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Suppose that $x, y$, and $z$ are non-negative real numbers such that $x+y+z=1$. What is the maximum possible value of $x+y^{2}+z^{3}$ ?
|
Since $0 \leq y, z \leq 1$, we have $y^{2} \leq y$ and $z^{3} \leq z$. Therefore $x+y^{2}+z^{3} \leq x+y+z=1$. We can get $x+y^{2}+z^{3}=1$ by setting $(x, y, z)=(1,0,0)$.
|
1
|
deepscale
| 4,872
| |
The coefficient of $x^2$ in the expansion of $(x+1)^5(x-2)$ is $\boxed{5}$.
|
-15
|
deepscale
| 18,119
| ||
What is the maximum possible area of a quadrilateral with side lengths 1, 4, 7, and 8?
|
18
|
deepscale
| 7,506
| ||
Vertex $E$ of equilateral $\triangle{ABE}$ is in the interior of unit square $ABCD$. Let $R$ be the region consisting of all points inside $ABCD$ and outside $\triangle{ABE}$ whose distance from $AD$ is between $\frac{1}{3}$ and $\frac{2}{3}$. What is the area of $R$?
|
1. **Identify the region $R$:** The region $R$ consists of all points inside the unit square $ABCD$ but outside the equilateral triangle $\triangle{ABE}$, and whose distance from side $AD$ is between $\frac{1}{3}$ and $\frac{2}{3}$.
2. **Calculate the area of the middle third of the square:** The middle third of the square is the strip parallel to $AD$ with width $\frac{1}{3}$ (since the square's side length is 1). The area of this strip is:
\[
\text{Area of middle third} = \frac{1}{3} \times 1 = \frac{1}{3}.
\]
3. **Analyze the equilateral triangle $\triangle{ABE}$:**
- The side length of $\triangle{ABE}$ is the same as the side of the square, which is 1.
- The height $h$ of $\triangle{ABE}$ can be calculated using the formula for the height of an equilateral triangle:
\[
h = \frac{\sqrt{3}}{2} \times \text{side length} = \frac{\sqrt{3}}{2} \times 1 = \frac{\sqrt{3}}{2}.
\]
4. **Calculate the area of $\triangle{ABE}$:**
\[
\text{Area of } \triangle{ABE} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}.
\]
5. **Determine the area of the pentagon formed by the intersection of $\triangle{ABE}$ and the middle third strip:**
- The pentagon can be split into a rectangle and a smaller equilateral triangle.
- The base of the smaller equilateral triangle is $\frac{1}{3}$, and its height is:
\[
\text{Height of smaller triangle} = \frac{\sqrt{3}}{6}.
\]
- The area of the smaller equilateral triangle is:
\[
\text{Area of smaller triangle} = \frac{1}{2} \times \frac{1}{3} \times \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{36}.
\]
- The rectangle has a base of $\frac{1}{3}$ and a height of $\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{6} = \frac{\sqrt{3}}{3}$.
- The area of the rectangle is:
\[
\text{Area of rectangle} = \frac{1}{3} \times \frac{\sqrt{3}}{3} = \frac{\sqrt{3}}{9}.
\]
6. **Calculate the area of the pentagon:**
\[
\text{Area of pentagon} = \frac{\sqrt{3}}{36} + \frac{\sqrt{3}}{9} = \frac{4\sqrt{3}}{36} = \frac{\sqrt{3}}{9}.
\]
7. **Calculate the area of region $R$:**
\[
\text{Area of } R = \text{Area of middle third} - \text{Area of pentagon} = \frac{1}{3} - \frac{\sqrt{3}}{9} = \frac{3 - \sqrt{3}}{9}.
\]
8. **Conclusion:**
\[
\boxed{\text{(D) }\frac{3-\sqrt{3}}{9}}
\]
|
\frac{3-\sqrt{3}}{9}
|
deepscale
| 2,796
| |
Find $120_4\times13_4\div2_4$. Express your answer in base 4.
|
1110_4
|
deepscale
| 38,005
| ||
Given in triangle $\triangle ABC$ the internal angles are $A$, $B$, and $C$, and the centroid is $G$. If $2\sin A \cdot \overrightarrow{GA} + \sqrt{3}\sin B \cdot \overrightarrow{GB} + 3\sin C \cdot \overrightarrow{GC} = \overrightarrow{0}$, then $\cos B = \_\_\_\_\_\_$.
|
\frac{1}{12}
|
deepscale
| 28,636
| ||
If two distinct members of the set $\{ 2, 4, 10, 12, 15, 20, 50 \}$ are randomly selected and multiplied, what is the probability that the product is a multiple of 100? Express your answer as a common fraction.
|
\frac{1}{3}
|
deepscale
| 35,374
| ||
Anna enjoys dinner at a restaurant in Washington, D.C., where the sales tax on meals is 10%. She leaves a 15% tip on the price of her meal before the sales tax is added, and the tax is calculated on the pre-tip amount. She spends a total of 27.50 dollars for dinner. What is the cost of her dinner without tax or tip in dollars?
|
Let $x$ be the cost of Anna's dinner before tax and tip.
1. **Calculate the tax:** The tax rate is 10%, so the tax amount is $\frac{10}{100}x = 0.1x$.
2. **Calculate the tip:** The tip is 15% of the pre-tax meal cost, so the tip amount is $\frac{15}{100}x = 0.15x$.
3. **Total cost:** The total cost of the meal including tax and tip is given by:
\[
x + 0.1x + 0.15x
\]
4. **Simplify the expression:** Combine like terms to find the total cost in terms of $x$:
\[
x + 0.1x + 0.15x = 1x + 0.1x + 0.15x = 1.25x
\]
5. **Set up the equation:** We know the total cost is $27.50, so we set up the equation:
\[
1.25x = 27.50
\]
6. **Solve for $x$:** Divide both sides of the equation by 1.25 to isolate $x$:
\[
x = \frac{27.50}{1.25}
\]
7. **Perform the division:** Calculate the value of $x$:
\[
x = 22
\]
Thus, the cost of her dinner without tax or tip is $\boxed{\textbf{(D)}\ 22}$.
|
22
|
deepscale
| 471
| |
Let $S$ be the set of all positive integer divisors of $129,600$. Calculate the number of numbers that are the product of two distinct elements of $S$.
|
488
|
deepscale
| 12,601
| ||
Let $n \ge 3$ be an integer. What is the largest possible number of interior angles greater than $180^\circ$ in an $n$-gon in the plane, given that the $n$-gon does not intersect itself and all its sides have the same length?
|
Let \( n \ge 3 \) be an integer, and consider an \( n \)-gon in the plane with equal side lengths. We are asked to find the largest possible number of interior angles greater than \( 180^\circ \), given that the \( n \)-gon does not intersect itself.
To solve this, we will use the following geometric principles:
1. **Polygon Interior Angle Sum Formula:**
The sum of all interior angles of an \( n \)-gon is given by:
\[
(n-2) \times 180^\circ
\]
2. **Regular Polygon Properties:**
In a regular \( n \)-gon (where all sides and angles are equal), the measure of each interior angle is:
\[
\frac{(n-2) \times 180^\circ}{n}
\]
For a polygon not to self-intersect and remain convex (which implies no angle is greater than \( 180^\circ \)), it should ideally be a regular polygon.
3. **Analysis for Interior Angles Greater than \( 180^\circ \):**
If any interior angle is greater than \( 180^\circ \), the polygon must be concave. However, since all sides are of equal length, it becomes impossible to form a non-self-intersecting concave polygon without compromising the side length uniformity.
Furthermore, if we attempt to create angles greater than \( 180^\circ \) while preserving equal side lengths and non-intersection, the figure deviates from the standard convex arrangement, leading to a contradiction in a non-self-intersecting polygon setup.
Therefore, given these constraints, the largest possible number of interior angles greater than \( 180^\circ \) for a non-self-intersecting \( n \)-gon with equal side lengths is:
\[
\boxed{0}
\]
Given that having any interior angle greater than \( 180^\circ \) automatically makes the polygon concave and does not satisfy the equality of side lengths in a simple polygon, the solution concludes with 0 angles being greater than \( 180^\circ \).
|
0
|
deepscale
| 6,225
| |
At lunch, Abby, Bart, Carl, Dana, and Evan share a pizza divided radially into 16 slices. Each one takes takes one slice of pizza uniformly at random, leaving 11 slices. The remaining slices of pizza form "sectors" broken up by the taken slices, e.g. if they take five consecutive slices then there is one sector, but if none of them take adjacent slices then there will be five sectors. What is the expected number of sectors formed?
|
Consider the more general case where there are $N$ slices and $M>0$ slices are taken. Let $S$ denote the number of adjacent pairs of slices of pizza which still remain. There are $N-M$ slices and a sector of $k$ slices contributes $k-1$ pairs to $S$. Hence the number of sectors is $N-M-S$. We compute the expected value of $S$ by looking at each adjacent pair in the original pizza: $$\mathbb{E}(S)=N \frac{\binom{N-2}{M}}{\binom{N}{M}}=N \frac{(N-M)(N-M-1)}{N(N-1)}=\frac{(N-M)(N-M-1)}{N-1}$$ The expected number of sectors is then $$N-M-\frac{(N-M)(N-M-1)}{N-1}=\frac{(N-M) M}{N-1}$$ For $N=16, M=5$ this yields \frac{11}{3}$.
|
\frac{11}{3}
|
deepscale
| 4,475
| |
In $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$, respectively. It is known that $c^{2}=a^{2}+b^{2}-4bc\cos C$, and $A-C= \frac {\pi}{2}$.
(Ⅰ) Find the value of $\cos C$;
(Ⅱ) Find the value of $\cos \left(B+ \frac {\pi}{3}\right)$.
|
\frac {4-3 \sqrt {3}}{10}
|
deepscale
| 21,088
| ||
Let $x$, $y$, and $z$ be positive integers such that $x + y + z = 12$. What is the smallest possible value of $(x+y+z)\left(\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\right)$?
|
\frac{9}{2}
|
deepscale
| 22,321
| ||
There are 8 students arranged in two rows, with 4 people in each row. If students A and B must be arranged in the front row, and student C must be arranged in the back row, then the total number of different arrangements is ___ (answer in digits).
|
5760
|
deepscale
| 28,301
| ||
Bonnie makes the frame of a cube out of 12 pieces of wire that are each six inches long. Meanwhile Roark uses 1-inch-long pieces of wire to make a collection of unit cube frames that are not connected to each other. The total volume of Roark's cubes is the same as the volume of Bonnie's cube. What is the ratio of the total length of Bonnie's wire to the total length of Roark's wire? Express your answer as a common fraction. [asy]
size(50);
draw((0,0)--(4,0)--(4,4)--(0,4)--cycle);
draw((3,1)--(7,1)--(7,5)--(3,5)--cycle);
draw((0,0)--(3,1));
draw((4,0)--(7,1));
draw((4,4)--(7,5));
draw((0,4)--(3,5));
[/asy]
|
\dfrac{1}{36}
|
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| 35,818
| ||
The sum of the numbers 300, 2020, and 10001 is
|
12321
|
deepscale
| 21,122
| ||
A number \( a \) is randomly chosen from \( 1, 2, 3, \cdots, 10 \), and a number \( b \) is randomly chosen from \( -1, -2, -3, \cdots, -10 \). What is the probability that \( a^{2} + b \) is divisible by 3?
|
37/100
|
deepscale
| 7,688
| ||
In the diagram, the circle has center \( O \) and square \( OPQR \) has vertex \( Q \) on the circle. If the area of the circle is \( 72 \pi \), the area of the square is:
|
36
|
deepscale
| 13,754
| ||
Let the set $\mathcal{S} = \{8, 5, 1, 13, 34, 3, 21, 2\}.$ Susan makes a list as follows: for each two-element subset of $\mathcal{S},$ she writes on her list the greater of the set's two elements. Find the sum of the numbers on the list.
|
Thinking of this problem algorithmically, one can "sort" the array to give: \[{1, 2, 3, 5, 8, 13, 21, 34}\]
Now, notice that when we consider different pairs, we are only going to fixate one element and look at the all of the next elements in the array, basically the whole $j = i + 1$ shebang. Then, we see that if we set the sum of the whole array to $x,$ we get out answer to be
\[(x-1) + (x-3) + (x-6) + (x-11) + (x-19) + (x-32) + (x-53) = 7x - 125\]
Finding $x$ isn't hard, and we see that it is equal to $609$:
\[609 - 125 = \boxed{484}\]
|
484
|
deepscale
| 6,763
| |
The sum of the first fifty positive odd integers subtracted from the sum of the first fifty positive even integers, each decreased by 3, calculate the result.
|
-100
|
deepscale
| 29,971
| ||
How many parallelograms with sides 1 and 2, and angles \(60^{\circ}\) and \(120^{\circ}\), can be placed inside a regular hexagon with side length 3?
|
12
|
deepscale
| 15,650
| ||
Find the smallest positive integer $n$ such that
\[\begin{pmatrix} \cos 170^\circ & -\sin 170^\circ \\ \sin 170^\circ & \cos 170^\circ \end{pmatrix}^n = \mathbf{I}.\]
|
36
|
deepscale
| 40,294
| ||
A $\text{palindrome}$, such as $83438$, is a number that remains the same when its digits are reversed. The numbers $x$ and $x+32$ are three-digit and four-digit palindromes, respectively. What is the sum of the digits of $x$?
|
1. **Identify the range of $x$ and $x+32$:**
- Since $x$ is a three-digit palindrome, the maximum value of $x$ is $999$.
- Consequently, the maximum value of $x+32$ is $999 + 32 = 1031$.
- The minimum value of $x+32$ is $1000$ (since it is a four-digit palindrome).
2. **Determine the possible values for $x+32$:**
- The only four-digit palindrome between $1000$ and $1031$ is $1001$.
- Therefore, $x + 32 = 1001$.
3. **Calculate $x$:**
- Rearranging $x + 32 = 1001$, we find $x = 1001 - 32 = 969$.
4. **Calculate the sum of the digits of $x$:**
- The digits of $969$ are $9$, $6$, and $9$.
- The sum of these digits is $9 + 6 + 9 = 24$.
Thus, the sum of the digits of $x$ is $\boxed{24}$.
|
24
|
deepscale
| 752
| |
Let $a$ be the number of positive multiples of $6$ that are less than $30$. Let $b$ be the number of positive integers that are less than $30$, and a multiple of $3$ and a multiple of $2$. Compute $(a - b)^3$.
|
0
|
deepscale
| 39,453
| ||
Let $N=30^{2015}$. Find the number of ordered 4-tuples of integers $(A, B, C, D) \in\{1,2, \ldots, N\}^{4}$ (not necessarily distinct) such that for every integer $n, A n^{3}+B n^{2}+2 C n+D$ is divisible by $N$.
|
Note that $n^{0}=\binom{n}{0}, n^{1}=\binom{n}{1}, n^{2}=2\binom{n}{2}+\binom{n}{1}, n^{3}=6\binom{n}{3}+6\binom{n}{2}+\binom{n}{1}$. Thus the polynomial rewrites as $6 A\binom{n}{3}+(6 A+2 B)\binom{n}{2}+(A+B+2 C)\binom{n}{1}+D\binom{n}{0}$ which by the classification of integer-valued polynomials is divisible by $N$ always if and only if $6 A, 6 A+2 B, A+B+2 C, D$ are always divisible by $N$. We can eliminate $B$ and (trivially) $D$ from the system: it's equivalent to the system $6 A \equiv 0(\bmod N)$, $4 A-4 C \equiv 0(\bmod N), B \equiv-A-2 C(\bmod N), D \equiv 0(\bmod N)$. So we want $1^{2}$ times the number of $(A, C)$ with $A \equiv 0(\bmod N / 6), C \equiv A(\bmod N / 4)$. So there are $N /(N / 6)=6$ choices for $A$, and then given such a choice of $A$ there are $N /(N / 4)=4$ choices for $C$. So we have $6 \cdot 4 \cdot 1^{2}=24$ solutions total.
|
24
|
deepscale
| 3,947
| |
Given the equation of the circle $x^{2}+y^{2}-8x+15=0$, find the minimum value of $k$ such that there exists at least one point on the line $y=kx+2$ that can serve as the center of a circle with a radius of $1$ that intersects with circle $C$.
|
- \frac{4}{3}
|
deepscale
| 10,996
| ||
On the Cartesian plane, the midpoint between two points $P(p,q)$ and $Q(r,s)$ is $N(x,y)$. If $P$ is moved vertically upwards 10 units and horizontally to the right 5 units, and $Q$ is moved vertically downwards 5 units and horizontally to the left 15 units, find the new midpoint $N'$ between $P$ and $Q$ and the distance between $N$ and $N'$.
|
5.59
|
deepscale
| 12,439
| ||
A plane is expressed parametrically by
\[\mathbf{v} = \begin{pmatrix} 1 + s - t \\ 2 - s \\ 3 - 2s + 2t \end{pmatrix}.\]Find the equation of the plane. Enter your answer in the form
\[Ax + By + Cz + D = 0,\]where $A,$ $B,$ $C,$ $D$ are integers such that $A > 0$ and $\gcd(|A|,|B|,|C|,|D|) = 1.$
|
2x + z - 5 = 0
|
deepscale
| 40,305
| ||
Point $P$ is inside triangle $\triangle ABC$. Line $AC$ intersects line $BP$ at $Q$, and line $AB$ intersects line $CP$ at $R$. Given that $AR = RB = CP$, and $CQ = PQ$, find $\angle BRC$.
|
120
|
deepscale
| 28,486
| ||
Given a geometric progression \( b_1, b_2, \ldots, b_{3000} \) with all positive terms and a total sum \( S \). It is known that if every term with an index that is a multiple of 3 (i.e., \( b_3, b_6, \ldots, b_{3000} \)) is increased by 50 times, the sum \( S \) increases by 10 times. How will \( S \) change if every term in an even position (i.e., \( b_2, b_4, \ldots, b_{3000} \)) is increased by 2 times?
|
\frac{11}{8}
|
deepscale
| 24,979
| ||
Find the maximum possible value of $H \cdot M \cdot M \cdot T$ over all ordered triples $(H, M, T)$ of integers such that $H \cdot M \cdot M \cdot T=H+M+M+T$.
|
If any of $H, M, T$ are zero, the product is 0. We can do better (examples below), so we may now restrict attention to the case when $H, M, T \neq 0$. When $M \in\{-2,-1,1,2\}$, a little casework gives all the possible $(H, M, T)=(2,1,4),(4,1,2),(-1,-2,1),(1,-2,-1)$. If $M=-2$, i.e. $H-4+T=4 H T$, then $-15=(4 H-1)(4 T-1)$, so $4 H-1 \in\{ \pm 1, \pm 3, \pm 5, \pm 15\}$ (only $-1,+3,-5,+15$ are possible) corresponding to $4 T-1 \in\{\mp 15, \mp 5, \mp 3, \mp 1\}$ (only $+15,-5,+3,-1$ are possible). But $H, T$ are nonzero, we can only have $4 H-1 \in\{+3,-5\}$, yielding $(-1,-2,1)$ and $(1,-2,-1)$. If $M=+2$, i.e. $H+4+T=4 H T$, then $17=(4 H-1)(4 T-1)$, so $4 H-1 \in\{ \pm 1, \pm 17\}$ (only $-1,-17$ are possible) corresponding to $4 T-1 \in\{ \pm 17, \pm 1\}$ (only $-17,-1$ are possible). But $H, T$ are nonzero, so there are no possibilities here. If $M=-1$, i.e. $H-2+T=H T$, then $-1=(H-1)(T-1)$, so we have $H-1 \in\{ \pm 1\}$ and $T-1 \in\{\mp 1\}$, neither of which is possible (as $H, T \neq 0)$. If $M=+1$, i.e. $H+2+T=H T$, then $3=(H-1)(T-1)$, so we have $H-1 \in\{ \pm 1, \pm 3\}$. Since $H, T \neq 0, H-1 \in\{+1,+3\}$, yielding $(2,1,4)$ and $(4,1,2)$. Now suppose there is such a triple $(H, M, T)$ for $|M| \geq 3$. The equation in the problem gives $\left(M^{2} H-\right.$ 1) $\left(M^{2} T-1\right)=2 M^{3}+1$. Note that since $H, T \neq 0,\left|2 M^{3}+1\right|=\left|M^{2} H-1\right| \cdot\left|M^{2} T-1\right| \geq \min \left(M^{2}-\right.$ $\left.1, M^{2}+1\right)^{2}=M^{4}-2 M^{2}+1>2|M|^{3}+1$ gives a contradiction.
|
8
|
deepscale
| 3,315
| |
Given sets $A=\{2,3,4\}$ and $B=\{a+2,a\}$, if $A \cap B = B$, find $A^cB$ ___.
|
\{3\}
|
deepscale
| 17,485
| ||
A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of $n$ if the the area of the small square is exactly $\frac1{1985}$.
|
Line Segment $DE = \frac{1}{n}$, so $EC = 1 - \frac{1}{n} = \frac{n-1}{n}$. Draw line segment $HE$ parallel to the corresponding sides of the small square, $HE$ has length $\frac{1}{\sqrt{1985}}$, as it is the same length as the sides of the square. Notice that $\triangle CEL$ is similar to $\triangle HDE$ by $AA$ similarity. Thus, $\frac{LC}{HE} = \frac{EC}{DE} = n-1$, so $LC = \frac{n-1}{\sqrt{1985}}$. Notice that $\triangle CEL$ is also similar to $\triangle CDF$ by $AA$ similarity. Thus, $\frac{FC}{EC} = \frac{DC}{LC}$, and the expression simplifies into a quadratic equation $n^2 - n - 992 = 0$. Solving this quadratic equation yields $n =\boxed{32}$.
|
32
|
deepscale
| 6,449
| |
Given the parametric equations of curve $C$ are
$$
\begin{cases}
x=3+ \sqrt {5}\cos \alpha \\
y=1+ \sqrt {5}\sin \alpha
\end{cases}
(\alpha \text{ is the parameter}),
$$
with the origin of the Cartesian coordinate system as the pole and the positive half-axis of $x$ as the polar axis, establish a polar coordinate system.
$(1)$ Find the polar equation of curve $C$;
$(2)$ If the polar equation of a line is $\sin \theta-\cos \theta= \frac {1}{\rho }$, find the length of the chord cut from curve $C$ by the line.
|
\sqrt {2}
|
deepscale
| 23,632
| ||
Given the sequence \(\{a_n\}\) with the sum of its first \(n\) terms denoted by \(S_n\), let \(T_n = \frac{S_1 + S_2 + \cdots + S_n}{n}\). \(T_n\) is called the "mean" of the sequence \(a_1, a_2, \cdots, a_n\). It is known that the "mean" of the sequence \(a_1, a_2, \cdots, a_{1005}\) is 2012. Determine the "mean" of the sequence \(-1, a_1, a_2, \cdots, a_{1005}\).
|
2009
|
deepscale
| 19,533
| ||
The non-negative difference between two numbers \(a\) and \(b\) is \(a-b\) or \(b-a\), whichever is greater than or equal to 0. For example, the non-negative difference between 24 and 64 is 40. In the sequence \(88, 24, 64, 40, 24, \ldots\), each number after the second is obtained by finding the non-negative difference between the previous 2 numbers. The sum of the first 100 numbers in this sequence is:
|
760
|
deepscale
| 27,923
| ||
Find the sum of squares of all distinct complex numbers $x$ satisfying the equation $0=4 x^{10}-7 x^{9}+5 x^{8}-8 x^{7}+12 x^{6}-12 x^{5}+12 x^{4}-8 x^{3}+5 x^{2}-7 x+4$
|
For convenience denote the polynomial by $P(x)$. Notice $4+8=7+5=12$ and that the consecutive terms $12 x^{6}-12 x^{5}+12 x^{4}$ are the leading terms of $12 \Phi_{14}(x)$, which is suggestive. Indeed, consider $\omega$ a primitive 14 -th root of unity; since $\omega^{7}=-1$, we have $4 \omega^{10}=-4 \omega^{3},-7 \omega^{9}=7 \omega^{2}$, and so on, so that $P(\omega)=12\left(\omega^{6}-\omega^{5}+\cdots+1\right)=12 \Phi_{14}(\omega)=0$. Dividing, we find $P(x)=\Phi_{14}(x)\left(4 x^{4}-3 x^{3}-2 x^{2}-3 x+4\right)$. This second polynomial is symmetric; since 0 is clearly not a root, we have $4 x^{4}-3 x^{3}-2 x^{2}-3 x+4=0 \Longleftrightarrow 4\left(x+\frac{1}{x}\right)^{2}-3\left(x+\frac{1}{x}\right)-10=0$. Setting $y=x+1 / x$ and solving the quadratic gives $y=2$ and $y=-5 / 4$ as solutions; replacing $y$ with $x+1 / x$ and solving the two resulting quadratics give the double root $x=1$ and the roots $(-5 \pm i \sqrt{39}) / 8$ respectively. Together with the primitive fourteenth roots of unity, these are all the roots of our polynomial. Explicitly, the roots are $e^{\pi i / 7}, e^{3 \pi i / 7}, e^{5 \pi i / 7}, e^{9 \pi i / 7}, e^{11 \pi i / 7}, e^{13 \pi i / 7}, 1,(-5 \pm i \sqrt{39}) / 8$. The sum of squares of the roots of unity (including 1) is just 0 by symmetry (or a number of other methods). The sum of the squares of the final conjugate pair is $\frac{2\left(5^{2}-39\right)}{8^{2}}=-\frac{14}{32}=-\frac{7}{16}$.
|
-\frac{7}{16}
|
deepscale
| 3,352
| |
If $f(x)$ is a monic quartic polynomial such that $f(-1)=-1$, $f(2)=-4$, $f(-3)=-9$, and $f(4)=-16$, find $f(1)$.
|
23
|
deepscale
| 36,949
| ||
Find the minimum value of
\[3x^2 + 3xy + y^2 - 3x + 3y + 9\]
over all real numbers $x$ and $y.$
|
\frac{45}{8}
|
deepscale
| 24,836
| ||
What is the smallest positive value of $m$ so that the equation $10x^2 - mx + 360 = 0$ has integral solutions with one root being a multiple of the other?
|
120
|
deepscale
| 9,524
| ||
When Alia was young, she could cycle 18 miles in 2 hours. Now, as an older adult, she walks 8 kilometers in 3 hours. Given that 1 mile is approximately 1.609 kilometers, determine how many minutes longer it takes for her to walk a kilometer now compared to when she was young.
|
18
|
deepscale
| 20,897
| ||
During the National Day holiday, a fruit company organized 20 cars to transport three types of fruits, $A$, $B$, and $C$, totaling 120 tons for sale in other places. It is required that all 20 cars be fully loaded, each car can only transport the same type of fruit, and each type of fruit must be transported by at least 3 cars. According to the information provided in the table below, answer the following questions:
| | $A$ | $B$ | $C$ |
|----------|------|------|------|
| Cargo Capacity per Car (tons) | 7 | 6 | 5 |
| Profit per ton of Fruit (yuan) | 1200 | 1800 | 1500 |
$(1)$ Let the number of cars transporting fruit $A$ be $x$, and the number of cars transporting fruit $B$ be $y$. Find the functional relationship between $y$ and $x$, and determine how many arrangements of cars are possible.
$(2)$ Let $w$ represent the profit obtained from sales. How should the cars be arranged to maximize the profit from this sale? Determine the maximum value of $w$.
|
198900
|
deepscale
| 20,293
| ||
A survey of $150$ teachers determined the following:
- $90$ had high blood pressure
- $60$ had heart trouble
- $50$ had diabetes
- $30$ had both high blood pressure and heart trouble
- $20$ had both high blood pressure and diabetes
- $10$ had both heart trouble and diabetes
- $5$ had all three conditions
What percent of the teachers surveyed had none of the conditions?
|
3.33\%
|
deepscale
| 32,125
| ||
In triangle $XYZ,$ $\angle X = 60^\circ,$ $\angle Y = 75^\circ,$ and $YZ = 6.$ Find $XZ.$
|
3\sqrt{2} + \sqrt{6}
|
deepscale
| 30,692
| ||
Let \( r(x) \) have a domain of \(\{ -2, -1, 0, 1 \}\) and a range of \(\{ 1, 3, 5, 7 \}\). Let \( s(x) \) have a domain of \(\{ 0, 1, 2, 3, 4, 5 \}\) and be defined by \( s(x) = 2x + 1 \). What is the sum of all possible values of \( s(r(x)) \)?
|
21
|
deepscale
| 17,076
| ||
Two mutually perpendicular chords \( AB \) and \( CD \) are drawn in a circle. Determine the distance between the midpoint of segment \( AD \) and the line \( BC \), given that \( BD = 6 \), \( AC = 12 \), and \( BC = 10 \). If necessary, round your answer to two decimal places.
|
2.5
|
deepscale
| 25,009
| ||
A triangle has vertices $A=(4,3)$, $B=(-4,-1)$, and $C=(9,-7)$. Calculate the equation of the bisector of $\angle A$ in the form $3x - by + c = 0$. Determine the value of $b+c$.
|
-6
|
deepscale
| 24,896
| ||
You have a length of string and 7 beads in the 7 colors of the rainbow. You place the beads on the string as follows - you randomly pick a bead that you haven't used yet, then randomly add it to either the left end or the right end of the string. What is the probability that, at the end, the colors of the beads are the colors of the rainbow in order? (The string cannot be flipped, so the red bead must appear on the left side and the violet bead on the right side.)
|
The threading method does not depend on the colors of the beads, so at the end all configurations are equally likely. Since there are $7!=5040$ configurations in total, the probability of any particular configuration is $\frac{1}{5040}$.
|
\frac{1}{5040}
|
deepscale
| 4,871
| |
Find the greatest common factor of 8! and 9!.
|
40320
|
deepscale
| 18,824
| ||
Distinct points $A$, $B$, $C$, and $D$ lie on a line, with $AB=BC=CD=1$. Points $E$ and $F$ lie on a second line, parallel to the first, with $EF=1$. A triangle with positive area has three of the six points as its vertices. How many possible values are there for the area of the triangle?
|
1. **Identify Possible Bases and Heights**:
- The points $A$, $B$, $C$, and $D$ are collinear with equal distances between consecutive points, i.e., $AB = BC = CD = 1$.
- Points $E$ and $F$ are also collinear on a different line parallel to the line containing $A$, $B$, $C$, and $D$, with $EF = 1$.
- The height of any triangle formed by choosing one side as the base on one line and the vertex on the other line is the perpendicular distance between these two parallel lines.
2. **Calculate Possible Base Lengths**:
- Since the lines are parallel and the distances between consecutive points on the same line are equal, the possible lengths of the bases (sides of the triangle on the same line) are:
- $AB = BC = CD = EF = 1$
- $AC = BD = 2$ (since $AC = AB + BC$ and $BD = BC + CD$)
- $AD = 3$ (since $AD = AB + BC + CD$)
3. **Determine Unique Triangle Areas**:
- The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
- Since the height is constant (distance between the two parallel lines), the area of the triangle is directly proportional to the length of the base.
- The unique base lengths available are $1$, $2$, and $3$.
4. **Conclusion**:
- Each unique base length corresponds to a unique area for the triangles formed.
- Therefore, there are exactly three possible different areas for the triangles that can be formed using these points.
Thus, the number of possible values for the area of the triangle is $\boxed{3}$.
|
3
|
deepscale
| 2,216
| |
Let $F_1 = (0,1)$ and $F_ 2= (4,1).$ Then the set of points $P$ such that
\[PF_1 + PF_2 = 6\]form an ellipse. The equation of this ellipse can be written as
\[\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1.\]Find $h + k + a + b.$
|
6 + \sqrt{5}
|
deepscale
| 37,447
| ||
How many paths are there from point $C$ to point $D$ on a grid, if every step must be either to the right or upwards, and the grid dimensions are now 7 steps to the right and 9 steps upward?
|
11440
|
deepscale
| 23,004
| ||
In the Cartesian coordinate system xOy, curve $C_1: x^2+y^2=1$ is given. Taking the origin O of the Cartesian coordinate system xOy as the pole and the positive half-axis of x as the polar axis, a polar coordinate system is established with the same unit length. It is known that the line $l: \rho(2\cos\theta-\sin\theta)=6$.
(1) After stretching all the x-coordinates and y-coordinates of points on curve $C_1$ by $\sqrt{3}$ and 2 times respectively, curve $C_2$ is obtained. Please write down the Cartesian equation of line $l$ and the parametric equation of curve $C_2$;
(2) Find a point P on curve $C_2$ such that the distance from point P to line $l$ is maximized, and calculate this maximum value.
|
2\sqrt{5}
|
deepscale
| 18,750
| ||
Let $S_n$ be the sum of the first $n$ terms of the difference sequence $\{a_n\}$, given that $a_2 + a_{12} = 24$ and $S_{11} = 121$.
(1) Find the general term formula for $\{a_n\}$.
(2) Let $b_n = \frac {1}{a_{n+1}a_{n+2}}$, and $T_n = b_1 + b_2 + \ldots + b_n$. If $24T_n - m \geq 0$ holds for all $n \in \mathbb{N}^*$, find the maximum value of the real number $m$.
|
m = \frac {3}{7}
|
deepscale
| 22,859
| ||
A sphere is inscribed in the tetrahedron whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The radius of the sphere is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$
|
Note that the center is obviously at some point $(r,r,r)$ since it must be equidistant from the x-y, x-z, and y-z planes. Now note that the point-plane distance formula is $\widehat{n}\cdot \vec{d}$ (the direction vector of the normal to the plane dotted with the distance between a point on the plane and the point we're interested in). By inspection we know plane $ABC$'s cartesian form is $2x+3y+6z=12$ thus its normal vector is $<\frac{2}{7},\frac{3}{7},\frac{6}{7}>.$ Let the "arbitrary point on the plane be $A.$ Then \[<\frac{2}{7},\frac{3}{7},\frac{6}{7}>\cdot<6-r,-r,-r>=r\] which implies \[2(6-r)-3r-6r=7r\] so $12=18r$ and $r=\frac{2}{3}$ giving us $\boxed{005}$ as the answer.
~Dhillonr25
|
5
|
deepscale
| 6,712
| |
Let the function $f(x)=\ln x- \frac{1}{2}ax^{2}-bx$.
$(1)$ When $a=b= \frac{1}{2}$, find the maximum value of the function $f(x)$;
$(2)$ Let $F(x)=f(x)+ \frac{1}{2}ax^{2}+bx+ \frac{a}{x}$, $(0 < x\leqslant 3)$, the slope of the tangent line at any point $P(x_{0},y_{0})$ on its graph is $k\leqslant \frac{1}{2}$ always holds, find the range of the real number $a$;
$(3)$ When $a=0$, $b=-1$, the equation $2mf(x)=x^{2}$ has a unique real solution, find the value of the positive number $m$.
|
\frac{1}{2}
|
deepscale
| 23,848
| ||
The areas of two squares are in the ratio $25:36$. What is the ratio of their perimeters? Express your answer in the form $a:b$.
|
5:6
|
deepscale
| 39,119
| ||
Given two points A and B on a number line, their distance is 2, and the distance between point A and the origin O is 3. Then, the sum of all possible distances between point B and the origin O equals to .
|
12
|
deepscale
| 21,491
| ||
Given that the broad money supply $\left(M2\right)$ balance was 2912000 billion yuan, express this number in scientific notation.
|
2.912 \times 10^{6}
|
deepscale
| 25,307
| ||
Solve for the largest value of $x$ such that $5(9x^2+9x+10) = x(9x-40).$ Express your answer as a simplified common fraction.
|
-\dfrac{10}{9}
|
deepscale
| 33,976
| ||
Let the function \( f(x) \) defined on \( (0, +\infty) \) satisfy \( f(x) > -\frac{3}{x} \) for any \( x \in (0, +\infty) \) and \( f\left(f(x) + \frac{3}{x}\right) = 2 \). Find \( f(5) \).
|
\frac{7}{5}
|
deepscale
| 29,538
| ||
Seventy percent of a train's passengers are men, and fifteen percent of those men are in the business class. What is the number of men in the business class if the train is carrying 300 passengers?
|
32
|
deepscale
| 32,553
| ||
If the inequality system $\left\{\begin{array}{l}{x-m＞0}\\{x-2＜0}\end{array}\right.$ has only one positive integer solution, then write down a value of $m$ that satisfies the condition: ______.
|
0.5
|
deepscale
| 22,236
| ||
In $\triangle ABC$, the sides opposite to angles A, B, and C are denoted as $a$, $b$, and $c$ respectively. Given that $a= \sqrt{2}$, $b=2$, and $\sin B - \cos B = \sqrt{2}$, find the measure of angle $A$.
|
\frac{\pi}{6}
|
deepscale
| 27,820
| ||
Given points $A(\cos\alpha, \sin\alpha)$ and $B(\cos\beta, \sin\beta)$, where $\alpha, \beta$ are acute angles, and that $|AB| = \frac{\sqrt{10}}{5}$:
(1) Find the value of $\cos(\alpha - \beta)$;
(2) If $\tan \frac{\alpha}{2} = \frac{1}{2}$, find the values of $\cos\alpha$ and $\cos\beta$.
|
\frac{24}{25}
|
deepscale
| 31,481
| ||
Let acute triangle $ABC$ have circumcenter $O$, and let $M$ be the midpoint of $BC$. Let $P$ be the unique point such that $\angle BAP=\angle CAM, \angle CAP=\angle BAM$, and $\angle APO=90^{\circ}$. If $AO=53, OM=28$, and $AM=75$, compute the perimeter of $\triangle BPC$.
|
The point $P$ has many well-known properties, including the property that $\angle BAP=\angle ACP$ and $\angle CAP=\angle BAP$. We prove this for completeness. Invert at $A$ with radius $\sqrt{AB \cdot AC}$ and reflect about the $A$-angle bisector. Let $P^{\prime}$ be the image of $P$. The angle conditions translate to - $P^{\prime}$ lies on line $AM$ - $P^{\prime}$ lies on the line parallel to $BC$ that passes through the reflection of $A$ about $BC$ (since $P$ lies on the circle with diameter $\overline{AO})$ In other words, $P^{\prime}$ is the reflection of $A$ about $M$. Then $BP^{\prime} \| AC$ and $CP^{\prime} \| AB$, so the circumcircles of $\triangle ABP$ and $\triangle ACP$ are tangent to $AC$ and $AB$, respectively. This gives the desired result. Extend $BP$ and $CP$ to meet the circumcircle of $\triangle ABC$ again at $B^{\prime}$ and $C^{\prime}$, respectively. Then $\angle C^{\prime}BA=\angle ACP=\angle BAP$, so $BC^{\prime} \| AP$. Similarly, $CB^{\prime} \| AP$, so $BCB^{\prime}C^{\prime}$ is an isosceles trapezoid. In particular, this means $B^{\prime}P=CP$, so $BP+PC=BB^{\prime}$. Now observe that $\angle ABP=\angle CAP=\angle BAM$, so if $AM$ meets the circumcircle of $\triangle ABC$ again at $A^{\prime}$, then $AA^{\prime}=BB^{\prime}$. Thus the perimeter of $\triangle BPC$ is $BP+PC+BC=BB^{\prime}+BC=AA^{\prime}+BC$. Now we compute. We have $$BC=2 \sqrt{AO^{2}-OM^{2}}=2 \sqrt{81 \cdot 25}=90$$ and Power of a Point gives $$MA^{\prime}=\frac{BM^{2}}{AM}=\frac{45^{2}}{75}=27$$ Thus $AA^{\prime}+BC=75+27+90=192$.
|
192
|
deepscale
| 3,548
| |
A tour group has three age categories of people, represented in a pie chart. The central angle of the sector corresponding to older people is $9^{\circ}$ larger than the central angle for children. The percentage of total people who are young adults is $5\%$ higher than the percentage of older people. Additionally, there are 9 more young adults than children. What is the total number of people in the tour group?
|
120
|
deepscale
| 9,670
| ||
\( P \) is the midpoint of the height \( V O \) of the regular quadrilateral pyramid \( V-ABCD \). If the distance from point \( P \) to the side faces is 3, and the distance to the base is 5, what is the volume of the regular quadrilateral pyramid?
|
750
|
deepscale
| 7,558
| ||
Find the integer $n$, $0 \le n \le 5$, such that \[n \equiv -3736 \pmod{6}.\]
|
2
|
deepscale
| 37,732
| ||
Given a sequence of positive terms $\{a_n\}$ with the sum of the first $n$ terms denoted as $S_n$, it satisfies the equation $2S_n = a_n^2 + a_n$ for all natural numbers $n$. Define a new sequence $\{c_n\}$ where $c_n = (-1)^n \frac{2a_n + 1}{2S_n}$. Find the sum of the first 2016 terms of the sequence $\{c_n\}$.
|
- \frac{2016}{2017}
|
deepscale
| 32,035
| ||
Let $ S $ be the set of all sides and diagonals of a regular hexagon. A pair of elements of $ S $ are selected at random without replacement. What is the probability that the two chosen segments have the same length?
|
\frac{17}{35}
|
deepscale
| 23,009
| ||
Using the digits 0, 1, 2, 3, 4, 5 to form numbers without repeating any digit. Calculate:
(1) How many six-digit numbers can be formed?
(2) How many three-digit numbers can be formed that contain at least one even number?
(3) How many three-digit numbers can be formed that are divisible by 3?
|
40
|
deepscale
| 18,564
| ||
One material particle entered the opening of a pipe, and after 6.8 minutes, a second particle entered the same opening. Upon entering the pipe, each particle immediately began linear motion along the pipe: the first particle moved uniformly at a speed of 5 meters per minute, while the second particle covered 3 meters in the first minute and in each subsequent minute covered 0.5 meters more than in the previous minute. How many minutes will it take for the second particle to catch up with the first?
|
17
|
deepscale
| 14,040
| ||
On the Cartesian plane, find the number of integer coordinate points (points where both x and y are integers) that satisfy the following system of inequalities:
\[
\begin{cases}
y \leq 3x, \\
y \geq \frac{1}{3}x, \\
x + y \leq 100.
\end{cases}
\]
|
2551
|
deepscale
| 16,135
| ||
How many natural numbers greater than 9 but less than 100 are relatively prime to 30?
|
24
|
deepscale
| 16,900
| ||
Given the set $M$ consisting of all functions $f(x)$ that satisfy the property: there exist real numbers $a$ and $k$ ($k \neq 0$) such that for all $x$ in the domain of $f$, $f(a+x) = kf(a-x)$. The pair $(a,k)$ is referred to as the "companion pair" of the function $f(x)$.
1. Determine whether the function $f(x) = x^2$ belongs to set $M$ and explain your reasoning.
2. If $f(x) = \sin x \in M$, find all companion pairs $(a,k)$ for the function $f(x)$.
3. If $(1,1)$ and $(2,-1)$ are both companion pairs of the function $f(x)$, where $f(x) = \cos(\frac{\pi}{2}x)$ for $1 \leq x < 2$ and $f(x) = 0$ for $x=2$. Find all zeros of the function $y=f(x)$ when $2014 \leq x \leq 2016$.
|
2016
|
deepscale
| 27,528
| ||
How many six-digit multiples of 27 have only 3, 6, or 9 as their digits?
|
Divide by 3. We now want to count the number of six-digit multiples of 9 that only have 1, 2, or 3 as their digits. Due to the divisibility rule for 9, we only need to consider when the digit sum is a multiple of 9. Note that $3 \cdot 6=18$ is the maximum digit sum. If the sum is 18, the only case is 333333. Otherwise, the digit sum is 9. The possibilities here, up to ordering of the digits, are 111222 and 111123. The first has $\binom{6}{3}=20$ cases, while the second has $6 \cdot 5=30$. Thus the final answer is $1+20+30=51$.
|
51
|
deepscale
| 5,011
| |
Simplify first, then evaluate: $(1-\frac{1}{a+1})÷\frac{a^2-2a+1}{a^2-1}$, where $a=-2$.
|
\frac{2}{3}
|
deepscale
| 18,800
| ||
Find the distance between the foci of the ellipse \[x^2 + 4y^2 = 400.\]
|
20\sqrt3
|
deepscale
| 37,532
| ||
Points $A$, $B$, and $C$ form an isosceles triangle with $AB = AC$. Points $M$, $N$, and $O$ are the midpoints of sides $AB$, $BC$, and $CA$ respectively. Find the number of noncongruent triangles that can be drawn using any three of these six points as vertices.
|
10
|
deepscale
| 32,819
| ||
The numbers $a, b, c, d$ belong to the interval $[-8.5,8.5]$. Find the maximum value of the expression $a + 2b + c + 2d - ab - bc - cd - da$.
|
306
|
deepscale
| 13,743
| ||
Leticia has a $9\times 9$ board. She says that two squares are *friends* is they share a side, if they are at opposite ends of the same row or if they are at opposite ends of the same column. Every square has $4$ friends on the board. Leticia will paint every square one of three colors: green, blue or red. In each square a number will be written based on the following rules:
- If the square is green, write the number of red friends plus twice the number of blue friends.
- If the square is red, write the number of blue friends plus twice the number of green friends.
- If the square is blue, write the number of green friends plus twice the number of red friends.
Considering that Leticia can choose the coloring of the squares on the board, find the maximum possible value she can obtain when she sums the numbers in all the squares.
|
486
|
deepscale
| 29,945
| ||
The diagram below shows a $4\times4$ rectangular array of points, each of which is $1$ unit away from its nearest neighbors.
[asy] unitsize(0.25inch); defaultpen(linewidth(0.7)); int i, j; for(i = 0; i < 4; ++i) for(j = 0; j < 4; ++j) dot(((real)i, (real)j)); [/asy]
Define a growing path to be a sequence of distinct points of the array with the property that the distance between consecutive points of the sequence is strictly increasing. Let $m$ be the maximum possible number of points in a growing path, and let $r$ be the number of growing paths consisting of exactly $m$ points. Find $mr$.
|
240
|
deepscale
| 35,095
| ||
Two cards are chosen at random from a standard 52-card deck. What is the probability that the first card is a spade and the second card is a king?
|
\frac{17}{884}
|
deepscale
| 32,361
| ||
Find all functions $f : \mathbb{R} \to \mathbb{R}$ such that
\[f(x(x + f(y))) = (x + y)f(x),\]
for all $x, y \in\mathbb{R}$.
|
Let's consider the functional equation \( f(x(x + f(y))) = (x + y)f(x) \) for all \( x, y \in \mathbb{R} \).
### Step 1: Test simple functions
First, let's test the simplest potential solutions.
1. **\( f(x) = 0 \):**
- Substituting \( f(x) = 0 \) into the equation gives:
\[
f(x(x + f(y))) = f(0) = 0, \quad (x + y)f(x) = 0
\]
- Both sides are equal for any \( x, y \in \mathbb{R} \).
- Thus, \( f(x) = 0 \) is a solution.
2. **\( f(x) = x \):**
- Substituting \( f(x) = x \) into the equation gives:
\[
f(x(x + f(y))) = f(x(x + y)) = x(x + y)
\]
- The right-hand side becomes:
\[
(x + y)f(x) = (x + y)x = x(x + y)
\]
- Both sides are equal for any \( x, y \in \mathbb{R} \).
- Thus, \( f(x) = x \) is another solution.
### Step 2: Prove that these are the only solutions
Let's analyze whether any other function could satisfy the condition.
#### Case Analysis
1. **\( x = 0 \):**
- Set \( x = 0 \) in the original equation:
\[
f(0(0 + f(y))) = 0 \cdot f(0)
\]
- Simplifies to \( f(0) = 0 \).
2. **Assume a non-trivial \( f \):**
- Assume there exists a function \( f \) other than the two tested functions, which means \( f(x) \neq 0 \) and \( f(x) \neq x \) for some \( x \).
- For \( f(x) = x \) to hold, any assumption leads back to either \( f(x) = 0 \) through function continuity implied by the symmetry of \( f \) or inherently linear functions as assumed initially.
### Conclusion
Having tested function forms and considered continuity and linearity constraints arising from the equation structure, we establish no other solutions exist. Therefore, the solutions are refined to:
\[
\boxed{f(x) = 0 \text{ and } f(x) = x}
\]
These are the only functions that satisfy the given functional equation for all real numbers \( x \) and \( y \).
|
f(x) = 0 \text{ and } f(x) = x.
|
deepscale
| 5,977
| |
A person has a probability of $\frac{1}{2}$ to hit the target in each shot. What is the probability of hitting the target 3 times out of 6 shots, with exactly 2 consecutive hits? (Answer with a numerical value)
|
\frac{3}{16}
|
deepscale
| 27,767
| ||
Given a pair of standard $8$-sided dice is rolled once. The sum of the numbers rolled, if it is a prime number, determines the diameter of a circle. Find the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference.
|
\frac{3}{64}
|
deepscale
| 26,571
| ||
Given the vectors $\overrightarrow{m}=(x,y)$ and $\overrightarrow{n}=(x-y)$, let $P$ be a moving point on the curve $\overrightarrow{m}\cdot \overrightarrow{n}=1 (x > 0)$. If the distance from point $P$ to the line $x-y+1=0$ is always greater than $\lambda$, find the maximum value of the real number $\lambda$.
|
\frac{\sqrt{2}}{2}
|
deepscale
| 23,047
| ||
Find all values of \( a \) for which the equation \( x^{2} + 2ax = 8a \) has two distinct integer roots. Record the product of all such \( a \), rounding to the nearest hundredth if necessary.
|
506.25
|
deepscale
| 10,338
| ||
A general gathers his troops. When he arranges them in groups of 2, one soldier is left over. When he arranges them in groups of 3, two soldiers are left over. When he arranges them in groups of 5, three soldiers are left over. If the general arranges his soldiers in groups of 30, how many soldiers will be left over?
|
23
|
deepscale
| 16,205
| ||
On a shelf, there are 4 different comic books, 5 different fairy tale books, and 3 different story books, all lined up in a row. If the fairy tale books cannot be separated from each other, and the comic books also cannot be separated from each other, how many different arrangements are there?
|
345600
|
deepscale
| 11,219
| ||
In a game of rock-paper-scissors with $n$ people, the following rules are used to determine a champion: (a) In a round, each person who has not been eliminated randomly chooses one of rock, paper, or scissors to play. (b) If at least one person plays rock, at least one person plays paper, and at least one person plays scissors, then the round is declared a tie and no one is eliminated. If everyone makes the same move, then the round is also declared a tie. (c) If exactly two moves are represented, then everyone who made the losing move is eliminated from playing in all further rounds (for example, in a game with 8 people, if 5 people play rock and 3 people play scissors, then the 3 who played scissors are eliminated). (d) The rounds continue until only one person has not been eliminated. That person is declared the champion and the game ends. If a game begins with 4 people, what is the expected value of the number of rounds required for a champion to be determined?
|
For each positive integer $n$, let $E_{n}$ denote the expected number of rounds required to determine a winner among $n$ people. Clearly, $E_{1}=0$. When $n=2$, on the first move, there is a $\frac{1}{3}$ probability that there is a tie, and a $\frac{2}{3}$ probability that a winner is determined. In the first case, the expected number of additional rounds needed is exactly $E_{2}$; in the second, it is $E_{1}$. Therefore, we get the relation $$E_{2}=\frac{1}{3}\left(E_{2}+1\right)+\frac{2}{3}\left(E_{1}+1\right)$$ from which it follows that $E_{2}=\frac{3}{2}$. Next, if $n=3$, with probability $\frac{1}{9}$ there is only one distinct play among the three players, and with probability $\frac{6}{27}=\frac{2}{9}$ all three players make different plays. In both of these cases, no players are eliminated. In all remaining situations, which occur with total probability $\frac{2}{3}$, two players make one play and the third makes a distinct play; with probability $\frac{1}{3}$ two players are eliminated and with probability $\frac{1}{3}$ one player is eliminated. This gives the relation $$E_{3}=\frac{1}{3}\left(E_{3}+1\right)+\frac{1}{3}\left(E_{2}+1\right)+\frac{1}{3}\left(E_{1}+1\right)$$ from which we find that $E_{3}=\frac{9}{4}$. Finally, suppose $n=4$. With probability $\frac{1}{27}$, all four players make the same play, and with probability $\frac{3 \cdot 6 \cdot 2}{81}=\frac{4}{9}$, two players make one play, and the other two players make the other two plays; in both cases no players are eliminated, with total probability $\frac{1}{27}+\frac{4}{9}=\frac{13}{27}$ over the two cases. With probability $\frac{6 \cdot 4}{81}=\frac{8}{27}$, three players make one play and the fourth makes another; thus, there is a probability of $\frac{4}{27}$ for exactly one player being eliminated and a probability of $\frac{4}{27}$ of three players being eliminated. Then, there is a remaining probability of $\frac{6 \cdot 3}{81}=\frac{2}{9}$, two players make one play and the other two players make another. Similar analysis from before yields $$E_{4}=\frac{13}{27}\left(E_{4}+1\right)+\frac{4}{27}\left(E_{3}+1\right)+\frac{2}{9}\left(E_{2}+1\right)+\frac{4}{27}\left(E_{1}+1\right)$$ so it follows that $E_{4}=\frac{45}{14}$.
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\frac{45}{14}
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deepscale
| 4,917
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Solve for $y$: $3y+7y = 282-8(y-3)$.
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17
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deepscale
| 38,913
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Consider the largest solution to the equation \[\log_{5x^3} 5 + \log_{25x^4} 5 = -1.\] Find the value of \( x^{10} \).
|
0.0000001024
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deepscale
| 28,562
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In a school there are \( n \) students, each with a different student number. Each student number is a positive factor of \( 60^{60} \), and the H.C.F. of any two student numbers is not a student number in the school. Find the greatest possible value of \( n \).
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3721
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deepscale
| 32,933
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Let $x,$ $y,$ and $z$ be positive real numbers such that $x + y + z = 1.$ Find the minimum value of
\[\frac{x + y}{xyz}.\]
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16
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deepscale
| 36,710
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Using systematic sampling to select a sample of size 20 from 180 students, the students are randomly numbered from 1 to 180. They are then divided into 20 groups in order of their number (group 1: numbers 1-9, group 2: numbers 10-18, ..., group 20: numbers 172-180). If the number drawn from group 20 is 176, what is the number drawn from group 3?
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23
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deepscale
| 20,493
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Given the function $f(x)=4\cos (ωx- \frac {π}{6})\sin (π-ωx)-\sin (2ωx- \frac {π}{2})$, where $ω > 0$.
(1) Find the range of the function $f(x)$.
(2) If $y=f(x)$ is an increasing function in the interval $[- \frac {3π}{2}, \frac {π}{2}]$, find the maximum value of $ω$.
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\frac{1}{6}
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deepscale
| 21,966
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Find the number of ways to distribute 4 pieces of candy to 12 children such that no two consecutive children receive candy.
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Since 4 pieces of candy are distributed, there must be exactly 8 children who do not receive any candy; since no two consecutive children do receive candy, the 8 who do not must consist of 4 groups of consecutive children. We divide into cases based on the sizes of these groups: - \{5,1,1,1\} : there are 12 places to begin the group of 5 children who do not receive any candy - \{4,2,1,1\} : there are 12 places to begin the group of 4 children who do not receive candy and then 3 choices for the group of 2 children which does not receive candy, for a total of 36 choices - \{3,3,1,1\} : these 8 children can either be bunched in the order 3,3,1,1, or in the order 3,1,3,1; the first has 12 positions in which to begin the first group of 3 non-candy receiving children and the second has 6 possibilities (due to symmetry), for a total of 18 - \{3,2,2,1\} : there are 12 places to begin the group of 3 children who do not receive candy and then 3 choices for the group of 1 child which does not receive candy, for a total of 36 choices - \{2,2,2,2\} : there are $12 / 4=3$ ways in which this can occur This gives a total of $12+36+18+36+3=105$
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105
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deepscale
| 4,941
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What is the least positive integer that satisfies the following conditions?
a) When divided by 2, the remainder is 1.
b) When divided by 3, the remainder is 2.
c) When divided by 4, the remainder is 3.
d) When divided by 5, the remainder is 4.
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59
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deepscale
| 37,804
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