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Simplify: $$\sqrt[3]{5488000}$$
|
176.4
| |
For the function $f(x)=a- \frac {2}{2^{x}+1}(a\in\mathbb{R})$
$(1)$ Determine the monotonicity of the function $f(x)$ and provide a proof;
$(2)$ If there exists a real number $a$ such that the function $f(x)$ is an odd function, find $a$;
$(3)$ For the $a$ found in $(2)$, if $f(x)\geqslant \frac {m}{2^{x}}$ holds true for all $x\in[2,3]$, find the maximum value of $m$.
|
\frac {12}{5}
| |
Natural numbers \( m \) and \( n \) are such that \( m > n \), \( m \) is not divisible by \( n \), and \( m \) has the same remainder when divided by \( n \) as \( m + n \) has when divided by \( m - n \).
Find the ratio \( m : n \).
|
5/2
| |
What is the largest $2$-digit prime factor of the integer $n = {200\choose 100}$?
|
61
|
Expanding the binomial coefficient, we get ${200 \choose 100}=\frac{200!}{100!100!}$. Let the required prime be $p$; then $10 \le p < 100$. If $p > 50$, then the factor of $p$ appears twice in the denominator. Thus, we need $p$ to appear as a factor at least three times in the numerator, so $3p<200$. The largest such prime is $\boxed{061}$, which is our answer.
|
The midpoints of the sides of a regular octagon \( ABCDEFGH \) are joined to form a smaller octagon. What fraction of the area of \( ABCDEFGH \) is enclosed by the smaller octagon?
|
\frac{1}{2}
| |
Given $\sqrt{2 + \frac{2}{3}} = 2\sqrt{\frac{2}{3}}, \sqrt{3 + \frac{3}{8}} = 3\sqrt{\frac{3}{8}}, \sqrt{4 + \frac{4}{15}} = 4\sqrt{\frac{4}{15}}\ldots$, if $\sqrt{6 + \frac{a}{b}} = 6\sqrt{\frac{a}{b}}$ (where $a,b$ are real numbers), please deduce $a = \_\_\_\_$, $b = \_\_\_\_$.
|
35
| |
Five friends earned $18, $23, $28, $35, and $45. If they split their earnings equally among themselves, how much will the friend who earned $45 need to give to the others?
|
15.2
| |
Given that $O$ is the coordinate origin, the complex numbers $z_1$ and $z_2$ correspond to the vectors $\overrightarrow{OZ_1}$ and $\overrightarrow{OZ_2}$, respectively. $\bar{z_1}$ is the complex conjugate of $z_1$. The vectors are represented as $\overrightarrow{OZ_1} = (10 - a^2, \frac{1}{a + 5})$ and $\overrightarrow{OZ_2} = (2a - 5, 2 - a)$, where $a \in \mathbb{R}$, and $(z_2 - z_1)$ is a purely imaginary number.
(1) Determine the quadrant in which the point corresponding to the complex number $\bar{z_1}$ lies in the complex plane.
(2) Calculate $|z_1 \cdot z_2|$.
|
\frac{\sqrt{130}}{8}
| |
Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac12$. Find the value of
\[\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}.\]
|
\frac{49}{58}
| |
Petya wrote a natural number \( A \) on the board. If you multiply it by 8, you get the square of a natural number. How many such three-digit numbers \( B \) exist for which \( A \cdot B \) is also a square of a natural number?
|
15
| |
Right triangle $ABC$ has side lengths $BC=6$, $AC=8$, and $AB=10$. A circle centered at $O$ is tangent to line $BC$ at $B$ and passes through $A$. A circle centered at $P$ is tangent to line $AC$ at $A$ and passes through $B$. What is $OP$?
|
\frac{35}{12}
|
1. **Identify the Geometry of the Problem:**
- Triangle $ABC$ is a right triangle with $AB$ as the hypotenuse, so $AB = 10$, $BC = 6$, and $AC = 8$.
- Circle centered at $O$ is tangent to $BC$ at $B$ and passes through $A$.
- Circle centered at $P$ is tangent to $AC$ at $A$ and passes through $B$.
2. **Properties of Circles and Tangency:**
- Since the circle with center $O$ is tangent to $BC$ at $B$ and passes through $A$, $O$ lies on the perpendicular bisector of $AB$ and $OB \perp BC$.
- Similarly, since the circle with center $P$ is tangent to $AC$ at $A$ and passes through $B$, $P$ lies on the perpendicular bisector of $AB$ and $PA \perp AC$.
3. **Geometric Configuration:**
- Let the intersection of $OB$ and $PA$ be point $D$. Since $OB \perp BC$ and $PA \perp AC$, and both $OB$ and $PA$ are perpendicular to sides of the right triangle $ABC$, $ACBD$ forms a rectangle.
4. **Midpoint and Similar Triangles:**
- Let $M$ be the midpoint of $AB$. Thus, $BM = \frac{AB}{2} = 5$.
- Since $O$ and $P$ are on the perpendicular bisector of $AB$, $M$, $O$, and $P$ are collinear with $OM \perp AB$.
- By similarity of triangles $\triangle MOB \sim \triangle CBA$ (by AA criterion, as $\angle OMB = \angle BAC = 90^\circ$ and $\angle MOB = \angle CBA$), we have:
\[
\frac{BO}{AB} = \frac{BM}{AC} \implies BO = \frac{BM \cdot AB}{AC} = \frac{5 \cdot 10}{8} = \frac{50}{8} = \frac{25}{4}.
\]
5. **Finding $OD$ and $OP$:**
- Since $D$ is the intersection of $OB$ and $PA$, and $ACBD$ is a rectangle, $BD = AC = 8$.
- Therefore, $OD = BD - BO = 8 - \frac{25}{4} = \frac{32}{4} - \frac{25}{4} = \frac{7}{4}$.
- By similarity of triangles $\triangle DOP \sim \triangle CBA$ (by AA criterion, as $\angle DOP = \angle BAC = 90^\circ$ and $\angle ODP = \angle ABC$), we have:
\[
\frac{OP}{AB} = \frac{DO}{BC} \implies OP = \frac{DO \cdot AB}{BC} = \frac{\frac{7}{4} \cdot 10}{6} = \frac{70}{24} = \frac{35}{12}.
\]
6. **Conclusion:**
- The distance $OP$ is $\boxed{\textbf{(C)}\ \frac{35}{12}}$. $\blacksquare$
|
Calculate $7 \cdot 7! + 5 \cdot 5! + 3 \cdot 3! + 3!$.
|
35904
| |
Find the remainder when $2x^6-x^4+4x^2-7$ is divided by $x^2+4x+3$.
|
-704x-706
| |
What is the arithmetic mean of all positive two-digit multiples of 7?
|
56
| |
An electronic clock always displays the date as an eight-digit number. For example, January 1, 2011, is displayed as 20110101. What is the last day of 2011 that can be evenly divided by 101? The date is displayed as $\overline{2011 \mathrm{ABCD}}$. What is $\overline{\mathrm{ABCD}}$?
|
1221
| |
Tom's favorite number is between $100$ and $150$. It is a multiple of $13$, but not a multiple of $3$. The sum of its digits is a multiple of $4$. What is Tom's favorite number?
|
143
| |
What is the result if you add the largest odd two-digit number to the smallest even three-digit number?
|
199
| |
Find the least three digit number that is equal to the sum of its digits plus twice the product of its digits.
|
397
| |
Let \( f(x) \) be a function defined on \( \mathbf{R} \). If \( f(x) + x^{2} \) is an odd function, and \( f(x) + 2^{x} \) is an even function, then the value of \( f(1) \) is ______.
|
-\frac{7}{4}
| |
Let $T$ be the set of all rational numbers $r$, $0<r<1$, that have a repeating decimal expansion in the form $0.efghefgh\ldots=0.\overline{efgh}$, where the digits $e$, $f$, $g$, and $h$ are not necessarily distinct. To write the elements of $T$ as fractions in lowest terms, how many different numerators are required?
|
6000
| |
The military kitchen needs 1000 jin of rice and 200 jin of millet for dinner. Upon arriving at the rice store, the quartermaster finds a promotion: "Rice is 1 yuan per jin, with 1 jin of millet given for every 10 jin purchased (fractions of 10 jins do not count); Millet is 2 yuan per jin, with 2 jins of rice given for every 5 jin purchased (fractions of 5 jins do not count)." How much money does the quartermaster need to spend to buy enough rice and millet for dinner?
|
1200
| |
Let $f(n) = \frac{x_1 + x_2 + \cdots + x_n}{n}$, where $n$ is a positive integer. If $x_k = (-1)^k, k = 1, 2, \cdots, n$, the set of possible values of $f(n)$ is:
|
$\{0, -\frac{1}{n}\}$
|
To find the set of possible values of $f(n)$, we first need to evaluate the sum $x_1 + x_2 + \cdots + x_n$ where $x_k = (-1)^k$ for $k = 1, 2, \ldots, n$.
1. **Expression for $x_k$:**
- $x_k = (-1)^k$ means that $x_k$ alternates between $-1$ and $1$ starting with $-1$ when $k$ is odd and $1$ when $k$ is even.
2. **Summing $x_k$:**
- If $n$ is odd, then there are $\frac{n+1}{2}$ terms of $-1$ (since the odd indices up to $n$ are $\frac{n+1}{2}$) and $\frac{n-1}{2}$ terms of $1$ (since the even indices up to $n$ are $\frac{n-1}{2}$).
- If $n$ is even, then there are $\frac{n}{2}$ terms of $-1$ and $\frac{n}{2}$ terms of $1$.
3. **Calculating the sum $S_n = x_1 + x_2 + \cdots + x_n$:**
- For $n$ odd:
\[
S_n = \left(\frac{n+1}{2}\right)(-1) + \left(\frac{n-1}{2}\right)(1) = -\frac{n+1}{2} + \frac{n-1}{2} = -1
\]
- For $n$ even:
\[
S_n = \left(\frac{n}{2}\right)(-1) + \left(\frac{n}{2}\right)(1) = -\frac{n}{2} + \frac{n}{2} = 0
\]
4. **Calculating $f(n) = \frac{S_n}{n}$:**
- For $n$ odd:
\[
f(n) = \frac{-1}{n}
\]
- For $n$ even:
\[
f(n) = \frac{0}{n} = 0
\]
5. **Set of possible values of $f(n)$:**
- From the calculations above, $f(n)$ can be either $0$ or $-\frac{1}{n}$ depending on whether $n$ is even or odd, respectively. However, since $n$ varies over all positive integers, $-\frac{1}{n}$ simplifies to just $-1$ when $n=1$ and approaches $0$ as $n$ increases. Thus, the set of possible values of $f(n)$ is $\{0, -\frac{1}{n}\}$.
### Conclusion:
The set of possible values of $f(n)$ is $\{0, -\frac{1}{n}\}$, which corresponds to choice $\boxed{\text{C}}$.
|
A pirate is searching for buried treasure on 6 islands. On each island, there is a $\frac{1}{4}$ chance that the island has buried treasure and no traps, a $\frac{1}{12}$ chance that the island has traps but no treasure, and a $\frac{2}{3}$ chance that the island has neither traps nor treasure. What is the probability that while searching all 6 islands, the pirate will encounter exactly 3 islands with treasure, and none with traps?
|
\frac{5}{54}
| |
Let the sequence $\left\{a_{i}\right\}_{i=0}^{\infty}$ be defined by $a_{0}=\frac{1}{2}$ and $a_{n}=1+\left(a_{n-1}-1\right)^{2}$. Find the product $$\prod_{i=0}^{\infty} a_{i}=a_{0} a_{1} a_{2}$$
|
\frac{2}{3}
|
Let $\left\{b_{i}\right\}_{i=0}^{\infty}$ be defined by $b_{n}=a_{n}-1$ and note that $b_{n}=b_{n-1}^{2}$. The infinite product is then $$\left(1+b_{0}\right)\left(1+b_{0}^{2}\right)\left(1+b_{0}^{4}\right) \ldots\left(1+b_{0}^{2^{k}}\right) \ldots$$ By the polynomial identity $$(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots\left(1+x^{2^{k}}\right) \cdots=1+x+x^{2}+x^{3}+\cdots=\frac{1}{1-x}$$ Our desired product is then simply $$\frac{1}{1-\left(a_{0}-1\right)}=\frac{2}{3}$$
|
The Rotokas alphabet, consisting of letters A, E, G, I, K, O, P, R, T, U, and V (note S is removed from the original alphabet), is used to design four-letter license plates. How many license plates are possible that start with E, end with O, cannot contain I, and must not have repeated letters?
|
56
| |
The median of the numbers 3, 7, x, 14, 20 is equal to the mean of those five numbers. Calculate the sum of all real numbers \( x \) for which this is true.
|
28
| |
Compute $\frac{x^8+12x^4+36}{x^4+6}$ when $x=5$.
|
631
| |
Let $(x^2+1)(2x+1)^9 = a_0 + a_1(x+2) + a_2(x+2)^2 + \ldots + a_{11}(x+2)^{11}$, then calculate the value of $a_0 + a_1 + a_2 + \ldots + a_{11}$.
|
-2
| |
In $\triangle ABC$, $AB = 86$, and $AC=97$. A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$. Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$?
|
61
|
1. **Assign Variables:**
Let $x$ represent $CX$, and let $y$ represent $BX$. Since the circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$, we have $AB = AX = 86$.
2. **Apply Stewart's Theorem:**
Stewart's Theorem states that for a point $X$ on side $BC$ of $\triangle ABC$, the following relation holds:
\[
AB^2 \cdot CX + AC^2 \cdot BX = BC \cdot (BX \cdot CX + AX^2).
\]
Plugging in the known values, we get:
\[
86^2 \cdot x + 97^2 \cdot y = (x+y) \cdot (xy + 86^2).
\]
Simplifying, we have:
\[
7396x + 9409y = xy(x+y) + 7396(x+y).
\]
Rearranging terms, we obtain:
\[
x^2y + xy^2 + 7396y = 9409y.
\]
3. **Simplify the Equation:**
Subtract $7396y$ from both sides:
\[
x^2y + xy^2 = 2013y.
\]
Since $y \neq 0$, we can divide both sides by $y$:
\[
x^2 + xy = 2013.
\]
4. **Factorize and Solve for $x$ and $y$:**
We know $x(x+y) = 2013$. The prime factors of $2013$ are $3$, $11$, and $61$. We need to find values of $x$ and $y$ such that $x < x+y$ and $x+y < 183$ (from the triangle inequality $BC < AB + AC = 183$).
5. **Check Possible Values:**
- If $x = 33$ and $x+y = 61$, then $y = 61 - 33 = 28$.
- Check if these values satisfy the triangle inequality:
\[
x + y = 33 + 28 = 61 < 183.
\]
- Check if $x(x+y) = 33 \cdot 61 = 2013$, which is true.
6. **Conclusion:**
Therefore, the length of $BC$ is $x+y = 61$.
Thus, the correct answer is $\boxed{\textbf{(D) }61}$.
|
The region consisting of all points in three-dimensional space within 3 units of line segment $\overline{AB}$ has volume $216\pi$. What is the length $\textit{AB}$?
|
20
|
1. **Understanding the Geometry**: The region described is a cylinder with hemispheres capping both ends. The radius of the cylinder and hemispheres is $3$ units.
2. **Volume of Hemispheres**: Each hemisphere has a radius of $3$ units. The volume $V$ of a sphere is given by $V = \frac{4}{3}\pi r^3$. Therefore, the volume of one hemisphere is half of this:
\[
V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3}\pi (3)^3 = \frac{1}{2} \times 36\pi = 18\pi
\]
Since there are two hemispheres, their combined volume is:
\[
V_{\text{total hemispheres}} = 2 \times 18\pi = 36\pi
\]
3. **Volume of Cylinder**: The total volume of the region is given as $216\pi$. Subtracting the volume of the hemispheres from the total volume gives the volume of the cylinder:
\[
V_{\text{cylinder}} = 216\pi - 36\pi = 180\pi
\]
4. **Calculating the Height of the Cylinder**: The volume $V$ of a cylinder is given by $V = \pi r^2 h$, where $r$ is the radius and $h$ is the height. Here, $r = 3$, so:
\[
180\pi = \pi (3)^2 h = 9\pi h
\]
Solving for $h$:
\[
h = \frac{180\pi}{9\pi} = 20
\]
5. **Conclusion**: The height of the cylinder, which corresponds to the length of the line segment $\overline{AB}$, is $20$ units.
Thus, the length of $\overline{AB}$ is $\boxed{\textbf{(D)}\ 20}$.
|
In triangle $\triangle ABC$, $a$, $b$, and $c$ are the opposite sides of angles $A$, $B$, and $C$ respectively, and $\dfrac{\cos B}{\cos C}=-\dfrac{b}{2a+c}$.
(1) Find the measure of angle $B$;
(2) If $b=\sqrt {13}$ and $a+c=4$, find the area of $\triangle ABC$.
|
\dfrac{3\sqrt{3}}{4}
| |
What is the ratio of the volume of a cube with edge length six inches to the volume of a cube with edge length one foot? Express your answer as a common fraction.
|
\frac{1}{8}
| |
How many digits are in the product $4^5 \cdot 5^{10}$?
|
11
|
To find the number of digits in the product $4^5 \cdot 5^{10}$, we first calculate each term separately and then use logarithmic properties to determine the number of digits in the product.
1. **Calculate $4^5$:**
\[
4^5 = (2^2)^5 = 2^{10} = 1024
\]
$4^5 = 1024$ has $4$ digits.
2. **Calculate $5^{10}$:**
\[
5^{10} = (5^2)^5 = 25^5
\]
To find the number of digits in $25^5$, we can estimate using logarithms:
\[
\log_{10}(25^5) = 5 \log_{10}(25) \approx 5 \times 1.39794 = 6.9897
\]
Since the number of digits $d$ in a number $n$ is given by $d = \lfloor \log_{10}(n) \rfloor + 1$, we have:
\[
d = \lfloor 6.9897 \rfloor + 1 = 6 + 1 = 7
\]
Therefore, $5^{10}$ has $7$ digits.
3. **Calculate the number of digits in $4^5 \cdot 5^{10}$:**
We use the property that the number of digits in the product of two numbers is the sum of the number of digits of each number minus one (if there is no carry in the most significant digit). However, a more accurate method is to use logarithms:
\[
\log_{10}(4^5 \cdot 5^{10}) = \log_{10}(4^5) + \log_{10}(5^{10})
\]
\[
= 10 \log_{10}(2) + 10 \log_{10}(5) = 10 (\log_{10}(2) + \log_{10}(5)) = 10 \log_{10}(10) = 10
\]
Therefore, the number of digits is:
\[
\lfloor 10 \rfloor + 1 = 10 + 1 = 11
\]
Thus, the product $4^5 \cdot 5^{10}$ has $\boxed{\textbf{(D)} 11}$ digits.
|
Compute $17^9 \div 17^7$.
|
289
| |
Consider a $4 \times 4$ grid of squares. Aziraphale and Crowley play a game on this grid, alternating turns, with Aziraphale going first. On Aziraphales turn, he may color any uncolored square red, and on Crowleys turn, he may color any uncolored square blue. The game ends when all the squares are colored, and Aziraphales score is the area of the largest closed region that is entirely red. If Aziraphale wishes to maximize his score, Crowley wishes to minimize it, and both players play optimally, what will Aziraphales score be?
|
\[ 6 \]
|
We claim that the answer is 6. On Aziraphale's first two turns, it is always possible for him to take 2 adjacent squares from the central four; without loss of generality, suppose they are the squares at $(1,1)$ and $(1,2)$. If allowed, Aziraphale's next turn will be to take one of the remaining squares in the center, at which point there will be seven squares adjacent to a red square, and so Aziraphale can guarantee at least two more adjacent red squares. After that, since the number of blue squares is always at most the number of red squares, Aziraphale can guarantee another adjacent red square, making his score at least 6. If, however, Crowley does not allow Aziraphale to attain another central red square - i.e. coloring the other two central squares blue - then Aziraphale will continue to take squares from the second row, $\operatorname{WLOG}(1,3)$. If Aziraphale is also allowed to take $(1,0)$, he will clearly attain at least 6 adjacent red squares as each red square in this row has two adjacent squares to it, and otherwise (if Crowley takes $(1,0)$), Aziraphale will take $(0,1)$ and guarantee a score of at least $4+\frac{4}{2}=6$ as there are 4 uncolored squares adjacent to a red one. Therefore, the end score will be at least 6. We now show that this is the best possible for Aziraphale; i.e. Crowley can always limit the score to 6. Crowley can play by the following strategy: if Aziraphale colors a square in the second row, Crowley will color the square below it, if Aziraphale colors a square in the third row, Crowley will color the square above it. Otherwise, if Aziraphale colors a square in the first or fourth rows, Crowley will color an arbitrary square in the same row. It is clear that the two "halves" of the board cannot be connected by red squares, and so the largest contiguous red region will occur entirely in one half of the grid, but then the maximum score is $4+\frac{4}{2}=6$. The optimal score is thus both at least 6 and at most 6, so it must be 6 as desired.
|
A triangular region is enclosed by the lines with equations $y = \frac{1}{2} x + 3$, $y = -2x + 6$ and $y = 1$. What is the area of the triangular region? Express your answer as a decimal to the nearest hundredth.
|
8.45
| |
Given \(x\) and \(y\) are real numbers, and:
\[
z_1 = x + (y + 2)i,
\]
\[
z_2 = (x - 2) + yi,
\]
with the condition \(\left|z_1\right| + \left|z_2\right| = 4\),
Find the maximum value of \(|x + y|\).
|
2\sqrt{2}
| |
If $f(1) = 3$, $f(2)= 12$, and $f(x) = ax^2 + bx + c$, what is the value of $f(3)$?
|
21
| |
There is a reservoir A and a town B connected by a river. When the reservoir does not release water, the water in the river is stationary; when the reservoir releases water, the water in the river flows at a constant speed. When the reservoir was not releasing water, speedboat M traveled for 50 minutes from A towards B and covered $\frac{1}{3}$ of the river's length. At this moment, the reservoir started releasing water, and the speedboat took only 20 minutes to travel another $\frac{1}{3}$ of the river's length. The driver then turned off the speedboat's engine and allowed it to drift with the current, taking $\quad$ minutes for the speedboat to reach B.
|
100/3
| |
Let $n$ be a three-digit integer with nonzero digits, not all of which are the same. Define $f(n)$ to be the greatest common divisor of the six integers formed by any permutation of $n$ s digits. For example, $f(123)=3$, because $\operatorname{gcd}(123,132,213,231,312,321)=3$. Let the maximum possible value of $f(n)$ be $k$. Find the sum of all $n$ for which $f(n)=k$.
|
5994
|
Let $n=\overline{a b c}$, and assume without loss of generality that $a \geq b \geq c$. We have $k \mid 100 a+10 b+c$ and $k \mid 100 a+10 c+b$, so $k \mid 9(b-c)$. Analogously, $k \mid 9(a-c)$ and $k \mid 9(a-b)$. Note that if $9 \mid n$, then 9 also divides any permutation of $n$ s digits, so $9 \mid f(n)$ as well; ergo, $f(n) \geq 9$, implying that $k \geq 9$. If $k$ is not a multiple of 3 , then we have $k \mid c-a \Longrightarrow k \leq c-a<9$, contradiction, so $3 \mid k$. Let $x=\min (a-b, b-c, a-c)$. If $x=1$, then we have $k \mid 9$, implying $k=9$ - irrelevant to our investigation. So we can assume $x \geq 2$. Note also that $x \leq 4$, as $2 x \leq(a-b)+(b-c)=a-c \leq 9-1$, and if $x=4$ we have $n=951 \Longrightarrow f(n)=3$. If $x=3$, then since $3|k| 100 a+10 b+c \Longrightarrow 3 \mid a+b+c$, we have $a \equiv b \equiv c(\bmod 3)$ (e.g. if $b-c=3$, then $b \equiv c(\bmod 3)$, so $a \equiv b \equiv c(\bmod 3)$ - the other cases are analogous). This gives us the possibilites $n=147,258,369$, which give $f(n)=3,3,9$ respectively. Hence we can conclude that $x=2$; therefore $k \mid 18$. We know also that $k \geq 9$, so either $k=9$ or $k=18$. If $k=18$, then all the digits of $n$ must be even, and $n$ must be a multiple of 9 ; it is clear that these are sufficient criteria. As $n$ 's digits are all even, the sum of them is also even, and hence their sum is 18. Since $a \geq b \geq c$, we have $a+b+c=18 \leq 3 a \Longrightarrow a \geq 6$, but if $a=6$ then $a=b=c=6$, contradicting the problem statement. Thus $a=8$, and this gives us the solutions $n=882,864$ along with their permutations. It remains to calculate the sum of the permutations of these solutions. In the $n=882$ case, each digit is either 8,8 , or 2 (one time each), and in the $n=864$ case, each digit is either 8,6 , or 4 (twice each). Hence the desired sum is $111(8+8+2)+111(8 \cdot 2+6 \cdot 2+4 \cdot 2)=111(54)=5994$.
|
Find the sum of the distinct prime factors of $7^7 - 7^4$.
|
31
| |
A school wishes to understand the psychological state of learning among its senior students and adopts a systematic sampling method to select 40 students out of 800 for a test. The students are randomly assigned numbers from 1 to 800 and then grouped. In the first group, number 18 is selected through simple random sampling. Among the 40 selected students, those with numbers in the range [1, 200] take test paper A, numbers in the range [201, 560] take test paper B, and the remaining students take test paper C. Calculate the number of students who take test paper C.
|
12
| |
In the Cartesian coordinate plane, a polar coordinate system is established with the origin as the pole and the non-negative half of the x-axis as the polar axis. It is known that point A has polar coordinates $$( \sqrt{2}, \frac{\pi}{4})$$, and the parametric equation of line $l$ is:
$$\begin{cases} x= \frac{3}{2} - \frac{\sqrt{2}}{2}t \\ y= \frac{1}{2} + \frac{\sqrt{2}}{2}t \end{cases}$$ (where $t$ is the parameter), and point A lies on line $l$.
(Ⅰ) Find the corresponding parameter $t$ of point A;
(Ⅱ) If the parametric equation of curve C is:
$$\begin{cases} x=2\cos\theta \\ y=\sin\theta \end{cases}$$ (where $\theta$ is the parameter), and line $l$ intersects curve C at points M and N, find the length of line segment |MN|.
|
\frac{4\sqrt{2}}{5}
| |
Given $x+x^{-1}=3$, calculate the value of $x^{ \frac {3}{2}}+x^{- \frac {3}{2}}$.
|
\sqrt{5}
| |
Petya cut an 8x8 square along the borders of the cells into parts of equal perimeter. It turned out that not all parts are equal. What is the maximum possible number of parts he could get?
|
21
| |
A man buys a house for $10,000 and rents it. He puts $12\frac{1}{2}\%$ of each month's rent aside for repairs and upkeep; pays $325 a year taxes and realizes $5\frac{1}{2}\%$ on his investment. The monthly rent (in dollars) is:
|
83.33
|
1. **Understanding the problem**: A man buys a house for $10,000 and aims to realize a $5\frac{1}{2}\%$ return on his investment. This means he wants to earn $5.5\%$ of $10,000$ annually:
\[
0.055 \times 10,000 = 550 \text{ dollars per year}
\]
2. **Accounting for taxes**: He also pays $325$ in taxes each year. Therefore, the total amount he needs to earn from rent to cover both the return on investment and taxes is:
\[
550 + 325 = 875 \text{ dollars per year}
\]
3. **Calculating monthly earnings**: To find out how much he needs to earn per month, divide the annual requirement by 12:
\[
\frac{875}{12} \approx 72.9167 \text{ dollars per month}
\]
This is the amount he needs after setting aside money for repairs and upkeep.
4. **Adjusting for repairs and upkeep**: The problem states that he sets aside $12\frac{1}{2}\%$ of each month's rent for repairs and upkeep. Since $12\frac{1}{2}\%$ is equivalent to $\frac{1}{8}$, he keeps $\frac{7}{8}$ of the rent. Therefore, the amount calculated above ($72.9167$ dollars) represents $\frac{7}{8}$ of the actual rent he charges. To find the full rent, we solve:
\[
\text{Full rent} = \frac{72.9167}{\frac{7}{8}} = 72.9167 \times \frac{8}{7} \approx 83.3333 \text{ dollars per month}
\]
5. **Conclusion**: The monthly rent that the man charges is approximately $83.33$. Thus, the correct answer is:
\[
\boxed{\textbf{(B)}\ \ 83.33}
\]
|
Given a sequence $\{a_n\}$ whose sum of the first $n$ terms is $S_n$, and the point $(n, \frac{S_n}{n})$ lies on the line $y= \frac{1}{2}x+ \frac{11}{2}$. The sequence $\{b_n\}$ satisfies $b_{n+2}-2b_{n+1}+b_n=0$ $(n\in{N}^*)$, and $b_3=11$, with the sum of the first $9$ terms being $153$.
$(1)$ Find the general formula for the sequences $\{a_n\}$ and $\{b_n\}$;
$(2)$ Let $c_n= \frac{3}{(2a_n-11)(2b_n-1)}$, and the sum of the first $n$ terms of the sequence $\{c_n\}$ be $T_n$. Find the maximum positive integer value of $k$ such that the inequality $T_n > \frac{k}{57}$ holds for all $n\in{N}^*$;
|
18
| |
Given that $\log_{10} \sin x + \log_{10} \cos x = -1$ and that $\log_{10} (\sin x + \cos x) = \frac{1}{2} (\log_{10} n - 1),$ find $n.$
|
12
|
Using the properties of logarithms, we can simplify the first equation to $\log_{10} \sin x + \log_{10} \cos x = \log_{10}(\sin x \cos x) = -1$. Therefore, \[\sin x \cos x = \frac{1}{10}.\qquad (*)\]
Now, manipulate the second equation. \begin{align*} \log_{10} (\sin x + \cos x) &= \frac{1}{2}(\log_{10} n - \log_{10} 10) \\ \log_{10} (\sin x + \cos x) &= \left(\log_{10} \sqrt{\frac{n}{10}}\right) \\ \sin x + \cos x &= \sqrt{\frac{n}{10}} \\ (\sin x + \cos x)^{2} &= \left(\sqrt{\frac{n}{10}}\right)^2 \\ \sin^2 x + \cos^2 x +2 \sin x \cos x &= \frac{n}{10} \\ \end{align*}
By the Pythagorean identities, $\sin ^2 x + \cos ^2 x = 1$, and we can substitute the value for $\sin x \cos x$ from $(*)$. $1 + 2\left(\frac{1}{10}\right) = \frac{n}{10} \Longrightarrow n = \boxed{012}$.
|
Let's define the distance between two numbers as the absolute value of their difference. It is known that the sum of the distances from twelve consecutive natural numbers to a certain number \(a\) is 358, and the sum of the distances from these same twelve numbers to another number \(b\) is 212. Find all possible values of \(a\), given that \(a + b = 114.5\).
|
\frac{190}{3}
| |
David drives from his home to the airport to catch a flight. He drives $35$ miles in the first hour, but realizes that he will be $1$ hour late if he continues at this speed. He increases his speed by $15$ miles per hour for the rest of the way to the airport and arrives $30$ minutes early. How many miles is the airport from his home?
|
210
|
Let's denote the total distance from David's home to the airport as $d$ miles. According to the problem, David drives the first 35 miles in one hour. If he continues at this speed for the entire journey, he would be 1 hour late. This means that the total time required to travel at 35 mph to be on time would be $t+1$ hours, where $t$ is the actual time he should take to reach on time.
1. **Calculate the total time if he continued at 35 mph:**
\[
d = 35(t+1)
\]
2. **David increases his speed to 50 mph for the remaining distance:**
After the first hour, the remaining distance is $d - 35$ miles. He increases his speed to $50$ mph and arrives 30 minutes early. Thus, the time taken for the remaining journey at 50 mph is $t - 1.5$ hours (since he is 30 minutes early, which is 0.5 hours).
\[
d - 35 = 50(t - 1.5)
\]
3. **Set up the equations:**
From the first equation:
\[
d = 35(t+1)
\]
From the second equation:
\[
d - 35 = 50(t - 1.5)
\]
4. **Substitute $d$ from the first equation into the second equation:**
\[
35(t+1) - 35 = 50(t - 1.5)
\]
\[
35t + 35 - 35 = 50t - 75
\]
\[
35t = 50t - 75
\]
\[
15t = 75
\]
\[
t = 5 \text{ hours}
\]
5. **Find the total distance $d$:**
\[
d = 35(t+1) = 35(5+1) = 35 \times 6 = 210 \text{ miles}
\]
Thus, the airport is $\boxed{210}$ miles from David's home. This corresponds to choice $\textbf{(C)}$.
|
What is the maximum number of rooks that can be placed in an \(8 \times 8 \times 8\) cube so that they do not attack each other?
|
64
| |
A rectangular piece of paper $A B C D$ is folded and flattened such that triangle $D C F$ falls onto triangle $D E F$, with vertex $E$ landing on side $A B$. Given that $\angle 1 = 22^{\circ}$, find $\angle 2$.
|
44
| |
Find all positive integer $ m$ if there exists prime number $ p$ such that $ n^m\minus{}m$ can not be divided by $ p$ for any integer $ n$.
|
m \neq 1
|
We are asked to find all positive integers \( m \) such that there exists a prime number \( p \) for which \( n^m - m \) is not divisible by \( p \) for any integer \( n \).
We claim that the answer is all \( m \neq 1 \).
First, consider \( m = 1 \). In this case, the expression becomes \( n - 1 \), which can clearly be a multiple of any prime \( p \) by choosing \( n \equiv 1 \pmod{p} \).
Now, consider \( m > 1 \). Let \( p \) be an arbitrary prime factor of \( m \). Write \( m = p^k l \), where \( \gcd(l, p) = 1 \). Assume that no prime \( q \) exists such that \( n^m - m \equiv 0 \pmod{q} \) has no solution for \( n \).
Consider the expression \( (p^k l)^{p-1} + (p^k l)^{p-2} + \cdots + p^k l + 1 \). Since the left-hand side is not congruent to \( 1 \pmod{p^{k+1}} \), we can choose \( q \) such that \( q \not\equiv 1 \pmod{p^{k+1}} \). We will show that this \( q \) leads to a contradiction.
First, note that the remainder of \( m^{p-1} + m^{p-2} + \cdots + 1 \) when divided by \( m - 1 \) is \( p \), which is relatively prime to \( m - 1 = p^k l - 1 \). Thus, \( \gcd(q, m - 1) = 1 \), so \( m \not\equiv 1 \pmod{q} \).
Since \( n^m \equiv m \pmod{q} \), we have \( n^{p^k l} \equiv p^k l \pmod{q} \), so \( n^{p^{k+1} l} \equiv (p^k l)^p \equiv 1 \pmod{q} \). Let the order of \( n \pmod{q} \) be \( x \). This means \( x \mid p^{k+1} l \). However, since \( n^{p^k l} \equiv p^k l \pmod{q} \), which is not congruent to \( 1 \pmod{q} \), we have that \( x \) is not a factor of \( p^k l \), so \( p^{k+1} \mid x \mid (q - 1) \), implying \( q \equiv 1 \pmod{p^{k+1}} \), which is a contradiction.
Thus, there exists a prime \( q \) such that \( q \) is not a factor of \( n^m - m \) for all integers \( n \).
The answer is: \boxed{m \neq 1}.
|
If the perimeter of a square is 28, what is the side length of the square?
|
7
|
Since a square has four equal sides, the side length of a square equals one-quarter of the perimeter of the square. Thus, the side length of a square with perimeter 28 is $28 \div 4 = 7$.
|
Using the digits 0, 1, 2, 3, and 4, how many even numbers can be formed without repeating any digits?
|
163
| |
The whole numbers from 1 to \( 2k \) are split into two equal-sized groups in such a way that any two numbers from the same group share no more than two distinct prime factors. What is the largest possible value of \( k \)?
|
44
| |
A rectangular box measures $a \times b \times c$, where $a$, $b$, and $c$ are integers and $1\leq a \leq b \leq c$. The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?
|
10
|
1. **Equating Volume and Surface Area**: Given a rectangular box with dimensions $a \times b \times c$, the volume is $abc$ and the surface area is $2(ab + bc + ca)$. Setting these equal gives:
\[
2(ab + bc + ca) = abc.
\]
Dividing both sides by $2abc$:
\[
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2}.
\]
2. **Constraints on $a$**: Since $\frac{1}{a} < \frac{1}{2}$, it follows that $a > 2$, so $a \geq 3$. Also, since $a \leq b \leq c$, we have $\frac{1}{a} \geq \frac{1}{b} \geq \frac{1}{c}$. Thus:
\[
\frac{1}{2} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \leq \frac{3}{a}.
\]
This implies $a \leq 6$. Therefore, $a$ can be $3, 4, 5,$ or $6$.
3. **Case Analysis**:
- **Case $a=3$**: We have $\frac{1}{b} + \frac{1}{c} = \frac{1}{6}$. Possible values for $b$ are those for which $b \geq 3$ and $\frac{1}{b} < \frac{1}{6}$, leading to $b \geq 7$. Checking each $b$ from $7$ to $12$, we find valid $(a, b, c)$ as $(3, 7, 42)$, $(3, 8, 24)$, $(3, 9, 18)$, $(3, 10, 15)$, $(3, 12, 12)$.
- **Case $a=4$**: We have $\frac{1}{b} + \frac{1}{c} = \frac{1}{4}$. Possible values for $b$ are those for which $b \geq 4$ and $\frac{1}{b} < \frac{1}{4}$, leading to $b \geq 5$. Checking each $b$ from $5$ to $8$, we find valid $(a, b, c)$ as $(4, 5, 20)$, $(4, 6, 12)$, $(4, 8, 8)$.
- **Case $a=5$**: We have $\frac{1}{b} + \frac{1}{c} = \frac{3}{10}$. Possible values for $b$ are those for which $b \geq 5$ and $\frac{1}{b} < \frac{3}{10}$, leading to $b \geq 5$. Checking each $b$ from $5$ to $6$, we find valid $(a, b, c)$ as $(5, 5, 10)$.
- **Case $a=6$**: We have $\frac{1}{b} + \frac{1}{c} = \frac{1}{6}$. Since $b \geq 6$, the only solution is $(6, 6, 6)$.
4. **Counting Solutions**: Adding up the solutions from each case, we have $5 + 3 + 1 + 1 = 10$ solutions.
Thus, the number of ordered triples $(a, b, c)$ that satisfy the given conditions is $\boxed{\textbf{(B)}\; 10}$.
|
Find the largest root of the equation $|\sin (2 \pi x) - \cos (\pi x)| = ||\sin (2 \pi x)| - |\cos (\pi x)|$, which belongs to the interval $\left(\frac{1}{4}, 2\right)$.
|
1.5
| |
Given the function $f(x)=\sqrt{3}\sin \omega x+\cos \omega x (\omega > 0)$, where the x-coordinates of the points where the graph of $f(x)$ intersects the x-axis form an arithmetic sequence with a common difference of $\frac{\pi}{2}$, determine the probability of the event "$g(x) \geqslant \sqrt{3}$" occurring, where $g(x)$ is the graph of $f(x)$ shifted to the left along the x-axis by $\frac{\pi}{6}$ units, when a number $x$ is randomly selected from the interval $[0,\pi]$.
|
\frac{1}{6}
| |
On New Year's Eve, Santa Claus gave the children the following task: by using all nine digits from 1 to 9 exactly once, insert either "+" or "-" between each pair of adjacent digits so that the result yields all possible two-digit prime numbers. How many such numbers can be obtained?
|
10
| |
Two parabolas are the graphs of the equations $y=3x^2+4x-5$ and $y=x^2+11$. Give all points where they intersect. List the points in order of increasing $x$-coordinate, separated by semicolons.
|
(-4, 27);(2, 15)
| |
How many ways are there to arrange numbers from 1 to 8 in circle in such way the adjacent numbers are coprime?
Note that we consider the case of rotation and turn over as distinct way.
|
72
| |
The line $y = b-x$ with $0 < b < 4$ intersects the $y$-axis at $P$ and the line $x=4$ at $S$. If the ratio of the area of triangle $QRS$ to the area of triangle $QOP$ is 9:25, what is the value of $b$? Express the answer as a decimal to the nearest tenth.
[asy]
draw((0,-3)--(0,5.5),Arrows);
draw((4,-3.5)--(4,5),Arrows);
draw((-2,0)--(6,0),Arrows);
draw((-2,4.5)--(6,-3.5),Arrows);
dot((0,0));
dot((2.5,0));
dot((4,0));
dot((4,-1.5));
dot((0,2.5));
label("O",(0,0),SW);
label("P",(0,2.5),NE);
label("Q",(2.5,0),NE);
label("R",(4,0),NE);
label("S",(4,-1.5),SW);
label("$y$-axis",(0,5.5),N);
label("$x=4$",(4,5),N);
label("$x$-axis",(6,0),E);
label("$y=b-x$",(6,-3.5),SE);
[/asy]
|
2.5
| |
What is the tens digit of the smallest positive integer that is divisible by each of 20, 16, and 2016?
|
8
|
We note that $20=2^{2} \cdot 5$ and $16=2^{4}$ and $2016=16 \cdot 126=2^{5} \cdot 3^{2} \cdot 7$. For an integer to be divisible by each of $2^{2} \cdot 5$, $2^{4}$, and $2^{5} \cdot 3^{2} \cdot 7$, it must include at least 5 factors of 2, at least 2 factors of 3, at least 1 factor of 5, and at least 1 factor of 7. The smallest such positive integer is $2^{5} \cdot 3^{2} \cdot 5^{1} \cdot 7^{1}=10080$. The tens digit of this integer is 8.
|
Square ABCD has its center at $(8,-8)$ and has an area of 4 square units. The top side of the square is horizontal. The square is then dilated with the dilation center at (0,0) and a scale factor of 2. What are the coordinates of the vertex of the image of square ABCD that is farthest from the origin? Give your answer as an ordered pair.
|
(18, -18)
| |
What is the minimum number of planes determined by $6$ points in space which are not all coplanar, and among which no three are collinear?
|
11
| |
Let $f(x) = e^x - ax + 3$ where $a \in \mathbb{R}$.
1. Discuss the monotonicity of the function $f(x)$.
2. If the minimum value of the function $f(x)$ on the interval $[1,2]$ is $4$, find the value of $a$.
|
e - 1
| |
If A, B, and C stand in a row, calculate the probability that A and B are adjacent.
|
\frac{2}{3}
| |
Find the number of permutations $x_1, x_2, x_3, x_4, x_5$ of numbers $1, 2, 3, 4, 5$ such that the sum of five products \[x_1x_2x_3 + x_2x_3x_4 + x_3x_4x_5 + x_4x_5x_1 + x_5x_1x_2\] is divisible by $3$.
|
80
|
Since $3$ is one of the numbers, a product with a $3$ in it is automatically divisible by $3,$ so WLOG $x_3=3,$ we will multiply by $5$ afterward since any of $x_1, x_2, \ldots, x_5$ would be $3,$ after some cancelation we see that now all we need to find is the number of ways that $x_5x_1(x_4+x_2)$ is divisible by $3,$ since $x_5x_1$ is never divisible by $3,$ now we just need to find the number of ways $x_4+x_2$ is divisible by $3.$ Note that $x_2$ and $x_4$ can be $(1, 2), (2, 1), (1, 5), (5, 1), (2, 4), (4, 2), (4, 5),$ or $(5, 4).$ We have $2$ ways to designate $x_1$ and $x_5$ for a total of $8 \cdot 2 = 16.$ So the desired answer is $16 \cdot 5=\boxed{080}.$
~math31415926535
~MathFun1000 (Rephrasing for clarity)
|
Circles of radius $3$ and $6$ are externally tangent to each other and are internally tangent to a circle of radius $9$. The circle of radius $9$ has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord.
[asy] pointpen = black; pathpen = black + linewidth(0.7); size(150); pair A=(0,0), B=(6,0), C=(-3,0), D=C+6*expi(acos(1/3)), F=B+3*expi(acos(1/3)), P=IP(F--F+3*(D-F),CR(A,9)), Q=IP(F--F+3*(F-D),CR(A,9)); D(CR(A,9)); D(CR(B,3)); D(CR(C,6)); D(P--Q); [/asy]
|
224
|
We label the points as following: the centers of the circles of radii $3,6,9$ are $O_3,O_6,O_9$ respectively, and the endpoints of the chord are $P,Q$. Let $A_3,A_6,A_9$ be the feet of the perpendiculars from $O_3,O_6,O_9$ to $\overline{PQ}$ (so $A_3,A_6$ are the points of tangency). Then we note that $\overline{O_3A_3} \parallel \overline{O_6A_6} \parallel \overline{O_9A_9}$, and $O_6O_9 : O_9O_3 = 3:6 = 1:2$. Thus, $O_9A_9 = \frac{2 \cdot O_6A_6 + 1 \cdot O_3A_3}{3} = 5$ (consider similar triangles). Applying the Pythagorean Theorem to $\triangle O_9A_9P$, we find that \[PQ^2 = 4(A_9P)^2 = 4[(O_9P)^2-(O_9A_9)^2] = 4[9^2-5^2] = \boxed{224}\]
[asy] pointpen = black; pathpen = black + linewidth(0.7); size(150); pair A=(0,0), B=(6,0), C=(-3,0), D=C+6*expi(acos(1/3)), F=B+3*expi(acos(1/3)),G=5*expi(acos(1/3)), P=IP(F--F+3*(D-F),CR(A,9)), Q=IP(F--F+3*(F-D),CR(A,9)); D(CR(D(MP("O_9",A)),9)); D(CR(D(MP("O_3",B)),3)); D(CR(D(MP("O_6",C)),6)); D(MP("P",P,NW)--MP("Q",Q,NE)); D((-9,0)--(9,0)); D(A--MP("A_9",G,N)); D(B--MP("A_3",F,N)); D(C--MP("A_6",D,N)); D(A--P); D(rightanglemark(A,G,P,12)); [/asy]
|
Find all ordered pairs of integers $(x, y)$ such that $3^{x} 4^{y}=2^{x+y}+2^{2(x+y)-1}$.
|
(0,1), (1,1), (2,2)
|
The right side is $2^{x+y}\left(1+2^{x+y-1}\right)$. If the second factor is odd, it needs to be a power of 3 , so the only options are $x+y=2$ and $x+y=4$. This leads to two solutions, namely $(1,1)$ and $(2,2)$. The second factor can also be even, if $x+y-1=0$. Then $x+y=1$ and $3^{x} 4^{y}=2+2$, giving $(0,1)$ as the only other solution.
|
Given a hyperbola with eccentricity $2$ and equation $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ $(a > 0, b > 0)$, the right focus $F_2$ of the hyperbola is the focus of the parabola $y^2 = 8x$. A line $l$ passing through point $F_2$ intersects the right branch of the hyperbola at two points $P$ and $Q$. $F_1$ is the left focus of the hyperbola. If $PF_1 \perp QF_1$, then find the slope of line $l$.
|
\dfrac{3\sqrt{7}}{7}
| |
In the coordinate plane, a parallelogram $O A B C$ is drawn such that its center is at the point $\left(\frac{19}{2}, \frac{15}{2}\right)$, and the points $A, B,$ and $C$ have natural number coordinates. Find the number of such parallelograms. (Here, $O$ denotes the origin - the point $(0,0)$; two parallelograms with the same set of vertices are considered the same, i.e., $OABC$ and $OCBA$ are considered the same parallelogram.)
|
126
| |
An ellipse has foci $(2, 2)$ and $(2, 6)$, and it passes through the point $(14, -3).$ Given this, we can write the equation of the ellipse in standard form as \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1,\]where $a, b, h, k$ are constants, and $a$ and $b$ are positive. Find the ordered quadruple $(a, b, h, k)$.
(Enter your answer as an ordered list, for example, "1, 3, -9, 2".)
|
(8\sqrt3, 14, 2, 4)
| |
How many of the first 512 smallest positive integers written in base 8 use the digit 5 or 6 (or both)?
|
296
| |
Given the function $f(x)=\sin x\cos x-\sqrt{3}\cos^{2}x$.
(1) Find the smallest positive period of $f(x)$;
(2) Find the maximum and minimum values of $f(x)$ when $x\in[0,\frac{\pi }{2}]$.
|
-\sqrt{3}
| |
Let $Q(x)=x^{2}+2x+3$, and suppose that $P(x)$ is a polynomial such that $P(Q(x))=x^{6}+6x^{5}+18x^{4}+32x^{3}+35x^{2}+22x+8$. Compute $P(2)$.
|
2
|
Note that $Q(-1)=2$. Therefore, $P(2)=P(Q(-1))=1-6+18-32+35-22+8=2$.
|
Let $a,b$ be positive reals such that $\frac{1}{a}+\frac{1}{b}\leq2\sqrt2$ and $(a-b)^2=4(ab)^3$ . Find $\log_a b$ .
|
-1
| |
The horizontal and vertical distances between adjacent points equal 1 unit. What is the area of triangle $ABC$?
|
\frac{1}{2}
|
To solve this problem, we need to first understand the positions of points $A$, $B$, $C$, and $D$ on the grid. However, the problem statement does not provide specific coordinates for these points, and the solution provided seems to assume a specific configuration without describing it. Let's assume a configuration based on the solution's description and calculate accordingly.
1. **Assume the positions of points on a grid:**
- Let $A$ be at $(0, 0)$.
- Let $D$ be at $(4, 0)$ (since $AD = 4$ units horizontally).
- Let $B$ be at $(0, 2)$ (since $BF = 2$ units vertically from $A$).
- Let $C$ be at $(3, 0)$ (since $CD = 3$ units horizontally).
2. **Calculate the area of $\triangle ADC$:**
- The base $AD = 4$ units.
- The height from $C$ to line $AD$ is $0$ units (since $C$ and $D$ are on the same horizontal line).
- Therefore, the area of $\triangle ADC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 0 = 0$.
3. **Calculate the area of $\triangle ABD$:**
- The base $AD = 4$ units.
- The height $BF = 2$ units (vertical distance from $B$ to line $AD$).
- Therefore, the area of $\triangle ABD = \frac{1}{2} \times 4 \times 2 = 4$.
4. **Calculate the area of $\triangle CBD$:**
- The base $CD = 3$ units.
- The height $BE = 1$ unit (vertical distance from $B$ to line $CD$).
- Therefore, the area of $\triangle CBD = \frac{1}{2} \times 3 \times 1 = \frac{3}{2}$.
5. **Calculate the area of $\triangle ABC$:**
- Since $\triangle ABC$ is part of $\triangle ADC$, and we subtract the areas of $\triangle ABD$ and $\triangle CBD$ from it:
- Area of $\triangle ABC = \text{Area of } \triangle ADC - \text{Area of } \triangle ABD - \text{Area of } \triangle CBD = 0 - 4 - \frac{3}{2}$.
- This calculation seems incorrect as it results in a negative area, which is not possible.
**Revisiting the problem:**
- The initial assumption about the positions of points or the calculation of the area of $\triangle ADC$ might be incorrect. The area of $\triangle ADC$ should not be zero if it indeed covers half of a $4 \times 3$ grid. Let's assume $\triangle ADC$ covers the entire rectangle formed by points $A$, $D$, and a point directly above $C$ at $(3, 3)$:
- The area of the rectangle is $4 \times 3 = 12$.
- Thus, the area of $\triangle ADC = \frac{1}{2} \times 12 = 6$.
**Correct calculation for the area of $\triangle ABC$:**
- Area of $\triangle ABC = \text{Area of } \triangle ADC - \text{Area of } \triangle ABD - \text{Area of } \triangle CBD = 6 - 4 - \frac{3}{2} = \frac{1}{2}$.
Therefore, the correct answer is $\boxed{B}$.
|
There are several teacups in the kitchen, some with handles and the others without handles. The number of ways of selecting two cups without a handle and three with a handle is exactly $1200$ . What is the maximum possible number of cups in the kitchen?
|
29
| |
Consider the geometric sequence $5$, $\dfrac{15}{4}$, $\dfrac{45}{16}$, $\dfrac{135}{64}$, $\ldots$. Find the tenth term of the sequence. Express your answer as a common fraction.
|
\frac{98415}{262144}
| |
In the complex plane, the points corresponding to the complex number $1+ \sqrt {3}i$ and $- \sqrt {3}+i$ are $A$ and $B$, respectively, with $O$ as the coordinate origin. Calculate the measure of $\angle AOB$.
|
\dfrac {\pi}{2}
| |
Given a set of sample data with $8$ numbers, the average is $8$, and the variance is $12$. Two unknown numbers are added to this set of sample data to form a new set of sample data. It is known that the average of the new sample data is $9$. Find the minimum value of the variance of the new sample data.
|
13.6
| |
A positive integer $n$ is known as an [i]interesting[/i] number if $n$ satisfies
\[{\ \{\frac{n}{10^k}} \} > \frac{n}{10^{10}} \]
for all $k=1,2,\ldots 9$.
Find the number of interesting numbers.
|
999989991
|
A positive integer \( n \) is known as an interesting number if \( n \) satisfies
\[
\left\{ \frac{n}{10^k} \right\} > \frac{n}{10^{10}}
\]
for all \( k = 1, 2, \ldots, 9 \), where \( \{ x \} \) denotes the fractional part of \( x \).
To determine the number of interesting numbers, we can use a computational approach to check each number \( n \) from 1 to \( 10^{10} - 1 \) to see if it satisfies the given condition for all \( k \).
The computational solution involves iterating through each number \( n \) and verifying the condition for each \( k \) from 1 to 9. If the condition holds for all \( k \), the number \( n \) is counted as an interesting number.
After running the computational check, the total number of interesting numbers is found to be 999989991.
The answer is: \boxed{999989991}.
|
What is the $y$-intercept of the line $y = x + 4$ after it is translated down 6 units?
|
-2
|
The line with equation $y = x + 4$ has a $y$-intercept of 4. When the line is translated 6 units downwards, all points on the line are translated 6 units down. This moves the $y$-intercept from 4 to $4 - 6 = -2$.
|
Randomly select $3$ out of $6$ small balls with the numbers $1$, $2$, $3$, $4$, $5$, and $6$, which are of the same size and material. The probability that exactly $2$ of the selected balls have consecutive numbers is ____.
|
\frac{3}{5}
| |
Given a set of data: $10.1$, $9.8$, $10$, $x$, $10.2$, the average of these data is $10$. Calculate the variance of this set of data.
|
0.02
| |
Given the function $y=\sin (2x+1)$, determine the direction and magnitude of the horizontal shift required to obtain this graph from the graph of the function $y=\sin 2x$.
|
\frac{1}{2}
| |
Let the operation $\#$ be defined as $\#(a, b, c) = b^2 - 4ac$, for all real numbers $a, b$ and $c$. What is the value of $\#(1, 2, 3)$?
|
-8
| |
For how many ordered pairs of positive integers $(a, b)$ such that $a \le 50$ is it true that $x^2 - ax + b$ has integer roots?
|
625
| |
A thin diverging lens with an optical power of $D_{p} = -6$ diopters is illuminated by a beam of light with a diameter $d_{1} = 10$ cm. On a screen positioned parallel to the lens, a light spot with a diameter $d_{2} = 20$ cm is observed. After replacing the thin diverging lens with a thin converging lens, the size of the spot on the screen remains unchanged. Determine the optical power $D_{c}$ of the converging lens.
|
18
| |
From the set \(\left\{-3, -\frac{5}{4}, -\frac{1}{2}, 0, \frac{1}{3}, 1, \frac{4}{5}, 2\right\}\), two numbers are drawn without replacement. Find the concept of the two numbers being the slopes of perpendicular lines.
|
3/28
| |
Calculate the volume of the solid bounded by the surfaces \(x + z = 6\), \(y = \sqrt{x}\), \(y = 2\sqrt{x}\), and \(z = 0\) using a triple integral.
|
\frac{48}{5} \sqrt{6}
| |
In $\triangle ABC$, it is known that $BC=1$, $B= \frac{\pi}{3}$, and the area of $\triangle ABC$ is $\sqrt{3}$. Determine the length of $AC$.
|
\sqrt{13}
| |
The function $f(x)$ satisfies
\[f(x + y) = f(x) f(y)\]for all real numbers $x$ and $y.$ If $f(2) = 3,$ find $f(6).$
|
27
| |
If $x$, $y$, and $z$ are positive with $xy=20\sqrt[3]{2}$, $xz = 35\sqrt[3]{2}$, and $yz=14\sqrt[3]{2}$, then what is $xyz$?
|
140
| |
A new website registered $2000$ people. Each of them invited $1000$ other registered people to be their friends. Two people are considered to be friends if and only if they have invited each other. What is the minimum number of pairs of friends on this website?
|
1000
|
Consider a new website with \(2000\) registered people. Each person invites \(1000\) other people from the group to be their friends. According to the rules, two people are actually considered friends if and only if they have invited each other.
We need to determine the minimum number of pairs of friends on this website.
To solve this problem, let's model the situation using graph theory, where each registered user is a vertex and each invitation is a directed edge. We want to identify the minimum number of mutual (bidirectional) edges, which represent pairs of friends.
### Analysis:
1. **Total Invitations**: Each of the \(2000\) people invites \(1000\) others, resulting in a total of \(2000 \times 1000 = 2000000\) directed invitations (edges).
2. **Mutual Friend Condition**: A mutual friendship is formed if, for any given pair of users \( (A, B) \), both users have invited each other. This means that if there is a directed edge from \(A\) to \(B\) and another directed edge from \(B\) to \(A\), then \(A\) and \(B\) are friends.
3. **Undirected Graph Formation**: We convert our directed edges into undirected edges when mutual invites occur, reducing the redundancy of counting two opposite directed edges as a single undirected edge (friendship).
4. **Balancing Invitations**: Each person's outgoing invitations count, which must be balanced with incoming invitations in the sense of closing needed mutual invitations to form friendships.
### Calculation:
By construction, if each person can potentially invite 1000 others, then:
- The maximum number of direct reciprocal (mutual) invitation pairs any user can be part of is constrained by their limiting outgoing or incoming invites.
For a minimal scenario (to minimize mutual friendships while satisfying conditions):
- Consider half of the \(2000\) people invite one set of \(1000\) people and the other half a different set.
- This partitioning under constraints leads to a scenario in which one set of \(1000\) complete friendships appear across the balanced invitations. Thus, half of \(2000\) (partition) gives us a direct calculation:
\[
\frac{2000}{2} = 1000
\]
Hence, under such an optimal (and edge-constrained) configuration, we determine that the minimum number of friendship pairs achievable by these invitations is:
\[
\boxed{1000}.
\]
|
Given that the point $P$ on the ellipse $\frac{x^{2}}{64} + \frac{y^{2}}{28} = 1$ is 4 units away from the left focus, find the distance from point $P$ to the right directrix.
|
16
| |
Estimate $N=\prod_{n=1}^{\infty} n^{n^{-1.25}}$. An estimate of $E>0$ will receive $\lfloor 22 \min (N / E, E / N)\rfloor$ points.
|
9000000
|
We approximate $\ln N=\sum_{n=1}^{\infty} \frac{\ln n}{n^{5 / 4}}$ with an integral as $\int_{1}^{\infty} \frac{\ln x}{x^{5 / 4}} d x =\left.\left(-4 x^{-1 / 4} \ln x-16 x^{-1 / 4}\right)\right|_{1} ^{\infty} =16$. Therefore $e^{16}$ is a good approximation. We can estimate $e^{16}$ by repeated squaring: $e \approx 2.72$, $e^{2} \approx 7.4$, $e^{4} \approx 55$, $e^{8} \approx 3000$, $e^{16} \approx 9000000$. The true value of $e^{16}$ is around 8886111, which is reasonably close to the value of $N$. Both $e^{16}$ and 9000000 would be worth 20 points.
|
Given vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, where $|\overrightarrow{a}|= \sqrt {2}$, $|\overrightarrow{b}|=2$, and $(\overrightarrow{a}-\overrightarrow{b})\perp \overrightarrow{a}$, calculate the angle between vector $\overrightarrow{a}$ and $\overrightarrow{b}$.
|
\frac{\pi}{4}
|
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