haowu89/deepthinking_deepscale · Datasets at Hugging Face

problem stringlengths 10 2.37k	original_solution stringclasses 890 values	answer stringlengths 0 253	source stringclasses 1 value	index int64 6 40.3k
The picture shows the same die in three different positions. When the die is rolled, what is the probability of rolling a 'YES'? A) \(\frac{1}{3}\) B) \(\frac{1}{2}\) C) \(\frac{5}{9}\) D) \(\frac{2}{3}\) E) \(\frac{5}{6}\)		\frac{1}{2}	deepscale	17,255
In the triangular pyramid \(ABCD\) with base \(ABC\), the lateral edges are pairwise perpendicular, \(DA = DB = 5, DC = 1\). A ray of light is emitted from a point on the base. After reflecting exactly once from each lateral face (the ray does not reflect from the edges), the ray hits a point on the pyramid's base. What is the minimum distance the ray could travel?		\frac{10 \sqrt{3}}{9}	deepscale	28,364
Compute the smallest positive integer $k$ such that 49 divides $\binom{2 k}{k}$.	The largest $a$ such that $7^{a} \left\lvert\,\binom{ 2 k}{k}\right.$ is equal to the number of carries when you add $k+k$ in base 7 , by Kummer's Theorem. Thus, we need two carries, so $2 k$ must have at least 3 digits in base 7 . Hence, $k \geq 25$. We know $k=25$ works because $25+25=34_{7}+34_{7}=101_{7}$ has two carries.	25	deepscale	4,409
The three sides of a triangle are 25, 39, and 40. Find the diameter of its circumscribed circle.		\frac{125}{3}	deepscale	22,122
In the regular quadrangular pyramid \(P-ABCD\), \(M\) and \(N\) are the midpoints of \(PA\) and \(PB\) respectively. If the tangent of the dihedral angle between a side face and the base is \(\sqrt{2}\), find the cosine of the angle between skew lines \(DM\) and \(AN\).		1/6	deepscale	14,138
A rectangular piece of paper with vertices $A B C D$ is being cut by a pair of scissors. The pair of scissors starts at vertex $A$, and then cuts along the angle bisector of $D A B$ until it reaches another edge of the paper. One of the two resulting pieces of paper has 4 times the area of the other piece. What is the ratio of the longer side of the original paper to the shorter side?	Without loss of generality, let $A B>A D$, and let $x=A D, y=A B$. Let the cut along the angle bisector of $\angle D A B$ meet $C D$ at $E$. Note that $A D E$ is a $45-45-90$ triangle, so $D E=A D=x$, and $E C=y-x$. Now, $[A D E]=\frac{x^{2}}{2}$, and $[A E C B]=x\left(y-\frac{x}{2}\right)=4[A D E]$. Equating and dividing both sides by $x$, we find that $2 x=y-\frac{x}{2}$, so $y / x=\frac{5}{2}$.	\frac{5}{2}	deepscale	4,934
Let $n > 2$ be an integer and let $\ell \in \{1, 2,\dots, n\}$. A collection $A_1,\dots,A_k$ of (not necessarily distinct) subsets of $\{1, 2,\dots, n\}$ is called $\ell$-large if $\|A_i\| \ge \ell$ for all $1 \le i \le k$. Find, in terms of $n$ and $\ell$, the largest real number $c$ such that the inequality \[ \sum_{i=1}^k\sum_{j=1}^k x_ix_j\frac{\|A_i\cap A_j\|^2}{\|A_i\|\cdot\|A_j\|}\ge c\left(\sum_{i=1}^k x_i\right)^2 \] holds for all positive integer $k$, all nonnegative real numbers $x_1,x_2,\dots,x_k$, and all $\ell$-large collections $A_1,A_2,\dots,A_k$ of subsets of $\{1,2,\dots,n\}$.	To solve the problem, we need to find the largest real number \( c \) such that the inequality \[ \sum_{i=1}^k \sum_{j=1}^k x_i x_j \frac{\|A_i \cap A_j\|^2}{\|A_i\| \cdot \|A_j\|} \ge c \left(\sum_{i=1}^k x_i\right)^2 \] holds for all positive integers \( k \), all nonnegative real numbers \( x_1, x_2, \dots, x_k \), and all \(\ell\)-large collections \( A_1, A_2, \dots, A_k \) of subsets of \(\{1, 2, \dots, n\}\). ### Step-by-Step Solution 1. Understanding the Constraints: Each \( A_i \) is a subset of \(\{1, 2, \dots, n\}\) with \(\|A_i\| \geq \ell\). The sets are \(\ell\)-large, meaning every set has at least \(\ell\) elements. 2. Expression Simplification: The expression on the left side of the inequality involves the squared size of the intersections \( \|A_i \cap A_j\| \) normalized by the sizes of \( \|A_i\| \) and \( \|A_j\| \). 3. Cauchy-Schwarz Application: To handle the sum of squares, we consider applying the Cauchy-Schwarz inequality in terms of sums and intersections: \[ \left(\sum_{i=1}^k x_i\right)^2 \leq k \sum_{i=1}^k x_i^2. \] 4. Bounding the Intersection Size: Since \( \|A_i\|, \|A_j\| \geq \ell \), the intersection \( \|A_i \cap A_j\| \) can be at most \(\min(\|A_i\|, \|A_j\|)\), but more typically involves sizing relative to \( n \), such as \(\|A_i \cap A_j\| \leq \ell\). 5. Finding \( c \): The challenge is finding a universal lower bound on the given expression. Consider setting boundaries based on specific configurations of \( A_i \) making the set sizes minimal at \(\ell\). Assume: \[ \|A_i\| = \ell \quad \text{for all } i, \] then we simplify the inequality's left side, using symmetry and the fact \( \|A_i \cap A_j\| \) can be estimated within strict bounds for large \( n \). The strategy is identifying the smallest reliable bound for: \[ \frac{\ell^2 - 2\ell + n}{n(n-1)}. \] ### Conclusion Thus, after considering possible configurations and analytic optimization, the bound for the largest real number \( c \) that satisfies the inequality for all valid configurations is: \[ \boxed{\frac{\ell^2 - 2\ell + n}{n(n-1)}}. \]	\frac{\ell^2 - 2\ell + n}{n(n-1)}	deepscale	5,985
Simplify and find the value of: $a^{2}-\left(a-\dfrac{2a}{a+1}\right)\div \dfrac{a^{2}-2a+1}{a^{2}-1}$, where $a$ is a solution of the equation $x^{2}-x-\dfrac{7}{2}=0$.		\dfrac{7}{2}	deepscale	23,529
When five students are lining up to take a photo, and two teachers join in, with the order of the five students being fixed, calculate the total number of ways for the two teachers to stand in line with the students for the photo.		42	deepscale	32,986
Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD}, BC=CD=43$, and $\overline{AD}\perp\overline{BD}$. Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$, and let $P$ be the midpoint of $\overline{BD}$. Given that $OP=11$, the length of $AD$ can be written in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$?	1. Identify the properties of the trapezoid: Given that $ABCD$ is a trapezoid with $\overline{AB}\parallel\overline{CD}$ and $BC=CD=43$. Also, $\overline{AD}\perp\overline{BD}$, which implies that $\triangle ABD$ is a right triangle. 2. Intersection and midpoint properties: Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$, and $P$ be the midpoint of $\overline{BD}$. Given $OP=11$. 3. Extend lines and identify congruent triangles: Extend $\overline{AD}$ and $\overline{BC}$ to meet at point $E$. Since $\overline{AD}\perp\overline{BD}$, $\angle BDE = 90^\circ$. By the properties of trapezoids and the congruence of triangles $\triangle BPC$ and $\triangle DPC$, we have $\angle DBE \cong \angle DBA$. By ASA (Angle-Side-Angle) congruence, $\triangle ABD \cong \triangle EBD$, and thus $AD = ED$. This implies $D$ is the midpoint of $\overline{AE}$. 4. Midpoint of $\overline{DE}$ and congruence: Let $M$ be the midpoint of $\overline{DE}$. Since $\triangle CME$ is congruent to $\triangle BPC$, we have $BC = CE$, making $C$ the midpoint of $\overline{BE}$. 5. Centroid and median properties: $\overline{AC}$ and $\overline{BD}$ are medians of $\triangle ABE$, making $O$ the centroid of $\triangle ABE$. The centroid divides each median in a 2:1 ratio, so $\frac{BO}{2} = DO = \frac{BD}{3}$. Given $P$ is the midpoint of $BD$, $DP = \frac{BD}{2}$. Given $OP = 11$, we have $DP - DO = 11$, leading to $\frac{BD}{2} - \frac{BD}{3} = 11 \implies \frac{BD}{6} = 11 \implies BD = 66$. 6. Calculate $AB$ and use the Pythagorean theorem: Since $\triangle ABD$ is similar to $\triangle CBP$ and $\triangle CPD$ by a factor of 2, $AB = 2 \cdot 43 = 86$. Applying the Pythagorean theorem in $\triangle ABD$, we have: \[ AB^2 - BD^2 = AD^2 \implies 86^2 - 66^2 = AD^2 \implies 7396 - 4356 = AD^2 \implies 3040 = AD^2 \implies AD = \sqrt{3040} = 4\sqrt{190} \] 7. Final answer: The length of $AD$ is $4\sqrt{190}$, where $m=4$ and $n=190$. Thus, $m+n = 4+190 = \boxed{\textbf{(D) }194}$. $\blacksquare$	194	deepscale	443
Given a $5 \times 5$ grid where the number in the $i$-th row and $j$-th column is denoted by \( a_{ij} \) (where \( a_{ij} \in \{0, 1\} \)), with the condition that \( a_{ij} = a_{ji} \) for \( 1 \leq i, j \leq 5 \). Calculate the total number of ways to fill the grid such that there are exactly five 1's in the grid.		326	deepscale	13,883
Determine the coefficient of the $x^5$ term in the expansion of $(x+1)(x^2-x-2)^3$.		-6	deepscale	22,395
A person's age at the time of their death was one 31st of their birth year. How old was this person in 1930?		39	deepscale	14,969
Find the maximum value of $\cos x + 2 \sin x,$ over all angles $x.$		\sqrt{5}	deepscale	39,817
How many triangles exist such that the lengths of the sides are integers not greater than 10?		125	deepscale	14,062
An integer $n>1$ is given . Find the smallest positive number $m$ satisfying the following conditions： for any set $\{a,b\}$ $\subset \{1,2,\cdots,2n-1\}$ ,there are non-negative integers $ x, y$ ( not all zero) such that $2n\|ax+by$ and $x+y\leq m.$	Given an integer \( n > 1 \), we aim to find the smallest positive number \( m \) satisfying the following conditions: for any set \(\{a, b\} \subset \{1, 2, \ldots, 2n-1\}\), there exist non-negative integers \( x \) and \( y \) (not both zero) such that \( 2n \mid ax + by \) and \( x + y \leq m \). To determine the smallest \( m \), we analyze the conditions: 1. Consider \( a = 1 \) and \( b = 2 \). If \( 2n \mid ax + by \), then: \[ 2n \leq x + 2y \leq 2(x + y) \leq 2m. \] This implies \( m \geq n \). 2. We now show that \( m \leq n \). - Case 1: If \(\gcd(a, 2n) > 1\) or \(\gcd(b, 2n) > 1\). Without loss of generality, assume \(\gcd(a, 2n) > 1\). Choose \( x = \frac{2n}{\gcd(a, 2n)} \) and \( y = 0 \). Then: \[ x + y = \frac{2n}{\gcd(a, 2n)} \leq \frac{2n}{2} = n. \] - Case 2: If \(\gcd(a, 2n) = 1\) and \(\gcd(b, 2n) = 1\). Let \( c \in [0, 2n-1] \) such that \( c \equiv ba^{-1} \pmod{2n} \). The equation \( ax + by \equiv 0 \pmod{2n} \) is equivalent to \( x + cy \equiv 0 \pmod{2n} \). Choose \( y = \left\lfloor \frac{2n}{c} \right\rfloor \) and \( x = 2n - c \left\lfloor \frac{2n}{c} \right\rfloor \). - Subcase 2.1: If \( 2 < c < n \), then: \[ x + y \leq c - 1 + \left\lfloor \frac{2n}{c} \right\rfloor < c - 1 + \frac{2n}{c} = \frac{(c-2)(c-n)}{c} + n + 1 < n + 1. \] Hence, \( x + y \leq n \). - Subcase 2.2: If \( c \geq n + 1 \), then: \[ x + y = 2n - (c-1) \left\lfloor \frac{2n}{c} \right\rfloor = 2n - (c-1) \leq 2n - (n+1-1) = n. \] Hence, \( x + y \leq n \). In conclusion, the smallest positive number \( m \) that satisfies the given conditions is \( n \). The answer is \(\boxed{n}\).	n	deepscale	2,924
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}	deepscale	24,408
Non-negative numbers \(a\) and \(b\) satisfy the equations \(a^2 + b^2 = 74\) and \(ab = 35\). What is the value of the expression \(a^2 - 12a + 54\)?		19	deepscale	12,082
Compute $$\sum_{k=1}^{2000} k(\lceil \log_{2}{k}\rceil- \lfloor\log_{2}{k} \rfloor).$$		1998953	deepscale	17,520
In right triangle $ABC$ with $\angle A = 90^\circ$, we have $AB =16$ and $BC = 24$. Find $\sin A$.		1	deepscale	36,256
Two integers have a sum of $26$. When two more integers are added to the first two, the sum is $41$. Finally, when two more integers are added to the sum of the previous $4$ integers, the sum is $57$. What is the minimum number of even integers among the $6$ integers?	1. Identify the sums at each stage: - Let the first two integers be $x$ and $y$. We know $x + y = 26$. - Let the next two integers added be $a$ and $b$. Then, $x + y + a + b = 41$. - Let the final two integers added be $m$ and $n$. Then, $x + y + a + b + m + n = 57$. 2. Calculate the sums of the additional integers: - From $x + y = 26$ to $x + y + a + b = 41$, the sum of $a$ and $b$ is $41 - 26 = 15$. - From $x + y + a + b = 41$ to $x + y + a + b + m + n = 57$, the sum of $m$ and $n$ is $57 - 41 = 16$. 3. Analyze the parity of the sums: - The sum $x + y = 26$ is even. This can be achieved with two even numbers or two odd numbers. - The sum $a + b = 15$ is odd. This requires one even and one odd integer (since the sum of two odd numbers or two even numbers is even). - The sum $m + n = 16$ is even. This can be achieved with two even numbers or two odd numbers. 4. Determine the minimum number of even integers: - For $x + y = 26$, we can choose both $x$ and $y$ to be odd (e.g., $x = 13$, $y = 13$), requiring $0$ even integers. - For $a + b = 15$, we must have one even integer (e.g., $a = 8$, $b = 7$). - For $m + n = 16$, we can choose both $m$ and $n$ to be odd (e.g., $m = 7$, $n = 9$), requiring $0$ even integers. 5. Conclusion: - The minimum number of even integers required among the six integers is $1$ (from the pair $a$ and $b$). Thus, the answer is $\boxed{\textbf{(A)}\ 1}$.	1	deepscale	2,103
A soccer ball kicked vertically upward reaches, after $t$ seconds, a height of $s$ meters where $s = 180t - 20t^2$. Find the maximum height reached by the ball.		405	deepscale	17,145
What is the value of $m$ if Tobias downloads $m$ apps, each app costs $\$ 2.00$ plus $10 \%$ tax, and he spends $\$ 52.80$ in total on these $m$ apps?	Since the tax rate is $10 \%$, then the tax on each $\$ 2.00$ app is $\$ 2.00 \times \frac{10}{100}=\$ 0.20$. Therefore, including tax, each app costs $\$ 2.00+\$ 0.20=\$ 2.20$. Since Tobias spends $\$ 52.80$ on apps, he downloads $\frac{\$ 52.80}{\$ 2.20}=24$ apps. Therefore, $m=24$.	24	deepscale	5,501
There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$, the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$, and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime integers. Find $p+q$.	Note that if $\tan \theta$ is positive, then $\theta$ is in the first or third quadrant, so $0^{\circ} < \theta < 90^{\circ} \pmod{180^{\circ}}$. Also notice that the only way $\tan{\left(2^{n}\theta\right)}$ can be positive for all $n$ that are multiples of $3$ is when $2^0\theta, 2^3\theta, 2^6\theta$, etc. are all the same value $\pmod{180^{\circ}}$. (This also must be true in order for $\theta$ to be unique.) This is the case if $8\theta = \theta \pmod{180^{\circ}}$, so $7\theta = 0^{\circ} \pmod{180^{\circ}}$. Therefore, the only possible values of $\theta$ between $0^{\circ}$ and $90^{\circ}$ are $\frac{180}{7}^{\circ}$, $\frac{360}{7}^{\circ}$, and $\frac{540}{7}^{\circ}$. $\frac{180}{7}^{\circ}$ does not work since $\tan(2 \cdot \frac{180}{7}^{\circ})$ is positive, and $\frac{360}{7}^{\circ}$ does not work because $\tan(4 \cdot \frac{360}{7}^{\circ})$ is positive. Thus, $\theta = \frac{540}{7}^{\circ}$, and a quick check verifies that it works. $540 + 7 = \boxed{547}$.	547	deepscale	7,265
Brachycephalus frogs have three toes on each foot and two fingers on each hand. The common frog has five toes on each foot and four fingers on each hand. Some Brachycephalus and common frogs are in a bucket. Each frog has all its fingers and toes. Between them they have 122 toes and 92 fingers. How many frogs are in the bucket? A 15 B 17 C 19 D 21 E 23		15	deepscale	11,543
A bar of chocolate is made of 10 distinguishable triangles as shown below. How many ways are there to divide the bar, along the edges of the triangles, into two or more contiguous pieces?	Every way to divide the bar can be described as a nonempty set of edges to break, with the condition that every endpoint of a broken edge is either on the boundary of the bar or connects to another broken edge. Let the center edge have endpoints $X$ and $Y$. We do casework on whether the center edge is broken. If the center edge is broken, then we just need some other edge connecting to $X$ to be broken, and some other edge connecting to $Y$ to be broken. We have $2^{5}$ choices for the edges connecting to $X$, of which 1 fails. Similarly, we have $2^{5}-1$ valid choices for the edges connecting to $Y$. This yields $\left(2^{5}-1\right)^{2}=961$ possibilities. If the center edge is not broken, then the only forbidden arrangements are those with exactly one broken edge at $X$ or those with exactly one broken edge at $Y$. Looking at just the edges connecting to $X$, we have 5 cases with exactly one broken edge. Thus, there are $2^{5}-5=27$ ways to break the edges connecting to $X$. Similarly there are 27 valid choices for the edges connecting to $Y$. This yields $27^{2}-1=728$ cases, once we subtract the situation where no edges are broken. The final answer is $961+728=1689$.	1689	deepscale	4,801
Ryan has 3 red lava lamps and 3 blue lava lamps. He arranges them in a row on a shelf randomly, and then randomly turns 3 of them on. What is the probability that the leftmost lamp is blue and off, and the rightmost lamp is red and on?		\dfrac{9}{100}	deepscale	34,810
A cone is perfectly fitted inside a cube such that the cone's base is one face of the cube and its vertex touches the opposite face. A sphere is inscribed in the same cube. Given that one edge of the cube is 8 inches, calculate: 1. The volume of the inscribed sphere. 2. The volume of the inscribed cone. Express your answer in terms of $\pi$.		\frac{128}{3}\pi	deepscale	9,492
Natural numbers \( a, b, c \) are chosen such that \( a < b < c \). It is also known that the system of equations \( 2x + y = 2025 \) and \( y = \|x - a\| + \|x - b\| + \|x - c\| \) has exactly one solution. Find the minimum possible value of \( c \).		1013	deepscale	31,607
What is the greatest common multiple of 8 and 12 that is less than 90?		72	deepscale	38,954
Remove all perfect squares from the sequence of positive integers \(1, 2, 3, \cdots\) to get a new sequence, and calculate the 2003rd term of this new sequence.		2047	deepscale	25,673
Three fair coins are to be tossed once. For each head that results, one fair die is to be rolled. If all three coins show heads, roll an additional fourth die. Determine the probability that the sum of the die rolls is odd.		\frac{7}{16}	deepscale	7,715
There exist $r$ unique nonnegative integers $n_1 > n_2 > \cdots > n_r$ and $r$ unique integers $a_k$ ($1\le k\le r$) with each $a_k$ either $1$ or $- 1$ such that\[a_13^{n_1} + a_23^{n_2} + \cdots + a_r3^{n_r} = 2008.\]Find $n_1 + n_2 + \cdots + n_r$.		21	deepscale	38,172
Steven subtracts the units digit from the tens digit for each two-digit number. He then finds the sum of all his answers. What is the value of Steven's sum?		45	deepscale	23,394
Alice conducted a survey among a group of students regarding their understanding of snakes. She found that $92.3\%$ of the students surveyed believed that snakes are venomous. Of the students who believed this, $38.4\%$ erroneously thought that all snakes are venomous. Knowing that only 31 students held this incorrect belief, calculate the total number of students Alice surveyed.		88	deepscale	29,235
If A and B can only undertake the first three tasks, while the other three can undertake all four tasks, calculate the total number of different selection schemes for the team leader group to select four people from five volunteers to undertake four different tasks.		72	deepscale	19,445
Convert the binary number $11011001_2$ to base 4.		3121_4	deepscale	8,715
A ball travels on a parabolic path in which the height (in feet) is given by the expression $-16t^2+32t+15$, where $t$ is the time after launch. What is the maximum height of the ball, in feet?		31	deepscale	34,454
How many 9 step paths are there from $E$ to $G$ which pass through $F$?[asy]size(4cm,4cm);int w=6;int h=5;int i;pen p=fontsize(9);for (i=0; i<h; ++i){draw((0,i) -- (w-1,i));}for (i=0; i<w; ++i){draw((i, 0)--(i,h-1));}label("G", (w-1,0), SE, p);label("E", (0,h-1), NW, p);label("F", (3,3), NE, p);[/asy]		40	deepscale	34,723
Let $ ABP, BCQ, CAR$ be three non-overlapping triangles erected outside of acute triangle $ ABC$. Let $ M$ be the midpoint of segment $ AP$. Given that $ \angle PAB \equal{} \angle CQB \equal{} 45^\circ$, $ \angle ABP \equal{} \angle QBC \equal{} 75^\circ$, $ \angle RAC \equal{} 105^\circ$, and $ RQ^2 \equal{} 6CM^2$, compute $ AC^2/AR^2$. [i]Zuming Feng.[/i]	Let \( ABP, BCQ, CAR \) be three non-overlapping triangles erected outside of acute triangle \( ABC \). Let \( M \) be the midpoint of segment \( AP \). Given that \( \angle PAB = \angle CQB = 45^\circ \), \( \angle ABP = \angle QBC = 75^\circ \), \( \angle RAC = 105^\circ \), and \( RQ^2 = 6CM^2 \), we aim to compute \( \frac{AC^2}{AR^2} \). Construct parallelogram \( CADP \). Claim: \( \triangle AQR \sim \triangle ADC \). Proof: Observe that \( \triangle BPA \sim \triangle BCQ \), hence \( \triangle BAQ \sim \triangle BPC \). Consequently, \[ \frac{AQ}{AD} = \frac{AQ}{CP} = \frac{BP}{BA} = \sqrt{\frac{3}{2}} = \frac{QR}{DC}. \] Since \( \angle RAC = 105^\circ \) and \( \angle QAD = \angle CPA + \angle QAP = 180^\circ - \angle (CP, AQ) = 180^\circ - \angle ABP = 105^\circ \), we can use SSA similarity (since \( 105^\circ > 90^\circ \)) to conclude that \( \triangle AQR \sim \triangle ADC \). Thus, it follows that \[ \frac{AC^2}{AR^2} = \frac{2}{3}. \] The answer is: \(\boxed{\frac{2}{3}}\).	\frac{2}{3}	deepscale	2,903
Petya and Vasya are playing the following game. Petya thinks of a natural number \( x \) with a digit sum of 2012. On each turn, Vasya chooses any natural number \( a \) and finds out the digit sum of the number \( \|x-a\| \) from Petya. What is the minimum number of turns Vasya needs to determine \( x \) with certainty?		2012	deepscale	22,092
If three different numbers are selected from 2, 3, 4, 5, 6 to be $a$, $b$, $c$ such that $N = abc + ab + bc + a - b - c$ reaches its maximum value, then this maximum value is.		167	deepscale	25,503
A shape is created by aligning five unit cubes in a straight line. Then, one additional unit cube is attached to the top of the second cube in the line and another is attached beneath the fourth cube in the line. Calculate the ratio of the volume to the surface area.		\frac{1}{4}	deepscale	28,060
A cube \(ABCDA_1B_1C_1D_1\) has edge length 1. Point \(M\) is taken on the side diagonal \(A_1D\), and point \(N\) is taken on \(CD_1\), such that the line segment \(MN\) is parallel to the diagonal plane \(A_1ACC_1\). Find the minimum length of \(MN\).		\frac{\sqrt{3}}{3}	deepscale	31,621
Rhombus $ABCD$ has side length $3$ and $\angle B = 110$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$? A) $0.81$ B) $1.62$ C) $2.43$ D) $2.16$ E) $3.24$		2.16	deepscale	31,849
Translate the graph of the function y = 2sin( $$\frac {π}{3}$$ - x) - cos( $$\frac {π}{6}$$ + x) by shifting it to the right by $$\frac {π}{4}$$ units. Determine the minimum value of the corresponding function.		-1	deepscale	18,241
On an island, there are knights, liars, and followers; each person knows who is who. All 2018 island residents were lined up and each was asked to answer "Yes" or "No" to the question: "Are there more knights than liars on the island?" The residents responded one by one in such a way that the others could hear. Knights always told the truth, liars always lied. Each follower answered the same as the majority of the preceding respondents, and if the "Yes" and "No" answers were split equally, they could give either answer. It turned out that there were exactly 1009 "Yes" answers. What is the maximum number of followers that could be among the island residents?		1009	deepscale	18,047
What is the median of the following list of $4040$ numbers? \[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\]	1. Identify the total number of terms and the position of the median: The list consists of $4040$ numbers, which includes all integers from $1$ to $2020$ and their squares. Since the list has an even number of terms, the median will be the average of the $2020$-th and $2021$-st terms. 2. Determine the range of perfect squares within the list: We need to find the largest perfect square less than or equal to $2020$. We calculate: \[ 44^2 = 1936 \quad \text{and} \quad 45^2 = 2025 \] Since $45^2$ is greater than $2020$, the perfect squares up to $44^2$ are all less than or equal to $2020$. 3. Count the number of terms up to $2020$: There are $2020$ integers from $1$ to $2020$, and there are $44$ perfect squares from $1^2$ to $44^2$. Thus, there are: \[ 2020 + 44 = 2064 \text{ terms less than or equal to } 2020. \] 4. Locate the $2020$-th and $2021$-st terms: Since $2064$ terms are less than or equal to $2020$, the $2020$-th and $2021$-st terms are among the integers and not among the squares of integers greater than $44$. Specifically, these terms are among the integers from $1$ to $2020$. 5. Calculate the $2020$-th and $2021$-st terms: The $2020$-th term is the last integer in the sequence from $1$ to $2020$, which is $2020$. The $2021$-st term is the first integer square greater than $44^2$, which is $45^2 = 2025$. 6. Correct the identification of the $2020$-th and $2021$-st terms: Since we have $2064$ terms up to $2020$, and we need the $2020$-th and $2021$-st terms, we actually need to consider the terms just before reaching $2020$ in the sequence. We calculate: \[ 2064 - 2020 = 44 \quad \text{and} \quad 2064 - 2021 = 43 \] Thus, the $2020$-th term is $2020 - 44 = 1976$ and the $2021$-st term is $2020 - 43 = 1977$. 7. Calculate the median: The median is the average of the $2020$-th and $2021$-st terms: \[ \text{Median} = \frac{1976 + 1977}{2} = 1976.5 \] Thus, the median of the list is $\boxed{\textbf{(C)}\ 1976.5}$.	1976.5	deepscale	1,727
At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?	1. Assume Total Number of Students: Let's assume there are 100 students at Typico High School for simplicity in calculation. 2. Students Who Like Dancing: - Given that 60% of the students like dancing. - Therefore, $60\% \times 100 = 60$ students like dancing. 3. Students Who Dislike Dancing: - The rest of the students dislike dancing, which is $100 - 60 = 40$ students. 4. Students Who Like Dancing but Say They Dislike It: - Of the students who like dancing, 20% say that they dislike it. - Therefore, $20\% \times 60 = 0.20 \times 60 = 12$ students like dancing but say they dislike it. 5. Students Who Dislike Dancing and Say So: - Of the students who dislike dancing, 90% say that they dislike it. - Therefore, $90\% \times 40 = 0.90 \times 40 = 36$ students dislike dancing and say so. 6. Total Students Who Say They Dislike Dancing: - This includes students who dislike dancing and say so, and students who like dancing but say they dislike it. - Total = Students who dislike and say so + Students who like but say they dislike = $36 + 12 = 48$ students. 7. Fraction of Students Who Say They Dislike Dancing but Actually Like It: - We need to find the fraction of students who say they dislike dancing but actually like it. - This is the number of students who like dancing but say they dislike it divided by the total number of students who say they dislike dancing. - Fraction = $\frac{\text{Students who like but say dislike}}{\text{Total who say dislike}} = \frac{12}{48} = \frac{1}{4}$. 8. Convert Fraction to Percentage: - To find the percentage, multiply the fraction by 100%. - Percentage = $\frac{1}{4} \times 100\% = 25\%$. Thus, the fraction of students who say they dislike dancing but actually like it is $\boxed{25\%}$, corresponding to choice $\textbf{(D)}$.	$25\%$	deepscale	927
Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number?	As shown above, we have $1000x+y=9xy$, so $1000/y=9-1/x$. $1000/y$ must be just a little bit smaller than 9, so we find $y=112$, $x=14$, and the solution is $\boxed{126}$.	126	deepscale	6,628
In the figure with circle $Q$, angle $KAT$ measures 42 degrees. What is the measure of minor arc $AK$ in degrees? [asy] import olympiad; size(150); defaultpen(linewidth(0.8)); dotfactor=4; draw(unitcircle); draw(dir(84)--(-1,0)--(1,0)); dot("$A$",(-1,0),W); dot("$K$",dir(84),NNE); dot("$T$",(1,0),E); dot("$Q$",(0,0),S); [/asy]		96	deepscale	36,301
Three tanks contain water. The number of litres in each is shown in the table: Tank A: 3600 L, Tank B: 1600 L, Tank C: 3800 L. Water is moved from each of Tank A and Tank C into Tank B so that each tank contains the same volume of water. How many litres of water are moved from Tank A to Tank B?	In total, the three tanks contain $3600 \mathrm{~L} + 1600 \mathrm{~L} + 3800 \mathrm{~L} = 9000 \mathrm{~L}$. If the water is divided equally between the three tanks, each will contain $\frac{1}{3} \cdot 9000 \mathrm{~L} = 3000 \mathrm{~L}$. Therefore, $3600 \mathrm{~L} - 3000 \mathrm{~L} = 600 \mathrm{~L}$ needs to be moved from Tank A to Tank B.	600	deepscale	5,369
Consider a month with 31 days, where the number of the month is a product of two distinct primes (e.g., July, represented as 7). Determine how many days in July are relatively prime to the month number.		27	deepscale	30,593
In the game of preference, each of the three players is dealt 10 cards, and two cards are placed in the kitty. How many different arrangements are possible in this game? (Consider possible distributions without accounting for which 10 cards go to each specific player.)		\frac{32!}{(10!)^3 \cdot 2! \cdot 3!}	deepscale	12,964
Given 4 distinct books that are to be distributed evenly between two students, find the probability that the Chinese language book and the Mathematics book are given to the same student.		\frac{1}{3}	deepscale	7,395
Find the largest positive number \( c \) such that for every natural number \( n \), the inequality \( \{n \sqrt{2}\} \geqslant \frac{c}{n} \) holds, where \( \{n \sqrt{2}\} = n \sqrt{2} - \lfloor n \sqrt{2} \rfloor \) and \( \lfloor x \rfloor \) denotes the integer part of \( x \). Determine the natural number \( n \) for which \( \{n \sqrt{2}\} = \frac{c}{n} \). (This problem appeared in the 30th International Mathematical Olympiad, 1989.)		\frac{1}{2\sqrt{2}}	deepscale	26,017
For any real number $x$, the symbol $\lfloor x \rfloor$ represents the integer part of $x$, that is, $\lfloor x \rfloor$ is the largest integer not exceeding $x$. Calculate the value of $\lfloor \log_{2}1 \rfloor + \lfloor \log_{2}2 \rfloor + \lfloor \log_{2}3 \rfloor + \lfloor \log_{2}4 \rfloor + \ldots + \lfloor \log_{2}1024 \rfloor$.		8204	deepscale	18,535
What is the last two digits of the decimal representation of $9^{8^{7^{\cdot^{\cdot^{\cdot^{2}}}}}}$ ?		21	deepscale	19,962
Find the number of integers between 1 and 2013 with the property that the sum of its digits equals 9.		101	deepscale	13,288
Calculate the volumes of the solids formed by rotating the regions bounded by the graphs of the functions around the y-axis. $$ y = \arcsin x, \quad y = \arccos x, \quad y = 0 $$		\frac{\pi}{2}	deepscale	16,048
In $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$, respectively. Given that $c\sin A= \sqrt {3}a\cos C$ and $(a-c)(a+c)=b(b-c)$, consider the function $f(x)=2\sin x\cos ( \frac {π}{2}-x)- \sqrt {3}\sin (π+x)\cos x+\sin ( \frac {π}{2}+x)\cos x$. (1) Find the period and the equation of the axis of symmetry of the function $y=f(x)$. (2) Find the value of $f(B)$.		\frac {5}{2}	deepscale	17,679
Given that $y=f\left(x\right)+x^{2}$ is an odd function, and $f\left(1\right)=1$, if $g\left(x\right)=f\left(x\right)+2$, then $g\left(-1\right)=$____.		-1	deepscale	26,429
Let $c = \frac{2\pi}{11}.$ What is the value of \[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?\]	1. Define the constant and simplify the expression: Let \( c = \frac{2\pi}{11} \). We need to evaluate: \[ \frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c} \] 2. Substitute \( c \) into the expression: \[ \frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}} \] 3. Use the properties of sine function: - \(\sin(x + 2\pi) = \sin(x)\) for periodicity. - \(\sin(2\pi - x) = \sin(x)\) for symmetry. - \(\sin(-x) = -\sin(x)\) for odd function property. Applying these properties: \[ \sin \frac{12\pi}{11} = \sin(2\pi - \frac{12\pi}{11}) = \sin \frac{10\pi}{11} \] \[ \sin \frac{18\pi}{11} = \sin(2\pi - \frac{18\pi}{11}) = \sin \frac{16\pi}{11} = \sin(2\pi - \frac{16\pi}{11}) = \sin \frac{4\pi}{11} \] \[ \sin \frac{24\pi}{11} = \sin(2\pi - \frac{24\pi}{11}) = \sin \frac{20\pi}{11} = \sin(2\pi - \frac{20\pi}{11}) = \sin \frac{2\pi}{11} \] \[ \sin \frac{30\pi}{11} = \sin(2\pi - \frac{30\pi}{11}) = \sin \frac{8\pi}{11} \] 4. Substitute back and simplify: \[ \frac{\sin \frac{6\pi}{11} \cdot \sin \frac{10\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{2\pi}{11} \cdot \sin \frac{8\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}} \] Notice that each term in the numerator matches a corresponding term in the denominator. Therefore, the entire expression simplifies to 1. 5. Conclude with the final answer: \[ \boxed{\textbf{(E)}\ 1} \]	1	deepscale	1,064
Triangle $PQR$ has positive integer side lengths with $PQ = PR$. Let $J$ be the intersection of the bisectors of $\angle Q$ and $\angle R$. Suppose $QJ = 10$. Find the smallest possible perimeter of $\triangle PQR$.		1818	deepscale	28,023
For some positive integer $k$, when 60 is divided by $k^2$, the remainder is 6. What is the remainder when 100 is divided by $k$?		1	deepscale	38,362
Today is December 19, 2010. What is the integer part of the sum $\frac{2010}{1000}+\frac{1219}{100}+\frac{27}{10}$?		16	deepscale	10,755
Given the new operation $n\heartsuit m=n^{3+m}m^{2+n}$, evaluate $\frac{2\heartsuit 4}{4\heartsuit 2}$.		\frac{1}{2}	deepscale	18,914
Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ with left and right foci $F_{1}$ and $F_{2}$ respectively, draw a line $l$ through the right focus that intersects the ellipse at points $P$ and $Q$. Find the maximum area of the inscribed circle of triangle $F_{1} PQ$.		\frac{9 \pi}{16}	deepscale	15,855
The water tank in the diagram below is in the shape of an inverted right circular cone. The radius of its base is 16 feet, and its height is 96 feet. The water in the tank is $25\%$ of the tank's capacity. The height of the water in the tank can be written in the form $a\sqrt[3]{b}$, where $a$ and $b$ are positive integers and $b$ is not divisible by a perfect cube greater than 1. What is $a+b$? [asy] size(150); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw(shift(0,96)yscale(0.5)Circle((0,0),16)); draw((-16,96)--(0,0)--(16,96)--(0,96)); draw(scale(0.75)shift(0,96)yscale(0.5)*Circle((0,0),16)); draw((-18,72)--(-20,72)--(-20,0)--(-18,0)); label("water's height",(-20,36),W); draw((20,96)--(22,96)--(22,0)--(20,0)); label("96'",(22,48),E); label("16'",(8,96),S); [/asy]		50	deepscale	36,275
What is $\frac{~\frac{2}{5}~}{\frac{3}{7}}$?		\frac{14}{15}	deepscale	38,768
The positive integers are arranged in rows and columns as shown below. \| Row 1 \| 1 \| \| Row 2 \| 2 \| 3 \| \| Row 3 \| 4 \| 5 \| 6 \| \| Row 4 \| 7 \| 8 \| 9 \| 10 \| \| Row 5 \| 11 \| 12 \| 13 \| 14 \| 15 \| \| Row 6 \| 16 \| 17 \| 18 \| 19 \| 20 \| 21 \| \| ... \| More rows continue to list the positive integers in order, with each new row containing one more integer than the previous row. How many integers less than 2000 are in the column that contains the number 2000?		16	deepscale	31,929
A real number \(x\) is chosen uniformly at random from the interval \([0,1000]\). Find the probability that \(\left\lfloor\frac{\left\lfloor\frac{x}{2.5}\right\rfloor}{2.5}\right\rfloor=\left\lfloor\frac{x}{6.25}\right\rfloor\).	Let \(y=\frac{x}{2.5}\), so \(y\) is chosen uniformly at random from [0,400]. Then we need \(\left\lfloor\frac{\lfloor y\rfloor}{2.5}\right\rfloor=\left\lfloor\frac{y}{2.5}\right\rfloor\). Let \(y=5a+b\), where \(0 \leq b<5\) and \(a\) is an integer. Then \(\left\lfloor\frac{\lfloor y\rfloor}{2.5}\right\rfloor=2a+\left\lfloor\frac{\lfloor b\rfloor}{2.5}\right\rfloor\) while \(\left\lfloor\frac{y}{2.5}\right\rfloor=2a+\left\lfloor\frac{b}{2.5}\right\rfloor\), so we need \(\left\lfloor\frac{\lfloor b\rfloor}{2.5}\right\rfloor=\left\lfloor\frac{b}{2.5}\right\rfloor\), where \(b\) is selected uniformly at random from \([0,5]\). This can be shown to always hold except for \(b \in[2.5,3)\), so the answer is \(1-\frac{0.5}{5}=\frac{9}{10}\).	\frac{9}{10}	deepscale	4,915
A batch of disaster relief supplies is loaded into 26 trucks. The trucks travel at a constant speed of \( v \) kilometers per hour directly to the disaster area. If the distance between the two locations is 400 kilometers and the distance between every two trucks must be at least \( \left(\frac{v}{20}\right)^{2} \) kilometers, how many hours will it take to transport all the supplies to the disaster area?		10	deepscale	31,063
The distance between two parallel lines $x-y=1$ and $2x-2y+3=0$ is ______.		\frac{5\sqrt{2}}{4}	deepscale	17,193
Adam and Simon start on bicycle trips from the same point at the same time. Adam travels east at 8mph and Simon travels south at 6mph. After how many hours are they 60 miles apart?		6	deepscale	38,640
Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$. What is the value of $\frac{x+3y}{3x-y}$?	1. Start with the given equation: \[ \frac{3x+y}{x-3y} = -2 \] 2. Cross-multiply to eliminate the fraction: \[ 3x + y = -2(x - 3y) \] 3. Distribute the -2 on the right-hand side: \[ 3x + y = -2x + 6y \] 4. Rearrange the equation to isolate terms involving \(x\) and \(y\) on opposite sides: \[ 3x + 2x = 6y - y \] \[ 5x = 5y \] 5. Simplify the equation: \[ x = y \] 6. Substitute \(x = y\) into the expression \(\frac{x+3y}{3x-y}\): \[ \frac{x + 3y}{3x - y} = \frac{y + 3y}{3y - y} \] \[ \frac{4y}{2y} \] 7. Simplify the fraction: \[ \frac{4y}{2y} = 2 \] 8. Thus, the value of \(\frac{x+3y}{3x-y}\) is: \[ \boxed{\textbf{(D)}\ 2} \]	2	deepscale	250
A cube with edge length 2 cm has a dot marked at the center of the top face. The cube is on a flat table and rolls without slipping, making a full rotation back to its initial orientation, with the dot back on top. Calculate the length of the path followed by the dot in terms of $\pi. $		2\sqrt{2}\pi	deepscale	28,866
At the round table, $10$ people are sitting, some of them are knights, and the rest are liars (knights always say pride, and liars always lie) . It is clear thath I have at least one knight and at least one liar. What is the largest number of those sitting at the table can say: ''Both of my neighbors are knights '' ? (A statement that is at least partially false is considered false.)	To solve this problem, we need to maximize the number of people at a round table who can truthfully say: "Both of my neighbors are knights." Considering the rules: - Knights always tell the truth. - Liars always lie. - At least one knight and one liar are present. Let's analyze the configuration of people around the table: 1. If a person is truthfully saying "Both of my neighbors are knights," the person themselves must be a knight, as knights tell the truth. 2. If the person making the statement is a liar, then one or both neighbors must not be knights (since liars lie). 3. Consider the maximum possible scenario where nine individuals are knights. With only one person left, that person must be a liar (since at least one liar is required). 4. Arrange the people such that the liar is strategically placed to disrupt the truth of the statement for only themselves and not for the rest. Let's denote: - \( K \) for a knight, - \( L \) for a liar. A possible arrangement that satisfies the conditions is: \( K, K, K, K, K, K, K, K, K, L \). - Here, each of the nine knights can truthfully say, "Both of my neighbors are knights," because they are flanked by knights on both sides. - The liar cannot truthfully make this statement as their claim would be "false," given at least one of their neighbors is a liar (themselves). Therefore, the maximum number of people who can truthfully say that both their neighbors are knights is: \[ \boxed{9} \]	9	deepscale	5,969
A square is inscribed in another square such that its vertices lie on the sides of the first square, and its sides form angles of $60^{\circ}$ with the sides of the first square. What fraction of the area of the given square is the area of the inscribed square?		4 - 2\sqrt{3}	deepscale	13,472
Misha made himself a homemade dartboard at the summer house. The round board is divided into sectors by circles - it can be used to throw darts. Points are awarded according to the number written in the sector, as indicated in the diagram. Misha threw 8 darts 3 times. The second time, he scored twice as many points as the first time, and the third time, he scored 1.5 times more points than the second time. How many points did he score the second time?		48	deepscale	14,722
Find the largest negative root of the equation $$ 4 \sin (3 x) + 13 \cos (3 x) = 8 \sin (x) + 11 \cos (x) $$		-0.1651	deepscale	25,066
A piece of string is cut in two at a point selected at random. The probability that the longer piece is at least x times as large as the shorter piece is	Let's consider a string of unit length (length = 1) for simplicity. We cut the string at a point $C$ chosen uniformly at random along the string. Let $C$ be the length of the shorter piece after the cut, and $1 - C$ be the length of the longer piece. We need to find the probability that the longer piece is at least $x$ times as large as the shorter piece. #### Step 1: Set up the inequality We need the longer piece to be at least $x$ times the length of the shorter piece. This can happen in two ways: 1. $1 - C \geq xC$ (if $C$ is the shorter piece) 2. $C \geq x(1 - C)$ (if $1 - C$ is the shorter piece) #### Step 2: Solve the inequalities 1. Solving $1 - C \geq xC$: \[ 1 - C \geq xC \implies 1 \geq (x + 1)C \implies C \leq \frac{1}{x + 1} \] This inequality is valid when $C$ is the shorter piece, which is true when $C \leq \frac{1}{2}$. 2. Solving $C \geq x(1 - C)$: \[ C \geq x - xC \implies C + xC \geq x \implies (1 + x)C \geq x \implies C \geq \frac{x}{x + 1} \] This inequality is valid when $1 - C$ is the shorter piece, which is true when $C \geq \frac{1}{2}$. #### Step 3: Determine the valid intervals 1. From $C \leq \frac{1}{x + 1}$, the valid interval is $[0, \frac{1}{x + 1}]$. 2. From $C \geq \frac{x}{x + 1}$, the valid interval is $[\frac{x}{x + 1}, 1]$. However, we need to consider the intersection of these intervals with the intervals where each piece is actually the shorter piece: - $C \leq \frac{1}{2}$ intersects with $[0, \frac{1}{x + 1}]$ if $\frac{1}{x + 1} \leq \frac{1}{2}$. - $C \geq \frac{1}{2}$ intersects with $[\frac{x}{x + 1}, 1]$ if $\frac{x}{x + 1} \geq \frac{1}{2}$. #### Step 4: Calculate the probability The probability that $C$ falls in either $[0, \frac{1}{x + 1}]$ or $[\frac{x}{x + 1}, 1]$ is the sum of the lengths of these intervals: \[ \text{Probability} = \left(\frac{1}{x + 1} - 0\right) + \left(1 - \frac{x}{x + 1}\right) = \frac{1}{x + 1} + \frac{1}{x + 1} = \frac{2}{x + 1} \] Thus, the probability that the longer piece is at least $x$ times as large as the shorter piece is $\boxed{\textbf{(E) }\frac{2}{x+1}}$.	\frac{2}{x+1}	deepscale	2,111
Let $P$ be a plane passing through the origin. When $\begin{pmatrix} 5 \\ 3 \\ 5 \end{pmatrix}$ is projected onto plane $P,$ the result is $\begin{pmatrix} 3 \\ 5 \\ 1 \end{pmatrix}.$ When $\begin{pmatrix} 4 \\ 0 \\ 7 \end{pmatrix}$ is projected onto plane $P,$ what is the result?		\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}	deepscale	39,819
Point $F$ is taken on the extension of side $AD$ of parallelogram $ABCD$. $BF$ intersects diagonal $AC$ at $E$ and side $DC$ at $G$. If $EF = 32$ and $GF = 24$, then $BE$ equals: [asy] size(7cm); pair A = (0, 0), B = (7, 0), C = (10, 5), D = (3, 5), F = (5.7, 9.5); pair G = intersectionpoints(B--F, D--C)[0]; pair E = intersectionpoints(A--C, B--F)[0]; draw(A--D--C--B--cycle); draw(A--C); draw(D--F--B); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$F$", F, N); label("$G$", G, NE); label("$E$", E, SE); //Credit to MSTang for the asymptote[/asy] $\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 16$		16	deepscale	36,077
How many whole numbers between 1 and 2000 do not contain the digit 2?		6560	deepscale	24,797
12 Smurfs are seated around a round table. Each Smurf dislikes the 2 Smurfs next to them, but does not dislike the other 9 Smurfs. Papa Smurf wants to form a team of 5 Smurfs to rescue Smurfette, who was captured by Gargamel. The team must not include any Smurfs who dislike each other. How many ways are there to form such a team?		36	deepscale	10,154
Given vectors $a=(2\cos \alpha,\sin ^{2}\alpha)$ and $b=(2\sin \alpha,t)$, where $\alpha\in\left( 0,\frac{\pi}{2} \right)$ and $t$ is a real number. $(1)$ If $a-b=\left( \frac{2}{5},0 \right)$, find the value of $t$; $(2)$ If $t=1$ and $a\cdot b=1$, find the value of $\tan \left( 2\alpha+\frac{\pi}{4} \right)$.		\frac{23}{7}	deepscale	32,841
On a table there are $100$ red and $k$ white buckets for which all of them are initially empty. In each move, a red and a white bucket is selected and an equal amount of water is added to both of them. After some number of moves, there is no empty bucket and for every pair of buckets that are selected together at least once during the moves, the amount of water in these buckets is the same. Find all the possible values of $k$ .		100	deepscale	20,334
Given in the polar coordinate system, point P moves on the curve $\rho^2\cos\theta-2\rho=0$, the minimum distance from point P to point $Q(1, \frac{\pi}{3})$ is \_\_\_\_\_\_.		\frac{3}{2}	deepscale	23,178
Given the equations $y = 2x^2$ and $y = 4x + c$, determine the value(s) of $c$ for which these two equations have precisely two identical solutions.		-2	deepscale	7,521
A television's price was discounted $40\%$ and then slashed another $40\%$ during a liquidation sale. By what percent has the original price been reduced?		64\%	deepscale	34,468
Determine the number of ways to select a sequence of 8 sets $A_{1}, A_{2}, \ldots, A_{8}$, such that each is a subset (possibly empty) of \{1,2\}, and $A_{m}$ contains $A_{n}$ if $m$ divides $n$.	Consider an arbitrary $x \in\{1,2\}$, and let us consider the number of ways for $x$ to be in some of the sets so that the constraints are satisfied. We divide into a few cases: - Case: $x \notin A_{1}$. Then $x$ cannot be in any of the sets. So there is one possibility. - Case: $x \in A_{1}$ but $x \notin A_{2}$. Then the only other sets that $x$ could be in are $A_{3}, A_{5}, A_{7}$, and $x$ could be in some collection of them. There are 8 possibilities in this case. - Case: $x \in A_{2}$. Then $x \in A_{1}$ automatically. There are 4 independent choices to be make here: (1) whether $x \in A_{5} ;(2)$ whether $x \in A_{7} ;(3)$ whether $x \in A_{3}$, and if yes, whether $x \in A_{6}$; (4) whether $x \in A_{4}$, and if yes, whether $x \in A_{8}$. There are $2 \times 2 \times 3 \times 3=36$ choices here. Therefore, there are $1+8+36=45$ ways to place $x$ into some of the sets. Since the choices for $x=1$ and $x=2$ are made independently, we see that the total number of possibilities is $45^{2}=2025$.	2025	deepscale	4,225
Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $\|a\|^2 + \|b\|^2 + \|c\|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$	As noted in the previous solutions, $a+b+c = 0$. Let $a = a_1+a_2 i$, $b = b_1+b_2 i$, $c = c_1+c_2 i$ and we have $a_1 + b_1 + c_1 = a_2 + b_2 + c_2 = 0$. Then the given $\|a\|^2 + \|b\|^2 + \|c\|^2 = 250$ translates to $\sum_{} ( {a_1}^2 + {a_2}^2 ) = 250.$ Note that in a right triangle, the sum of the squares of the three sides is equal to two times the square of the hypotenuse, by the pythagorean theorem. Thus, we have \[2h^2 = (a_1 - b_1)^2 + (a_2 - b_2)^2 + (b_1 - c_1)^2 + (b_2 - c_2)^2 + (a_1 - c_1)^2 + (a_2 - c_2)^2\] \[= 2 \left( \sum_{} ( {a_1}^2 + {a_2}^2 ) \right) - 2 \left( \sum_{cyc} a_1 b_1 + \sum_{cyc} a_2 b_2 \right)\] \[= 500 - \left( (a_1 + b_1 + c_1)^2 + (a_2 + b_2 + c_2)^2 - \sum_{cyc} ( {a_1}^2 + {a_2}^2 ) \right)\] \[= 500 - (0^2 + 0^2 - 250)\] so $h^2 = \boxed{375}$ and we may conclude. ~ rzlng	375	deepscale	7,044
Ivan wanted to buy nails. In one store, where 100 grams of nails cost 180 rubles, he couldn't buy the required amount because he was short 1430 rubles. Then he went to another store where 100 grams cost 120 rubles. He bought the required amount and received 490 rubles in change. How many kilograms of nails did Ivan buy?		3.2	deepscale	9,107
Consider a regular polygon with $2^n$ sides, for $n \ge 2$ , inscribed in a circle of radius $1$ . Denote the area of this polygon by $A_n$ . Compute $\prod_{i=2}^{\infty}\frac{A_i}{A_{i+1}}$		\frac{2}{\pi}	deepscale	24,251
Consider a city grid with intersections labeled A, B, C, and D. Assume a student walks from intersection A to intersection B every morning, always walking along the designated paths and only heading east or south. The student passes through intersections C and D along the way. The intersections are placed such that A to C involves 3 eastward moves and 2 southward moves, and C to D involves 2 eastward moves and 1 southward move, and finally from D to B requires 1 eastward move and 2 southward moves. Each morning, at each intersection where he has a choice, he randomly chooses whether to go east or south with probability $\frac{1}{2}$. Determine the probability that the student walks through C, and then D on any given morning. A) $\frac{15}{77}$ B) $\frac{10}{462}$ C) $\frac{120}{462}$ D) $\frac{3}{10}$ E) $\frac{64}{462}$		\frac{15}{77}	deepscale	21,145
In a game of Chomp, two players alternately take bites from a 5-by-7 grid of unit squares. To take a bite, a player chooses one of the remaining squares, then removes ("eats") all squares in the quadrant defined by the left edge (extended upward) and the lower edge (extended rightward) of the chosen square. For example, the bite determined by the shaded square in the diagram would remove the shaded square and the four squares marked by $\times.$ (The squares with two or more dotted edges have been removed form the original board in previous moves.) The object of the game is to make one's opponent take the last bite. The diagram shows one of the many subsets of the set of 35 unit squares that can occur during the game of Chomp. How many different subsets are there in all? Include the full board and empty board in your count.	By drawing possible examples of the subset, one can easily see that making one subset is the same as dividing the game board into two parts. One can also see that it is the same as finding the shortest route from the upper left hand corner to the lower right hand corner; Such a route would require 5 lengths that go down, and 7 that go across, with the shape on the right "carved" out by the path a possible subset. Therefore, the total number of such paths is $\binom{12}{5}=\boxed{792}$	792	deepscale	6,562
What is the number of degrees in the smaller angle formed by the hour and minute hands of a clock at 8:15? Express your answer as a decimal to the nearest tenth. [asy] size(200); draw(Circle((0,0),5),linewidth(1.2)); pair[] mins; for(int i = 0; i < 60; ++i){ mins[i] = 4.5dir(-6i + 90); dot(mins[i]); } for(int i = 1; i <= 12; ++i){ label((string)i,mins[5i % 60],dir(-30i - 90)); } fill(Circle((0,0),0.25)); [/asy]		157.5	deepscale	36,316
A certain department store sells suits and ties, with each suit priced at $1000$ yuan and each tie priced at $200 yuan. During the "National Day" period, the store decided to launch a promotion offering two discount options to customers.<br/>Option 1: Buy one suit and get one tie for free;<br/>Option 2: Pay 90% of the original price for both the suit and the tie.<br/>Now, a customer wants to buy 20 suits and $x$ ties $\left(x > 20\right)$.<br/>$(1)$ If the customer chooses Option 1, the payment will be ______ yuan (expressed as an algebraic expression in terms of $x$). If the customer chooses Option 2, the payment will be ______ yuan (expressed as an algebraic expression in terms of $x$).<br/>$(2)$ If $x=30$, calculate and determine which option is more cost-effective at this point.<br/>$(3)$ When $x=30$, can you come up with a more cost-effective purchasing plan? Please describe your purchasing method.		21800	deepscale	23,283
There is a positive integer s such that there are s solutions to the equation $64sin^2(2x)+tan^2(x)+cot^2(x)=46$ in the interval $(0,\frac{\pi}{2})$ all of the form $\frac{m_k}{n_k}\pi$ where $m_k$ and $n_k$ are relatively prime positive integers, for $k = 1, 2, 3, . . . , s$ . Find $(m_1 + n_1) + (m_2 + n_2) + (m_3 + n_3) + · · · + (m_s + n_s)$ .		100	deepscale	27,997

The picture shows the same die in three different positions. When the die is rolled, what is the probability of rolling a 'YES'? A) $\frac{1}{3}$ B) $\frac{1}{2}$ C) $\frac{5}{9}$ D) $\frac{2}{3}$ E) $\frac{5}{6}$

\frac{1}{2}

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17,255

In the triangular pyramid $ABCD$ with base $ABC$, the lateral edges are pairwise perpendicular, $DA = DB = 5, DC = 1$. A ray of light is emitted from a point on the base. After reflecting exactly once from each lateral face (the ray does not reflect from the edges), the ray hits a point on the pyramid's base. What is the minimum distance the ray could travel?

\frac{10 \sqrt{3}}{9}

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28,364

Compute the smallest positive integer $k$ such that 49 divides $\binom{2 k}{k}$.

The largest $a$ such that $7^{a} \left\lvert\,\binom{ 2 k}{k}\right.$ is equal to the number of carries when you add $k+k$ in base 7 , by Kummer's Theorem. Thus, we need two carries, so $2 k$ must have at least 3 digits in base 7 . Hence, $k \geq 25$. We know $k=25$ works because $25+25=34_{7}+34_{7}=101_{7}$ has two carries.

25

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4,409

The three sides of a triangle are 25, 39, and 40. Find the diameter of its circumscribed circle.

\frac{125}{3}

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22,122

In the regular quadrangular pyramid $P-ABCD$, $M$ and $N$ are the midpoints of $PA$ and $PB$ respectively. If the tangent of the dihedral angle between a side face and the base is $\sqrt{2}$, find the cosine of the angle between skew lines $DM$ and $AN$.

1/6

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14,138

A rectangular piece of paper with vertices $A B C D$ is being cut by a pair of scissors. The pair of scissors starts at vertex $A$, and then cuts along the angle bisector of $D A B$ until it reaches another edge of the paper. One of the two resulting pieces of paper has 4 times the area of the other piece. What is the ratio of the longer side of the original paper to the shorter side?

Without loss of generality, let $A B>A D$, and let $x=A D, y=A B$. Let the cut along the angle bisector of $\angle D A B$ meet $C D$ at $E$. Note that $A D E$ is a $45-45-90$ triangle, so $D E=A D=x$, and $E C=y-x$. Now, $[A D E]=\frac{x^{2}}{2}$, and $[A E C B]=x\left(y-\frac{x}{2}\right)=4[A D E]$. Equating and dividing both sides by $x$, we find that $2 x=y-\frac{x}{2}$, so $y / x=\frac{5}{2}$.

\frac{5}{2}

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4,934

Let $n > 2$ be an integer and let $\ell \in \{1, 2,\dots, n\}$. A collection $A_1,\dots,A_k$ of (not necessarily distinct) subsets of $\{1, 2,\dots, n\}$ is called $\ell$-large if $|A_i| \ge \ell$ for all $1 \le i \le k$. Find, in terms of $n$ and $\ell$, the largest real number $c$ such that the inequality \[ \sum_{i=1}^k\sum_{j=1}^k x_ix_j\frac{|A_i\cap A_j|^2}{|A_i|\cdot|A_j|}\ge c\left(\sum_{i=1}^k x_i\right)^2 \] holds for all positive integer $k$, all nonnegative real numbers $x_1,x_2,\dots,x_k$, and all $\ell$-large collections $A_1,A_2,\dots,A_k$ of subsets of $\{1,2,\dots,n\}$.

To solve the problem, we need to find the largest real number $ c $ such that the inequality \[ \sum_{i=1}^k \sum_{j=1}^k x_i x_j \frac{|A_i \cap A_j|^2}{|A_i| \cdot |A_j|} \ge c \left(\sum_{i=1}^k x_i\right)^2 \] holds for all positive integers $ k $, all nonnegative real numbers $ x_1, x_2, \dots, x_k $, and all $\ell$-large collections $ A_1, A_2, \dots, A_k $ of subsets of $\{1, 2, \dots, n\}$. ### Step-by-Step Solution 1. **Understanding the Constraints**: Each $ A_i $ is a subset of $\{1, 2, \dots, n\}$ with $|A_i| \geq \ell$. The sets are $\ell$-large, meaning every set has at least $\ell$ elements. 2. **Expression Simplification**: The expression on the left side of the inequality involves the squared size of the intersections $ |A_i \cap A_j| $ normalized by the sizes of $ |A_i| $ and $ |A_j| $. 3. **Cauchy-Schwarz Application**: To handle the sum of squares, we consider applying the Cauchy-Schwarz inequality in terms of sums and intersections: \[ \left(\sum_{i=1}^k x_i\right)^2 \leq k \sum_{i=1}^k x_i^2. \] 4. **Bounding the Intersection Size**: Since $ |A_i|, |A_j| \geq \ell $, the intersection $ |A_i \cap A_j| $ can be at most $\min(|A_i|, |A_j|)$, but more typically involves sizing relative to $ n $, such as $|A_i \cap A_j| \leq \ell$. 5. **Finding $ c $**: The challenge is finding a universal lower bound on the given expression. Consider setting boundaries based on specific configurations of $ A_i $ making the set sizes minimal at $\ell$. Assume: \[ |A_i| = \ell \quad \text{for all } i, \] then we simplify the inequality's left side, using symmetry and the fact $ |A_i \cap A_j| $ can be estimated within strict bounds for large $ n $. The strategy is identifying the smallest reliable bound for: \[ \frac{\ell^2 - 2\ell + n}{n(n-1)}. \] ### Conclusion Thus, after considering possible configurations and analytic optimization, the bound for the largest real number $ c $ that satisfies the inequality for all valid configurations is: \[ \boxed{\frac{\ell^2 - 2\ell + n}{n(n-1)}}. \]

\frac{\ell^2 - 2\ell + n}{n(n-1)}

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5,985

Simplify and find the value of: $a^{2}-\left(a-\dfrac{2a}{a+1}\right)\div \dfrac{a^{2}-2a+1}{a^{2}-1}$, where $a$ is a solution of the equation $x^{2}-x-\dfrac{7}{2}=0$.

\dfrac{7}{2}

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23,529

When five students are lining up to take a photo, and two teachers join in, with the order of the five students being fixed, calculate the total number of ways for the two teachers to stand in line with the students for the photo.

42

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32,986

Trapezoid $ABCD$ has $\overline{AB}\parallel\overline{CD}, BC=CD=43$, and $\overline{AD}\perp\overline{BD}$. Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$, and let $P$ be the midpoint of $\overline{BD}$. Given that $OP=11$, the length of $AD$ can be written in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. What is $m+n$?

1. **Identify the properties of the trapezoid**: Given that $ABCD$ is a trapezoid with $\overline{AB}\parallel\overline{CD}$ and $BC=CD=43$. Also, $\overline{AD}\perp\overline{BD}$, which implies that $\triangle ABD$ is a right triangle. 2. **Intersection and midpoint properties**: Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$, and $P$ be the midpoint of $\overline{BD}$. Given $OP=11$. 3. **Extend lines and identify congruent triangles**: Extend $\overline{AD}$ and $\overline{BC}$ to meet at point $E$. Since $\overline{AD}\perp\overline{BD}$, $\angle BDE = 90^\circ$. By the properties of trapezoids and the congruence of triangles $\triangle BPC$ and $\triangle DPC$, we have $\angle DBE \cong \angle DBA$. By ASA (Angle-Side-Angle) congruence, $\triangle ABD \cong \triangle EBD$, and thus $AD = ED$. This implies $D$ is the midpoint of $\overline{AE}$. 4. **Midpoint of $\overline{DE}$ and congruence**: Let $M$ be the midpoint of $\overline{DE}$. Since $\triangle CME$ is congruent to $\triangle BPC$, we have $BC = CE$, making $C$ the midpoint of $\overline{BE}$. 5. **Centroid and median properties**: $\overline{AC}$ and $\overline{BD}$ are medians of $\triangle ABE$, making $O$ the centroid of $\triangle ABE$. The centroid divides each median in a 2:1 ratio, so $\frac{BO}{2} = DO = \frac{BD}{3}$. Given $P$ is the midpoint of $BD$, $DP = \frac{BD}{2}$. Given $OP = 11$, we have $DP - DO = 11$, leading to $\frac{BD}{2} - \frac{BD}{3} = 11 \implies \frac{BD}{6} = 11 \implies BD = 66$. 6. **Calculate $AB$ and use the Pythagorean theorem**: Since $\triangle ABD$ is similar to $\triangle CBP$ and $\triangle CPD$ by a factor of 2, $AB = 2 \cdot 43 = 86$. Applying the Pythagorean theorem in $\triangle ABD$, we have: \[ AB^2 - BD^2 = AD^2 \implies 86^2 - 66^2 = AD^2 \implies 7396 - 4356 = AD^2 \implies 3040 = AD^2 \implies AD = \sqrt{3040} = 4\sqrt{190} \] 7. **Final answer**: The length of $AD$ is $4\sqrt{190}$, where $m=4$ and $n=190$. Thus, $m+n = 4+190 = \boxed{\textbf{(D) }194}$. $\blacksquare$

194

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443

Given a $5 \times 5$ grid where the number in the $i$-th row and $j$-th column is denoted by $ a_{ij} $ (where $ a_{ij} \in \{0, 1\} $), with the condition that $ a_{ij} = a_{ji} $ for $ 1 \leq i, j \leq 5 $. Calculate the total number of ways to fill the grid such that there are exactly five 1's in the grid.

326

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13,883

Determine the coefficient of the $x^5$ term in the expansion of $(x+1)(x^2-x-2)^3$.

-6

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22,395

A person's age at the time of their death was one 31st of their birth year. How old was this person in 1930?

39

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14,969

Find the maximum value of $\cos x + 2 \sin x,$ over all angles $x.$

\sqrt{5}

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39,817

How many triangles exist such that the lengths of the sides are integers not greater than 10?

125

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14,062

An integer $n>1$ is given . Find the smallest positive number $m$ satisfying the following conditions： for any set $\{a,b\}$ $\subset \{1,2,\cdots,2n-1\}$ ,there are non-negative integers $ x, y$ ( not all zero) such that $2n|ax+by$ and $x+y\leq m.$

Given an integer $ n > 1 $, we aim to find the smallest positive number $ m $ satisfying the following conditions: for any set $\{a, b\} \subset \{1, 2, \ldots, 2n-1\}$, there exist non-negative integers $ x $ and $ y $ (not both zero) such that $ 2n \mid ax + by $ and $ x + y \leq m $. To determine the smallest $ m $, we analyze the conditions: 1. Consider $ a = 1 $ and $ b = 2 $. If $ 2n \mid ax + by $, then: \[ 2n \leq x + 2y \leq 2(x + y) \leq 2m. \] This implies $ m \geq n $. 2. We now show that $ m \leq n $. - **Case 1**: If $\gcd(a, 2n) > 1$ or $\gcd(b, 2n) > 1$. Without loss of generality, assume $\gcd(a, 2n) > 1$. Choose $ x = \frac{2n}{\gcd(a, 2n)} $ and $ y = 0 $. Then: \[ x + y = \frac{2n}{\gcd(a, 2n)} \leq \frac{2n}{2} = n. \] - **Case 2**: If $\gcd(a, 2n) = 1$ and $\gcd(b, 2n) = 1$. Let $ c \in [0, 2n-1] $ such that $ c \equiv ba^{-1} \pmod{2n} $. The equation $ ax + by \equiv 0 \pmod{2n} $ is equivalent to $ x + cy \equiv 0 \pmod{2n} $. Choose $ y = \left\lfloor \frac{2n}{c} \right\rfloor $ and $ x = 2n - c \left\lfloor \frac{2n}{c} \right\rfloor $. - **Subcase 2.1**: If $ 2 < c < n $, then: \[ x + y \leq c - 1 + \left\lfloor \frac{2n}{c} \right\rfloor < c - 1 + \frac{2n}{c} = \frac{(c-2)(c-n)}{c} + n + 1 < n + 1. \] Hence, $ x + y \leq n $. - **Subcase 2.2**: If $ c \geq n + 1 $, then: \[ x + y = 2n - (c-1) \left\lfloor \frac{2n}{c} \right\rfloor = 2n - (c-1) \leq 2n - (n+1-1) = n. \] Hence, $ x + y \leq n $. In conclusion, the smallest positive number $ m $ that satisfies the given conditions is $ n $. The answer is $\boxed{n}$.

n

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2,924

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}

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24,408

Non-negative numbers $a$ and $b$ satisfy the equations $a^2 + b^2 = 74$ and $ab = 35$. What is the value of the expression $a^2 - 12a + 54$?

19

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12,082

Compute $$\sum_{k=1}^{2000} k(\lceil \log_{2}{k}\rceil- \lfloor\log_{2}{k} \rfloor).$$

1998953

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17,520

In right triangle $ABC$ with $\angle A = 90^\circ$, we have $AB =16$ and $BC = 24$. Find $\sin A$.

1

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36,256

Two integers have a sum of $26$. When two more integers are added to the first two, the sum is $41$. Finally, when two more integers are added to the sum of the previous $4$ integers, the sum is $57$. What is the minimum number of even integers among the $6$ integers?

1. **Identify the sums at each stage:** - Let the first two integers be $x$ and $y$. We know $x + y = 26$. - Let the next two integers added be $a$ and $b$. Then, $x + y + a + b = 41$. - Let the final two integers added be $m$ and $n$. Then, $x + y + a + b + m + n = 57$. 2. **Calculate the sums of the additional integers:** - From $x + y = 26$ to $x + y + a + b = 41$, the sum of $a$ and $b$ is $41 - 26 = 15$. - From $x + y + a + b = 41$ to $x + y + a + b + m + n = 57$, the sum of $m$ and $n$ is $57 - 41 = 16$. 3. **Analyze the parity of the sums:** - The sum $x + y = 26$ is even. This can be achieved with two even numbers or two odd numbers. - The sum $a + b = 15$ is odd. This requires one even and one odd integer (since the sum of two odd numbers or two even numbers is even). - The sum $m + n = 16$ is even. This can be achieved with two even numbers or two odd numbers. 4. **Determine the minimum number of even integers:** - For $x + y = 26$, we can choose both $x$ and $y$ to be odd (e.g., $x = 13$, $y = 13$), requiring $0$ even integers. - For $a + b = 15$, we must have one even integer (e.g., $a = 8$, $b = 7$). - For $m + n = 16$, we can choose both $m$ and $n$ to be odd (e.g., $m = 7$, $n = 9$), requiring $0$ even integers. 5. **Conclusion:** - The minimum number of even integers required among the six integers is $1$ (from the pair $a$ and $b$). Thus, the answer is $\boxed{\textbf{(A)}\ 1}$.

1

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2,103

A soccer ball kicked vertically upward reaches, after $t$ seconds, a height of $s$ meters where $s = 180t - 20t^2$. Find the maximum height reached by the ball.

405

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17,145

What is the value of $m$ if Tobias downloads $m$ apps, each app costs $\$ 2.00$ plus $10 \%$ tax, and he spends $\$ 52.80$ in total on these $m$ apps?

Since the tax rate is $10 \%$, then the tax on each $\$ 2.00$ app is $\$ 2.00 \times \frac{10}{100}=\$ 0.20$. Therefore, including tax, each app costs $\$ 2.00+\$ 0.20=\$ 2.20$. Since Tobias spends $\$ 52.80$ on apps, he downloads $\frac{\$ 52.80}{\$ 2.20}=24$ apps. Therefore, $m=24$.

24

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5,501

There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$, the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$, and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime integers. Find $p+q$.

Note that if $\tan \theta$ is positive, then $\theta$ is in the first or third quadrant, so $0^{\circ} < \theta < 90^{\circ} \pmod{180^{\circ}}$. Also notice that the only way $\tan{\left(2^{n}\theta\right)}$ can be positive for all $n$ that are multiples of $3$ is when $2^0\theta, 2^3\theta, 2^6\theta$, etc. are all the same value $\pmod{180^{\circ}}$. (This also must be true in order for $\theta$ to be unique.) This is the case if $8\theta = \theta \pmod{180^{\circ}}$, so $7\theta = 0^{\circ} \pmod{180^{\circ}}$. Therefore, the only possible values of $\theta$ between $0^{\circ}$ and $90^{\circ}$ are $\frac{180}{7}^{\circ}$, $\frac{360}{7}^{\circ}$, and $\frac{540}{7}^{\circ}$. $\frac{180}{7}^{\circ}$ does not work since $\tan(2 \cdot \frac{180}{7}^{\circ})$ is positive, and $\frac{360}{7}^{\circ}$ does not work because $\tan(4 \cdot \frac{360}{7}^{\circ})$ is positive. Thus, $\theta = \frac{540}{7}^{\circ}$, and a quick check verifies that it works. $540 + 7 = \boxed{547}$.

547

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7,265

Brachycephalus frogs have three toes on each foot and two fingers on each hand. The common frog has five toes on each foot and four fingers on each hand. Some Brachycephalus and common frogs are in a bucket. Each frog has all its fingers and toes. Between them they have 122 toes and 92 fingers. How many frogs are in the bucket? A 15 B 17 C 19 D 21 E 23

15

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11,543

A bar of chocolate is made of 10 distinguishable triangles as shown below. How many ways are there to divide the bar, along the edges of the triangles, into two or more contiguous pieces?

Every way to divide the bar can be described as a nonempty set of edges to break, with the condition that every endpoint of a broken edge is either on the boundary of the bar or connects to another broken edge. Let the center edge have endpoints $X$ and $Y$. We do casework on whether the center edge is broken. If the center edge is broken, then we just need some other edge connecting to $X$ to be broken, and some other edge connecting to $Y$ to be broken. We have $2^{5}$ choices for the edges connecting to $X$, of which 1 fails. Similarly, we have $2^{5}-1$ valid choices for the edges connecting to $Y$. This yields $\left(2^{5}-1\right)^{2}=961$ possibilities. If the center edge is not broken, then the only forbidden arrangements are those with exactly one broken edge at $X$ or those with exactly one broken edge at $Y$. Looking at just the edges connecting to $X$, we have 5 cases with exactly one broken edge. Thus, there are $2^{5}-5=27$ ways to break the edges connecting to $X$. Similarly there are 27 valid choices for the edges connecting to $Y$. This yields $27^{2}-1=728$ cases, once we subtract the situation where no edges are broken. The final answer is $961+728=1689$.

1689

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4,801

Ryan has 3 red lava lamps and 3 blue lava lamps. He arranges them in a row on a shelf randomly, and then randomly turns 3 of them on. What is the probability that the leftmost lamp is blue and off, and the rightmost lamp is red and on?

\dfrac{9}{100}

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34,810

A cone is perfectly fitted inside a cube such that the cone's base is one face of the cube and its vertex touches the opposite face. A sphere is inscribed in the same cube. Given that one edge of the cube is 8 inches, calculate: 1. The volume of the inscribed sphere. 2. The volume of the inscribed cone. Express your answer in terms of $\pi$.

\frac{128}{3}\pi

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9,492

Natural numbers $ a, b, c $ are chosen such that $ a < b < c $. It is also known that the system of equations $ 2x + y = 2025 $ and $ y = |x - a| + |x - b| + |x - c| $ has exactly one solution. Find the minimum possible value of $ c $.

1013

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31,607

What is the greatest common multiple of 8 and 12 that is less than 90?

72

deepscale

38,954

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

2047

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25,673

Three fair coins are to be tossed once. For each head that results, one fair die is to be rolled. If all three coins show heads, roll an additional fourth die. Determine the probability that the sum of the die rolls is odd.

\frac{7}{16}

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7,715

There exist $r$ unique nonnegative integers $n_1 > n_2 > \cdots > n_r$ and $r$ unique integers $a_k$ ($1\le k\le r$) with each $a_k$ either $1$ or $- 1$ such that\[a_13^{n_1} + a_23^{n_2} + \cdots + a_r3^{n_r} = 2008.\]Find $n_1 + n_2 + \cdots + n_r$.

21

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38,172

Steven subtracts the units digit from the tens digit for each two-digit number. He then finds the sum of all his answers. What is the value of Steven's sum?

45

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23,394

Alice conducted a survey among a group of students regarding their understanding of snakes. She found that $92.3\%$ of the students surveyed believed that snakes are venomous. Of the students who believed this, $38.4\%$ erroneously thought that all snakes are venomous. Knowing that only 31 students held this incorrect belief, calculate the total number of students Alice surveyed.

88

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29,235

If A and B can only undertake the first three tasks, while the other three can undertake all four tasks, calculate the total number of different selection schemes for the team leader group to select four people from five volunteers to undertake four different tasks.

72

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19,445

Convert the binary number $11011001_2$ to base 4.

3121_4

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8,715

A ball travels on a parabolic path in which the height (in feet) is given by the expression $-16t^2+32t+15$, where $t$ is the time after launch. What is the maximum height of the ball, in feet?

31

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34,454

How many 9 step paths are there from $E$ to $G$ which pass through $F$?[asy]size(4cm,4cm);int w=6;int h=5;int i;pen p=fontsize(9);for (i=0; i<h; ++i){draw((0,i) -- (w-1,i));}for (i=0; i<w; ++i){draw((i, 0)--(i,h-1));}label("G", (w-1,0), SE, p);label("E", (0,h-1), NW, p);label("F", (3,3), NE, p);[/asy]

40

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34,723

Let $ ABP, BCQ, CAR$ be three non-overlapping triangles erected outside of acute triangle $ ABC$. Let $ M$ be the midpoint of segment $ AP$. Given that $ \angle PAB \equal{} \angle CQB \equal{} 45^\circ$, $ \angle ABP \equal{} \angle QBC \equal{} 75^\circ$, $ \angle RAC \equal{} 105^\circ$, and $ RQ^2 \equal{} 6CM^2$, compute $ AC^2/AR^2$. [i]Zuming Feng.[/i]

Let $ ABP, BCQ, CAR $ be three non-overlapping triangles erected outside of acute triangle $ ABC $. Let $ M $ be the midpoint of segment $ AP $. Given that $ \angle PAB = \angle CQB = 45^\circ $, $ \angle ABP = \angle QBC = 75^\circ $, $ \angle RAC = 105^\circ $, and $ RQ^2 = 6CM^2 $, we aim to compute $ \frac{AC^2}{AR^2} $. Construct parallelogram $ CADP $. **Claim:** $ \triangle AQR \sim \triangle ADC $. **Proof:** Observe that $ \triangle BPA \sim \triangle BCQ $, hence $ \triangle BAQ \sim \triangle BPC $. Consequently, \[ \frac{AQ}{AD} = \frac{AQ}{CP} = \frac{BP}{BA} = \sqrt{\frac{3}{2}} = \frac{QR}{DC}. \] Since $ \angle RAC = 105^\circ $ and $ \angle QAD = \angle CPA + \angle QAP = 180^\circ - \angle (CP, AQ) = 180^\circ - \angle ABP = 105^\circ $, we can use SSA similarity (since $ 105^\circ > 90^\circ $) to conclude that $ \triangle AQR \sim \triangle ADC $. Thus, it follows that \[ \frac{AC^2}{AR^2} = \frac{2}{3}. \] The answer is: $\boxed{\frac{2}{3}}$.

\frac{2}{3}

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2,903

Petya and Vasya are playing the following game. Petya thinks of a natural number $ x $ with a digit sum of 2012. On each turn, Vasya chooses any natural number $ a $ and finds out the digit sum of the number $ |x-a| $ from Petya. What is the minimum number of turns Vasya needs to determine $ x $ with certainty?

2012

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22,092

If three different numbers are selected from 2, 3, 4, 5, 6 to be $a$, $b$, $c$ such that $N = abc + ab + bc + a - b - c$ reaches its maximum value, then this maximum value is.

167

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25,503

A shape is created by aligning five unit cubes in a straight line. Then, one additional unit cube is attached to the top of the second cube in the line and another is attached beneath the fourth cube in the line. Calculate the ratio of the volume to the surface area.

\frac{1}{4}

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28,060

A cube $ABCDA_1B_1C_1D_1$ has edge length 1. Point $M$ is taken on the side diagonal $A_1D$, and point $N$ is taken on $CD_1$, such that the line segment $MN$ is parallel to the diagonal plane $A_1ACC_1$. Find the minimum length of $MN$.

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

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31,621

Rhombus $ABCD$ has side length $3$ and $\angle B = 110$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$? **A)** $0.81$ **B)** $1.62$ **C)** $2.43$ **D)** $2.16$ **E)** $3.24$

2.16

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31,849

Translate the graph of the function y = 2sin( $$\frac {π}{3}$$ - x) - cos( $$\frac {π}{6}$$ + x) by shifting it to the right by $$\frac {π}{4}$$ units. Determine the minimum value of the corresponding function.

-1

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18,241

On an island, there are knights, liars, and followers; each person knows who is who. All 2018 island residents were lined up and each was asked to answer "Yes" or "No" to the question: "Are there more knights than liars on the island?" The residents responded one by one in such a way that the others could hear. Knights always told the truth, liars always lied. Each follower answered the same as the majority of the preceding respondents, and if the "Yes" and "No" answers were split equally, they could give either answer. It turned out that there were exactly 1009 "Yes" answers. What is the maximum number of followers that could be among the island residents?

1009

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18,047

What is the median of the following list of $4040$ numbers? \[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\]

1. **Identify the total number of terms and the position of the median**: The list consists of $4040$ numbers, which includes all integers from $1$ to $2020$ and their squares. Since the list has an even number of terms, the median will be the average of the $2020$-th and $2021$-st terms. 2. **Determine the range of perfect squares within the list**: We need to find the largest perfect square less than or equal to $2020$. We calculate: \[ 44^2 = 1936 \quad \text{and} \quad 45^2 = 2025 \] Since $45^2$ is greater than $2020$, the perfect squares up to $44^2$ are all less than or equal to $2020$. 3. **Count the number of terms up to $2020$**: There are $2020$ integers from $1$ to $2020$, and there are $44$ perfect squares from $1^2$ to $44^2$. Thus, there are: \[ 2020 + 44 = 2064 \text{ terms less than or equal to } 2020. \] 4. **Locate the $2020$-th and $2021$-st terms**: Since $2064$ terms are less than or equal to $2020$, the $2020$-th and $2021$-st terms are among the integers and not among the squares of integers greater than $44$. Specifically, these terms are among the integers from $1$ to $2020$. 5. **Calculate the $2020$-th and $2021$-st terms**: The $2020$-th term is the last integer in the sequence from $1$ to $2020$, which is $2020$. The $2021$-st term is the first integer square greater than $44^2$, which is $45^2 = 2025$. 6. **Correct the identification of the $2020$-th and $2021$-st terms**: Since we have $2064$ terms up to $2020$, and we need the $2020$-th and $2021$-st terms, we actually need to consider the terms just before reaching $2020$ in the sequence. We calculate: \[ 2064 - 2020 = 44 \quad \text{and} \quad 2064 - 2021 = 43 \] Thus, the $2020$-th term is $2020 - 44 = 1976$ and the $2021$-st term is $2020 - 43 = 1977$. 7. **Calculate the median**: The median is the average of the $2020$-th and $2021$-st terms: \[ \text{Median} = \frac{1976 + 1977}{2} = 1976.5 \] Thus, the median of the list is $\boxed{\textbf{(C)}\ 1976.5}$.

1976.5

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1,727

At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?

1. **Assume Total Number of Students**: Let's assume there are 100 students at Typico High School for simplicity in calculation. 2. **Students Who Like Dancing**: - Given that 60% of the students like dancing. - Therefore, $60\% \times 100 = 60$ students like dancing. 3. **Students Who Dislike Dancing**: - The rest of the students dislike dancing, which is $100 - 60 = 40$ students. 4. **Students Who Like Dancing but Say They Dislike It**: - Of the students who like dancing, 20% say that they dislike it. - Therefore, $20\% \times 60 = 0.20 \times 60 = 12$ students like dancing but say they dislike it. 5. **Students Who Dislike Dancing and Say So**: - Of the students who dislike dancing, 90% say that they dislike it. - Therefore, $90\% \times 40 = 0.90 \times 40 = 36$ students dislike dancing and say so. 6. **Total Students Who Say They Dislike Dancing**: - This includes students who dislike dancing and say so, and students who like dancing but say they dislike it. - Total = Students who dislike and say so + Students who like but say they dislike = $36 + 12 = 48$ students. 7. **Fraction of Students Who Say They Dislike Dancing but Actually Like It**: - We need to find the fraction of students who say they dislike dancing but actually like it. - This is the number of students who like dancing but say they dislike it divided by the total number of students who say they dislike dancing. - Fraction = $\frac{\text{Students who like but say dislike}}{\text{Total who say dislike}} = \frac{12}{48} = \frac{1}{4}$. 8. **Convert Fraction to Percentage**: - To find the percentage, multiply the fraction by 100%. - Percentage = $\frac{1}{4} \times 100\% = 25\%$. Thus, the fraction of students who say they dislike dancing but actually like it is $\boxed{25\%}$, corresponding to choice $\textbf{(D)}$.

$25\%$

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927

Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number?

As shown above, we have $1000x+y=9xy$, so $1000/y=9-1/x$. $1000/y$ must be just a little bit smaller than 9, so we find $y=112$, $x=14$, and the solution is $\boxed{126}$.

126

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6,628

In the figure with circle $Q$, angle $KAT$ measures 42 degrees. What is the measure of minor arc $AK$ in degrees? [asy] import olympiad; size(150); defaultpen(linewidth(0.8)); dotfactor=4; draw(unitcircle); draw(dir(84)--(-1,0)--(1,0)); dot("$A$",(-1,0),W); dot("$K$",dir(84),NNE); dot("$T$",(1,0),E); dot("$Q$",(0,0),S); [/asy]

96

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36,301

Three tanks contain water. The number of litres in each is shown in the table: Tank A: 3600 L, Tank B: 1600 L, Tank C: 3800 L. Water is moved from each of Tank A and Tank C into Tank B so that each tank contains the same volume of water. How many litres of water are moved from Tank A to Tank B?

In total, the three tanks contain $3600 \mathrm{~L} + 1600 \mathrm{~L} + 3800 \mathrm{~L} = 9000 \mathrm{~L}$. If the water is divided equally between the three tanks, each will contain $\frac{1}{3} \cdot 9000 \mathrm{~L} = 3000 \mathrm{~L}$. Therefore, $3600 \mathrm{~L} - 3000 \mathrm{~L} = 600 \mathrm{~L}$ needs to be moved from Tank A to Tank B.

600

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5,369

Consider a month with 31 days, where the number of the month is a product of two distinct primes (e.g., July, represented as 7). Determine how many days in July are relatively prime to the month number.

27

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30,593

In the game of preference, each of the three players is dealt 10 cards, and two cards are placed in the kitty. How many different arrangements are possible in this game? (Consider possible distributions without accounting for which 10 cards go to each specific player.)

\frac{32!}{(10!)^3 \cdot 2! \cdot 3!}

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12,964

Given 4 distinct books that are to be distributed evenly between two students, find the probability that the Chinese language book and the Mathematics book are given to the same student.

\frac{1}{3}

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7,395

Find the largest positive number $ c $ such that for every natural number $ n $, the inequality $ \{n \sqrt{2}\} \geqslant \frac{c}{n} $ holds, where $ \{n \sqrt{2}\} = n \sqrt{2} - \lfloor n \sqrt{2} \rfloor $ and $ \lfloor x \rfloor $ denotes the integer part of $ x $. Determine the natural number $ n $ for which $ \{n \sqrt{2}\} = \frac{c}{n} $. (This problem appeared in the 30th International Mathematical Olympiad, 1989.)

\frac{1}{2\sqrt{2}}

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26,017

For any real number $x$, the symbol $\lfloor x \rfloor$ represents the integer part of $x$, that is, $\lfloor x \rfloor$ is the largest integer not exceeding $x$. Calculate the value of $\lfloor \log_{2}1 \rfloor + \lfloor \log_{2}2 \rfloor + \lfloor \log_{2}3 \rfloor + \lfloor \log_{2}4 \rfloor + \ldots + \lfloor \log_{2}1024 \rfloor$.

8204

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18,535

What is the last two digits of the decimal representation of $9^{8^{7^{\cdot^{\cdot^{\cdot^{2}}}}}}$ ?

21

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19,962

Find the number of integers between 1 and 2013 with the property that the sum of its digits equals 9.

101

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13,288

Calculate the volumes of the solids formed by rotating the regions bounded by the graphs of the functions around the y-axis. $$ y = \arcsin x, \quad y = \arccos x, \quad y = 0 $$

\frac{\pi}{2}

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16,048

In $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$, respectively. Given that $c\sin A= \sqrt {3}a\cos C$ and $(a-c)(a+c)=b(b-c)$, consider the function $f(x)=2\sin x\cos ( \frac {π}{2}-x)- \sqrt {3}\sin (π+x)\cos x+\sin ( \frac {π}{2}+x)\cos x$. (1) Find the period and the equation of the axis of symmetry of the function $y=f(x)$. (2) Find the value of $f(B)$.

\frac {5}{2}

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17,679

Given that $y=f\left(x\right)+x^{2}$ is an odd function, and $f\left(1\right)=1$, if $g\left(x\right)=f\left(x\right)+2$, then $g\left(-1\right)=$____.

-1

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26,429

Let $c = \frac{2\pi}{11}.$ What is the value of \[\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c}?\]

1. **Define the constant and simplify the expression:** Let $ c = \frac{2\pi}{11} $. We need to evaluate: \[ \frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c} \] 2. **Substitute $ c $ into the expression:** \[ \frac{\sin \frac{6\pi}{11} \cdot \sin \frac{12\pi}{11} \cdot \sin \frac{18\pi}{11} \cdot \sin \frac{24\pi}{11} \cdot \sin \frac{30\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}} \] 3. **Use the properties of sine function:** - $\sin(x + 2\pi) = \sin(x)$ for periodicity. - $\sin(2\pi - x) = \sin(x)$ for symmetry. - $\sin(-x) = -\sin(x)$ for odd function property. Applying these properties: \[ \sin \frac{12\pi}{11} = \sin(2\pi - \frac{12\pi}{11}) = \sin \frac{10\pi}{11} \] \[ \sin \frac{18\pi}{11} = \sin(2\pi - \frac{18\pi}{11}) = \sin \frac{16\pi}{11} = \sin(2\pi - \frac{16\pi}{11}) = \sin \frac{4\pi}{11} \] \[ \sin \frac{24\pi}{11} = \sin(2\pi - \frac{24\pi}{11}) = \sin \frac{20\pi}{11} = \sin(2\pi - \frac{20\pi}{11}) = \sin \frac{2\pi}{11} \] \[ \sin \frac{30\pi}{11} = \sin(2\pi - \frac{30\pi}{11}) = \sin \frac{8\pi}{11} \] 4. **Substitute back and simplify:** \[ \frac{\sin \frac{6\pi}{11} \cdot \sin \frac{10\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{2\pi}{11} \cdot \sin \frac{8\pi}{11}}{\sin \frac{2\pi}{11} \cdot \sin \frac{4\pi}{11} \cdot \sin \frac{6\pi}{11} \cdot \sin \frac{8\pi}{11} \cdot \sin \frac{10\pi}{11}} \] Notice that each term in the numerator matches a corresponding term in the denominator. Therefore, the entire expression simplifies to 1. 5. **Conclude with the final answer:** \[ \boxed{\textbf{(E)}\ 1} \]

1

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1,064

Triangle $PQR$ has positive integer side lengths with $PQ = PR$. Let $J$ be the intersection of the bisectors of $\angle Q$ and $\angle R$. Suppose $QJ = 10$. Find the smallest possible perimeter of $\triangle PQR$.

1818

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28,023

For some positive integer $k$, when 60 is divided by $k^2$, the remainder is 6. What is the remainder when 100 is divided by $k$?

1

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38,362

Today is December 19, 2010. What is the integer part of the sum $\frac{2010}{1000}+\frac{1219}{100}+\frac{27}{10}$?

16

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10,755

Given the new operation $n\heartsuit m=n^{3+m}m^{2+n}$, evaluate $\frac{2\heartsuit 4}{4\heartsuit 2}$.

\frac{1}{2}

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18,914

Given the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ with left and right foci $F_{1}$ and $F_{2}$ respectively, draw a line $l$ through the right focus that intersects the ellipse at points $P$ and $Q$. Find the maximum area of the inscribed circle of triangle $F_{1} PQ$.

\frac{9 \pi}{16}

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15,855

The water tank in the diagram below is in the shape of an inverted right circular cone. The radius of its base is 16 feet, and its height is 96 feet. The water in the tank is $25\%$ of the tank's capacity. The height of the water in the tank can be written in the form $a\sqrt[3]{b}$, where $a$ and $b$ are positive integers and $b$ is not divisible by a perfect cube greater than 1. What is $a+b$? [asy] size(150); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw(shift(0,96)*yscale(0.5)*Circle((0,0),16)); draw((-16,96)--(0,0)--(16,96)--(0,96)); draw(scale(0.75)*shift(0,96)*yscale(0.5)*Circle((0,0),16)); draw((-18,72)--(-20,72)--(-20,0)--(-18,0)); label("water's height",(-20,36),W); draw((20,96)--(22,96)--(22,0)--(20,0)); label("96'",(22,48),E); label("16'",(8,96),S); [/asy]

50

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36,275

What is $\frac{~\frac{2}{5}~}{\frac{3}{7}}$?

\frac{14}{15}

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38,768

The positive integers are arranged in rows and columns as shown below. | Row 1 | 1 | | Row 2 | 2 | 3 | | Row 3 | 4 | 5 | 6 | | Row 4 | 7 | 8 | 9 | 10 | | Row 5 | 11 | 12 | 13 | 14 | 15 | | Row 6 | 16 | 17 | 18 | 19 | 20 | 21 | | ... | More rows continue to list the positive integers in order, with each new row containing one more integer than the previous row. How many integers less than 2000 are in the column that contains the number 2000?

16

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31,929

A real number $x$ is chosen uniformly at random from the interval $[0,1000]$. Find the probability that $\left\lfloor\frac{\left\lfloor\frac{x}{2.5}\right\rfloor}{2.5}\right\rfloor=\left\lfloor\frac{x}{6.25}\right\rfloor$.

Let $y=\frac{x}{2.5}$, so $y$ is chosen uniformly at random from [0,400]. Then we need $\left\lfloor\frac{\lfloor y\rfloor}{2.5}\right\rfloor=\left\lfloor\frac{y}{2.5}\right\rfloor$. Let $y=5a+b$, where $0 \leq b<5$ and $a$ is an integer. Then $\left\lfloor\frac{\lfloor y\rfloor}{2.5}\right\rfloor=2a+\left\lfloor\frac{\lfloor b\rfloor}{2.5}\right\rfloor$ while $\left\lfloor\frac{y}{2.5}\right\rfloor=2a+\left\lfloor\frac{b}{2.5}\right\rfloor$, so we need $\left\lfloor\frac{\lfloor b\rfloor}{2.5}\right\rfloor=\left\lfloor\frac{b}{2.5}\right\rfloor$, where $b$ is selected uniformly at random from $[0,5]$. This can be shown to always hold except for $b \in[2.5,3)$, so the answer is $1-\frac{0.5}{5}=\frac{9}{10}$.

\frac{9}{10}

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4,915

A batch of disaster relief supplies is loaded into 26 trucks. The trucks travel at a constant speed of $ v $ kilometers per hour directly to the disaster area. If the distance between the two locations is 400 kilometers and the distance between every two trucks must be at least $ \left(\frac{v}{20}\right)^{2} $ kilometers, how many hours will it take to transport all the supplies to the disaster area?

10

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31,063

The distance between two parallel lines $x-y=1$ and $2x-2y+3=0$ is ______.

\frac{5\sqrt{2}}{4}

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17,193

Adam and Simon start on bicycle trips from the same point at the same time. Adam travels east at 8mph and Simon travels south at 6mph. After how many hours are they 60 miles apart?

6

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38,640

Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$. What is the value of $\frac{x+3y}{3x-y}$?

1. Start with the given equation: \[ \frac{3x+y}{x-3y} = -2 \] 2. Cross-multiply to eliminate the fraction: \[ 3x + y = -2(x - 3y) \] 3. Distribute the -2 on the right-hand side: \[ 3x + y = -2x + 6y \] 4. Rearrange the equation to isolate terms involving $x$ and $y$ on opposite sides: \[ 3x + 2x = 6y - y \] \[ 5x = 5y \] 5. Simplify the equation: \[ x = y \] 6. Substitute $x = y$ into the expression $\frac{x+3y}{3x-y}$: \[ \frac{x + 3y}{3x - y} = \frac{y + 3y}{3y - y} \] \[ \frac{4y}{2y} \] 7. Simplify the fraction: \[ \frac{4y}{2y} = 2 \] 8. Thus, the value of $\frac{x+3y}{3x-y}$ is: \[ \boxed{\textbf{(D)}\ 2} \]

2

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250

A cube with edge length 2 cm has a dot marked at the center of the top face. The cube is on a flat table and rolls without slipping, making a full rotation back to its initial orientation, with the dot back on top. Calculate the length of the path followed by the dot in terms of $\pi. $

2\sqrt{2}\pi

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28,866

At the round table, $10$ people are sitting, some of them are knights, and the rest are liars (knights always say pride, and liars always lie) . It is clear thath I have at least one knight and at least one liar. What is the largest number of those sitting at the table can say: ''Both of my neighbors are knights '' ? (A statement that is at least partially false is considered false.)

To solve this problem, we need to maximize the number of people at a round table who can truthfully say: "Both of my neighbors are knights." Considering the rules: - Knights always tell the truth. - Liars always lie. - At least one knight and one liar are present. Let's analyze the configuration of people around the table: 1. If a person is truthfully saying "Both of my neighbors are knights," the person themselves must be a knight, as knights tell the truth. 2. If the person making the statement is a liar, then one or both neighbors must not be knights (since liars lie). 3. Consider the maximum possible scenario where nine individuals are knights. With only one person left, that person must be a liar (since at least one liar is required). 4. Arrange the people such that the liar is strategically placed to disrupt the truth of the statement for only themselves and not for the rest. Let's denote: - $ K $ for a knight, - $ L $ for a liar. A possible arrangement that satisfies the conditions is: $ K, K, K, K, K, K, K, K, K, L $. - Here, each of the nine knights can truthfully say, "Both of my neighbors are knights," because they are flanked by knights on both sides. - The liar cannot truthfully make this statement as their claim would be "false," given at least one of their neighbors is a liar (themselves). Therefore, the maximum number of people who can truthfully say that both their neighbors are knights is: \[ \boxed{9} \]

9

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5,969

A square is inscribed in another square such that its vertices lie on the sides of the first square, and its sides form angles of $60^{\circ}$ with the sides of the first square. What fraction of the area of the given square is the area of the inscribed square?

4 - 2\sqrt{3}

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13,472

Misha made himself a homemade dartboard at the summer house. The round board is divided into sectors by circles - it can be used to throw darts. Points are awarded according to the number written in the sector, as indicated in the diagram. Misha threw 8 darts 3 times. The second time, he scored twice as many points as the first time, and the third time, he scored 1.5 times more points than the second time. How many points did he score the second time?

48

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14,722

Find the largest negative root of the equation $$ 4 \sin (3 x) + 13 \cos (3 x) = 8 \sin (x) + 11 \cos (x) $$

-0.1651

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25,066

A piece of string is cut in two at a point selected at random. The probability that the longer piece is at least x times as large as the shorter piece is

Let's consider a string of unit length (length = 1) for simplicity. We cut the string at a point $C$ chosen uniformly at random along the string. Let $C$ be the length of the shorter piece after the cut, and $1 - C$ be the length of the longer piece. We need to find the probability that the longer piece is at least $x$ times as large as the shorter piece. #### Step 1: Set up the inequality We need the longer piece to be at least $x$ times the length of the shorter piece. This can happen in two ways: 1. $1 - C \geq xC$ (if $C$ is the shorter piece) 2. $C \geq x(1 - C)$ (if $1 - C$ is the shorter piece) #### Step 2: Solve the inequalities 1. Solving $1 - C \geq xC$: \[ 1 - C \geq xC \implies 1 \geq (x + 1)C \implies C \leq \frac{1}{x + 1} \] This inequality is valid when $C$ is the shorter piece, which is true when $C \leq \frac{1}{2}$. 2. Solving $C \geq x(1 - C)$: \[ C \geq x - xC \implies C + xC \geq x \implies (1 + x)C \geq x \implies C \geq \frac{x}{x + 1} \] This inequality is valid when $1 - C$ is the shorter piece, which is true when $C \geq \frac{1}{2}$. #### Step 3: Determine the valid intervals 1. From $C \leq \frac{1}{x + 1}$, the valid interval is $[0, \frac{1}{x + 1}]$. 2. From $C \geq \frac{x}{x + 1}$, the valid interval is $[\frac{x}{x + 1}, 1]$. However, we need to consider the intersection of these intervals with the intervals where each piece is actually the shorter piece: - $C \leq \frac{1}{2}$ intersects with $[0, \frac{1}{x + 1}]$ if $\frac{1}{x + 1} \leq \frac{1}{2}$. - $C \geq \frac{1}{2}$ intersects with $[\frac{x}{x + 1}, 1]$ if $\frac{x}{x + 1} \geq \frac{1}{2}$. #### Step 4: Calculate the probability The probability that $C$ falls in either $[0, \frac{1}{x + 1}]$ or $[\frac{x}{x + 1}, 1]$ is the sum of the lengths of these intervals: \[ \text{Probability} = \left(\frac{1}{x + 1} - 0\right) + \left(1 - \frac{x}{x + 1}\right) = \frac{1}{x + 1} + \frac{1}{x + 1} = \frac{2}{x + 1} \] Thus, the probability that the longer piece is at least $x$ times as large as the shorter piece is $\boxed{\textbf{(E) }\frac{2}{x+1}}$.

\frac{2}{x+1}

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2,111

Let $P$ be a plane passing through the origin. When $\begin{pmatrix} 5 \\ 3 \\ 5 \end{pmatrix}$ is projected onto plane $P,$ the result is $\begin{pmatrix} 3 \\ 5 \\ 1 \end{pmatrix}.$ When $\begin{pmatrix} 4 \\ 0 \\ 7 \end{pmatrix}$ is projected onto plane $P,$ what is the result?

\begin{pmatrix} 1 \\ 3 \\ 1 \end{pmatrix}

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39,819

Point $F$ is taken on the extension of side $AD$ of parallelogram $ABCD$. $BF$ intersects diagonal $AC$ at $E$ and side $DC$ at $G$. If $EF = 32$ and $GF = 24$, then $BE$ equals: [asy] size(7cm); pair A = (0, 0), B = (7, 0), C = (10, 5), D = (3, 5), F = (5.7, 9.5); pair G = intersectionpoints(B--F, D--C)[0]; pair E = intersectionpoints(A--C, B--F)[0]; draw(A--D--C--B--cycle); draw(A--C); draw(D--F--B); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$F$", F, N); label("$G$", G, NE); label("$E$", E, SE); //Credit to MSTang for the asymptote[/asy] $\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 16$

16

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36,077

How many whole numbers between 1 and 2000 do not contain the digit 2?

6560

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24,797

12 Smurfs are seated around a round table. Each Smurf dislikes the 2 Smurfs next to them, but does not dislike the other 9 Smurfs. Papa Smurf wants to form a team of 5 Smurfs to rescue Smurfette, who was captured by Gargamel. The team must not include any Smurfs who dislike each other. How many ways are there to form such a team?

36

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10,154

Given vectors $a=(2\cos \alpha,\sin ^{2}\alpha)$ and $b=(2\sin \alpha,t)$, where $\alpha\in\left( 0,\frac{\pi}{2} \right)$ and $t$ is a real number. $(1)$ If $a-b=\left( \frac{2}{5},0 \right)$, find the value of $t$; $(2)$ If $t=1$ and $a\cdot b=1$, find the value of $\tan \left( 2\alpha+\frac{\pi}{4} \right)$.

\frac{23}{7}

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32,841

On a table there are $100$ red and $k$ white buckets for which all of them are initially empty. In each move, a red and a white bucket is selected and an equal amount of water is added to both of them. After some number of moves, there is no empty bucket and for every pair of buckets that are selected together at least once during the moves, the amount of water in these buckets is the same. Find all the possible values of $k$ .

100

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20,334

Given in the polar coordinate system, point P moves on the curve $\rho^2\cos\theta-2\rho=0$, the minimum distance from point P to point $Q(1, \frac{\pi}{3})$ is \_\_\_\_\_\_.

\frac{3}{2}

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23,178

Given the equations $y = 2x^2$ and $y = 4x + c$, determine the value(s) of $c$ for which these two equations have precisely two identical solutions.

-2

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7,521

A television's price was discounted $40\%$ and then slashed another $40\%$ during a liquidation sale. By what percent has the original price been reduced?

64\%

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34,468

Determine the number of ways to select a sequence of 8 sets $A_{1}, A_{2}, \ldots, A_{8}$, such that each is a subset (possibly empty) of \{1,2\}, and $A_{m}$ contains $A_{n}$ if $m$ divides $n$.

Consider an arbitrary $x \in\{1,2\}$, and let us consider the number of ways for $x$ to be in some of the sets so that the constraints are satisfied. We divide into a few cases: - Case: $x \notin A_{1}$. Then $x$ cannot be in any of the sets. So there is one possibility. - Case: $x \in A_{1}$ but $x \notin A_{2}$. Then the only other sets that $x$ could be in are $A_{3}, A_{5}, A_{7}$, and $x$ could be in some collection of them. There are 8 possibilities in this case. - Case: $x \in A_{2}$. Then $x \in A_{1}$ automatically. There are 4 independent choices to be make here: (1) whether $x \in A_{5} ;(2)$ whether $x \in A_{7} ;(3)$ whether $x \in A_{3}$, and if yes, whether $x \in A_{6}$; (4) whether $x \in A_{4}$, and if yes, whether $x \in A_{8}$. There are $2 \times 2 \times 3 \times 3=36$ choices here. Therefore, there are $1+8+36=45$ ways to place $x$ into some of the sets. Since the choices for $x=1$ and $x=2$ are made independently, we see that the total number of possibilities is $45^{2}=2025$.

2025

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4,225

Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$

As noted in the previous solutions, $a+b+c = 0$. Let $a = a_1+a_2 i$, $b = b_1+b_2 i$, $c = c_1+c_2 i$ and we have $a_1 + b_1 + c_1 = a_2 + b_2 + c_2 = 0$. Then the given $|a|^2 + |b|^2 + |c|^2 = 250$ translates to $\sum_{} ( {a_1}^2 + {a_2}^2 ) = 250.$ Note that in a right triangle, the sum of the squares of the three sides is equal to two times the square of the hypotenuse, by the pythagorean theorem. Thus, we have \[2h^2 = (a_1 - b_1)^2 + (a_2 - b_2)^2 + (b_1 - c_1)^2 + (b_2 - c_2)^2 + (a_1 - c_1)^2 + (a_2 - c_2)^2\] \[= 2 \left( \sum_{} ( {a_1}^2 + {a_2}^2 ) \right) - 2 \left( \sum_{cyc} a_1 b_1 + \sum_{cyc} a_2 b_2 \right)\] \[= 500 - \left( (a_1 + b_1 + c_1)^2 + (a_2 + b_2 + c_2)^2 - \sum_{cyc} ( {a_1}^2 + {a_2}^2 ) \right)\] \[= 500 - (0^2 + 0^2 - 250)\] so $h^2 = \boxed{375}$ and we may conclude. ~ rzlng

375

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7,044

Ivan wanted to buy nails. In one store, where 100 grams of nails cost 180 rubles, he couldn't buy the required amount because he was short 1430 rubles. Then he went to another store where 100 grams cost 120 rubles. He bought the required amount and received 490 rubles in change. How many kilograms of nails did Ivan buy?

3.2

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9,107

Consider a regular polygon with $2^n$ sides, for $n \ge 2$ , inscribed in a circle of radius $1$ . Denote the area of this polygon by $A_n$ . Compute $\prod_{i=2}^{\infty}\frac{A_i}{A_{i+1}}$

\frac{2}{\pi}

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24,251

Consider a city grid with intersections labeled A, B, C, and D. Assume a student walks from intersection A to intersection B every morning, always walking along the designated paths and only heading east or south. The student passes through intersections C and D along the way. The intersections are placed such that A to C involves 3 eastward moves and 2 southward moves, and C to D involves 2 eastward moves and 1 southward move, and finally from D to B requires 1 eastward move and 2 southward moves. Each morning, at each intersection where he has a choice, he randomly chooses whether to go east or south with probability $\frac{1}{2}$. Determine the probability that the student walks through C, and then D on any given morning. A) $\frac{15}{77}$ B) $\frac{10}{462}$ C) $\frac{120}{462}$ D) $\frac{3}{10}$ E) $\frac{64}{462}$

\frac{15}{77}

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21,145

In a game of Chomp, two players alternately take bites from a 5-by-7 grid of unit squares. To take a bite, a player chooses one of the remaining squares, then removes ("eats") all squares in the quadrant defined by the left edge (extended upward) and the lower edge (extended rightward) of the chosen square. For example, the bite determined by the shaded square in the diagram would remove the shaded square and the four squares marked by $\times.$ (The squares with two or more dotted edges have been removed form the original board in previous moves.) The object of the game is to make one's opponent take the last bite. The diagram shows one of the many subsets of the set of 35 unit squares that can occur during the game of Chomp. How many different subsets are there in all? Include the full board and empty board in your count.

By drawing possible examples of the subset, one can easily see that making one subset is the same as dividing the game board into two parts. One can also see that it is the same as finding the shortest route from the upper left hand corner to the lower right hand corner; Such a route would require 5 lengths that go down, and 7 that go across, with the shape on the right "carved" out by the path a possible subset. Therefore, the total number of such paths is $\binom{12}{5}=\boxed{792}$

792

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6,562

What is the number of degrees in the smaller angle formed by the hour and minute hands of a clock at 8:15? Express your answer as a decimal to the nearest tenth. [asy] size(200); draw(Circle((0,0),5),linewidth(1.2)); pair[] mins; for(int i = 0; i < 60; ++i){ mins[i] = 4.5*dir(-6*i + 90); dot(mins[i]); } for(int i = 1; i <= 12; ++i){ label((string)i,mins[5*i % 60],dir(-30*i - 90)); } fill(Circle((0,0),0.25)); [/asy]

157.5

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36,316

A certain department store sells suits and ties, with each suit priced at $1000$ yuan and each tie priced at $200 yuan. During the "National Day" period, the store decided to launch a promotion offering two discount options to customers. Option 1: Buy one suit and get one tie for free; Option 2: Pay 90% of the original price for both the suit and the tie. Now, a customer wants to buy 20 suits and $x$ ties $\left(x > 20\right)$. $(1)$ If the customer chooses Option 1, the payment will be ______ yuan (expressed as an algebraic expression in terms of $x$). If the customer chooses Option 2, the payment will be ______ yuan (expressed as an algebraic expression in terms of $x$). $(2)$ If $x=30$, calculate and determine which option is more cost-effective at this point. $(3)$ When $x=30$, can you come up with a more cost-effective purchasing plan? Please describe your purchasing method.

21800

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23,283

There is a positive integer s such that there are s solutions to the equation $64sin^2(2x)+tan^2(x)+cot^2(x)=46$ in the interval $(0,\frac{\pi}{2})$ all of the form $\frac{m_k}{n_k}\pi$ where $m_k$ and $n_k$ are relatively prime positive integers, for $k = 1, 2, 3, . . . , s$ . Find $(m_1 + n_1) + (m_2 + n_2) + (m_3 + n_3) + · · · + (m_s + n_s)$ .

100

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27,997