[ 1 = m(6m+1) \implies 6m^2 + m - 1 = 0 \] 4. **Solve the quadratic equation**: The quadratic equation $6m^2 + m - 1 = 0$ can be solved using the quadratic formula: \[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \quad \text{where } a = 6, b = 1, c = -1 \] \[ m = \frac{-1 \pm \sqrt{1 + 24}}{12} = \frac{-1 \pm 5}{12} \] \[ m = \frac{4}{12} = \frac{1}{3}, \quad m = \frac{-6}{12} = -\frac{1}{2} \] 5. **Sum of all possible values of $m$**: Using Vieta's formulas, the sum of the roots of the equation $6m^2 + m - 1 = 0$ is given by $-\frac{b}{a} = -\frac{1}{6}$. Thus, the sum of all possible values of $m$ is $\boxed{\textbf{(B)} - \!\frac {1}{6}}$.

Huahua is writing letters to Yuanyuan with a pen. When she finishes the 3rd pen refill, she is working on the 4th letter; when she finishes the 5th letter, the 4th pen refill is not yet used up. If Huahua uses the same amount of ink for each letter, how many pen refills does she need to write 16 letters?

13

Students guess that Norb's age is $24, 28, 30, 32, 36, 38, 41, 44, 47$, and $49$. Norb says, "At least half of you guessed too low, two of you are off by one, and my age is a prime number." How old is Norb?

37

1. **Identify the conditions given by Norb:** - At least half of the guesses are too low. - Two guesses are off by one. - Norb's age is a prime number. 2. **List of guesses:** $24, 28, 30, 32, 36, 38, 41, 44, 47, 49$. 3. **Determine possible values for Norb's age ($x$) based on guesses off by one:** - If two guesses are off by one, then $x-1$ and $x+1$ must both be in the list. - Possible pairs $(x-1, x+1)$ from the list are: - $(28, 30)$, hence $x = 29$ - $(30, 32)$, hence $x = 31$ - $(36, 38)$, hence $x = 37$ - $(47, 49)$, hence $x = 48$ 4. **Check which of these values are prime numbers:** - $29$ is prime. - $31$ is prime. - $37$ is prime. - $48$ is not prime. So, possible values for $x$ are $29$, $31$, and $37$. 5. **Check the condition that at least half of the guesses are too low:** - If $x = 29$, the guesses too low are $24, 28$. This is not half. - If $x = 31$, the guesses too low are $24, 28, 30$. This is not half. - If $x = 37$, the guesses too low are $24, 28, 30, 32, 36$. This is exactly half of the total guesses (5 out of 10). 6. **Conclusion:** - The only value that satisfies all conditions (prime number, two guesses off by one, and at least half of the guesses are too low) is $37$. Thus, Norb's age is $\boxed{\textbf{(C)}\ 37}$.

Tim's quiz scores were 85, 87, 92, 94, 78, and 96. Calculate his mean score and find the range of his scores.

18

Find the only positive real number $x$ for which $\displaystyle \frac{x-4}{9} = \frac{4}{x-9}$.

13

Evaluate the product \[\left(\sqrt{5}+\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{6}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).\]

104

We group the first and last factors as well as the two middle factors, then apply the difference of squares repeatedly: \begin{align*} \left(\left(\sqrt{6} + \sqrt{7}\right)^2 - \sqrt{5}^2\right)\left(\sqrt{5}^2 - \left(\sqrt{6} - \sqrt{7}\right)^2\right) &= \left(13 + 2\sqrt{42} - 5\right)\left(5 - \left(13 - 2\sqrt{42}\right)\right) \\ &= \left(2\sqrt{42} + 8\right)\left(2\sqrt{42} - 8\right) \\ &= \left(2\sqrt{42}\right)^2 - 8^2 \\ &= \boxed{104}. \end{align*} ~Azjps (Solution) ~MRENTHUSIASM (Revision)

Find the smallest positive integer $n$ such that the polynomial $(x+1)^{n}-1$ is "divisible by $x^{2}+1$ modulo 3", or more precisely, either of the following equivalent conditions holds: there exist polynomials $P, Q$ with integer coefficients such that $(x+1)^{n}-1=\left(x^{2}+1\right) P(x)+3 Q(x)$; or more conceptually, the remainder when (the polynomial) $(x+1)^{n}-1$ is divided by (the polynomial) $x^{2}+1$ is a polynomial with (integer) coefficients all divisible by 3.

8

We have $(x+1)^{2}=x^{2}+2 x+1 \equiv 2 x,(x+1)^{4} \equiv(2 x)^{2} \equiv-4 \equiv-1$, and $(x+1)^{8} \equiv(-1)^{2}=1$. So the order $n$ divides 8, as $x+1$ and $x^{2}+1$ are relatively prime polynomials modulo 3 (or more conceptually, in $\mathbb{F}_{3}[x]$ ), but cannot be smaller by our computations of the 2 nd and 4 th powers.

The distances between the points on a line are given as $2, 4, 5, 7, 8, k, 13, 15, 17, 19$. Determine the value of $k$.

12

The ship decided to determine the depth of the ocean at its location. The signal sent by the echo sounder was received on the ship 8 seconds later. The speed of sound in water is 1.5 km/s. Determine the depth of the ocean.

6000

Given the numbers 1, 3, 5 and 2, 4, 6, calculate the total number of different three-digit numbers that can be formed when arranging these numbers on three cards.

48

In triangle $\triangle ABC$, $A=60^{\circ}$, $a=\sqrt{6}$, $b=2$. $(1)$ Find $\angle B$; $(2)$ Find the area of $\triangle ABC$.

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

Quadrilateral $PQRS$ is a square. A circle with center $S$ has arc $PXC$. A circle with center $R$ has arc $PYC$. If $PQ = 3$ cm, what is the total number of square centimeters in the football-shaped area of regions II and III combined? Express your answer as a decimal to the nearest tenth.

5.1

The equation $ \sin^2 x + \sin^2 3x + \sin^2 5x + \sin^2 6x = \frac{81}{32} $ can be reduced to the equivalent equation \[ \cos ax \cos bx \cos cx = 0, \] for some positive integers $ a, $ $ b, $ and $ c. $ Find $ a + b + c. $

14

Suppose $X$ and $Y$ are digits in base $d > 8$ such that $\overline{XY}_d + \overline{XX}_d = 234_d$. Find $X_d - Y_d$ in base $d$.

-2

Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio $3: 2: 1,$what is the least possible total for the number of bananas?

408

Let $A,B,C$ be the fraction of bananas taken by the first, second, and third monkeys respectively. Then we have the system of equations \[\frac{3}{4}A+\frac{3}{8}B+\frac{11}{24}C=\frac{1}{2}\] \[\frac{1}{8}A+\frac{1}{4}B+\frac{11}{24}C=\frac{1}{3}\] \[\frac{1}{8}A+\frac{3}{8}B+\frac{2}{24}C=\frac{1}{6}.\] Solve this your favorite way to get that \[(A,B,C)=\left( \frac{11}{51}, \frac{13}{51}, \frac{9}{17} \right).\] We need the amount taken by the first and second monkeys to be divisible by 8 and the third by 24 (but for the third, we already have divisibility by 9). Thus our minimum is $8 \cdot 51 = \boxed{408}.$ ~Dhillonr25

Denote by $\mathbb{Z}^2$ the set of all points $(x,y)$ in the plane with integer coordinates. For each integer $n \geq 0$, let $P_n$ be the subset of $\mathbb{Z}^2$ consisting of the point $(0,0)$ together with all points $(x,y)$ such that $x^2 + y^2 = 2^k$ for some integer $k \leq n$. Determine, as a function of $n$, the number of four-point subsets of $P_n$ whose elements are the vertices of a square.

5n+1

The answer is $5n+1$. We first determine the set $P_n$. Let $Q_n$ be the set of points in $\mathbb{Z}^2$ of the form $(0, \pm 2^k)$ or $(\pm 2^k, 0)$ for some $k \leq n$. Let $R_n$ be the set of points in $\mathbb{Z}^2$ of the form $(\pm 2^k, \pm 2^k)$ for some $k \leq n$ (the two signs being chosen independently). We prove by induction on $n$ that \[ P_n = \{(0,0)\} \cup Q_{\lfloor n/2 \rfloor} \cup R_{\lfloor (n-1)/2 \rfloor}. \] We take as base cases the straightforward computations \begin{align*} P_0 &= \{(0,0), (\pm 1, 0), (0, \pm 1)\} \\ P_1 &= P_0 \cup \{(\pm 1, \pm 1)\}. \end{align*} For $n \geq 2$, it is clear that $\{(0,0)\} \cup Q_{\lfloor n/2 \rfloor} \cup R_{\lfloor (n-1)/2 \rfloor} \subseteq P_n$, so it remains to prove the reverse inclusion. For $(x,y) \in P_n$, note that $x^2 + y^2 \equiv 0 \pmod{4}$; since every perfect square is congruent to either 0 or 1 modulo 4, $x$ and $y$ must both be even. Consequently, $(x/2, y/2) \in P_{n-2}$, so we may appeal to the induction hypothesis to conclude. We next identify all of the squares with vertices in $P_n$. In the following discussion, let $(a,b)$ and $(c,d)$ be two opposite vertices of a square, so that the other two vertices are \[ \left( \frac{a-b+c+d}{2}, \frac{a+b-c+d}{2} \right) \] and \[ \left( \frac{a+b+c-d}{2}, \frac{-a+b+c+d}{2} \right). \] \begin{itemize} \item Suppose that $(a,b) = (0,0)$. Then $(c,d)$ may be any element of $P_n$ not contained in $P_0$. The number of such squares is $4n$. \item Suppose that $(a,b), (c,d) \in Q_k$ for some $k$. There is one such square with vertices \[ \{(0, 2^k), (0, 2^{-k}), (2^k, 0), (2^{-k}, 0)\} \] for $k = 0,\dots,\lfloor \frac{n}{2} \rfloor$, for a total of $\lfloor \frac{n}{2} \rfloor + 1$. To show that there are no others, by symmetry it suffices to rule out the existence of a square with opposite vertices $(a,0)$ and $(c,0)$ where $a > \left| c \right|$. The other two vertices of this square would be $((a+c)/2, (a-c)/2)$ and $((a+c)/2, (-a+c)/2)$. These cannot belong to any $Q_k$, or be equal to $(0,0)$, because $|a+c|, |a-c| \geq a - |c| > 0$ by the triangle inequality. These also cannot belong to any $R_k$ because $(a + |c|)/2 > (a - |c|)/2$. (One can also phrase this argument in geometric terms.) \item Suppose that $(a,b), (c,d) \in R_k$ for some $k$. There is one such square with vertices \[ \{(2^k, 2^k), (2^k, -2^k), (-2^k, 2^k), (-2^k, -2^k)\} \] for $k=0,\dots, \lfloor \frac{n-1}{2} \rfloor$, for a total of $\lfloor \frac{n+1}{2} \rfloor$. To show that there are no others, we may reduce to the previous case: rotating by an angle of $\frac{\pi}{4}$ and then rescaling by a factor of $\sqrt{2}$ would yield a square with two opposite vertices in some $Q_k$ not centered at $(0,0)$, which we have already ruled out. \item It remains to show that we cannot have $(a,b) \in Q_k$ and $(c,d) \in R_k$ for some $k$. By symmetry, we may reduce to the case where $(a,b) = (0, 2^k)$ and $(c,d) = (2^\ell, \pm 2^\ell)$. If $d>0$, then the third vertex $(2^{k-1}, 2^{k-1} + 2^\ell)$ is impossible. If $d<0$, then the third vertex $(-2^{k-1}, 2^{k-1} - 2^\ell)$ is impossible. \end{itemize} Summing up, we obtain \[ 4n + \left\lfloor \frac{n}{2} \right\rfloor + 1 + \left\lfloor \frac{n+1}{2} \right\rfloor = 5n+1 \] squares, proving the claim.

Seven students count from 1 to 1000 as follows: Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000. Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers. Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers. Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers. Finally, George says the only number that no one else says. What number does George say?

365

1. **Alice's Numbers:** Alice says all numbers except those that are the middle number in each consecutive group of three numbers. This means Alice skips numbers of the form $3n - 1$ for $n = 1, 2, 3, \ldots, 333$. The numbers Alice skips are: \[ 2, 5, 8, \ldots, 998 \] which are calculated as: \[ 3 \times 1 - 1, 3 \times 2 - 1, 3 \times 3 - 1, \ldots, 3 \times 333 - 1 \] 2. **Barbara's Numbers:** Barbara says the numbers that Alice doesn't say, but she also skips the middle number in each consecutive group of three numbers among these. The numbers Barbara skips are those of the form $3(3n - 1) - 1$ for $n = 1, 2, 3, \ldots$. Calculating a few terms: \[ 3(3 \times 1 - 1) - 1 = 5, \quad 3(3 \times 2 - 1) - 1 = 14, \quad 3(3 \times 3 - 1) - 1 = 23, \ldots \] 3. **Candice's Numbers:** Candice says the numbers that neither Alice nor Barbara says, but also skips the middle number in each consecutive group of three numbers. The numbers Candice skips are those of the form $3(3(3n - 1) - 1) - 1$ for $n = 1, 2, 3, \ldots$. Calculating a few terms: \[ 3(3(3 \times 1 - 1) - 1) - 1 = 14, \quad 3(3(3 \times 2 - 1) - 1) - 1 = 41, \ldots \] 4. **Continuing the Pattern:** Following the same pattern for Debbie, Eliza, and Fatima, each skips numbers of the form $3(\text{previous pattern}) - 1$. 5. **George's Number:** George says the only number that no one else says. Following the pattern to George, the first number he says is calculated as: \[ 3(3(3(3(3(3 \times 1 - 1) - 1) - 1) - 1) - 1) - 1 = 365 \] Checking the next number George would say: \[ 3(3(3(3(3(3 \times 2 - 1) - 1) - 1) - 1) - 1) - 1 > 1000 \] Thus, the only number George says within the range 1 to 1000 is 365. $\boxed{\textbf{(C)}\ 365}$

Let $ a_{n} = 1 + 2 + \cdots + n $, where $ n \in \mathbf{N}_{+} $, and $ S_{m} = a_{1} + a_{2} + \cdots + a_{m} $, $ m = 1, 2, \cdots, m $. Find the number of values among $ S_{1}, S_{2}, \cdots, S_{2017} $ that are divisible by 2 but not by 4.

252

If $\frac{1}{4} + 4\left(\frac{1}{2013} + \frac{1}{x}\right) = \frac{7}{4}$, find the value of $1872 + 48 \times \left(\frac{2013 x}{x + 2013}\right)$.

2000

The natural number $a$ is divisible by 35 and has 75 different divisors, including 1 and $a$. Find the smallest such $a$.

490000

How many three-digit positive integers $x$ satisfy $3874x+481\equiv 1205 \pmod{23}$?

40

(In the preliminaries of optimal method and experimental design) When using the 0.618 method to find the optimal amount to add in an experiment, if the current range of excellence is $[628, 774]$ and the good point is 718, then the value of the addition point for the current experiment is ________.

684

Remove all perfect squares from the sequence of positive integers $1, 2, 3, \cdots$ to get a new sequence. What is the 2003rd term of this new sequence?

2048

Simplify $\frac{\sqrt{2}}{\sqrt{3}} \cdot \frac{\sqrt{4}}{\sqrt{5}} \cdot \frac{\sqrt{6}}{\sqrt{7}}$ and rationalize the denominator of the resulting fraction.

\frac{4\sqrt{35}}{35}

Given $\triangle ABC$, where the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively, and it satisfies $a\cos 2C+2c\cos A\cos C+a+b=0$. $(1)$ Find the size of angle $C$; $(2)$ If $b=4\sin B$, find the maximum value of the area $S$ of $\triangle ABC$.

\sqrt{3}

If a die is rolled, event $ A = \{1, 2, 3\} $ consists of rolling one of the faces 1, 2, or 3. Similarly, event $ B = \{1, 2, 4\} $ consists of rolling one of the faces 1, 2, or 4. The die is rolled 10 times. It is known that event $ A $ occurred exactly 6 times. a) Find the probability that under this condition, event $ B $ did not occur at all. b) Find the expected value of the random variable $ X $, which represents the number of occurrences of event $ B $.

\frac{16}{3}

The graph of the function $y=\sin(\omega x+ \frac {5\pi}{6})$ where $0<\omega<\pi$ intersects with the coordinate axes at points closest to the origin, which are $(0, \frac {1}{2})$ and $( \frac {1}{2}, 0)$. Determine the axis of symmetry of this graph closest to the y-axis.

-1

Let $ p, q, r, s $ be distinct real numbers such that the roots of $ x^2 - 12px - 13q = 0 $ are $ r $ and $ s $, and the roots of $ x^2 - 12rx - 13s = 0 $ are $ p $ and $ q $. Find the value of $ p + q + r + s $.

-13

Given two lines $l_{1}$: $(a-1)x+2y+1=0$, $l_{2}$: $x+ay+1=0$, find the value of $a$ that satisfies the following conditions: $(1) l_{1} \parallel l_{2}$ $(2) l_{1} \perp l_{2}$

\frac{1}{3}

How many three-digit whole numbers have at least one 7 or at least one 9 as digits?

452

The Bank of Springfield's Super High Yield savings account compounds annually at a rate of one percent. If Lisa invests 1000 dollars in one of these accounts, then how much interest will she earn after five years? (Give your answer to the nearest dollar.)

51

In square $ABCD$, points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA}$, respectively, so that $AE=AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD}$, respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$. See the figure below. Triangle $AEH$, quadrilateral $BFIE$, quadrilateral $DHJG$, and pentagon $FCGJI$ each has area $1.$ What is $FI^2$? [asy] real x=2sqrt(2); real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); real z=2sqrt(8-4sqrt(2)); pair A, B, C, D, E, F, G, H, I, J; A = (0,0); B = (4,0); C = (4,4); D = (0,4); E = (x,0); F = (4,y); G = (y,4); H = (0,x); I = F + z * dir(225); J = G + z * dir(225); draw(A--B--C--D--A); draw(H--E); draw(J--G^^F--I); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$E$", E, S); dot("$F$", F, dir(0)); dot("$G$", G, N); dot("$H$", H, W); dot("$I$", I, SW); dot("$J$", J, SW); [/asy] $\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2$

8-4\sqrt{2}

There are 100 positive integers written on a board. At each step, Alex composes 50 fractions using each number written on the board exactly once, brings these fractions to their irreducible form, and then replaces the 100 numbers on the board with the new numerators and denominators to create 100 new numbers. Find the smallest positive integer $n{}$ such that regardless of the values of the initial 100 numbers, after $n{}$ steps Alex can arrange to have on the board only pairwise coprime numbers.

99

To solve this problem, we aim to find the smallest positive integer $ n $ such that after $ n $ steps, the 100 numbers on the board are all pairwise coprime regardless of their initial values. ### Key Observations 1. **Irreducible Fractions**: At each step, Alex forms 50 fractions out of the 100 numbers. Each fraction $\frac{a}{b}$ is reduced to its irreducible form $\frac{p}{q}$, where $\gcd(p, q) = 1$. 2. **Numerators and Denominators**: The new set of numbers on the board after each step are the numerators and denominators of these 50 irreducible fractions. 3. **Pairwise Coprimeness**: For numbers to be pairwise coprime, each pair of numbers has a greatest common divisor of 1. ### Strategy - **Step Progression**: As we progress with the steps, fractions are reduced to irreducible form, potentially introducing many coprime pairs. However, we need them to all become pairwise coprime eventually. - **Minimizing the Steps**: To get the numbers pairwise coprime, consider the worst-case scenario: Starting with 100 numbers where no two numbers are coprime. ### Execution Analyzing each pair of numbers: - Each step incorporates forming pairs that guarantee at least one pair becomes coprime. By the nature of reduction to irreducible form, this iteration slowly increases the number of coprime pairs among the set. - After the first step, observe that some coprime pairs necessarily occur due to the fraction reduction process. - Due to the properties of the Euclidean algorithm, composed during the fraction process, this coprime nature spreads as the steps progress. - After 99 steps, according to the pigeonhole principle and the iterative application of number theory principles, all numbers can be arranged to be pairwise coprime. Thus, regardless of initial values, the minimum number of steps required to achieve pairwise coprimeness among the 100 numbers is: \[ \boxed{99} \]

The figure may be folded along the lines shown to form a number cube. Three number faces come together at each corner of the cube. What is the largest sum of three numbers whose faces come together at a corner?

14

To solve this problem, we need to understand how a cube is formed and how the numbers on the faces are arranged. In a standard die, opposite faces sum up to 7. This means: - If one face shows 1, the opposite face shows 6. - If one face shows 2, the opposite face shows 5. - If one face shows 3, the opposite face shows 4. Given this arrangement, no two of the highest numbers (6, 5, 4) can be adjacent because they are on opposite faces. Therefore, we need to find the highest possible sum of three numbers that can meet at a vertex. 1. **Check the possibility of the sum of 6, 5, and 4**: - These numbers are on opposite faces, so they cannot meet at a vertex. 2. **Check the possibility of the sum of 6, 5, and 3**: - Since 6 and 5 are on opposite faces, they cannot be adjacent, but 6 and 3 can be adjacent, and 5 and 3 can also be adjacent. - The numbers 6, 5, and 3 can meet at a vertex because they are not on directly opposite faces. 3. **Calculate the sum of 6, 5, and 3**: \[ 6 + 5 + 3 = 14 \] 4. **Verify if there is a higher possible sum**: - The next highest possible combinations would involve numbers less than 3 since 4 is opposite to 3 and cannot be at the same vertex with 6 or 5. - Any combination involving numbers less than 3 will result in a sum less than 14. Since no other combination of three numbers that can meet at a vertex exceeds the sum of 14, the largest sum of three numbers whose faces come together at a corner of the cube is 14. Thus, the answer is $\boxed{\text{D}}$.

If $n$ is $1$ less than a multiple of $50$, then what is the remainder when $n^2+2n+3$ is divided by $50$?

2

A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are $6.1$ cm, $8.2$ cm and $9.7$ cm. What is the area of the square in square centimeters?

36

Given a number $\overline{abcd}$ , where $a$ , $b$ , $c$ , and $d$ , represent the digits of $\overline{abcd}$ , find the minimum value of \[\frac{\overline{abcd}}{a+b+c+d}\] where $a$ , $b$ , $c$ , and $d$ are distinct <details><summary>Answer</summary>$\overline{abcd}=1089$ , minimum value of $\dfrac{\overline{abcd}}{a+b+c+d}=60.5$</details>

60.5

If $x+y=\frac{7}{13}$ and $x-y=\frac{1}{91}$, what is the value of $x^2-y^2$? Express your answer as a common fraction.

\frac{1}{169}

One of the following four-digit numbers is not divisible by 4: 3544, 3554, 3564, 3572, 3576. What is the product of the units digit and the tens digit of that number?

20

The average of $x+6$, $6x+2$, and $2x+7$ is $4x-7$. What is $x$?

12

The nine horizontal and nine vertical lines on an $8\times8$ checkerboard form $r$ rectangles, of which $s$ are squares. The number $s/r$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

125

To determine the two horizontal sides of a rectangle, we have to pick two of the horizontal lines of the checkerboard, or ${9\choose 2} = 36$. Similarly, there are ${9\choose 2}$ ways to pick the vertical sides, giving us $r = 1296$ rectangles. For $s$, there are $8^2$ unit squares, $7^2$ of the $2\times2$ squares, and so on until $1^2$ of the $8\times 8$ squares. Using the sum of squares formula, that gives us $s=1^2+2^2+\cdots+8^2=\dfrac{(8)(8+1)(2\cdot8+1)}{6}=12*17=204$. Thus $\frac sr = \dfrac{204}{1296}=\dfrac{17}{108}$, and $m+n=\boxed{125}$.

Yan is somewhere between his office and a concert hall. To get to the concert hall, he can either walk directly there, or walk to his office and then take a scooter to the concert hall. He rides 5 times as fast as he walks, and both choices take the same amount of time. What is the ratio of Yan's distance from his office to his distance from the concert hall?

\frac{2}{3}

Ana and Luíza train every day for the Big Race that will take place at the end of the year at school, each running at the same speed. The training starts at point $A$ and ends at point $B$, which are $3000 \mathrm{~m}$ apart. They start at the same time, but when Luíza finishes the race, Ana still has $120 \mathrm{~m}$ to reach point $B$. Yesterday, Luíza gave Ana a chance: "We will start at the same time, but I will start some meters before point $A$ so that we arrive together." How many meters before point $A$ should Luíza start?

125

In triangle $ABC$, $AB = 3$, $BC = 4$, $AC = 5$, and $BD$ is the angle bisector from vertex $B$. If $BD = k \sqrt{2}$, then find $k$.

\frac{12}{7}

Given $2^{3x} = 128$, calculate the value of $2^{-x}$.

\frac{1}{2^{\frac{7}{3}}}

Triangle $\triangle ABC$ in the figure has area $10$. Points $D, E$ and $F$, all distinct from $A, B$ and $C$, are on sides $AB, BC$ and $CA$ respectively, and $AD = 2, DB = 3$. If triangle $\triangle ABE$ and quadrilateral $DBEF$ have equal areas, then that area is

6

1. **Identify Key Relationships**: Given that $AD = 2$, $DB = 3$, and $[ABE] = [DBEF]$, where $[X]$ denotes the area of figure $X$. 2. **Area of $\triangle ABC$**: The area of $\triangle ABC$ is given as $10$. 3. **Ratio of Segments on $AB$**: Since $AD = 2$ and $DB = 3$, the total length of $AB = AD + DB = 2 + 3 = 5$. Thus, $AD:DB = 2:3$. 4. **Area of $\triangle ABE$ Using Similarity**: Since $D$ and $E$ are points on $AB$ and $BC$ respectively, and $[ABE] = [DBEF]$, we consider the implications of the areas being equal. Let $G$ be the intersection of $AE$ and $DF$. We have: \[ [DBEF] = [DBEG] + [EFG] \quad \text{and} \quad [ABE] = [ABEG] + [ADG]. \] Since $[DBEF] = [ABE]$, it follows that $[DBEG] + [EFG] = [ABEG] + [ADG]$. Given $[DBEG] = [ABEG]$, we deduce $[EFG] = [ADG]$. 5. **Implication of Equal Areas**: The equality $[EFG] = [ADG]$ implies that $[ADF] = [AFE]$ because $[ADF] = [ADG] + [AGF]$ and $[AFE] = [EFG] + [AGF]$. This equality suggests that $AF \parallel DE$. 6. **Similarity of Triangles**: Since $AF \parallel DE$, $\triangle DBE \sim \triangle ABC$. The ratio of similarity is the same as the ratio of the corresponding sides on $AB$, which is $\frac{BD}{BA} = \frac{3}{5}$. 7. **Area of $\triangle ABE$**: Since $\triangle ABE$ is similar to $\triangle ABC$ with a ratio of $\frac{3}{5}$, the area of $\triangle ABE$ is: \[ [ABE] = \left(\frac{3}{5}\right)^2 \times [ABC] = \frac{9}{25} \times 10 = 3.6. \] However, this calculation seems incorrect as it contradicts the given areas are equal and the choice options. We need to re-evaluate the area calculation considering the linear ratio for height, not the square of the ratio (since area scales with the square of the linear dimensions, but here we consider the height directly proportional, not the square). Correcting the area calculation: \[ [ABE] = \frac{3}{5} \times [ABC] = \frac{3}{5} \times 10 = 6. \] 8. **Conclusion**: The area of $\triangle ABE$ (and hence $[DBEF]$) is $\boxed{6}$. $\blacksquare$

In the final stage of a professional bowling tournament, the competition between the top five players is conducted as follows: the fifth and fourth place players compete first, the loser gets the 5th place prize; the winner competes with the third place player, the loser gets the 4th place prize; the winner competes with the second place player, the loser gets the 3rd place prize; the winner competes with the first place player, the loser gets the 2nd place prize, and the winner gets the 1st place prize. How many different possible outcomes of the prize distribution are there?

16

The $\textit{arithmetic derivative}$ $D(n)$ of a positive integer $n$ is defined via the following rules: - $D(1) = 0$ ; - $D(p)=1$ for all primes $p$ ; - $D(ab)=D(a)b+aD(b)$ for all positive integers $a$ and $b$ . Find the sum of all positive integers $n$ below $1000$ satisfying $D(n)=n$ .

31

Let point $P$ be a point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 (a > b > 0)$. Let $F_1$ and $F_2$ respectively be the left and right foci of the ellipse, and let $I$ be the incenter of $\triangle PF_1F_2$. If $S_{\triangle IPF_1} + S_{\triangle IPF_2} = 2S_{\triangle IF_1F_2}$, then the eccentricity of the ellipse is ______.

\frac{1}{2}

Given $f(x) = x^2$ and $g(x) = |x - 1|$, let $f_1(x) = g(f(x))$, $f_{n+1}(x) = g(f_n(x))$, calculate the number of solutions to the equation $f_{2015}(x) = 1$.

2017

Pirate Bob shares his treasure with Pirate Sam in a peculiar manner. Bob first declares, ``One for me, one for you,'' keeping one coin for himself and starting Sam's pile with one coin. Then Bob says, ``Two for me, and two for you,'' adding two more coins to his pile but updating Sam's total to two coins. This continues until Bob says, ``$x$ for me, $x$ for you,'' at which he takes $x$ more coins and makes Sam's total $x$ coins in total. After all coins are distributed, Pirate Bob has exactly three times as many coins as Pirate Sam. Find out how many gold coins they have between them?

20

There are six clearly distinguishable frogs sitting in a row. Two are green, three are red, and one is blue. Green frogs refuse to sit next to the red frogs, for they are highly poisonous. In how many ways can the frogs be arranged?

24

Let $x,$ $y,$ and $z$ be positive real numbers such that $x + y + z = 1.$ Find the maximum value of $x^3 y^2 z.$

\frac{1}{432}

Compute: $9 \cdot \frac{1}{13} \cdot 26.$

18

Given that a square $S_1$ has an area of $25$, the area of the square $S_3$ constructed by bisecting the sides of $S_2$ is formed by the points of bisection of $S_2$.

6.25

On the lateral side $ CD $ of the trapezoid $ ABCD (AD \parallel BC) $, point $ M $ is marked. From vertex $ A $, a perpendicular $ AH $ is dropped onto segment $ BM $. It turns out that $ AD = HD $. Find the length of segment $ AD $ if it is known that $ BC = 16 $, $ CM = 8 $, and $ MD = 9 $.

18

Consider the case when all numbers are equal. $\frac{5}{4} n + \frac{5}{4} = n$. If the first number is -5, then all numbers will be equal to -5. The same applies to all cases where the first number is equal to $-5 + 1024n$, $n \in \mathbb{Z}$.

-5

Big Al, the ape, ate 100 bananas from May 1 through May 5. Each day he ate six more bananas than on the previous day. How many bananas did Big Al eat on May 5?

32

1. **Identify the sequence type**: Big Al's banana consumption forms an arithmetic sequence because he eats six more bananas each day than the previous day. 2. **Define the terms of the sequence**: Let $a$ be the number of bananas Big Al ate on May 1. Then, the number of bananas he ate on subsequent days can be expressed as: - May 2: $a + 6$ - May 3: $a + 12$ - May 4: $a + 18$ - May 5: $a + 24$ 3. **Set up the equation for the sum of the sequence**: The sum of an arithmetic sequence can be calculated using the formula: \[ S = \frac{n}{2} \times (\text{first term} + \text{last term}) \] where $n$ is the number of terms. Here, $n = 5$, the first term is $a$, and the last term is $a + 24$. The total number of bananas eaten over the five days is given as 100. Therefore, we have: \[ \frac{5}{2} \times (a + (a + 24)) = 100 \] 4. **Simplify and solve for $a$**: \[ \frac{5}{2} \times (2a + 24) = 100 \] \[ 5a + 60 = 100 \] \[ 5a = 40 \] \[ a = 8 \] 5. **Calculate the number of bananas eaten on May 5**: Since $a = 8$, the number of bananas eaten on May 5 is: \[ a + 24 = 8 + 24 = 32 \] 6. **Conclusion**: Big Al ate $\boxed{32}$ bananas on May 5, which corresponds to choice $\textbf{(D)}\ 32$.

Given the coefficient of determination R^2 for four different regression models, where the R^2 values are 0.98, 0.67, 0.85, and 0.36, determine which model has the best fitting effect.

0.98

Find all values of $x$ that satisfy $x=1-x+x^{2}-x^{3}+x^{4}-x^{5}+\cdots$ (be careful; this is tricky).

x=\frac{-1+\sqrt{5}}{2}

Multiplying both sides by $1+x$ gives $(1+x) x=1$, or $x=\frac{-1 \pm \sqrt{5}}{2}$. However, the series only converges for $|x|<1$, so only the answer $x=\frac{-1+\sqrt{5}}{2}$ makes sense.

What is the sum of the proper divisors of 256?

255

Find all functions $f : \mathbb{N}\rightarrow \mathbb{N}$ satisfying following condition: \[f(n+1)>f(f(n)), \quad \forall n \in \mathbb{N}.\]

{f(n)=n}

We are tasked with finding all functions $ f: \mathbb{N} \rightarrow \mathbb{N} $ that satisfy the condition: \[ f(n+1) > f(f(n)), \quad \forall n \in \mathbb{N}. \] To solve this problem, let us first analyze the condition given: \[ f(n+1) > f(f(n)). \] This inequality implies that the function $ f $ must order the values of its output in a certain way: the value of $ f(n+1) $ must always be greater than the value of $ f(f(n)) $. ### Testing Simple Functions #### Assume $ f(n) = n $ Substituting $ f(n) = n $ into the inequality gives: \[ f(n+1) = n+1 \quad \text{and} \quad f(f(n)) = f(n) = n. \] The inequality becomes: \[ n+1 > n, \] which is obviously true. Thus, $ f(n) = n $ satisfies the condition. ### Considering Other Types of Functions Next, let us consider whether other non-linear or non-identity functions might satisfy this inequality. Suppose $ f $ is a strictly increasing function with $ f(1) = 1 $: - For $ n = 1 $, we have: \[ f(2) > f(f(1)) = f(1) = 1. \] - Also, for higher $ n $, the condition consistently requires the output to be greater than the iterated value. Since any deviation from $ f(n) = n $ leads to failure under $ f(n+1) > f(f(n)) $ if $ f $ does not grow at the rate of identity (for instance, significant jumps or non-monotonic behaviors), it must preserve the identity nature to maintain all properties aligned. Thus, any form of function where $ f(n) \neq n $ leads to contradictions or failure in maintaining strict inequality for all $ n $. Given these arguments, the only function that meets the criteria is: \[ f(n) = n. \] Therefore, the solution to the functional equation under the given condition is: \[ \boxed{f(n) = n}. \]

A certain university needs $40L$ of helium gas to make balloon decorations for its centennial celebration. The chemistry club voluntarily took on this task. The club's equipment can produce a maximum of $8L$ of helium gas per day. According to the plan, the club must complete the production within 30 days. Upon receiving the task, the club members immediately started producing helium gas at a rate of $xL$ per day. It is known that the cost of raw materials for producing $1L$ of helium gas is $100$ yuan. If the daily production of helium gas is less than $4L$, the additional cost per day is $W_1=4x^2+16$ yuan. If the daily production of helium gas is greater than or equal to $4L$, the additional cost per day is $W_2=17x+\frac{9}{x}-3$ yuan. The production cost consists of raw material cost and additional cost. $(1)$ Write the relationship between the total cost $W$ (in yuan) and the daily production $x$ (in $L$). $(2)$ When the club produces how many liters of helium gas per day, the total cost is minimized? What is the minimum cost?

4640

Given that $x$ is a root of the equation $x^{2}+x-6=0$, simplify $\frac{x-1}{\frac{2}{{x-1}}-1}$ and find its value.

\frac{8}{3}

For how many integer values of $a$ does the equation $$x^2 + ax + 12a = 0$$ have integer solutions for $x$?

16

Given a circle $x^2 + (y-1)^2 = 1$ with its tangent line $l$, which intersects the positive x-axis at point A and the positive y-axis at point B. Determine the y-intercept of the tangent line $l$ when the distance AB is minimized.

\frac{3+\sqrt{5}}{2}

For each even positive integer $x$, let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square.

899

First note that $g(k)=1$ if $k$ is odd and $2g(k/2)$ if $k$ is even. so $S_n=\sum_{k=1}^{2^{n-1}}g(2k). = \sum_{k=1}^{2^{n-1}}2g(k) = 2\sum_{k=1}^{2^{n-1}}g(k) = 2\sum_{k=1}^{2^{n-2}} g(2k-1)+g(2k).$ $2k-1$ must be odd so this reduces to $2\sum_{k=1}^{2^{n-2}}1+g(2k) = 2(2^{n-2}+\sum_{k=1}^{2^n-2}g(2k)).$ Thus $S_n=2(2^{n-2}+S_{n-1})=2^{n-1}+2S_{n-1}.$ Further noting that $S_0=1$ we can see that $S_n=2^{n-1}\cdot (n-1)+2^n\cdot S_0=2^{n-1}\cdot (n-1)+2^{n-1}\cdot2=2^{n-1}\cdot (n+1).$ which is the same as above. To simplify the process of finding the largest square $S_n$ we can note that if $n-1$ is odd then $n+1$ must be exactly divisible by an odd power of $2$. However, this means $n+1$ is even but it cannot be. Thus $n-1$ is even and $n+1$ is a large even square. The largest even square $< 1000$ is $900$ so $n+1= 900 => n= \boxed{899}$

Charlie folds an $\frac{17}{2}$-inch by 11-inch piece of paper in half twice, each time along a straight line parallel to one of the paper's edges. What is the smallest possible perimeter of the piece after two such folds?

$\frac{39}{2}$

Note that a piece of paper is folded in half, one pair of opposite sides is preserved and the other pair is halved. Hence, the net effect on the perimeter is to decrease it by one of the side lengths. Hence, the original perimeter is $2\left(\frac{17}{2}\right)+2 \cdot 11=39$ and by considering the cases of folding twice along one edge or folding once along each edge, one can see that this perimeter can be decreased by at most $11+\frac{17}{2}=\frac{39}{2}$. Hence, the minimal perimeter is $\frac{39}{2}$.

Given the natural numbers $1,2,3,\ldots,10,11,12$, divide them into two groups such that the quotient of the product of all numbers in the first group by the product of all numbers in the second group is an integer and takes on the smallest possible value. What is this quotient?

231

What is the smallest positive integer with exactly 10 positive integer divisors?

48

A, B, C, D, and E are five students who obtained the top 5 positions (no ties) in a math competition. When taking a photo, they stood in a line and each of them made the following statement: A said: The two students next to me have rankings lower than mine; B said: The two students next to me have rankings adjacent to mine; C said: All students to my right (at least one) have higher rankings than mine; D said: All students to my left (at least one) have lower rankings than mine; E said: I am standing in second from the right. Given that all their statements are true, determine the five-digit number $\overline{\mathrm{ABCDE}}$.

23514

The area of the closed region formed by the line $y = nx$ and the curve $y = x^2$ is \_\_\_\_\_\_ when the binomial coefficients of the third and fourth terms in the expansion of $(x - \frac{2}{x})^n$ are equal.

\frac{125}{6}

The inverse of $f(x) = \frac{2x-1}{x+5}$ may be written in the form $f^{-1}(x)=\frac{ax+b}{cx+d}$, where $a$, $b$, $c$, and $d$ are real numbers. Find $a/c$.

-5

A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode (most frequent value). Let $D$ be the difference between the mode and the arithmetic mean of the sample. What is the largest possible value of $\lfloor D\rfloor$? (For real $x$, $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.)

947

With the same reasoning as Solution 1, in order to get largest possible value of D, we can construct that our set of numbers as $\underbrace{1,1,1...1,}_\text{n times}\underbrace{2,2,2...2,}_\text{n times}\underbrace{3,3,3...3,}_\text{n times}........\underbrace{1000,1000,1000....}_\text{n+1 times}$ And, we need to find the value of n that makes the sum as low as possible. And, we can create a formula to make it easier. It isn't hard to find the sum. The numbers which are not 1000, average to $\frac{120}{2n}$ or $\frac{60}{n}$, and there are $120-n$ of them. So, they sum to $\frac{60}{n}(120-n)$. And, the sum of the numbers that are 1000 is $1000(n+1)$ so, our total sum gets us $1000(n+1)+120/2n(120-n)$ We want to minimize it, since the mode will always be 1000. And, testing the values n = 1, n = 2, n = 3, n = 4, we get these results. $n = 1: 2000+60*119 = 9140$ $n = 2: 3000+30*118 = 6540$ $n = 3: 4000+20*117 = 6340$ $n = 4: 5000+15*116 = 6740$ And, as n grows larger and larger from 4, the values will start increasing. Thus, the lowest possible sum is 6340. Dividing by 121, the lowest possible mean is 52.396...., and thus, the highest possible value of $D$ is 947.604, and the floor of that is $\boxed{947}$ - AlexLikeMath Notes ^ In fact, when $n = 2,3,4,5$ (which some simple testing shows that the maximum will occur around), it turns out that $\frac{121-n}{n-1}$ is an integer anyway, so indeed $k = \left\lfloor \frac{121-n}{n-1} \right\rfloor = \frac{121-n}{n-1}$.

On a sphere, there are four points A, B, C, and D satisfying $AB=1$, $BC=\sqrt{3}$, $AC=2$. If the maximum volume of tetrahedron D-ABC is $\frac{\sqrt{3}}{2}$, then the surface area of this sphere is _______.

\frac{100\pi}{9}

James is six years older than Louise. Eight years from now, James will be four times as old as Louise was four years before now. What is the sum of their current ages?

26

Suppose $ a, b $, and $ c $ are real numbers with $ a < b < 0 < c $. Let $ f(x) $ be the quadratic function $ f(x) = (x-a)(x-c) $ and $ g(x) $ be the cubic function $ g(x) = (x-a)(x-b)(x-c) $. Both $ f(x) $ and $ g(x) $ have the same $ y $-intercept of -8 and $ g(x) $ passes through the point $ (-a, 8) $. Determine the value of $ c $.

\frac{8}{3}