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There are 2020 quadratic equations written on the board:
$$
\begin{gathered}
2020 x^{2}+b x+2021=0 \\
2019 x^{2}+b x+2020=0 \\
2018 x^{2}+b x+2019=0 \\
\ldots \\
x^{2}+b x+2=0
\end{gathered}
$$
(each subsequent equation is obtained from the previous one by decreasing the leading coefficient and the constant term by one unit). Find the product of the roots of all the equations written on the board, given that each equation has two real roots.
|
2021
| |
In a triangle $ABC$ , let $I$ denote the incenter. Let the lines $AI,BI$ and $CI$ intersect the incircle at $P,Q$ and $R$ , respectively. If $\angle BAC = 40^o$ , what is the value of $\angle QPR$ in degrees ?
|
20
| |
A sphere is inscribed in a right cone with base radius \(15\) cm and height \(30\) cm. Find the radius \(r\) of the sphere, which can be expressed as \(b\sqrt{d} - b\) cm. What is the value of \(b + d\)?
|
12.5
| |
Compute the sum \[\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \cdots + \lfloor \sqrt{25} \rfloor.\]
|
75
| |
Given complex numbers \( z \) and \( \omega \) satisfying the following two conditions:
1. \( z + \omega + 3 = 0 \);
2. \( |z|, 2, |\omega| \) form an arithmetic sequence.
Is there a maximum value for \( \cos(\arg z - \arg \omega) \)? If so, find it.
|
\frac{1}{8}
| |
Given the curve $\frac{y^{2}}{b} - \frac{x^{2}}{a} = 1 (a \cdot b \neq 0, a \neq b)$ and the line $x + y - 2 = 0$, the points $P$ and $Q$ intersect at the curve and line, and $\overrightarrow{OP} \cdot \overrightarrow{OQ} = 0 (O$ is the origin$), then the value of $\frac{1}{b} - \frac{1}{a}$ is $\_\_\_\_\_\_\_\_\_$.
|
\frac{1}{2}
| |
Find, with proof, the least integer $N$ such that if any $2016$ elements are removed from the set $\{1, 2,...,N\}$ , one can still find $2016$ distinct numbers among the remaining elements with sum $N$ .
|
\boxed{6097392}
|
Since any $2016$ elements are removed, suppose we remove the integers from $1$ to $2016$ . Then the smallest possible sum of $2016$ of the remaining elements is \[2017+2018+\cdots + 4032 = 1008 \cdot 6049 = 6097392\] so clearly $N\ge 6097392$ . We will show that $N=6097392$ works.
$\vspace{0.2 in}$
$\{1,2\cdots 6097392\}$ contain the integers from $1$ to $6048$ , so pair these numbers as follows:
\[1, 6048\]
\[2, 6047\]
\[3, 6046\]
\[\cdots\]
\[3024, 3025\]
When we remove any $2016$ integers from the set $\{1,2,\cdots N\}$ , clearly we can remove numbers from at most $2016$ of the $3024$ pairs above, leaving at least $1008$ complete pairs. To get a sum of $N$ , simply take these $1008$ pairs, all of which sum to $6049$ . The sum of these $2016$ elements is $1008 \cdot 6049 = 6097392$ , as desired.
$\vspace{0.2 in}$
We have shown that $N$ must be at least $6097392$ , and that this value is attainable. Therefore our answer is $\boxed{6097392}$ .
The problems on this page are copyrighted by the Mathematical Association of America 's American Mathematics Competitions .
|
Find the number of integers from 1 to 1000 inclusive that give the same remainder when divided by 11 and by 12.
|
87
| |
Arrange the numbers 3, 4, 5, 6 into two natural numbers A and B, so that the product A×B is maximized. Find the value of A×B.
|
\left( 3402 \right)
| |
The minimum value of the function $f(x) = \cos^2 x + \sin x$ is given by $\frac{-1 + \sqrt{2}}{2}$.
|
-1
| |
Given that in the geometric sequence $\{a_n\}$ where all terms are positive, $a_1a_3=16$ and $a_3+a_4=24$, find the value of $a_5$.
|
32
| |
What is the area of the region defined by the inequality $|3x-18|+|2y+7| \le 3$?
|
3
|
1. **Understanding the given inequality**: The inequality provided is $|3x-18|+|2y+7|\le3$. This inequality represents the sum of the absolute values of two linear expressions and is less than or equal to 3.
2. **Translation of coordinates**: To simplify the inequality, we translate the coordinates. Let's define new variables $u$ and $v$ such that $u = 3x - 18$ and $v = 2y + 7$. Then, the inequality becomes $|u| + |v| \leq 3$.
3. **Reverting back to $x$ and $y$**: The transformations are $u = 3x - 18$ and $v = 2y + 7$. Solving for $x$ and $y$, we get:
\[ x = \frac{u + 18}{3}, \quad y = \frac{v - 7}{2}. \]
Setting $u = 0$ and $v = 0$ gives the new center of the region:
\[ x = \frac{18}{3} = 6, \quad y = \frac{-7}{2}. \]
Thus, the center of the region in the original coordinates is $(6, -\frac{7}{2})$.
4. **Understanding the region's shape**: The inequality $|u| + |v| \leq 3$ describes a diamond (or rhombus) centered at the origin in the $uv$-plane. The vertices of this diamond in the $uv$-plane are at $(\pm 3, 0)$ and $(0, \pm 3)$.
5. **Scaling back to $x$ and $y$ coordinates**: Since $u = 3x - 18$ and $v = 2y + 7$, the vertices translate to:
\[ x = \frac{3 \pm 18}{3} = 7 \text{ or } 5, \quad y = \frac{3 - 7}{2} = -2 \text{ or } \frac{-3 - 7}{2} = -5. \]
However, the correct scaling for the vertices in terms of $x$ and $y$ should be:
\[ x = \frac{\pm 3 + 18}{3} = 7 \text{ or } 5, \quad y = \frac{\pm 3 - 7}{2} = -2 \text{ or } -5. \]
This gives vertices at $(7, -2)$, $(5, -2)$, $(6, -1)$, and $(6, -3)$.
6. **Calculating the area of the diamond**: The lengths of the diagonals of the diamond are $2$ units (horizontal) and $3$ units (vertical), as each side of the inequality contributes half its maximum value to the diagonal. The area $A$ of a rhombus is given by:
\[ A = \frac{1}{2} \times \text{diagonal}_1 \times \text{diagonal}_2 = \frac{1}{2} \times 2 \times 3 = 3. \]
Thus, the area of the region defined by the original inequality is $\boxed{3}$.
|
Let $\triangle PQR$ be a right triangle with $Q$ as the right angle. A circle with diameter $QR$ intersects side $PR$ at point $S$. If the area of $\triangle PQR$ is $200$ and $PR = 40$, find the length of $QS$.
|
10
| |
Find the value of $\dfrac{2\cos 10^\circ - \sin 20^\circ }{\sin 70^\circ }$.
|
\sqrt{3}
| |
Given the complex number $z=(2m^{2}+3m-2)+(m^{2}+m-2)i$ where $(m\in\mathbb{R})$, find the value of $m$ under the following conditions:
$(1) z$ is a real number;
$(2) z$ is an imaginary number;
$(3) z$ is a pure imaginary number;
$(4) z=0$.
|
-2
| |
Distribute 5 students into dormitories A, B, and C, with each dormitory having at least 1 and at most 2 students. Among these, the number of different ways to distribute them without student A going to dormitory A is \_\_\_\_\_\_.
|
60
| |
Determine the smallest integer $B$ such that there exist several consecutive integers, including $B$, that add up to 2024.
|
-2023
| |
Vasya, Petya, and Kolya are in the same class. Vasya always lies in response to any question, Petya alternates between lying and telling the truth, and Kolya lies in response to every third question but tells the truth otherwise. One day, each of them was asked six consecutive times how many students are in their class. The responses were "Twenty-five" five times, "Twenty-six" six times, and "Twenty-seven" seven times. Can we determine the actual number of students in their class based on their answers?
|
27
| |
Find all values of $x$ such that
\[3^x + 4^x + 5^x = 6^x.\]
|
3
| |
Simplify and write the result as a common fraction: $$\sqrt[4]{\sqrt[3]{\sqrt{\frac{1}{65536}}}}$$
|
\frac{1}{\sqrt[3]{4}}
| |
Given the vertices of a regular 100-sided polygon \( A_{1}, A_{2}, A_{3}, \ldots, A_{100} \), in how many ways can three vertices be selected such that they form an obtuse triangle?
|
117600
| |
In trapezoid $ABCD$, the sides $AB$ and $CD$ are equal. The perimeter of $ABCD$ is
|
34
|
1. **Identify the Shape and Given Information**: We are given a trapezoid $ABCD$ where $AB$ and $CD$ are equal in length. We are also given that the perimeter of the trapezoid is one of the options provided.
2. **Assumption of Rectangle Presence**: The solution assumes the presence of a rectangle within the trapezoid, where the horizontal bases are $8$ units each. This assumption needs clarification or justification, which is not provided in the initial problem statement. However, let's proceed with this assumption for the sake of solving the problem.
3. **Calculation of Excess Length on the Bottom Base**: The solution states that the bottom base $CD$ is $16$ units in length, and the top base $AB$ is $8$ units. The excess length on the bottom base is then $16 - 8 = 8$ units.
4. **Division of Excess Length**: Since $AB$ and $CD$ are equal in length and the trapezoid is symmetric, the excess length of $8$ units is equally divided into two parts of $4$ units each on either side of the rectangle.
5. **Use of the Pythagorean Theorem**: The solution uses the Pythagorean theorem to determine the lengths of the non-horizontal sides $BC$ and $AD$. Given that the excess lengths form right triangles with the sides of the rectangle, and assuming these are $3-4-5$ triangles, the sides $BC$ and $AD$ (hypotenuses) are each $5$ units.
6. **Calculation of Perimeter**: The perimeter of trapezoid $ABCD$ is the sum of all its sides:
\[
AB + BC + CD + DA = 8 + 5 + 16 + 5 = 34 \text{ units}
\]
7. **Conclusion**: The perimeter of trapezoid $ABCD$ is $34$ units.
Thus, the correct answer is $\boxed{\text{(D)}\ 34}$.
|
Solve for $x$: $5(3x + 2) - 2 = -2(1 - 7x)$.
|
-10
| |
The value of the quadratic polynomial $a(x^3 - x^2 + 3x) + b(2x^2 + x) + x^3 - 5$ when $x = 2$ is $-17$. What is the value of this polynomial when $x = -2$?
|
-1
| |
There exist positive integers $A,B$ and $C$, with no common factor greater than $1$, such that
\[A \log_{200} 5 + B \log_{200} 2 = C.\]What is $A + B + C$?
|
6
|
1. **Express the equation using logarithm properties**:
Given the equation:
\[
A \log_{200} 5 + B \log_{200} 2 = C
\]
We can use the change of base formula for logarithms, $\log_b a = \frac{\log_k a}{\log_k b}$, to rewrite the logarithms in terms of base 10 (or any common base, but base 10 is standard):
\[
A \frac{\log 5}{\log 200} + B \frac{\log 2}{\log 200} = C
\]
Since $\log 200 = \log (2^3 \cdot 5^2) = 3 \log 2 + 2 \log 5$, the equation becomes:
\[
A \frac{\log 5}{3 \log 2 + 2 \log 5} + B \frac{\log 2}{3 \log 2 + 2 \log 5} = C
\]
2. **Simplify the equation**:
Multiply through by $3 \log 2 + 2 \log 5$ to clear the denominators:
\[
A \log 5 + B \log 2 = C (3 \log 2 + 2 \log 5)
\]
This can be rewritten using the properties of logarithms:
\[
\log(5^A) + \log(2^B) = \log(200^C)
\]
Simplifying further using $\log(ab) = \log a + \log b$:
\[
\log(5^A \cdot 2^B) = \log(200^C)
\]
Since $\log x = \log y$ implies $x = y$, we have:
\[
5^A \cdot 2^B = 200^C
\]
Expanding $200^C$:
\[
5^A \cdot 2^B = (2^3 \cdot 5^2)^C = 2^{3C} \cdot 5^{2C}
\]
3. **Equating the powers of the same bases**:
\[
5^A = 5^{2C} \quad \text{and} \quad 2^B = 2^{3C}
\]
Thus, $A = 2C$ and $B = 3C$.
4. **Determine the values of $A, B, C$**:
Since $A, B, C$ are relatively prime, the smallest value for $C$ that keeps $A$ and $B$ integers and maintains the gcd condition is $C = 1$. Therefore, $A = 2 \times 1 = 2$ and $B = 3 \times 1 = 3$.
5. **Calculate $A + B + C$**:
\[
A + B + C = 2 + 3 + 1 = 6
\]
Thus, the answer is $\boxed{6}$. $\mathrm{(A)}$
|
Given an arithmetic sequence $\{a\_n\}$ with a common difference $d > 0$, and $a\_2$, $a\_5-1$, $a\_{10}$ form a geometric sequence. If $a\_1=5$, and $S\_n$ represents the sum of the first $n$ terms of the sequence, find the minimum value of $\frac{2S\_n+n+32}{a\_n+1}$.
|
\frac{20}{3}
| |
Given the function $f(x)$ defined on the interval $[-2011, 2011]$ and satisfying $f(x_1+x_2) = f(x_1) + f(x_2) - 2011$ for any $x_1, x_2 \in [-2011, 2011]$, and $f(x) > 2011$ when $x > 0$, determine the value of $M+N$.
|
4022
| |
Given the function $f(x) = \sqrt{2}\cos(2x - \frac{\pi}{4})$, where $x \in \mathbb{R}$,
1. Find the smallest positive period of the function $f(x)$ and its intervals of monotonically increasing values.
2. Find the minimum and maximum values of the function $f(x)$ on the interval $\left[-\frac{\pi}{8}, \frac{\pi}{2}\right]$.
|
-1
| |
Express this sum as a common fraction: $0.\overline{7} + 0.\overline{13}$
|
\frac{10}{11}
| |
For each positive integer $ n$, let $ c(n)$ be the largest real number such that
\[ c(n) \le \left| \frac {f(a) \minus{} f(b)}{a \minus{} b}\right|\]
for all triples $ (f, a, b)$ such that
--$ f$ is a polynomial of degree $ n$ taking integers to integers, and
--$ a, b$ are integers with $ f(a) \neq f(b)$.
Find $ c(n)$.
[i]Shaunak Kishore.[/i]
|
\frac{1}{L_n}
|
For each positive integer \( n \), let \( c(n) \) be the largest real number such that
\[
c(n) \le \left| \frac{f(a) - f(b)}{a - b} \right|
\]
for all triples \( (f, a, b) \) such that:
- \( f \) is a polynomial of degree \( n \) taking integers to integers, and
- \( a, b \) are integers with \( f(a) \neq f(b) \).
To find \( c(n) \), we claim that \( c(n) = \frac{1}{L_n} \), where \( L_n = \text{lcm}(1, 2, 3, \ldots, n) \).
First, note that any polynomial \( f(X) \) that maps the integers to the integers can be represented as:
\[
f(X) = c_0 + c_1 \binom{X}{1} + c_2 \binom{X}{2} + \cdots + c_n \binom{X}{n}.
\]
### Lemma 1
\( L_n \cdot \frac{\binom{a}{n} - \binom{b}{n}}{a - b} \in \mathbb{Z} \).
**Proof:**
Consider the polynomial \( g(X) = \binom{X + b}{n} - \binom{b}{n} \). This polynomial can be written as:
\[
g(X) = d_1 \binom{X}{1} + \cdots + d_n \binom{X}{n}.
\]
Using the identity \( \frac{1}{X} \binom{X}{n} = \frac{1}{n} \binom{X-1}{n-1} \), the denominator of \( \frac{g(X)}{X} \) must have size at most \( L_n \). Thus, \( L_n \cdot \frac{g(X)}{X} \in \mathbb{Z} \), proving the lemma. \( \blacksquare \)
Now, consider:
\[
T = \frac{f(a) - f(b)}{a - b} = \sum_{k=0}^n c_k \frac{\binom{a}{k} - \binom{b}{k}}{a - b}.
\]
In particular, for each prime \( p \),
\[
v_p \left( c_k \frac{\binom{a}{k} - \binom{b}{k}}{a - b} \right) \ge -v_p(L_k) \ge -v_p(L_n),
\]
so \( v_p(T) \ge -v_p(L_n) \). Therefore, \( T \cdot L_n \in \mathbb{Z} \). If \( T \neq 0 \), then \( T \ge \frac{1}{L_n} \), establishing a lower bound on \( c(n) \).
To show that this lower bound is attainable, consider a suitable choice of \( c_i \) such that:
\[
\frac{f(N!) - f(0)}{N!} = \frac{1}{L_n}
\]
for large \( N \). Note that:
\[
\frac{\binom{N!}{k} - \binom{0}{k}}{N! - 0} = \frac{\binom{N!}{k}}{N!} = \frac{\binom{N! - 1}{k - 1}}{k}.
\]
No prime less than or equal to \( k \) divides \( \binom{N! - 1}{k - 1} \), as the expression can be written as \( \prod_{i=1}^{k-1} \frac{N! - i}{i} \) and \( \gcd \left( \frac{N! - i}{i}, L_k \right) = 1 \) for large \( N \) and \( k \le n \). Therefore:
\[
\frac{f(N!) - f(0)}{N! - 0} = \sum_{k=0}^n \frac{c_k t_k}{k}
\]
for \( \gcd(t_k, k) = 1 \) fixed and some \( c_k \). By Bézout's identity, we can choose suitable \( c_i \) such that the expression equals \( \frac{1}{L_n} \).
Thus, we conclude that:
\[
c(n) = \frac{1}{L_n}.
\]
The answer is: \boxed{\frac{1}{L_n}}.
|
Two particles move along the edges of a square $ABCD$ with \[A \Rightarrow B \Rightarrow C \Rightarrow D \Rightarrow A,\] starting simultaneously and moving at the same speed. One starts at vertex $A$, and the other starts at the midpoint of side $CD$. The midpoint of the line segment joining the two particles traces out a path that encloses a region $R$. What is the ratio of the area of $R$ to the area of square $ABCD$?
A) $\frac{1}{16}$
B) $\frac{1}{12}$
C) $\frac{1}{9}$
D) $\frac{1}{6}$
E) $\frac{1}{4}$
|
\frac{1}{4}
| |
Vasya thought of a four-digit number and wrote down the product of each pair of its adjacent digits on the board. After that, he erased one product, and the numbers 20 and 21 remained on the board. What is the smallest number Vasya could have in mind?
|
3745
| |
The cafe "Burattino" operates 6 days a week with a day off on Mondays. Kolya made two statements: "from April 1 to April 20, the cafe worked 18 days" and "from April 10 to April 30, the cafe also worked 18 days." It is known that he made a mistake once. How many days did the cafe work from April 1 to April 27?
|
23
| |
For odd primes $p$, let $f(p)$ denote the smallest positive integer $a$ for which there does not exist an integer $n$ satisfying $p \mid n^{2}-a$. Estimate $N$, the sum of $f(p)^{2}$ over the first $10^{5}$ odd primes $p$. An estimate of $E>0$ will receive $\left\lfloor 22 \min (N / E, E / N)^{3}\right\rfloor$ points.
|
2266067
|
Note that the smallest quadratic nonresidue $a$ is always a prime, because if $a=b c$ with $b, c>1$ then one of $b$ and $c$ is also a quadratic nonresidue. We apply the following heuristic: if $p_{1}$, $p_{2}, \ldots$ are the primes in increasing order, then given a "uniform random prime" $q$, the values of $\left(\frac{p_{1}}{q}\right),\left(\frac{p_{2}}{q}\right), \ldots$ are independent and are 1 with probability $\frac{1}{2}$ and -1 with probability $\frac{1}{2}$. Of course, there is no such thing as a uniform random prime. More rigorously, for any $n$, the joint distributions of $\left(\frac{p_{1}}{q}\right), \ldots,\left(\frac{p_{n}}{q}\right)$ where $q$ is a uniform random prime less than $N$ converges in distribution to $n$ independent coin flips between 1 and -1 as $N \rightarrow \infty$. For ease of explanation, we won't adopt this more formal view, but it is possible to make the following argument rigorous by looking at primes $q<N$ and sending $N \rightarrow \infty$. Given any $n$, the residue of $q \bmod n$ is uniform over the $\varphi(n)$ residues $\bmod n$ that are relatively prime to $n$. By quadratic reciprocity, conditioned on either $q \equiv 1(\bmod 4)$ or $q \equiv 3(\bmod 4)$, exactly half of the nonzero residues $\bmod p_{n}$ satisfy $\left(\frac{p_{n}}{q}\right)=1$ and exactly half satisfy $\left(\frac{p_{n}}{q}\right)=-1$ for odd $p_{n}$ (the case of $p_{n}=2$ is slightly different and one must look mod 8, but the result is the same). The residue of $q \bmod 8, p_{2}, p_{3}, \ldots, p_{n}$ are independent as these are pairwise relatively prime, yielding our heuristic. Thus, we may model our problem of finding the smallest quadratic nonresidue with the following process: independent fair coins are flipped for each prime, and we take the smallest prime that flipped heads. We can estimate the expected value of $f(p)^{2}$ as $\sum_{n=1}^{\infty} \frac{p_{n}^{2}}{2^{n}}$. Looking at the first few terms gives us $\frac{2^{2}}{2}+\frac{3^{2}}{4}+\frac{5^{2}}{8}+\frac{7^{2}}{16}+\frac{11^{2}}{32}+\frac{13^{2}}{64}+\frac{17^{2}}{128}+\frac{19^{2}}{256}+\frac{23^{2}}{512}+\frac{29^{2}}{1024} \approx 22$. The terms after this decay rapidly, so a good approximation is $E=22 \cdot 10^{5}$, good enough for 20 points. The more inaccurate $E=20 \cdot 10^{5}$ earns 15 points. This Python code computes the exact answer: ``` def smallest_nqr(p): for a in range(1,p): if pow(a, (p-1)//2,p)==p-1: return a ```import sympyprint(sum([smallest_nqr(p)**2 for p in sympy.ntheory.primerange(3,sympy.prime(10**5+2))])) Remark. In 1961, Erdős showed that as $N \rightarrow \infty$, the average value of $f(p)$ over odd primes $p<N$ will converge to $\sum_{n=1}^{\infty} \frac{p_{n}}{2^{n}} \approx 3.675$.
|
Egor wrote a number on the board and encrypted it according to the rules of letter puzzles (different letters correspond to different digits, identical letters - identical digits). The result was the word "GWATEMALA". How many different numbers could Egor have originally written if his number was divisible by 5?
|
114240
| |
Let \[f(x) =
\begin{cases}
x^2+2 &\text{if } x<n, \\
2x+5 &\text{if }x\ge{n}.
\end{cases}
\]If the graph $y=f(x)$ is continuous, find the sum of all possible values of $n$.
|
2
| |
Given three rays $AB$, $BC$, $BB_{1}$ are not coplanar, and the diagonals of quadrilaterals $BB_{1}A_{1}A$ and $BB_{1}C_{1}C$ bisect each other, and $\overrightarrow{AC_{1}}=x\overrightarrow{AB}+2y\overrightarrow{BC}+3z\overrightarrow{CC_{1}}$, find the value of $x+y+z$.
|
\frac{11}{6}
| |
Eight congruent copies of the parabola \( y = x^2 \) are arranged symmetrically around a circle such that each vertex is tangent to the circle, and each parabola is tangent to its two neighbors. Find the radius of the circle. Assume that one of the tangents to the parabolas corresponds to the line \( y = x \tan(45^\circ) \).
|
\frac{1}{4}
| |
Shelby drives her scooter at a speed of $30$ miles per hour if it is not raining, and $20$ miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of $16$ miles in $40$ minutes. How many minutes did she drive in the rain?
|
24
|
1. **Define Variables:**
Let $x$ be the time (in minutes) Shelby drove in the rain. Therefore, the time she drove when it was not raining is $40 - x$ minutes.
2. **Convert Speeds to Miles per Minute:**
- Shelby's speed in non-rainy conditions is $30$ miles per hour. Converting this to miles per minute:
\[
\frac{30 \text{ miles}}{60 \text{ minutes}} = \frac{1}{2} \text{ miles per minute}
\]
- Shelby's speed in rainy conditions is $20$ miles per hour. Converting this to miles per minute:
\[
\frac{20 \text{ miles}}{60 \text{ minutes}} = \frac{1}{3} \text{ miles per minute}
\]
3. **Set Up the Distance Equation:**
The total distance Shelby drove is the sum of the distances she drove in each weather condition:
\[
\left(\frac{1}{2} \text{ miles per minute}\right) \cdot (40 - x) \text{ minutes} + \left(\frac{1}{3} \text{ miles per minute}\right) \cdot x \text{ minutes} = 16 \text{ miles}
\]
4. **Simplify and Solve the Equation:**
\[
\frac{1}{2}(40 - x) + \frac{1}{3}x = 16
\]
Multiply through by 6 to clear the fractions:
\[
3(40 - x) + 2x = 96
\]
\[
120 - 3x + 2x = 96
\]
\[
120 - x = 96
\]
\[
x = 24
\]
5. **Conclusion:**
Shelby drove in the rain for $24$ minutes.
Thus, the answer is $\boxed{\textbf{(C)}\ 24}$.
|
Let positive integers \( a, b, c, d \) satisfy \( a > b > c > d \) and \( a+b+c+d=2004 \), \( a^2 - b^2 + c^2 - d^2 = 2004 \). Find the minimum value of \( a \).
|
503
| |
Solve for $x$: $x = \dfrac{35}{6-\frac{2}{5}}$.
|
\frac{25}{4}
| |
We call a positive integer $N$ [i]contagious[/i] if there are $1000$ consecutive non-negative integers such that the sum of all their digits is $N$. Find all contagious positive integers.
|
\{13500, 13501, 13502, \ldots\}
|
To determine which positive integers \( N \) are contagious, we consider 1000 consecutive non-negative integers and the sum of all their digits equating to \( N \).
Let the consecutive integers be \( x, x+1, x+2, \ldots, x+999 \). We need to calculate the sum of the digits of these 1000 numbers.
Let's start by considering the simple case where \( x = 0 \). The consecutive integers then are \( 0, 1, 2, \ldots, 999 \). The sum of all the digits in these numbers can be computed by considering the contribution from each place (units, tens, hundreds):
1. **Units Place:** Each digit from 0 to 9 appears 100 times across the 1000 numbers (as every complete set of 100 numbers repeats the digit series in the units place). Thus, the sum of these digits is:
\[
100 \times (0 + 1 + 2 + \ldots + 9) = 100 \times 45 = 4500.
\]
2. **Tens Place:** Each digit from 0 to 9 appears 100 times in the tens place (similar to the units place). Therefore, the contribution from the tens place is:
\[
10 \times (100 \times (0 + 1 + 2 + \ldots + 9)) = 10 \times 4500 = 45000.
\]
3. **Hundreds Place:** From numbers 0 to 999, digits from 0 to 9 appear 100 times in the hundreds place. Therefore, the sum from the hundreds place is:
\[
100 \times (0 + 1 + 2 + \ldots + 9) \times 100 = 450000.
\]
Adding these contributions together, we find the total digit sum of the numbers 0 to 999:
\[
4500 + 45000 + 450000 = 499500.
\]
Now, imagine shifting \( x \) from 0 to any other starting point up to 999. Each unit increase in \( x \) effectively cycles the number sequence, maintaining the same digit sum pattern shifted across different numbers.
Therefore, the sum of digits still repeats every 1000 numbers. Adjusting the sequence by \( k \) units (\( x \rightarrow x+k \)), the digit sum \( N \) is shifted by the sum of first \( k \) digits which follows:
\[
\sum_{i=0}^{k-1} ((i+1) \bmod 10) \leq 45k.
\]
If \( N = \text{original digit sum} + p \) for some integer \( p \), then each increase \( k \times 45 \) will match this pattern starting with:
\[
N = 13500 + p.
\]
Hence, since the structure repeats every 1000 numbers and variations just shift the start by up to 9 per cycle:
\[
\boxed{\{13500, 13501, 13502, \ldots\}}
\]
Thus all positive integers starting from 13500 and beyond are confirmed contagious numbers.
|
Calculate the sum of the repeating decimals $0.\overline{2}$, $0.\overline{02}$, and $0.\overline{0002}$ as a common fraction.
|
\frac{224422}{9999}
| |
In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?
|
0
|
1. **Identify the fixed elements and the problem requirements:**
- Points $A$ and $B$ are fixed in the plane and are $10$ units apart.
- We need to find point $C$ such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units.
2. **Set up a coordinate system:**
- Place $A$ at $(0,0)$ and $B$ at $(10,0)$. This simplifies calculations and any valid configuration of points can be rotated or translated to this setup.
3. **Express the area condition:**
- The area of $\triangle ABC$ is given by $\frac{1}{2} \times \text{base} \times \text{height} = 100$ square units.
- Here, the base $AB = 10$ units, so the height from $C$ to line $AB$ must satisfy $\frac{1}{2} \times 10 \times h = 100 \Rightarrow h = 20$.
- Therefore, the $y$-coordinate of $C$ must be $\pm 20$ to satisfy the area condition.
4. **Analyze the perimeter condition:**
- The perimeter of $\triangle ABC$ is $AB + AC + BC = 50$ units.
- Since $AB = 10$, we need $AC + BC = 40$ units.
5. **Calculate distances $AC$ and $BC$:**
- If $C$ is at $(x, 20)$ or $(x, -20)$, then:
- $AC = \sqrt{(x-0)^2 + (20-0)^2} = \sqrt{x^2 + 400}$
- $BC = \sqrt{(x-10)^2 + (20-0)^2} = \sqrt{(x-10)^2 + 400}$
6. **Check if there exists a point $C$ such that $AC + BC = 40$:**
- We need to find $x$ such that $\sqrt{x^2 + 400} + \sqrt{(x-10)^2 + 400} = 40$.
- To simplify, consider the case where $C$ is symmetrically placed above the midpoint of $AB$, i.e., at $(5, 20)$:
- $AC = BC = \sqrt{(5-0)^2 + (20-0)^2} = \sqrt{25 + 400} = \sqrt{425}$
- The perimeter in this case is $2\sqrt{425} + 10$.
7. **Calculate the minimal perimeter:**
- $2\sqrt{425} + 10 \approx 2 \times 20.62 + 10 = 41.24 + 10 = 51.24$ units, which is greater than $50$ units.
8. **Conclusion:**
- Since even in the symmetric case where the perimeter should be minimal, the perimeter exceeds $50$ units, there are no points $C$ that satisfy both the area and perimeter conditions simultaneously.
Thus, the answer is $\boxed{\textbf{(A) }0}$. $\blacksquare$
|
Points \(A = (2,8)\), \(B = (2,2)\), and \(C = (6,2)\) lie in the first quadrant and are vertices of triangle \(ABC\). Point \(D=(a,b)\) is also in the first quadrant, and together with \(A\), \(B\), and \(C\), forms quadrilateral \(ABCD\). The quadrilateral formed by joining the midpoints of \(\overline{AB}\), \(\overline{BC}\), \(\overline{CD}\), and \(\overline{DA}\) is a square. Additionally, the diagonal of this square has the same length as the side \(\overline{AB}\) of triangle \(ABC\). Find the sum of the coordinates of point \(D\).
A) 12
B) 14
C) 15
D) 16
E) 18
|
14
| |
In a convex polygon with 1992 sides, the minimum number of interior angles that are not acute is:
|
1989
| |
From the numbers $1,2,3, \cdots, 2014$, select 315 different numbers (order does not matter) to form an arithmetic sequence. Among these, the number of ways to form an arithmetic sequence that includes the number 1 is ___. The total number of ways to form an arithmetic sequence is ___.
|
5490
| |
A can of soup can feed $3$ adults or $5$ children. If there are $5$ cans of soup and $15$ children are fed, then how many adults would the remaining soup feed?
|
6
|
1. **Determine the number of cans needed to feed 15 children**:
Given that one can of soup can feed 5 children, we calculate the number of cans needed to feed 15 children as follows:
\[
\text{Number of cans} = \frac{15 \text{ children}}{5 \text{ children per can}} = 3 \text{ cans}
\]
2. **Calculate the remaining cans of soup**:
There are initially 5 cans of soup. After using 3 cans to feed the children, the remaining number of cans is:
\[
5 \text{ cans} - 3 \text{ cans} = 2 \text{ cans}
\]
3. **Determine how many adults can be fed with the remaining soup**:
Since each can of soup can feed 3 adults, the number of adults that can be fed with 2 cans is:
\[
3 \text{ adults per can} \times 2 \text{ cans} = 6 \text{ adults}
\]
Thus, the remaining soup can feed $\boxed{\textbf{(B)}\ 6}$ adults.
|
Integers $x$ and $y$ with $x>y>0$ satisfy $x+y+xy=104$. What is $x$?
|
34
| |
Mafia is a game where there are two sides: The village and the Mafia. Every night, the Mafia kills a person who is sided with the village. Every day, the village tries to hunt down the Mafia through communication, and at the end of every day, they vote on who they think the mafia are.**p6.** Patrick wants to play a game of mafia with his friends. If he has $10$ friends that might show up to play, each with probability $1/2$ , and they need at least $5$ players and a narrator to play, what is the probability that Patrick can play?**p7.** At least one of Kathy and Alex is always mafia. If there are $2$ mafia in a game with $6$ players, what is the probability that both Kathy and Alex are mafia?**p8.** Eric will play as mafia regardless of whether he is randomly selected to be mafia or not, and Euhan will play as the town regardless of what role he is selected as. If there are $2$ mafia and $6$ town, what is the expected value of the number of people playing as mafia in a random game with Eric and Euhan?**p9.** Ben is trying to cheat in mafia. As a mafia, he is trying to bribe his friend to help him win the game with his spare change. His friend will only help him if the change he has can be used to form at least $25$ different values. What is the fewest number of coins he can have to achieve this added to the fewest possible total value of those coins? He can only use pennies, nickels, dimes, and quarters.**p10.** Sammy, being the very poor mafia player he is, randomly shoots another player whenever he plays as the vigilante. What is the probability that the player he shoots is also not shot by the mafia nor saved by the doctor, if they both select randomly in a game with $8$ people? There are $2$ mafia, and they cannot select a mafia to be killed, and the doctor can save anyone.
PS. You should use hide for answers.
|
319/512
| |
In an apartment building, each floor in every entrance has the same number of apartments (more than one). Additionally, each entrance has the same number of floors. The number of floors is more than the number of apartments per floor, but less than the number of entrances. How many floors are in the building if there are a total of $715$ apartments?
|
11
| |
Three coplanar circles intersect as shown. What is the maximum number of points on the circles that a line passing through all three circles can touch?
[asy]import graph;
draw(Circle((-9,9),15));
draw(Circle((0,-9),15));
draw(Circle((9,9),15));
[/asy]
|
6
| |
Given that $0 < x < \frac{\pi}{2}$ and $\sin(2x - \frac{\pi}{4}) = -\frac{\sqrt{2}}{10}$, find the value of $\sin x + \cos x$.
|
\frac{2\sqrt{10}}{5}
| |
Suppose $a_i, b_i, c_i, i=1,2,\cdots ,n$, are $3n$ real numbers in the interval $\left [ 0,1 \right ].$ Define $$S=\left \{ \left ( i,j,k \right ) |\, a_i+b_j+c_k<1 \right \}, \; \; T=\left \{ \left ( i,j,k \right ) |\, a_i+b_j+c_k>2 \right \}.$$ Now we know that $\left | S \right |\ge 2018,\, \left | T \right |\ge 2018.$ Try to find the minimal possible value of $n$.
|
18
|
Suppose \( a_i, b_i, c_i \) for \( i = 1, 2, \ldots, n \) are \( 3n \) real numbers in the interval \([0, 1]\). Define the sets
\[
S = \{ (i, j, k) \mid a_i + b_j + c_k < 1 \}
\]
and
\[
T = \{ (i, j, k) \mid a_i + b_j + c_k > 2 \}.
\]
We are given that \( |S| \geq 2018 \) and \( |T| \geq 2018 \). We aim to find the minimal possible value of \( n \).
To establish a lower bound for \( n \), consider the projections of the sets \( S \) and \( T \) onto the coordinate planes. Note that \( S_{xy} \cap T_{xy} = \emptyset \), meaning that no pair \((a_i, b_j)\) can simultaneously satisfy \( a_i + b_j + c_k < 1 \) and \( a_i + b_j + c_k > 2 \) for any \( c_k \).
Thus, we have the inequalities:
\[
|S_{xy}| + |T_{xy}| \leq n^2, \quad |S_{yz}| + |T_{yz}| \leq n^2, \quad |S_{zx}| + |T_{zx}| \leq n^2.
\]
Applying the Projection Inequality and Hölder's Inequality, we obtain:
\[
2 \cdot 2018^{2/3} \leq |S|^{2/3} + |T|^{2/3} \leq |S_{xy}|^{1/3} \cdot |S_{yz}|^{1/3} \cdot |S_{zx}|^{1/3} + |T_{xy}|^{1/3} \cdot |T_{yz}|^{1/3} \cdot |T_{zx}|^{1/3} \leq (|S_{xy}| + |T_{xy}|)^{1/3} (|S_{yz}| + |T_{yz}|)^{1/3} (|S_{zx}| + |T_{zx}|)^{1/3} \leq n^2.
\]
Solving for \( n \), we get:
\[
2 \cdot 2018^{2/3} \leq n^2 \implies n \geq \sqrt{2} \cdot 2018^{1/3} \approx 17.8.
\]
Thus, the minimal possible value of \( n \) is:
\[
n \geq 18.
\]
The answer is: \boxed{18}.
|
In a certain region of the planet, seismic activity was studied. 80 percent of all days were quiet. The predictions of the devices promised a calm environment in 64 out of 100 cases, and in 70 percent of all cases where the day was calm, the predictions of the devices came true. What percentage of days with increased seismic activity are those in which the predictions did not match reality?
|
40
| |
In this final problem, a ball is again launched from the vertex of an equilateral triangle with side length 5. In how many ways can the ball be launched so that it will return again to a vertex for the first time after 2009 bounces?
|
502
|
We will use the same idea as in the previous problem. We first note that every vertex of a triangle can be written uniquely in the form $a(5,0)+b\left(\frac{5}{2}, \frac{5 \sqrt{3}}{2}\right)$, where $a$ and $b$ are non-negative integers. Furthermore, if a ball ends at $a(5,0)+b\left(\frac{5}{2}, \frac{5 \sqrt{3}}{2}\right)$, then it bounces off of a wall $2(a+b)-3$ times. Therefore, the possible directions that you can launch the ball in correspond to solutions to $2(a+b)-3=2009$, or $a+b=1006$. However, if $a$ and $b$ have a common factor, say, $k$, then the ball will pass through the vertex corresponding to $\frac{a}{k}$ and $\frac{b}{k}$ before it passes through the vertex corresponding to $a$ and $b$. Therefore, we must discount all such pairs $a, b$. This corresponds to when $a$ is even or $a=503$, so after removing these we are left with 502 remaining possible values of $a$, hence 502 possible directions in which to launch the ball.
|
At a certain meeting, a total of \(12k\) people attended. Each person greeted exactly \(3k+6\) other people. For any two individuals, the number of people who greeted both of them is the same. How many people attended the meeting?
|
36
| |
Let $a$ and $b$ be positive real numbers such that $a + 3b = 2.$ Find the minimum value of
\[\frac{2}{a} + \frac{4}{b}.\]
|
14
| |
What is the greatest common divisor of all the numbers $7^{n+2} + 8^{2n+1}$ for $n \in \mathbb{N}$?
|
57
| |
How many pairs of two-digit positive integers have a difference of 50?
|
40
| |
A small class of nine boys are to change their seating arrangement by drawing their new seat numbers from a box. After the seat change, what is the probability that there is only one pair of boys who have switched seats with each other and only three boys who have unchanged seats?
|
1/32
| |
Problem
Steve is piling $m\geq 1$ indistinguishable stones on the squares of an $n\times n$ grid. Each square can have an arbitrarily high pile of stones. After he finished piling his stones in some manner, he can then perform stone moves, defined as follows. Consider any four grid squares, which are corners of a rectangle, i.e. in positions $(i, k), (i, l), (j, k), (j, l)$ for some $1\leq i, j, k, l\leq n$ , such that $i<j$ and $k<l$ . A stone move consists of either removing one stone from each of $(i, k)$ and $(j, l)$ and moving them to $(i, l)$ and $(j, k)$ respectively,j or removing one stone from each of $(i, l)$ and $(j, k)$ and moving them to $(i, k)$ and $(j, l)$ respectively.
Two ways of piling the stones are equivalent if they can be obtained from one another by a sequence of stone moves.
How many different non-equivalent ways can Steve pile the stones on the grid?
|
\[
\binom{n+m-1}{m}^{2}
\]
|
Let the number of stones in row $i$ be $r_i$ and let the number of stones in column $i$ be $c_i$ . Since there are $m$ stones, we must have $\sum_{i=1}^n r_i=\sum_{i=1}^n c_i=m$
Lemma 1: If any $2$ pilings are equivalent, then $r_i$ and $c_i$ are the same in both pilings $\forall i$ .
Proof: We suppose the contrary. Note that $r_i$ and $c_i$ remain invariant after each move, therefore, if any of the $r_i$ or $c_i$ are different, they will remain different.
Lemma 2: Any $2$ pilings with the same $r_i$ and $c_i$ $\forall i$ are equivalent.
Proof: Suppose piling 1 and piling 2 not the same piling. Call a stone in piling 1 wrong if the stone occupies a position such that there are more stones in that position in piling 1 than piling 2. Similarly define a wrong stone in piling 2. Let a wrong stone be at $(a, b)$ in piling 1. Since $c_b$ is the same for both pilings, we must have a wrong stone in piling 2 at column b, say at $(c, b)$ , such that $c\not = a$ . Similarly, we must have a wrong stone in piling 1 at row c, say at $(c, d)$ where $d \not = b$ . Clearly, making the move $(a,b);(c,d) \implies (c,b);(a,d)$ in piling 1 decreases the number of wrong stones in piling 1. Therefore, the number of wrong stones in piling 1 must eventually be $0$ after a sequence of moves, so piling 1 and piling 2 are equivalent.
Lemma 3: Given the sequences $g_i$ and $h_i$ such that $\sum_{i=1}^n g_i=\sum_{i=1}^n h_i=m$ and $g_i, h_i\geq 0 \forall i$ , there is always a piling that satisfies $r_i=g_i$ and $c_i=h_i$ $\forall i$ .
Proof: We take the lowest $i$ , $j$ , such that $g_i, h_j >0$ and place a stone at $(i, j)$ , then we subtract $g_i$ and $h_j$ by $1$ each, until $g_i$ and $h_i$ become $0$ $\forall i$ , which will happen when $m$ stones are placed, because $\sum_{i=1}^n g_i$ and $\sum_{i=1}^n h_i$ are both initially $m$ and decrease by $1$ after each stone is placed. Note that in this process $r_i+g_i$ and $c_i+h_i$ remains invariant, thus, the final piling satisfies the conditions above.
By the above lemmas, the number of ways to pile is simply the number of ways to choose the sequences $r_i$ and $c_i$ such that $\sum_{i=1}^n r_i=\sum_{i=1}^n c_i=m$ and $r_i, c_i \geq 0 \forall i$ . By stars and bars, the number of ways is $\binom{n+m-1}{m}^{2}$ .
Solution by Shaddoll
|
How many strictly positive integers less than or equal to 120 are not divisible by 3, 5, or 7?
|
54
| |
A $3 \times 3$ square is partitioned into $9$ unit squares. Each unit square is painted either white or black with each color being equally likely, chosen independently and at random. The square is then rotated $90^{\circ}$ clockwise about its center, and every white square in a position formerly occupied by a black square is painted black. The colors of all other squares are left unchanged. What is the probability the grid is now entirely black?
|
\frac{49}{512}
|
1. **Identify the invariant**: The center square remains unchanged after a $90^\circ$ rotation. Therefore, for the entire grid to be black after the process, the center square must initially be black. The probability of the center square being black is $\frac{1}{2}$.
2. **Consider the rotation effect on other squares**: The rotation affects the positions of the other eight squares. Specifically, each corner square rotates into the position of another corner square, and each edge square rotates into the position of another edge square.
3. **Probability for edge squares**: There are four edge squares, and each pair of opposite edge squares (top-bottom and left-right) must be black to ensure they remain black after rotation. The probability that a specific pair of opposite edge squares is black is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. Since there are two independent pairs (top-bottom and left-right), the probability that both pairs are black is $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$.
4. **Probability for corner squares**: Similarly, there are four corner squares, and each pair of diagonally opposite corner squares must be black. The probability that one pair of diagonally opposite corners is black is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$. Since the corners rotate among themselves, the probability that all corners are black (ensuring they remain black after rotation) is $\frac{1}{16}$.
5. **Combine probabilities**: The probability that the center, all edge squares, and all corner squares are black is the product of their independent probabilities:
\[
\left(\frac{1}{2}\right) \times \left(\frac{1}{16}\right) \times \left(\frac{1}{16}\right) = \frac{1}{512}
\]
6. **Adjust for overcounting**: We need to consider the cases where the configuration of black squares is such that after rotation, any white square that moves into a black square's position gets painted black, resulting in an all-black grid. This includes configurations where not all squares were initially black but became black due to the painting rule. This adjustment is complex and requires careful counting or symmetry arguments, which were not fully detailed in the initial solution.
7. **Final probability**: The final probability, considering all configurations that result in an all-black grid after rotation and painting, is given as $\boxed{\textbf{(A)}\ \frac{49}{512}}$. This result implies that there are 49 configurations out of 512 possible initial configurations that satisfy the condition of the problem.
This solution involves a deeper analysis of the rotation and painting effects, ensuring that all cases where the grid ends up all black are counted, including those where the painting rule changes the color of some squares.
|
Simplify the fraction by rationalizing the denominator: $$\frac{4}{\sqrt{108}+2\sqrt{12}+2\sqrt{27}}.$$
|
\frac{\sqrt{3}}{12}
| |
The difference of the roots of $x^2-7x-9=0$ is:
|
\sqrt{85}
|
1. **Identify the coefficients of the quadratic equation**: Given the quadratic equation $x^2 - 7x - 9 = 0$, we can identify the coefficients as $a = 1$, $b = -7$, and $c = -9$.
2. **Apply Vieta's formulas**: Vieta's formulas tell us that for a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots $r_1 + r_2 = -\frac{b}{a}$ and the product of the roots $r_1r_2 = \frac{c}{a}$. Plugging in our coefficients:
- $r_1 + r_2 = -\frac{-7}{1} = 7$
- $r_1r_2 = \frac{-9}{1} = -9$
3. **Calculate the square of the difference of the roots**: We know from algebra that $(r_1 - r_2)^2 = (r_1 + r_2)^2 - 4r_1r_2$. Substituting the values from Vieta's formulas:
- $(r_1 - r_2)^2 = 7^2 - 4(-9)$
- $(r_1 - r_2)^2 = 49 + 36$
- $(r_1 - r_2)^2 = 85$
4. **Find the difference of the roots**: To find $r_1 - r_2$, we take the square root of $(r_1 - r_2)^2$:
- $r_1 - r_2 = \sqrt{85}$
5. **Conclusion**: The difference of the roots of the quadratic equation $x^2 - 7x - 9 = 0$ is $\sqrt{85}$.
Thus, the correct answer is $\boxed{\textbf{(E)}\ \sqrt{85}}$.
|
Find all natural numbers \( n \) such that, when writing the numbers \( n^3 \) and \( n^4 \) side by side in decimal notation, each of the ten digits appears exactly once.
(Former Yugoslavia Mathematical Olympiad, 1983)
|
18
| |
For the Shanghai World Expo, 20 volunteers were recruited, with each volunteer assigned a unique number from 1 to 20. If four individuals are to be selected randomly from this group and divided into two teams according to their numbers, with the smaller numbers in one team and the larger numbers in another, what is the total number of ways to ensure that both volunteers number 5 and number 14 are selected and placed on the same team?
|
21
| |
A magician and their assistant plan to perform a trick. The spectator writes a sequence of $N$ digits on a board. The magician's assistant then covers two adjacent digits with a black dot. Next, the magician enters and has to guess both covered digits (including the order in which they are arranged). What is the smallest $N$ for which the magician and the assistant can arrange the trick so that the magician can always correctly guess the covered digits?
|
101
| |
The diagonal of a particular square is 5 inches. The diameter of a particular circle is also 5 inches. By how many square inches is the area of the circle greater than the area of square? Express your answer as a decimal to the nearest tenth. [asy]
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
draw((2,0)--(0,2));
draw(circle((4.5,1),1.414));
draw((2+3.5,0)--(0+3.5,2));
[/asy]
|
7.1
| |
Sanitation workers plan to plant 7 trees in a row on one side of a road, choosing only from plane trees and willow trees. What is the number of planting methods where no two adjacent trees are both willows?
|
34
| |
Each of the lateral edges of a pyramid is equal to 269/32. The base of the pyramid is a triangle with sides 13, 14, 15. Find the volume of the pyramid.
|
483/8
| |
The number of terms in an A.P. (Arithmetic Progression) is even. The sum of the odd and even-numbered terms are 24 and 30, respectively. If the last term exceeds the first by 10.5, the number of terms in the A.P. is
|
8
|
1. **Define the variables:**
Let $a$ be the first term, $n$ be the number of terms, and $d$ be the common difference in the arithmetic progression (A.P.). The last term of the A.P. can be expressed as $a + (n-1)d$.
2. **Set up the equation for the difference between the last and first term:**
Given that the last term exceeds the first by 10.5, we can write:
\[
a + (n-1)d - a = 10.5
\]
Simplifying this, we get:
\[
(n-1)d = 10.5
\]
3. **Use the information about the sums of odd and even-numbered terms:**
The sum of the odd-numbered terms is 24 and the sum of the even-numbered terms is 30. Since the number of terms $n$ is even, there are $\frac{n}{2}$ odd-numbered terms and $\frac{n}{2}$ even-numbered terms.
The sum of an arithmetic sequence can be calculated using the formula:
\[
S = \frac{\text{number of terms}}{2} \times (\text{first term} + \text{last term})
\]
For the odd-numbered terms, the first term is $a$ and the last term is $a + (n-2)d$. Thus, the sum of the odd-numbered terms is:
\[
\frac{n}{2} \times (a + a + (n-2)d) = 24
\]
Simplifying, we get:
\[
n(a + (n-2)d) = 48
\]
For the even-numbered terms, the first term is $a + d$ and the last term is $a + (n-1)d$. Thus, the sum of the even-numbered terms is:
\[
\frac{n}{2} \times (a + d + a + (n-1)d) = 30
\]
Simplifying, we get:
\[
n(a + (n-1)d + d) = 60
\]
4. **Solve the system of equations:**
From $(n-1)d = 10.5$, we can express $d$ as:
\[
d = \frac{10.5}{n-1}
\]
Substituting $d$ into the equation $n(a + (n-2)d) = 48$ and $n(a + (n-1)d + d) = 60$, we need to solve these equations simultaneously. However, there seems to be an error in the original solution's manipulation of these equations. Let's correct this:
From $(n-1)d = 10.5$, we have:
\[
d = \frac{10.5}{n-1}
\]
Substituting $d$ into $n(a + (n-2)d) = 48$ and simplifying, we find:
\[
n(a + (n-2)\frac{10.5}{n-1}) = 48
\]
This equation needs to be solved along with the similar equation for the even-numbered terms. However, the original solution seems to have simplified the process by assuming the sums of the terms directly relate to $d$ and $n$. Let's recheck the calculation:
\[
(n-1)d = 10.5 \quad \text{and} \quad n \cdot d = 12
\]
Solving these, we find $d = 1.5$ and $n = 8$.
5. **Conclusion:**
Thus, the number of terms in the arithmetic sequence is $\boxed{\textbf{(E)}\ 8}$.
|
Given \(\alpha \in (0, \pi)\), if \(\sin \alpha + \cos \alpha = \frac{\sqrt{3}}{3}\), then \(\cos^2 \alpha - \sin^2 \alpha = \)
|
-\frac{\sqrt{5}}{3}
| |
Compute \((1+i^{-100}) + (2+i^{-99}) + (3+i^{-98}) + \cdots + (101+i^0) + (102+i^1) + \cdots + (201+i^{100})\).
|
20302
| |
Given two real numbers \( p > 1 \) and \( q > 1 \) such that \( \frac{1}{p} + \frac{1}{q} = 1 \) and \( pq = 9 \), what is \( q \)?
|
\frac{9 + 3\sqrt{5}}{2}
| |
Five different products, A, B, C, D, and E, are to be arranged in a row on a shelf. Products A and B must be placed together, while products C and D cannot be placed together. How many different arrangements are possible?
|
36
| |
In a convex 1950-gon, all diagonals are drawn. They divide it into polygons. Consider the polygon with the largest number of sides. What is the greatest number of sides it can have?
|
1949
| |
Cards numbered 1 and 2 must be placed into the same envelope, and three cards are left to be placed into two remaining envelopes, find the total number of different methods.
|
18
| |
In how many different ways can a chess king move from square $e1$ to square $h5$, if it is only allowed to move one square to the right, upward, or diagonally right-upward?
|
129
| |
Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number?
|
126
|
As shown above, we have $1000x+y=9xy$, so $1000/y=9-1/x$. $1000/y$ must be just a little bit smaller than 9, so we find $y=112$, $x=14$, and the solution is $\boxed{126}$.
|
Evaluate the expression: $\left(2\left(3\left(2\left(3\left(2\left(3 \times (2+1) \times 2\right)+2\right)\times 2\right)+2\right)\times 2\right)+2\right)$.
|
5498
| |
The diagram shows five circles of the same radius touching each other. A square is drawn so that its vertices are at the centres of the four outer circles. What is the ratio of the area of the shaded parts of the circles to the area of the unshaded parts of the circles?
|
2:3
| |
A block of iron solidifies from molten iron, and its volume reduces by $\frac{1}{34}$. Then, if this block of iron melts back into molten iron (with no loss in volume), by how much does its volume increase?
|
\frac{1}{33}
| |
Let $z$ be a complex number. In the complex plane, the distance from $z$ to 1 is 2 , and the distance from $z^{2}$ to 1 is 6 . What is the real part of $z$ ?
|
\frac{5}{4}
|
Note that we must have $|z-1|=2$ and \left|z^{2}-1\right|=6$, so $|z+1|=\frac{\left|z^{2}-1\right|}{|z-1|}=3$. Thus, the distance from $z$ to 1 in the complex plane is 2 and the distance from $z$ to -1 in the complex plane is 3 . Thus, $z, 1,-1$ form a triangle with side lengths $2,3,3$. The area of a triangle with sides $2,2,3$ can be computed to be \frac{3 \sqrt{7}}{4}$ by standard techniques, so the length of the altitude from $z$ to the real axis is \frac{3 \sqrt{7}}{4} \cdot \frac{2}{2}=\frac{3 \sqrt{7}}{4}$. The distance between 1 and the foot from $z$ to the real axis is \sqrt{2^{2}-\left(\frac{3 \sqrt{7}}{4}\right)^{2}}=\frac{1}{4}$ by the Pythagorean Theorem. It is clear that $z$ has positive imaginary part as the distance from $z$ to -1 is greater than the distance from $z$ to 1 , so the distance from 0 to the foot from $z$ to the real axis is $1+\frac{1}{4}=\frac{5}{4}$. This is exactly the real part of $z$ that we are trying to compute.
|
A room measures 16 feet by 12 feet and includes a column with a square base of 2 feet on each side. Find the area in square inches of the floor that remains uncovered by the column.
|
27,072
| |
In triangle $ABC$, it is known that $BC = 1$, $\angle B = \frac{\pi}{3}$, and the area of $\triangle ABC$ is $\sqrt{3}$. The length of $AC$ is __________.
|
\sqrt{13}
| |
Given a parallelogram \\(ABCD\\) where \\(AD=2\\), \\(∠BAD=120^{\\circ}\\), and point \\(E\\) is the midpoint of \\(CD\\), if \\( \overrightarrow{AE} \cdot \overrightarrow{BD}=1\\), then \\( \overrightarrow{BD} \cdot \overrightarrow{BE}=\\) \_\_\_\_\_\_.
|
13
| |
Find the largest value of $t$ such that \[\frac{13t^2 - 34t + 12}{3t - 2 } + 5t = 6t - 1.\]
|
\frac{5}{2}
| |
A rectangle that is inscribed in a larger rectangle (with one vertex on each side) is called unstuck if it is possible to rotate (however slightly) the smaller rectangle about its center within the confines of the larger. Of all the rectangles that can be inscribed unstuck in a 6 by 8 rectangle, the smallest perimeter has the form $\sqrt{N}\,$, for a positive integer $N\,$. Find $N\,$.
|
448
|
Note that the diagonal of the rectangle with minimum perimeter must have the diagonal along the middle segment of length 8 of the rectangle (any other inscribed rectangle can be rotated a bit, then made smaller; this one can't because then the rectangle cannot be inscribed since its longest diagonal is less than 8 in length). Then since a rectangle must have right angles, we draw a circle of radius 4 around the center of the rectangle. Picking the two midpoints on the sides of length 6 and opposite intersection points on the segments of length 8, we form a rectangle. Let $a$ and $b$ be the sides of the rectangle. Then $ab = 3(8) = 24$ since both are twice the area of the same right triangle, and $a^2+b^2 = 64$. So $(a+b)^2 = 64+2(24) = 112$, so $2(a+b) = \sqrt{\boxed{448}}$.
|
What is the largest base-4 number that has four digits? Express your answer in base 10.
|
255
| |
Parallelogram $ABCD$ has $AB=CD=6$ and $BC=AD=10$ , where $\angle ABC$ is obtuse. The circumcircle of $\triangle ABD$ intersects $BC$ at $E$ such that $CE=4$ . Compute $BD$ .
|
4\sqrt{6}
| |
Given a triangle $ABC$ with $\angle B = \frac{\pi}{3}$,
(I) if $AB=8\sqrt{3}$ and $AC=12$, find the area of $\triangle ABC$;
(II) if $AB=4$ and $\vec{BM} = \vec{MN} = \vec{NC}$ with $AN=2\sqrt{3}BM$, find the length of $AM$.
|
\sqrt{13}
| |
Given positive integers $n, k$ such that $n\ge 4k$, find the minimal value $\lambda=\lambda(n,k)$ such that for any positive reals $a_1,a_2,\ldots,a_n$, we have
\[ \sum\limits_{i=1}^{n} {\frac{{a}_{i}}{\sqrt{{a}_{i}^{2}+{a}_{{i}+{1}}^{2}+{\cdots}{{+}}{a}_{{i}{+}{k}}^{2}}}}
\le \lambda\]
Where $a_{n+i}=a_i,i=1,2,\ldots,k$
|
n - k
|
Given positive integers \( n \) and \( k \) such that \( n \geq 4k \), we aim to find the minimal value \( \lambda = \lambda(n, k) \) such that for any positive reals \( a_1, a_2, \ldots, a_n \), the following inequality holds:
\[
\sum_{i=1}^{n} \frac{a_i}{\sqrt{a_i^2 + a_{i+1}^2 + \cdots + a_{i+k}^2}} \leq \lambda,
\]
where \( a_{n+i} = a_i \) for \( i = 1, 2, \ldots, k \).
To determine the minimal value of \( \lambda \), consider the construction where \( a_i = q^i \) for \( 0 < q < 1 \) and let \( q \to 0 \). Then, for \( 1 \leq i \leq n-k \),
\[
\frac{a_i}{\sqrt{a_i^2 + a_{i+1}^2 + \cdots + a_{i+k}^2}} = \frac{1}{\sqrt{1 + q + \cdots + q^k}} \to 1.
\]
For \( n-k < i \leq n \),
\[
\frac{a_i}{\sqrt{a_i^2 + a_{i+1}^2 + \cdots + a_{i+k}^2}} = \frac{q^{i-1}}{\sqrt{q^{2(i-1)} + \cdots + q^{2(n-1)} + 1 + q^2 + \cdots + q^{2(i+k-n-1)}}} \to 0.
\]
Thus,
\[
\sum_{i=1}^n \frac{a_i}{\sqrt{a_i^2 + a_{i+1}^2 + \cdots + a_{i+k}^2}} \to n-k,
\]
implying that \( \lambda \geq n-k \).
To prove that \( \lambda = n-k \) is indeed the minimal value, we consider the case when \( n = 4 \) and \( k = 1 \). Squaring both sides, we need to show:
\[
\frac{a_1^2}{a_1^2 + a_2^2} + \frac{a_2^2}{a_2^2 + a_3^2} + \frac{a_3^2}{a_3^2 + a_4^2} + \frac{a_4^2}{a_4^2 + a_1^2} + \frac{2a_1a_2}{\sqrt{(a_1^2 + a_2^2)(a_2^2 + a_3^2)}} + \frac{2a_2a_3}{\sqrt{(a_2^2 + a_3^2)(a_3^2 + a_4^2)}} + \frac{2a_3a_4}{\sqrt{(a_3^2 + a_4^2)(a_4^2 + a_1^2)}} + \frac{2a_4a_1}{\sqrt{(a_4^2 + a_1^2)(a_1^2 + a_2^2)}} + \frac{2a_1a_3}{\sqrt{(a_1^2 + a_3^2)(a_3^2 + a_4^2)}} + \frac{2a_2a_4}{\sqrt{(a_2^2 + a_3^2)(a_4^2 + a_1^2)}} \leq 9.
\]
Using the Cauchy-Schwarz inequality and other properties of binomial coefficients, we can generalize this result for \( n = 4k \) and prove by induction for \( n > 4k \).
Therefore, the minimal value \( \lambda \) is:
\[
\lambda = n - k.
\]
The answer is: \boxed{n - k}.
|
Azmi has four blocks, each in the shape of a rectangular prism and each with dimensions $2 imes 3 imes 6$. She carefully stacks these four blocks on a flat table to form a tower that is four blocks high. What is the number of possible heights for this tower?
|
14
|
The height of each block is 2, 3 or 6. Thus, the total height of the tower of four blocks is the sum of the four heights, each of which equals 2, 3 or 6. If 4 blocks have height 6, the total height equals $4 imes 6=24$. If 3 blocks have height 6, the fourth block has height 3 or 2. Therefore, the possible heights are $3 imes 6+3=21$ and $3 imes 6+2=20$. If 2 blocks have height 6, the third and fourth blocks have height 3 or 2. Therefore, the possible heights are $2 imes 6+3+3=18$ and $2 imes 6+3+2=17$ and $2 imes 6+2+2=16$. If 1 block has height 6, the second, third and fourth blocks have height 3 or 2. Therefore, the possible heights are $6+3+3+3=15$ and $6+3+3+2=14$ and $6+3+2+2=13$ and $6+2+2+2=12$. If no blocks have height 6, the possible heights are $3+3+3+3=12$ and $3+3+3+2=11$ and $3+3+2+2=10$ and $3+2+2+2=9$ and $2+2+2+2=8$. The possible heights are thus $24, 21, 20, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8$. There are 14 possible heights.
|
Is the following number rational or irrational?
$$
\sqrt[3]{2016^{2} + 2016 \cdot 2017 + 2017^{2} + 2016^{3}} ?
$$
|
2017
| |
Elective 4-4: Coordinate System and Parametric Equations
In the Cartesian coordinate system $xOy$, the parametric equation of line $l_1$ is $\begin{cases}x=2+t \\ y=kt\end{cases}$ (where $t$ is the parameter), and the parametric equation of line $l_2$ is $\begin{cases}x=-2+m \\ y= \frac{m}{k}\end{cases}$ (where $m$ is the parameter). Let the intersection point of $l_1$ and $l_2$ be $P$. When $k$ changes, the trajectory of $P$ is curve $C$.
(1) Write the general equation of $C$;
(2) Establish a polar coordinate system with the origin as the pole and the positive half-axis of $x$ as the polar axis. Let line $l_3: \rho(\cos \theta +\sin \theta)− \sqrt{2} =0$, and $M$ be the intersection point of $l_3$ and $C$. Find the polar radius of $M$.
|
\sqrt{5}
| |
A small town has fewer than 6000 inhabitants. We know that there are $10\%$ more girls than boys among the children, and $15\%$ more men than women among the adults. There are $20\%$ more children than adults in the town.
How many people live in the town?
|
3311
| |
Given point \( A(2,0) \), point \( B \) lies on the curve \( y = \sqrt{1 - x^2} \), and the triangle \( \triangle ABC \) is an isosceles right triangle with \( A \) as the right angle vertex. Determine the maximum value of \( |OC| \).
|
2\sqrt{2} + 1
| |
Evaluate $x^2y^3z$ if $x = \frac13$, $y = \frac23$, and $z = -9$.
|
-\frac{8}{27}
|
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