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|---|---|---|---|---|---|
How many square units are in the area of the triangle whose vertices are the $x$ and $y$ intercepts of the curve $y = (x-3)^2 (x+2)$?
|
45
|
deepscale
| 34,305
| ||
In the rectangular coordinate system xOy, an ellipse C is given by the equation $$\frac {x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}=1$$ ($$a>b>0$$), with left and right foci $$F_1$$ and $$F_2$$, respectively. The left vertex's coordinates are ($$-\sqrt {2}$$, 0), and point M lies on the ellipse C such that the perimeter of $$\triangle MF_1F_2$$ is $$2\sqrt {2}+2$$.
(1) Find the equation of the ellipse C;
(2) A line l passes through $$F_1$$ and intersects ellipse C at A and B, satisfying |$$\overrightarrow {OA}+2 \overrightarrow {OB}$$|=|$$\overrightarrow {BA}- \overrightarrow {OB}$$|, find the area of $$\triangle ABO$$.
|
\frac {2\sqrt {3}}{5}
|
deepscale
| 29,270
| ||
If $f(a)=a-2$ and $F(a,b)=b^2+a$, then $F(3,f(4))$ is:
|
1. **Calculate $f(4)$**: Given the function $f(a) = a - 2$, substitute $a = 4$:
\[
f(4) = 4 - 2 = 2
\]
2. **Evaluate $F(3, f(4))$**: With $f(4) = 2$, we need to find $F(3, 2)$. The function $F(a, b) = b^2 + a$ is given, so substitute $a = 3$ and $b = 2$:
\[
F(3, 2) = 2^2 + 3 = 4 + 3 = 7
\]
3. **Conclusion**: The value of $F(3, f(4))$ is $7$. Therefore, the correct answer is:
\[
\boxed{\textbf{(C)}\ 7}
\]
|
7
|
deepscale
| 143
| |
Simplify first, then evaluate: $\left(\frac{x}{x-1}-1\right) \div \frac{{x}^{2}-1}{{x}^{2}-2x+1}$, where $x=\sqrt{5}-1$.
|
\frac{\sqrt{5}}{5}
|
deepscale
| 19,035
| ||
Three workshops A, B, and C of a factory produced the same kind of product, with quantities of 120, 60, and 30, respectively. To determine if there are significant differences in product quality, a stratified sampling method was used to take a sample of size n for inspection, with 2 samples taken from workshop B.
(I) How many samples should be taken from workshops A and C, and what is the sample size n?
(II) Let the n samples be denoted by $A_1$, $A_2$, ..., $A_n$. Now, randomly select 2 samples from these.
(i) List all possible sampling outcomes.
(ii) Let M be the event "the 2 selected samples come from different workshops." Calculate the probability of event M occurring.
|
\frac{2}{3}
|
deepscale
| 20,572
| ||
In the polar coordinate system, the curve $C\_1$: $ρ=2\cos θ$, and the curve $C\_2$: $ρ\sin ^{2}θ=4\cos θ$. Establish a rectangular coordinate system $(xOy)$ with the pole as the coordinate origin and the polar axis as the positive semi-axis $x$. The parametric equation of the curve $C$ is $\begin{cases} x=2+ \frac {1}{2}t \ y= \frac {\sqrt {3}}{2}t\end{cases}$ ($t$ is the parameter).
(I) Find the rectangular coordinate equations of $C\_1$ and $C\_2$;
(II) The curve $C$ intersects $C\_1$ and $C\_2$ at four distinct points, arranged in order along $C$ as $P$, $Q$, $R$, and $S$. Find the value of $||PQ|-|RS||$.
|
\frac {11}{3}
|
deepscale
| 29,819
| ||
There are 6 locked suitcases and 6 keys for them. However, it is unknown which key opens which suitcase. What is the minimum number of attempts needed to ensure that all suitcases are opened? How many attempts are needed if there are 10 suitcases and 10 keys?
|
45
|
deepscale
| 21,641
| ||
A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola $x^2 - y^2 = 2000^2$?
|
98
|
deepscale
| 38,134
| ||
Given that $\tan{\alpha} = 3$, calculate:
1. $$\frac{4\sin{\alpha} - \cos{\alpha}}{3\sin{\alpha} + 5\cos{\alpha}}$$
2. $$\sin{\alpha} \cdot \cos{\alpha}$$
|
\frac{3}{10}
|
deepscale
| 8,315
| ||
Given that the positive real numbers $x$ and $y$ satisfy the equation $x + 2y = 1$, find the minimum value of $\frac{y}{2x} + \frac{1}{y}$.
|
2 + \sqrt{2}
|
deepscale
| 10,109
| ||
The diagram shows the miles traveled by cyclists Clara and David. After five hours, how many more miles has Clara cycled than David?
[asy]
/* Modified AMC8 1999 #4 Problem */
draw((0,0)--(6,0)--(6,4.5)--(0,4.5)--cycle);
for(int x=0; x <= 6; ++x) {
for(real y=0; y <=4.5; y+=0.9) {
dot((x, y));
}
}
draw((0,0)--(5,3.6)); // Clara's line
draw((0,0)--(5,2.7)); // David's line
label(rotate(36)*"David", (3,1.35));
label(rotate(36)*"Clara", (3,2.4));
label(scale(0.75)*rotate(90)*"MILES", (-1, 2.25));
label(scale(0.75)*"HOURS", (3, -1));
label(scale(0.85)*"90", (0, 4.5), W);
label(scale(0.85)*"72", (0, 3.6), W);
label(scale(0.85)*"54", (0, 2.7), W);
label(scale(0.85)*"36", (0, 1.8), W);
label(scale(0.85)*"18", (0, 0.9), W);
label(scale(0.86)*"1", (1, 0), S);
label(scale(0.86)*"2", (2, 0), S);
label(scale(0.86)*"3", (3, 0), S);
label(scale(0.86)*"4", (4, 0), S);
label(scale(0.86)*"5", (5, 0), S);
label(scale(0.86)*"6", (6, 0), S);
[/asy]
|
18
|
deepscale
| 31,812
| ||
How many positive integer divisors of $3003^{3003}$ are divisible by exactly 3003 positive integers?
|
24
|
deepscale
| 9,175
| ||
In $\triangle ABC$, the sides corresponding to angles $A$, $B$, and $C$ are denoted as $a$, $b$, and $c$ respectively. It is known that $B= \frac{\pi}{4}$ and $\cos A-\cos 2A=0$.
$(1)$ Find the angle $C$.
$(2)$ If $b^{2}+c^{2}=a-bc+2$, find the area of $\triangle ABC$.
|
1- \frac{ \sqrt{3}}{3}
|
deepscale
| 7,779
| ||
The diagram shows a right-angled triangle \( ACD \) with a point \( B \) on the side \( AC \). The sides of triangle \( ABD \) have lengths 3, 7, and 8. What is the area of triangle \( BCD \)?
|
2\sqrt{3}
|
deepscale
| 24,118
| ||
Determine how many five-letter words can be formed such that they begin and end with the same letter and the third letter is always a vowel.
|
87880
|
deepscale
| 31,600
| ||
Let $n$ be a fixed positive integer. Determine the smallest possible rank of an $n \times n$ matrix that has zeros along the main diagonal and strictly positive real numbers off the main diagonal.
|
For $n=1$ the only matrix is (0) with rank 0. For $n=2$ the determinant of such a matrix is negative, so the rank is 2. We show that for all $n \geq 3$ the minimal rank is 3. Notice that the first three rows are linearly independent. Suppose that some linear combination of them, with coefficients $c_{1}, c_{2}, c_{3}$, vanishes. Observe that from the first column one deduces that $c_{2}$ and $c_{3}$ either have opposite signs or both zero. The same applies to the pairs $\left(c_{1}, c_{2}\right)$ and $\left(c_{1}, c_{3}\right)$. Hence they all must be zero. It remains to give an example of a matrix of rank (at most) 3. For example, the matrix $\left((i-j)^{2}\right)_{i, j=1}^{n}$ is the sum of three matrices of rank 1, so its rank cannot exceed 3.
|
3
|
deepscale
| 5,168
| |
If the line $(m+2)x+3y+3=0$ is parallel to the line $x+(2m-1)y+m=0$, then the real number $m=$ \_\_\_\_\_\_.
|
-\frac{5}{2}
|
deepscale
| 32,148
| ||
The graphs of four functions, labelled (2) through (5), are shown below. Note that the domain of function (3) is $$\{-5,-4,-3,-2,-1,0,1,2\}.$$ Find the product of the labels of the functions which are invertible. [asy]
size(8cm);
defaultpen(linewidth(.7pt)+fontsize(8pt));
import graph;
picture pic1,pic2,pic3,pic4;
draw(pic1,(-8,0)--(8,0),Arrows(4));
draw(pic1,(0,-8)--(0,8),Arrows(4));
draw(pic2,(-8,0)--(8,0),Arrows(4));
draw(pic2,(0,-8)--(0,8),Arrows(4));
draw(pic3,(-8,0)--(8,0),Arrows(4));
draw(pic3,(0,-8)--(0,8),Arrows(4));
draw(pic4,(-8,0)--(8,0),Arrows(4));
draw(pic4,(0,-8)--(0,8),Arrows(4));
real f(real x) {return x^2-2x;}
real h(real x) {return -atan(x);}
real k(real x) {return 4/x;}
real x;
draw(pic1,graph(f,-2,4),Arrows(4));
draw(pic3,graph(h,-8,8),Arrows(4));
draw(pic4,graph(k,-8,-0.125*4),Arrows(4));
draw(pic4,graph(k,0.125*4,8),Arrows(4));
dot(pic2,(-5,3)); dot(pic2,(-4,5)); dot(pic2,(-3,1)); dot(pic2,(-2,0));
dot(pic2,(-1,2)); dot(pic2,(0,-4)); dot(pic2,(1,-3)); dot(pic2,(2,-2));
label(pic1,"(2)",(0,-9));
label(pic2,"(3)",(0,-9));
label(pic3,"(4)",(0,-9));
label(pic4,"(5)",(0,-9));
add(pic1);
add(shift(20)*pic2);
add(shift(0,-20)*pic3);
add(shift(20,-20)*pic4);
[/asy]
|
60
|
deepscale
| 33,780
| ||
What is the largest perfect square factor of 1512?
|
36
|
deepscale
| 37,740
| ||
Given a linear function \( f(x) \). It is known that the distance between the points of intersection of the graphs \( y = x^2 - 1 \) and \( y = f(x) + 1 \) is \( 3\sqrt{10} \), and the distance between the points of intersection of the graphs \( y = x^2 \) and \( y = f(x) + 3 \) is \( 3\sqrt{14} \). Find the distance between the points of intersection of the graphs \( y = x^2 \) and \( y = f(x) \).
|
3\sqrt{2}
|
deepscale
| 15,626
| ||
Find the remainder when $x^5-x^4-x^3+x^2+x$ is divided by $(x^2-4)(x+1)$.
|
-8x^2+13x+20
|
deepscale
| 36,732
| ||
Evaluate the polynomial \[ x^3 - 2 x^2 - 8 x + 4, \]where $x$ is the positive number such that $x^2 - 2x - 8 = 0$.
|
4
|
deepscale
| 36,876
| ||
Solve for $x$: $$ \frac{1}{2} - \frac{1}{3} = \frac{1}{x}.$$
|
6
|
deepscale
| 33,253
| ||
For any integer $n>1$, the number of prime numbers greater than $n!+1$ and less than $n!+n$ is:
$\text{(A) } 0\quad\qquad \text{(B) } 1\quad\\ \text{(C) } \frac{n}{2} \text{ for n even, } \frac{n+1}{2} \text{ for n odd}\quad\\ \text{(D) } n-1\quad \text{(E) } n$
|
deepscale
| 38,209
| |||
Let \( f: \mathbb{N} \rightarrow \mathbb{N} \) be a function satisfying \( f(m+n) \geq f(m) + f(f(n)) - 1 \) for all \( m, n \in \mathbb{N} \). What values can \( f(2019) \) take?
|
2019
|
deepscale
| 27,624
| ||
In the interval [1, 6], three different integers are randomly selected. The probability that these three numbers are the side lengths of an obtuse triangle is ___.
|
\frac{1}{4}
|
deepscale
| 29,041
| ||
An ellipse and a hyperbola have the same foci $F\_1(-c,0)$, $F\_2(c,0)$. One endpoint of the ellipse's minor axis is $B$, and the line $F\_1B$ is parallel to one of the hyperbola's asymptotes. If the eccentricities of the ellipse and hyperbola are $e\_1$ and $e\_2$, respectively, find the minimum value of $3e\_1^2+e\_2^2$.
|
2\sqrt{3}
|
deepscale
| 19,100
| ||
Given an arithmetic sequence $\{a_n\}$, where $a_n \in \mathbb{N}^*$, and $S_n = \frac{1}{8}(a_n+2)^2$. If $b_n = \frac{1}{2}a_n - 30$, find the minimum value of the sum of the first \_\_\_\_\_\_ terms of the sequence $\{b_n\}$.
|
15
|
deepscale
| 20,483
| ||
Henry decides one morning to do a workout, and he walks $\frac{3}{4}$ of the way from his home to his gym. The gym is $2$ kilometers away from Henry's home. At that point, he changes his mind and walks $\frac{3}{4}$ of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks $\frac{3}{4}$ of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked $\frac{3}{4}$ of the distance toward either the gym or home from the point where he last changed his mind, he will get very close to walking back and forth between a point $A$ kilometers from home and a point $B$ kilometers from home. What is $|A-B|$?
|
1. **Define the sequence of positions**: Let $A$ be the point closer to Henry’s home, and $B$ be the point closer to the gym. Define $(a_n)$ to be the position of Henry after $2n$ walks (even steps, returning towards home), and $(b_n)$ to be the position of Henry after $2n - 1$ walks (odd steps, going towards the gym).
2. **Initial positions**:
- After the first walk towards the gym, Henry walks $\frac{3}{4}$ of the 2 km, so $b_1 = \frac{3}{4} \times 2 = \frac{3}{2}$.
- Then, he walks $\frac{3}{4}$ of the way back towards home from $\frac{3}{2}$ km, which is $\frac{3}{4} \times (\frac{3}{2} - 2) = \frac{3}{4} \times (-\frac{1}{2}) = -\frac{3}{8}$. Thus, $a_1 = 2 - \frac{3}{8} = \frac{13}{8}$.
3. **Recursive relations**:
- For $a_n$: Henry walks $\frac{3}{4}$ of the distance from $b_{n-1}$ to 2 km (towards home), then $\frac{1}{4}$ of this distance from $b_{n-1}$. Thus, $a_n = b_{n-1} - \frac{3}{4}(2 - b_{n-1}) = \frac{1}{4}b_{n-1} + \frac{3}{2}$.
- For $b_n$: Henry walks $\frac{3}{4}$ of the distance from $a_{n-1}$ to 2 km (towards the gym), then $\frac{1}{4}$ of this distance from $a_{n-1}$. Thus, $b_n = a_{n-1} + \frac{3}{4}(2 - a_{n-1}) = \frac{1}{4}a_{n-1} + \frac{3}{2}$.
4. **Solving the recursive relations**:
- We solve for $a_n$ and $b_n$ by finding a fixed point. Assume $a_n = k$ and $b_n = k$ for some $k$. Then, $k = \frac{1}{4}k + \frac{3}{2}$, solving for $k$ gives $k = 2$.
- However, this fixed point does not match the initial conditions, so we need to refine our approach.
5. **Finding the limits**:
- We observe that $a_n$ and $b_n$ converge to certain values as $n \to \infty$. We calculate these limits by considering the recursive relations and the fact that the changes in position decrease geometrically.
- $\lim_{n \to \infty} a_n = \frac{2}{5}$ and $\lim_{n \to \infty} b_n = \frac{8}{5}$.
6. **Calculate $|A-B|$**:
- $|A-B| = \left|\frac{2}{5} - \frac{8}{5}\right| = \left|-\frac{6}{5}\right| = \frac{6}{5}$.
Thus, the final answer is $\boxed{\textbf{(C) } 1 \frac{1}{5}}$.
|
1 \frac{1}{5}
|
deepscale
| 2,889
| |
There are 11 seats, and now we need to arrange for 2 people to sit down. It is stipulated that the middle seat (the 6th seat) cannot be occupied, and the two people must not sit next to each other. How many different seating arrangements are there?
|
84
|
deepscale
| 12,656
| ||
Find the sum of all roots of the equation:
$$
\begin{gathered}
\sqrt{2 x^{2}-2024 x+1023131} + \sqrt{3 x^{2}-2025 x+1023132} + \sqrt{4 x^{2}-2026 x+1023133} = \\
= \sqrt{x^{2}-x+1} + \sqrt{2 x^{2}-2 x+2} + \sqrt{3 x^{2}-3 x+3}
\end{gathered}
$$
|
2023
|
deepscale
| 14,300
| ||
If $x, y, z \in \mathbb{R}$ are solutions to the system of equations $$ \begin{cases}
x - y + z - 1 = 0
xy + 2z^2 - 6z + 1 = 0
\end{cases} $$ what is the greatest value of $(x - 1)^2 + (y + 1)^2$ ?
|
11
|
deepscale
| 10,166
| ||
If \[f(x) =
\begin{cases}
x^2-4 &\quad \text{if } x \ge -4, \\
x + 3 &\quad \text{otherwise},
\end{cases}
\]then for how many values of $x$ is $f(f(x)) = 5$?
|
5
|
deepscale
| 34,361
| ||
A wooden cube, whose edges are one centimeter long, rests on a horizontal surface. Illuminated by a point source of light that is $x$ centimeters directly above an upper vertex, the cube casts a shadow on the horizontal surface. The area of the shadow, which does not include the area beneath the cube is 48 square centimeters. Find the greatest integer that does not exceed $1000x$.
|
166
|
deepscale
| 35,901
| ||
One hundred and one of the squares of an $n\times n$ table are colored blue. It is known that there exists a unique way to cut the table to rectangles along boundaries of its squares with the following property: every rectangle contains exactly one blue square. Find the smallest possible $n$ .
|
101
|
deepscale
| 31,120
| ||
Points $R$, $S$ and $T$ are vertices of an equilateral triangle, and points $X$, $Y$ and $Z$ are midpoints of its sides. How many noncongruent triangles can be drawn using any three of these six points as vertices?
|
To solve this problem, we need to determine how many noncongruent triangles can be formed using any three of the six points $R$, $S$, $T$, $X$, $Y$, and $Z$, where $X$, $Y$, and $Z$ are the midpoints of the sides of equilateral triangle $\triangle RST$.
#### Step 1: Identify possible triangles
We start by noting that there are $\binom{6}{3} = 20$ ways to choose any three points from the six available. However, not all sets of three points will form a triangle (some may be collinear), and some triangles will be congruent to others.
#### Step 2: Analyze cases of triangles
We categorize the triangles based on their congruence to specific reference triangles:
**Case 1: Triangles congruent to $\triangle RST$**
- Only $\triangle RST$ itself fits this description.
**Case 2: Triangles congruent to $\triangle SYZ$**
- These triangles are formed by choosing one vertex from $\triangle RST$ and two midpoints from the sides not touching the chosen vertex. The triangles are $\triangle SYZ$, $\triangle RXY$, $\triangle TXZ$, and $\triangle XYZ$.
**Case 3: Triangles congruent to $\triangle RSX$**
- These triangles are formed by choosing two vertices from $\triangle RST$ and one midpoint from the side connecting these two vertices. The triangles are $\triangle RSX$, $\triangle TSX$, $\triangle STY$, $\triangle RTY$, $\triangle RSZ$, and $\triangle RTZ$.
**Case 4: Triangles congruent to $\triangle SYX$**
- These triangles are formed by choosing one vertex from $\triangle RST$ and two midpoints, one from each of the other two sides. The triangles are $\triangle SYX$, $\triangle SZX$, $\triangle TYZ$, $\triangle TYX$, $\triangle RXZ$, and $\triangle RYZ$.
#### Step 3: Count non-congruent triangles
From the analysis:
- Case 1 contributes 1 triangle.
- Case 2 contributes 4 triangles.
- Case 3 contributes 6 triangles.
- Case 4 contributes 6 triangles.
Adding these, we have $1 + 4 + 6 + 6 = 17$ triangles. However, we initially counted 20 possible sets of points. The remaining 3 sets ($SYR$, $RXT$, $TZS$) are collinear and do not form triangles.
#### Conclusion:
We have identified 4 non-congruent types of triangles: one from Case 1, one from Case 2, one from Case 3, and one from Case 4. Therefore, the number of noncongruent triangles that can be drawn using any three of these six points as vertices is $\boxed{4}$.
|
4
|
deepscale
| 530
| |
What is the sum of the exponents of the prime factors of the square root of the largest perfect square that divides $12!$?
|
1. **Identify the prime factors of $12!$ and their exponents:**
We start by finding the exponents of the prime factors of $12!$. For any prime $p$, the exponent of $p$ in $n!$ is given by:
\[
\sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor
\]
where $\left\lfloor x \right\rfloor$ denotes the floor function, or the greatest integer less than or equal to $x$.
2. **Calculate the exponents for each prime factor:**
- For $p = 2$:
\[
\left\lfloor \frac{12}{2} \right\rfloor + \left\lfloor \frac{12}{2^2} \right\rfloor + \left\lfloor \frac{12}{2^3} \right\rfloor = 6 + 3 + 1 = 10
\]
- For $p = 3$:
\[
\left\lfloor \frac{12}{3} \right\rfloor + \left\lfloor \frac{12}{3^2} \right\rfloor = 4 + 1 = 5
\]
- For $p = 5$:
\[
\left\lfloor \frac{12}{5} \right\rfloor = 2
\]
- Primes greater than $5$ contribute at most once, so they do not affect the largest perfect square.
3. **Determine the largest perfect square that divides $12!$:**
The exponents in the prime factorization of this square must be even. Therefore, we adjust the exponents:
- For $2$, the exponent remains $10$ (already even).
- For $3$, reduce $5$ to $4$ (the largest even number less than $5$).
- For $5$, the exponent remains $2$ (already even).
The prime factorization of the largest perfect square is:
\[
2^{10} \cdot 3^4 \cdot 5^2
\]
4. **Find the square root of this perfect square:**
Halving the exponents of the prime factorization gives:
\[
2^{5} \cdot 3^{2} \cdot 5^{1}
\]
5. **Sum the exponents of the prime factors of the square root:**
The exponents are $5$, $2$, and $1$. Their sum is:
\[
5 + 2 + 1 = 8
\]
6. **Conclusion:**
The sum of the exponents of the prime factors of the square root of the largest perfect square that divides $12!$ is $\boxed{\textbf{(C)}\ 8}$.
|
8
|
deepscale
| 583
| |
How many functions $f:\{0,1\}^{3} \rightarrow\{0,1\}$ satisfy the property that, for all ordered triples \left(a_{1}, a_{2}, a_{3}\right) and \left(b_{1}, b_{2}, b_{3}\right) such that $a_{i} \geq b_{i}$ for all $i, f\left(a_{1}, a_{2}, a_{3}\right) \geq f\left(b_{1}, b_{2}, b_{3}\right)$?
|
Consider the unit cube with vertices $\{0,1\}^{3}$. Let $O=(0,0,0), A=(1,0,0), B=(0,1,0), C=(0,0,1)$, $D=(0,1,1), E=(1,0,1), F=(1,1,0)$, and $P=(1,1,1)$. We want to find a function $f$ on these vertices such that $f(1, y, z) \geq f(0, y, z)$ (and symmetric representations). For instance, if $f(A)=1$, then $f(E)=f(F)=f(P)=1$ as well, and if $f(D)=1$, then $f(P)=1$ as well. We group the vertices into four levels: $L_{0}=\{O\}, L_{1}=\{A, B, C\}, L_{2}=\{D, E, F\}$, and $L_{3}=\{P\}$. We do casework on the lowest level of a 1 in a function. - If the 1 is in $L_{0}$, then $f$ maps everything to 1, for a total of 1 way. - If the 1 is in $L_{1}$, then $f(O)=0$. If there are 31 's in $L_{1}$, then everything but $O$ must be mapped to 1, for 1 way. If there are 21 's in $L_{1}$, then $f\left(L_{2}\right)=f\left(L_{3}\right)=1$, and there are 3 ways to choose the 21 's in $L_{1}$, for a total of 3 ways. If there is one 1, then WLOG $f(A)=1$. Then $f(E)=f(F)=f(P)=1$, and $f(D)$ equals either 0 or 1. There are $3 \cdot 2=6$ ways to do this. In total, there are $1+3+6=10$ ways for the lowest 1 to be in $L_{1}$. - If the lowest 1 is in $L_{2}$, then $f(O)=f\left(L_{1}\right)=0$. If there are 31 's in $L_{2}$, there is one way to make $f$. If there are 21 's, then we can pick the 21 's in 3 ways. Finally, if there is one 1, then we pick this 1 in 3 ways. There are $1+3+3=7$ ways. - The lowest 1 is in $L_{3}$. There is 1 way. - There are no 1's. Then $f$ sends everything to 0. There is 1 way. In total, there are $1+10+7+1+1=20$ total $f^{\prime}$ 's.
|
20
|
deepscale
| 4,417
| |
A sequence $a_{1}, a_{2}, a_{3}, \ldots$ of positive reals satisfies $a_{n+1}=\sqrt{\frac{1+a_{n}}{2}}$. Determine all $a_{1}$ such that $a_{i}=\frac{\sqrt{6}+\sqrt{2}}{4}$ for some positive integer $i$.
|
Clearly $a_{1}<1$, or else $1 \leq a_{1} \leq a_{2} \leq a_{3} \leq \ldots$ We can therefore write $a_{1}=\cos \theta$ for some $0<\theta<90^{\circ}$. Note that $\cos \frac{\theta}{2}=\sqrt{\frac{1+\cos \theta}{2}}$, and $\cos 15^{\circ}=$ $\frac{\sqrt{6}+\sqrt{2}}{4}$. Hence, the possibilities for $a_{1}$ are $\cos 15^{\circ}, \cos 30^{\circ}$, and $\cos 60^{\circ}$, which are $\frac{\sqrt{2}+\sqrt{6}}{2}, \frac{\sqrt{3}}{2}$, and $\frac{1}{2}$.
|
\frac{\sqrt{2}+\sqrt{6}}{2}, \frac{\sqrt{3}}{2}, \frac{1}{2}
|
deepscale
| 3,312
| |
How many polynomials of the form $x^5 + ax^4 + bx^3 + cx^2 + dx + 2020$, where $a$, $b$, $c$, and $d$ are real numbers, have the property that whenever $r$ is a root, so is $\frac{-1+i\sqrt{3}}{2} \cdot r$? (Note that $i=\sqrt{-1}$)
|
1. **Identify the form of the polynomial and the transformation of roots:**
Let $P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + 2020$, where $a$, $b$, $c$, and $d$ are real numbers. We are given that if $r$ is a root, then $\frac{-1+i\sqrt{3}}{2} \cdot r$ is also a root. We recognize $\frac{-1+i\sqrt{3}}{2}$ as a complex cube root of unity, denoted by $\omega = e^{2\pi i / 3}$, using Euler's formula $e^{ix} = \cos(x) + i\sin(x)$.
2. **Analyze the implications for the roots:**
Since $\omega = e^{2\pi i / 3}$, and $\omega^3 = 1$, the roots of the polynomial must cycle among $r$, $r\omega$, and $r\omega^2$. Given that $P(x)$ is a fifth-degree polynomial, it can have exactly five roots (counting multiplicity).
3. **Determine the structure of the polynomial:**
If $r$ is a root, then $r\omega$ and $r\omega^2$ must also be roots. This accounts for three roots. We need two more roots to satisfy the degree of the polynomial. If we denote another root by $w$, then $w$, $w\omega$, and $w\omega^2$ must also be roots. However, this would imply six roots, which is not possible for a fifth-degree polynomial. Therefore, $w$ must coincide with one of $r$, $r\omega$, or $r\omega^2$.
4. **Formulate the polynomial with the given roots:**
The polynomial can be expressed as $P(x) = (x-r)^m(x-r\omega)^n(x-r\omega^2)^p$ where $m+n+p = 5$. The coefficients of the polynomial must be real, which requires that the roots $r\omega$ and $r\omega^2$ (which are complex conjugates) have the same multiplicity, i.e., $n = p$.
5. **Calculate the possible distributions of root multiplicities:**
Since $m+n+p = 5$ and $n = p$, we have two cases:
- Case 1: $m = 1$, $n = p = 2$. This gives the polynomial $(x-r)(x-r\omega)^2(x-r\omega^2)^2$.
- Case 2: $m = 3$, $n = p = 1$. This gives the polynomial $(x-r)^3(x-r\omega)(x-r\omega^2)$.
6. **Conclude the number of possible polynomials:**
Each case corresponds to a distinct polynomial, as the multiplicities of the roots differ. Therefore, there are exactly two polynomials that satisfy the given conditions.
Thus, the answer is $\boxed{\textbf{(C) } 2}$.
|
2
|
deepscale
| 2,167
| |
Evaluate the expression $\sqrt{25\sqrt{15\sqrt{9}}}$.
|
5\sqrt{15}
|
deepscale
| 25,982
| ||
In the expansion of $(x^{2}+1)^{2}(x-1)^{6}$, the coefficient of $x^{5}$ is ____.
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-52
|
deepscale
| 21,539
| ||
Compute the sum:
\[\sum_{1 \le a < b < c} \frac{1}{3^a 5^b 7^c}.\]
(The sum is taken over all triples $(a,b,c)$ of positive integers such that $1 \le a < b < c.$)
|
\frac{1}{21216}
|
deepscale
| 19,348
| ||
At 7:00, five sheep, designated as A, B, C, D, and E, have distances to Wolf Castle forming an arithmetic sequence with a common difference of 20 meters. At 8:00, these same five sheep have distances to Wolf Castle forming another arithmetic sequence, but with a common difference of 30 meters, and their order has changed to B, E, C, A, D. Find how many more meters the fastest sheep can run per hour compared to the slowest sheep.
|
140
|
deepscale
| 10,771
| ||
In the diagram, the smaller circles touch the larger circle and touch each other at the center of the larger circle. The radius of the larger circle is $6.$ What is the area of the shaded region?
[asy]
size(100);
import graph;
filldraw(Circle((0,0),2),mediumgray);
filldraw(Circle((-1,0),1),white);
filldraw(Circle((1,0),1),white);
[/asy]
|
18\pi
|
deepscale
| 35,682
| ||
Points $A(3,5)$ and $B(7,10)$ are the endpoints of a diameter of a circle graphed in a coordinate plane. How many square units are in the area of the circle? Express your answer in terms of $\pi$.
|
\frac{41\pi}{4}
|
deepscale
| 34,533
| ||
Let the sum of the first $n$ terms of an arithmetic sequence $\{a_n\}$ be $S_n$, and it satisfies $S_{2016} > 0$, $S_{2017} < 0$. For any positive integer $n$, we have $|a_n| \geqslant |a_k|$. Determine the value of $k$.
|
1009
|
deepscale
| 26,148
| ||
Given three people, A, B, and C, they stand in a row in any random order. What is the probability that A stands in front of B and C does not stand in front of A?
|
\frac{1}{3}
|
deepscale
| 29,920
| ||
To open the safe, you need to enter a code — a number consisting of seven digits: twos and threes. The safe will open if there are more twos than threes, and the code is divisible by both 3 and 4. Create a code that opens the safe.
|
2222232
|
deepscale
| 10,205
| ||
Ava and Tiffany participate in a knockout tournament consisting of a total of 32 players. In each of 5 rounds, the remaining players are paired uniformly at random. In each pair, both players are equally likely to win, and the loser is knocked out of the tournament. The probability that Ava and Tiffany play each other during the tournament is $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Compute $100 a+b$.
|
Each match eliminates exactly one player, so exactly $32-1=31$ matches are played, each of which consists of a different pair of players. Among the $\binom{32}{2}=\frac{32 \cdot 31}{2}=496$ pairs of players, each pair is equally likely to play each other at some point during the tournament. Therefore, the probability that Ava and Tiffany form one of the 31 pairs of players that play each other is $\frac{31}{496}=\frac{1}{16}$, giving an answer of $100 \cdot 1+16=116$.
|
116
|
deepscale
| 3,842
| |
What is the largest quotient that can be formed using two numbers chosen from the set $\{ -24, -3, -2, 1, 2, 8 \}$?
|
To find the largest quotient formed using two numbers from the set $\{-24, -3, -2, 1, 2, 8\}$, we need to consider the absolute values of the numbers and the signs to maximize the quotient $\frac{a}{b}$.
1. **Maximizing the Quotient**:
- The quotient $\frac{a}{b}$ is maximized when $a$ is maximized and $b$ is minimized in absolute value.
- Additionally, to ensure the quotient is positive and thus eligible for being "largest", both $a$ and $b$ should either be positive or negative.
2. **Case 1: Both $a$ and $b$ are positive**:
- Choose $a = 8$ (maximum positive number in the set).
- Choose $b = 1$ (minimum positive number in the set).
- Compute the quotient: $\frac{8}{1} = 8$.
3. **Case 2: Both $a$ and $b$ are negative**:
- Choose $a = -24$ (maximum negative number in absolute value in the set).
- Choose $b = -2$ (minimum negative number in absolute value in the set).
- Compute the quotient: $\frac{-24}{-2} = 12$.
4. **Comparison of Results**:
- From Case 1, the quotient is $8$.
- From Case 2, the quotient is $12$.
- Since $12 > 8$, the largest possible quotient is $12$.
Thus, the largest quotient that can be formed using two numbers from the given set is $\boxed{12}$, corresponding to choice $\text{(D)}$.
|
12
|
deepscale
| 608
| |
A point $(x, y)$ is selected uniformly at random from the unit square $S=\{(x, y) \mid 0 \leq x \leq 1,0 \leq y \leq 1\}$. If the probability that $(3x+2y, x+4y)$ is in $S$ is $\frac{a}{b}$, where $a, b$ are relatively prime positive integers, compute $100a+b$.
|
Under the transformation $(x, y) \mapsto(3x+2y, x+4y), S$ is mapped to a parallelogram with vertices $(0,0),(3,1),(5,5)$, and $(2,4)$. Using the shoelace formula, the area of this parallelogram is 10. The intersection of the image parallelogram and $S$ is the quadrilateral with vertices $(0,0),\left(1, \frac{1}{3}\right),(1,1)$, and $\left(\frac{1}{2}, 1\right)$. To get this quadrilateral, we take away a right triangle with legs 1 and $\frac{1}{2}$ and a right triangle with legs 1 and $\frac{1}{3}$ from the unit square. So the quadrilateral has area $1-\frac{1}{2} \cdot \frac{1}{2}-\frac{1}{2} \cdot \frac{1}{3}=\frac{7}{12}$. Then the fraction of the image parallelogram that lies within $S$ is $\frac{7}{12}=\frac{7}{120}$, which is the probability that a point stays in $S$ after the mapping.
|
820
|
deepscale
| 5,005
| |
Suppose $f(x)=\frac{3}{2-x}$. If $g(x)=\frac{1}{f^{-1}(x)}+9$, find $g(3)$.
|
10
|
deepscale
| 33,577
| ||
In triangle \( \triangle ABC \), \( \angle BAC = 90^\circ \), \( AC = AB = 4 \), and point \( D \) is inside \( \triangle ABC \) such that \( AD = \sqrt{2} \). Find the minimum value of \( BD + CD \).
|
2\sqrt{10}
|
deepscale
| 29,846
| ||
Find $x$ if $\log_x32 = \dfrac{5}{2}$.
|
4
|
deepscale
| 33,404
| ||
Dr. Math's four-digit house number $WXYZ$ contains no zeroes and can be split into two different two-digit primes ``$WX$'' and ``$YZ$'' where the digits $W$, $X$, $Y$, and $Z$ are not necessarily distinct. If each of the two-digit primes is less than 60, how many such house numbers are possible?
|
156
|
deepscale
| 11,092
| ||
(Ⅰ) Find the equation of the line that passes through the intersection point of the two lines $2x-3y-3=0$ and $x+y+2=0$, and is perpendicular to the line $3x+y-1=0$.
(Ⅱ) Given the equation of line $l$ in terms of $x$ and $y$ as $mx+y-2(m+1)=0$, find the maximum distance from the origin $O$ to the line $l$.
|
2 \sqrt {2}
|
deepscale
| 24,087
| ||
Given an integer coefficient polynomial
$$
f(x)=x^{5}+a_{1} x^{4}+a_{2} x^{3}+a_{3} x^{2}+a_{4} x+a_{5}
$$
If $f(\sqrt{3}+\sqrt{2})=0$ and $f(1)+f(3)=0$, then $f(-1)=$ $\qquad$
|
24
|
deepscale
| 8,170
| ||
Let $a,b,c$ be the roots of $x^3-9x^2+11x-1=0$, and let $s=\sqrt{a}+\sqrt{b}+\sqrt{c}$. Find $s^4-18s^2-8s$.
|
-37
|
deepscale
| 37,027
| ||
Sandy plans to paint one wall in her bedroom. The wall is 9 feet high and 12 feet long. There is a 2-foot by 4-foot area on that wall that she will not have to paint due to the window. How many square feet will she need to paint?
|
100
|
deepscale
| 38,931
| ||
Three shepherds met on a large road, each driving their respective herds. Jack says to Jim:
- If I give you 6 pigs for one horse, your herd will have twice as many heads as mine.
And Dan remarks to Jack:
- If I give you 14 sheep for one horse, your herd will have three times as many heads as mine.
Jim, in turn, says to Dan:
- And if I give you 4 cows for one horse, your herd will become 6 times larger than mine.
The deals did not take place, but can you still say how many heads of livestock were in the three herds?
|
39
|
deepscale
| 15,635
| ||
Given an isosceles triangle $PQR$ with $PQ=QR$ and the angle at the vertex $108^\circ$. Point $O$ is located inside the triangle $PQR$ such that $\angle ORP=30^\circ$ and $\angle OPR=24^\circ$. Find the measure of the angle $\angle QOR$.
|
126
|
deepscale
| 29,964
| ||
At the namesake festival, 45 Alexanders, 122 Borises, 27 Vasily, and several Gennady attended. At the beginning of the festival, all of them lined up so that no two people with the same name stood next to each other. What is the minimum number of Gennadys that could have attended the festival?
|
49
|
deepscale
| 7,468
| ||
Points \( M, N, P, Q \) are taken on the diagonals \( D_1A, A_1B, B_1C, C_1D \) of the faces of cube \( ABCD A_1B_1C_1D_1 \) respectively, such that:
\[ D_1M: D_1A = BA_1: BN = B_1P: B_1C = DQ: DC_1 = \mu, \]
and the lines \( MN \) and \( PQ \) are mutually perpendicular. Find \( \mu \).
|
\frac{1}{\sqrt{2}}
|
deepscale
| 12,280
| ||
Given that $x$ and $y$ are distinct nonzero real numbers such that $x - \tfrac{2}{x} = y - \tfrac{2}{y}$, determine the product $xy$.
|
-2
|
deepscale
| 20,562
| ||
A regular triangular pyramid \(SABC\) is given, with the edge of its base equal to 1. Medians of the lateral faces are drawn from the vertices \(A\) and \(B\) of the base \(ABC\), and these medians do not intersect. It is known that the edges of a certain cube lie on the lines containing these medians. Find the length of the lateral edge of the pyramid.
|
\frac{\sqrt{6}}{2}
|
deepscale
| 12,446
| ||
A function $f: \N\rightarrow\N$ is circular if for every $p\in\N$ there exists $n\in\N,\ n\leq{p}$ such that $f^n(p)=p$ ( $f$ composed with itself $n$ times) The function $f$ has repulsion degree $k>0$ if for every $p\in\N$ $f^i(p)\neq{p}$ for every $i=1,2,\dots,\lfloor{kp}\rfloor$ . Determine the maximum repulsion degree can have a circular function.**Note:** Here $\lfloor{x}\rfloor$ is the integer part of $x$ .
|
1/2
|
deepscale
| 26,244
| ||
What is the largest number, all of whose digits are either 5, 3, or 1, and whose digits add up to $15$?
|
555
|
deepscale
| 16,864
| ||
Find the sum of all positive integers $n$ such that $1+2+\cdots+n$ divides $15\left[(n+1)^{2}+(n+2)^{2}+\cdots+(2 n)^{2}\right]$
|
We can compute that $1+2+\cdots+n=\frac{n(n+1)}{2}$ and $(n+1)^{2}+(n+2)^{2}+\cdots+(2 n)^{2}=\frac{2 n(2 n+1)(4 n+1)}{6}-\frac{n(n+1)(2 n+1)}{6}=\frac{n(2 n+1)(7 n+1)}{6}$, so we need $\frac{15(2 n+1)(7 n+1)}{3(n+1)}=\frac{5(2 n+1)(7 n+1)}{n+1}$ to be an integer. The remainder when $(2 n+1)(7 n+1)$ is divided by $(n+1)$ is 6, so after long division we need $\frac{30}{n+1}$ to be an integer. The solutions are one less than a divisor of 30 so the answer is $$1+2+4+5+9+14+29=64$$
|
64
|
deepscale
| 4,574
| |
1. How many four-digit numbers with no repeated digits can be formed using the digits 1, 2, 3, 4, 5, 6, 7, and the four-digit number must be even?
2. How many five-digit numbers with no repeated digits can be formed using the digits 0, 1, 2, 3, 4, 5, and the five-digit number must be divisible by 5? (Answer with numbers)
|
216
|
deepscale
| 23,218
| ||
Let $AD,BF$ and ${CE}$ be the altitudes of $\vartriangle ABC$. A line passing through ${D}$ and parallel to ${AB}$intersects the line ${EF}$at the point ${G}$. If ${H}$ is the orthocenter of $\vartriangle ABC$, find the angle ${\angle{CGH}}$.
|
Consider triangle \(\triangle ABC\) with altitudes \(AD\), \(BF\), and \(CE\). The orthocenter of the triangle is denoted by \(H\). A line through \(D\) that is parallel to \(AB\) intersects line \(EF\) at point \(G\).
To find the angle \(\angle CGH\), follow these steps:
1. **Identify the orthocenter \(H\):**
Since \(AD\), \(BF\), and \(CE\) are altitudes of \(\triangle ABC\), they meet at the orthocenter \(H\) of the triangle.
2. **Analyze parallelism:**
The line passing through \(D\) and parallel to \(AB\), when intersecting \(EF\) at \(G\), means that \(DG \parallel AB\).
3. **Use properties of cyclic quadrilaterals:**
The points \(A\), \(B\), \(C\), and \(H\) are concyclic in the circumcircle. The key insight is noticing the properties of angles formed by such a configuration:
- Since \(AB \parallel DG\), the angles \(\angle DAG = \angle ABC\) are equal.
- Consider the quadrilateral \(BFDG\): since it is cyclic, the opposite angles are supplementary.
4. **Calculate \(\angle CGH\):**
- Since \(H\) lies on the altitude \(AD\), \(\angle CDH = 90^\circ\).
- Now observe the angles formed at \(G\):
- Since \(DG \parallel AB\) and both are perpendicular to \(CE\), we have \(\angle CGH = 90^\circ\).
5. **Conclusion:**
This analysis ensures that \(\angle CGH\) is a right angle since \(G\) is where the line parallel to \(AB\) meets \(EF\) and forms perpendicularity with \(CE\).
Thus, the angle \(\angle CGH\) is:
\[
\boxed{90^\circ}
\]
|
90^\circ
|
deepscale
| 6,104
| |
8. Andrey likes all numbers that are not divisible by 3, and Tanya likes all numbers that do not contain digits that are divisible by 3.
a) How many four-digit numbers are liked by both Andrey and Tanya?
b) Find the total sum of the digits of all such four-digit numbers.
|
14580
|
deepscale
| 24,807
| ||
Given $sn(α+ \frac {π}{6})= \frac {1}{3}$, and $\frac {π}{3} < α < \pi$, find $\sin ( \frac {π}{12}-α)$.
|
- \frac {4+ \sqrt {2}}{6}
|
deepscale
| 17,337
| ||
Five points, no three of which are collinear, are given. Calculate the least possible value of the number of convex polygons whose some corners are formed by these five points.
|
16
|
deepscale
| 7,718
| ||
In rectangle $ABCD$, $AB = 4$ and $BC = 8$. The rectangle is folded so that points $B$ and $D$ coincide, forming the pentagon $ABEFC$. What is the length of segment $EF$? Express your answer in simplest radical form.
|
4\sqrt{5}
|
deepscale
| 32,953
| ||
From the numbers $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, the probability of randomly selecting two different numbers such that both numbers are odd is $\_\_\_\_\_\_\_\_\_$, and the probability that the product of the two numbers is even is $\_\_\_\_\_\_\_\_\_$.
|
\frac{13}{18}
|
deepscale
| 20,793
| ||
The device consists of three independently operating elements. The probabilities of failure-free operation of the elements (over time $t$) are respectively: $p_{1}=0.7$, $p_{2}=0.8$, $p_{3}=0.9$. Find the probabilities that over time $t$, the following will occur:
a) All elements operate without failure;
b) Two elements operate without failure;
c) One element operates without failure;
d) None of the elements operate without failure.
|
0.006
|
deepscale
| 29,987
| ||
In this square array of 16 dots, four dots are to be chosen at random. What is the probability that the four dots will be collinear? Express your answer as a common fraction.
[asy]
size(59);
for(int i = 0; i<4; ++i)
for(int j = 0; j<4; ++j)
dot((i,j),linewidth(7));
[/asy]
|
\frac{1}{182}
|
deepscale
| 35,350
| ||
Find the distance between the vertices of the hyperbola $9x^2 + 54x - y^2 + 10y + 55 = 0.$
|
\frac{2}{3}
|
deepscale
| 36,366
| ||
Find $3 \cdot 5 \cdot 7 + 15 \div 3.$
|
110
|
deepscale
| 38,820
| ||
Define an ordered triple $(A, B, C)$ of sets to be minimally intersecting if $|A \cap B| = |B \cap C| = |C \cap A| = 1$ and $A \cap B \cap C = \emptyset$. For example, $(\{1,2\},\{2,3\},\{1,3,4\})$ is a minimally intersecting triple. Let $N$ be the number of minimally intersecting ordered triples of sets for which each set is a subset of $\{1,2,3,4,5,6,7\}$. Find the remainder when $N$ is divided by $1000$.
Note: $|S|$ represents the number of elements in the set $S$.
|
Let each pair of two sets have one element in common. Label the common elements as $x$, $y$, $z$. Set $A$ will have elements $x$ and $y$, set $B$ will have $y$ and $z$, and set $C$ will have $x$ and $z$. There are $7 \cdot 6 \cdot 5 = 210$ ways to choose values of $x$, $y$ and $z$. There are $4$ unpicked numbers, and each number can either go in the first set, second set, third set, or none of them. Since we have $4$ choices for each of $4$ numbers, that gives us $4^4 = 256$.
Finally, $256 \cdot 210 = 53760$, so the answer is $\boxed{760}$.
|
760
|
deepscale
| 6,977
| |
The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land ACD?
|
1. **Identify the larger triangle and its dimensions**: The problem describes a larger triangle with vertices at points A, B, and D. The base of this triangle (AD) is along Main Street and measures 6 miles. The height of this triangle (from point B to line AD) is along Brown Road and measures 3 miles.
2. **Calculate the area of triangle ABD**: The area of a triangle is given by the formula:
\[
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}
\]
Substituting the known values:
\[
\text{Area of } \triangle ABD = \frac{1}{2} \times 6 \times 3 = 9 \text{ square miles}
\]
3. **Identify the smaller triangle and its dimensions**: Triangle ABC is within triangle ABD. The base of triangle ABC (AC) is along Aspen Road and measures 3 miles. The height of this triangle (from point B to line AC) is also along Brown Road and measures 3 miles.
4. **Calculate the area of triangle ABC**: Using the same area formula:
\[
\text{Area of } \triangle ABC = \frac{1}{2} \times 3 \times 3 = 4.5 \text{ square miles}
\]
5. **Determine the area of triangle ACD**: Since triangle ACD is the remaining part of triangle ABD after removing triangle ABC, its area is:
\[
\text{Area of } \triangle ACD = \text{Area of } \triangle ABD - \text{Area of } \triangle ABC = 9 - 4.5 = 4.5 \text{ square miles}
\]
Thus, the area of the plot of land ACD is $\boxed{\textbf{(C)}\ 4.5}$ square miles.
|
4.5
|
deepscale
| 2,361
| |
A conference center is setting up chairs in rows for a seminar. Each row can seat $13$ chairs, and currently, there are $169$ chairs set up. They want as few empty seats as possible but need to maintain complete rows. If $95$ attendees are expected, how many chairs should be removed?
|
65
|
deepscale
| 9,335
| ||
For a positive integer \( n \), let \( x_n \) be the real root of the equation \( n x^{3} + 2 x - n = 0 \). Define \( a_n = \left[ (n+1) x_n \right] \) (where \( [x] \) denotes the greatest integer less than or equal to \( x \)) for \( n = 2, 3, \ldots \). Then find \( \frac{1}{1005} \left( a_2 + a_3 + a_4 + \cdots + a_{2011} \right) \).
|
2013
|
deepscale
| 13,562
| ||
Let $A,$ $B,$ $C$ be the angles of a triangle. Evaluate
\[\begin{vmatrix} \sin^2 A & \cot A & 1 \\ \sin^2 B & \cot B & 1 \\ \sin^2 C & \cot C & 1 \end{vmatrix}.\]
|
0
|
deepscale
| 40,311
| ||
Given three real numbers \(p, q,\) and \(r\) such that
\[ p+q+r=5 \quad \text{and} \quad \frac{1}{p+q}+\frac{1}{q+r}+\frac{1}{p+r}=9 \]
What is the value of the expression
\[ \frac{r}{p+q}+\frac{p}{q+r}+\frac{q}{p+r} ? \]
|
42
|
deepscale
| 11,653
| ||
How many ways are there to allocate a group of 8 friends among a basketball team, a soccer team, a track team, and the option of not participating in any sports? Each team, including the non-participant group, could have anywhere from 0 to 8 members. Assume the friends are distinguishable.
|
65536
|
deepscale
| 22,046
| ||
Sixteen is 64$\%$ of what number?
|
25
|
deepscale
| 33,247
| ||
Let $A,$ $B,$ and $C$ be constants such that the equation \[\frac{(x+B)(Ax+28)}{(x+C)(x+7)} = 2\]has infinitely many solutions for $x.$ For these values of $A,$ $B,$ and $C,$ it turns out that there are only finitely many values of $x$ which are not solutions to the equation. Find the sum of these values of $x.$
|
-21
|
deepscale
| 36,462
| ||
If you want to form a rectangular prism with a surface area of 52, what is the minimum number of small cubes with edge length 1 needed?
|
16
|
deepscale
| 9,884
| ||
Let \( [x] \) denote the greatest integer not exceeding \( x \), e.g., \( [\pi]=3 \), \( [5.31]=5 \), and \( [2010]=2010 \). Given \( f(0)=0 \) and \( f(n)=f\left(\left[\frac{n}{2}\right]\right)+n-2\left[\frac{n}{2}\right] \) for any positive integer \( n \). If \( m \) is a positive integer not exceeding 2010, find the greatest possible value of \( f(m) \).
|
10
|
deepscale
| 16,003
| ||
In the drawing, there is a grid composed of 25 small equilateral triangles.
How many rhombuses can be formed from two adjacent small triangles?
|
30
|
deepscale
| 32,130
| ||
Find all values of $r$ such that $\lfloor r \rfloor + r = 16.5$.
|
8.5
|
deepscale
| 34,497
| ||
Square A has side lengths each measuring $x$ inches. Square B has side lengths each measuring $4x$ inches. What is the ratio of the area of the smaller square to the area of the larger square? Express your answer as a common fraction.
|
\frac{1}{16}
|
deepscale
| 39,565
| ||
Let $N$ be the number of ordered triples $(A,B,C)$ of integers satisfying the conditions (a) $0\le A<B<C\le99$, (b) there exist integers $a$, $b$, and $c$, and prime $p$ where $0\le b<a<c<p$, (c) $p$ divides $A-a$, $B-b$, and $C-c$, and (d) each ordered triple $(A,B,C)$ and each ordered triple $(b,a,c)$ form arithmetic sequences. Find $N$.
|
From condition (d), we have $(A,B,C)=(B-D,B,B+D)$ and $(b,a,c)=(a-d,a,a+d)$. Condition $\text{(c)}$ states that $p\mid B-D-a$, $p | B-a+d$, and $p\mid B+D-a-d$. We subtract the first two to get $p\mid-d-D$, and we do the same for the last two to get $p\mid 2d-D$. We subtract these two to get $p\mid 3d$. So $p\mid 3$ or $p\mid d$. The second case is clearly impossible, because that would make $c=a+d>p$, violating condition $\text{(b)}$. So we have $p\mid 3$, meaning $p=3$. Condition $\text{(b)}$ implies that $(b,a,c)=(0,1,2)$ or $(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})$. Now we return to condition $\text{(c)}$, which now implies that $(A,B,C)\equiv(-2,0,2)\pmod{3}$. Now, we set $B=3k$ for increasing positive integer values of $k$. $B=0$ yields no solutions. $B=3$ gives $(A,B,C)=(1,3,5)$, giving us $1$ solution. If $B=6$, we get $2$ solutions, $(4,6,8)$ and $(1,6,11)$. Proceeding in the manner, we see that if $B=48$, we get 16 solutions. However, $B=51$ still gives $16$ solutions because $C_\text{max}=2B-1=101>100$. Likewise, $B=54$ gives $15$ solutions. This continues until $B=96$ gives one solution. $B=99$ gives no solution. Thus, $N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}$.
|
272
|
deepscale
| 7,075
| |
In cube \( ABCD A_{1} B_{1} C_{1} D_{1} \), with an edge length of 6, points \( M \) and \( N \) are the midpoints of edges \( AB \) and \( B_{1} C_{1} \) respectively. Point \( K \) is located on edge \( DC \) such that \( D K = 2 K C \). Find:
a) The distance from point \( N \) to line \( AK \);
b) The distance between lines \( MN \) and \( AK \);
c) The distance from point \( A_{1} \) to the plane of triangle \( MNK \).
|
\frac{66}{\sqrt{173}}
|
deepscale
| 21,826
| ||
Nine chairs in a row are to be occupied by six students and Professors Alpha, Beta and Gamma. These three professors arrive before the six students and decide to choose their chairs so that each professor will be between two students. In how many ways can Professors Alpha, Beta and Gamma choose their chairs?
|
1. **Identify the constraints for the professors' seating**: Professors Alpha, Beta, and Gamma must each be seated between two students. This means they cannot occupy the first or last chair in the row, as these positions do not allow a student to be seated on both sides.
2. **Determine the possible seats for the professors**: Since the first and last chairs are not options for the professors, they can only choose from the remaining seven chairs (positions 2 through 8).
3. **Adjust the problem to account for the spacing requirement**: Each professor must be seated with at least one student between them. This introduces a spacing requirement that effectively reduces the number of available choices. If we place a professor in a chair, we must leave at least one chair empty between them and the next professor.
4. **Transform the problem into a simpler counting problem**: Consider the arrangement of chairs and the requirement that there must be at least one empty chair between each professor. We can think of this as first placing the professors and then filling in the gaps with students. If we place a professor, we eliminate the chair they sit in and the next chair (to maintain the gap). This transforms our choice of 7 chairs into a choice among fewer effective positions.
5. **Calculate the number of ways to choose positions for the professors**: We can simplify the problem by imagining that each time we place a professor, we remove an additional chair from our options to maintain the required gap. After placing one professor and removing the next chair for the gap, we have 5 effective chairs left from which to choose the positions of the remaining two professors. The number of ways to choose 3 positions from these 5 effective chairs is given by the combination formula $\binom{5}{3}$.
6. **Account for the arrangement of the professors**: Once the positions are chosen, the three professors can be arranged in these positions in $3!$ (factorial of 3) different ways.
7. **Calculate the total number of valid arrangements**:
\[
\binom{5}{3} \cdot 3! = 10 \cdot 6 = 60
\]
8. **Conclude with the final answer**:
\[
\boxed{60 \text{ (C)}}
\]
|
60
|
deepscale
| 1,399
| |
How many prime numbers are between 30 and 40?
|
2
|
deepscale
| 38,618
| ||
A geometric progression with 10 terms starts with the first term as 2 and has a common ratio of 3. Calculate the sum of the new geometric progression formed by taking the reciprocal of each term in the original progression.
A) $\frac{29523}{59049}$
B) $\frac{29524}{59049}$
C) $\frac{29525}{59049}$
D) $\frac{29526}{59049}$
|
\frac{29524}{59049}
|
deepscale
| 23,514
| ||
In a regular 2019-gon, numbers are placed at the vertices such that the sum of the numbers in any nine consecutive vertices is 300. It is known that the number at the 19th vertex is 19, and the number at the 20th vertex is 20. What number is at the 2019th vertex?
|
61
|
deepscale
| 14,781
|
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