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What is the greatest number of consecutive integers whose sum is $45?$
|
90
|
1. **Understanding the Problem:**
We need to find the greatest number of consecutive integers that sum up to $45$. These integers can be positive, negative, or zero.
2. **Exploring Small Cases:**
- If we consider only positive integers starting from $1$, the sum of the first few consecutive integers is:
\[
1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}
\]
Setting this equal to $45$ and solving for $n$:
\[
\frac{n(n+1)}{2} = 45 \implies n^2 + n - 90 = 0
\]
Solving this quadratic equation, we find $n = 9$ as a solution. However, this does not necessarily mean it is the maximum number of terms.
3. **Considering Negative and Zero Integers:**
- We can also include negative integers and zero. For example, the sequence $-44, -43, \ldots, 44, 45$ sums to $45$ because all pairs $(-k, k)$ for $k = 1$ to $44$ cancel each other out, leaving only $45$.
- The total number of integers in this sequence is $45 - (-44) + 1 = 90$.
4. **Generalizing the Sequence:**
- Let's consider a general sequence of consecutive integers $a, a+1, \ldots, a+(N-1)$, where $N$ is the number of terms. The sum of these terms can be expressed as:
\[
S = a + (a+1) + \cdots + (a+N-1) = Na + \frac{(N-1)N}{2}
\]
Setting $S = 45$, we get:
\[
Na + \frac{(N-1)N}{2} = 45
\]
Rearranging, we find:
\[
2Na + (N-1)N = 90 \implies N(2a + N - 1) = 90
\]
5. **Maximizing $N$:**
- We need to maximize $N$ such that $N(2a + N - 1) = 90$. Since $N$ must be a divisor of $90$, we consider the divisors of $90$: $1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90$.
- The largest divisor is $90$, which corresponds to the sequence $-44, -43, \ldots, 44, 45$.
6. **Conclusion:**
- The greatest number of consecutive integers whose sum is $45$ is $90$.
Thus, the answer is $\boxed{\textbf{(D) } 90}$.
|
A charity sells $140$ benefit tickets for a total of $2001$. Some tickets sell for full price (a whole dollar amount), and the rest sells for half price. How much money is raised by the full-price tickets?
|
$782
|
Let's denote the number of full-price tickets as $f$ and the number of half-price tickets as $h$. Let $p$ be the price of a full-price ticket. Then, the price of a half-price ticket is $\frac{p}{2}$. The total number of tickets sold is $140$, and the total revenue is $2001$ dollars. We can set up the following equations based on this information:
1. **Total number of tickets:**
\[ f + h = 140 \]
2. **Total revenue:**
\[ f \cdot p + h \cdot \frac{p}{2} = 2001 \]
We can solve these equations step-by-step:
#### Step 1: Express $h$ in terms of $f$
From equation (1), we have:
\[ h = 140 - f \]
#### Step 2: Substitute $h$ in the revenue equation
Substituting the expression for $h$ from Step 1 into equation (2), we get:
\[ f \cdot p + (140 - f) \cdot \frac{p}{2} = 2001 \]
#### Step 3: Simplify the equation
Expanding and simplifying the equation from Step 2, we have:
\[ f \cdot p + 70p - \frac{f \cdot p}{2} = 2001 \]
\[ 2f \cdot p + 140p - f \cdot p = 4002 \] (Multiplying through by 2 to clear the fraction)
\[ f \cdot p + 140p = 4002 \]
#### Step 4: Solve for $p$
Rearranging the equation from Step 3, we get:
\[ f \cdot p = 4002 - 140p \]
Since $p$ must be a whole number, and $f \cdot p$ must also be a whole number, we need to find a value of $p$ such that $4002 - 140p$ is a multiple of $p$. We can test integer values of $p$ that are factors of $4002$.
#### Step 5: Factorize $4002$ and test possible values of $p$
The prime factorization of $4002$ is:
\[ 4002 = 2 \cdot 3 \cdot 23 \cdot 29 \]
We test divisors of $4002$ that are reasonable for the price of a ticket. We find that $p = 23$ works because:
\[ f \cdot p = 4002 - 140 \cdot 23 = 4002 - 3220 = 782 \]
#### Step 6: Verify the number of full-price tickets
Using $p = 23$, we find $f$:
\[ f = \frac{782}{23} = 34 \]
#### Conclusion
The amount raised by the full-price tickets is $f \cdot p = 34 \cdot 23 = \boxed{\textbf{(A) } \textdollar 782}$.
|
Suppose \(AB = 1\), and the slanted segments form an angle of \(45^\circ\) with \(AB\). There are \(n\) vertices above \(AB\).
What is the length of the broken line?
|
\sqrt{2}
| |
Let $ABCD$ be a convex quadrilateral such that $AB + BC = 2021$ and $AD = CD$ . We are also given that $\angle ABC = \angle CDA = 90^o$ . Determine the length of the diagonal $BD$ .
|
\frac{2021 \sqrt{2}}{2}
| |
(1) Find the domain of the function $y= \sqrt {\sin x}+ \sqrt { \frac{1}{2}-\cos x}$.
(2) Find the range of the function $y=\cos ^{2}x-\sin x$, where $x\in\left[-\frac {\pi}{4}, \frac {\pi}{4}\right]$.
|
\frac{1-\sqrt{2}}{2}
| |
For what real values of $K$ does $x = K^2 (x-1)(x-2)$ have real roots?
|
all
|
1. **Rewrite the given equation**: Start by rewriting the equation $x = K^2 (x-1)(x-2)$ in a standard quadratic form:
\[
x = K^2x^2 - 3K^2x + 2K^2.
\]
Rearranging terms, we get:
\[
K^2x^2 - (3K^2 + 1)x + 2K^2 = 0.
\]
2. **Identify coefficients**: In the quadratic equation $ax^2 + bx + c = 0$, identify $a$, $b$, and $c$:
\[
a = K^2, \quad b = -(3K^2 + 1), \quad c = 2K^2.
\]
3. **Calculate the discriminant**: The discriminant $\Delta$ of a quadratic equation, which determines the nature of the roots, is given by:
\[
\Delta = b^2 - 4ac.
\]
Substituting the values of $a$, $b$, and $c$, we have:
\[
\Delta = (-(3K^2 + 1))^2 - 4(K^2)(2K^2).
\]
Simplifying further:
\[
\Delta = (9K^4 + 6K^2 + 1) - 8K^4 = K^4 + 6K^2 + 1.
\]
4. **Analyze the discriminant**: We need to check when $\Delta \geq 0$:
\[
K^4 + 6K^2 + 1 \geq 0.
\]
This is a quadratic in terms of $K^2$. Let $u = K^2$. Then the inequality becomes:
\[
u^2 + 6u + 1 \geq 0.
\]
The discriminant of this quadratic in $u$ is:
\[
\Delta_u = 6^2 - 4 \cdot 1 \cdot 1 = 36 - 4 = 32.
\]
Since $\Delta_u > 0$, the quadratic has two real roots. However, since $u = K^2 \geq 0$, we are only interested in non-negative values of $u$. The quadratic $u^2 + 6u + 1$ is always positive because it opens upwards (coefficient of $u^2$ is positive) and the vertex is negative:
\[
u = -\frac{b}{2a} = -\frac{6}{2} = -3.
\]
Since the vertex is at $u = -3$, the quadratic is positive for all $u \geq 0$.
5. **Conclusion**: Since $K^4 + 6K^2 + 1 \geq 0$ for all real $K$, the original quadratic equation has real roots for all real values of $K$.
\[
\boxed{\textbf{(E)}\ \text{all}}
\]
|
Simplify: $$\sqrt[3]{5488000}$$
|
176.4
| |
The sum of the base-$10$ logarithms of the divisors of $6^n$ is $540$. What is $n$?
A) 9
B) 10
C) 11
D) 12
E) 13
|
10
| |
There are 4 distinct balls and 4 distinct boxes. All the balls need to be placed into the boxes.
(1) If exactly one box remains empty, how many different arrangements are possible?
(2) If exactly two boxes remain empty, how many different arrangements are possible?
|
84
| |
What is the area of the portion of the circle defined by \(x^2 - 10x + y^2 = 9\) that lies above the \(x\)-axis and to the left of the line \(y = x-5\)?
|
4.25\pi
| |
What percent of square $EFGH$ is shaded? All angles in the diagram are right angles. [asy]
import graph;
defaultpen(linewidth(0.7));
xaxis(0,8,Ticks(1.0,NoZero));
yaxis(0,8,Ticks(1.0,NoZero));
fill((0,0)--(2,0)--(2,2)--(0,2)--cycle);
fill((3,0)--(5,0)--(5,5)--(0,5)--(0,3)--(3,3)--cycle);
fill((6,0)--(7,0)--(7,7)--(0,7)--(0,6)--(6,6)--cycle);
label("$E$",(0,0),SW);
label("$F$",(0,7),N);
label("$G$",(7,7),NE);
label("$H$",(7,0),E);
[/asy]
|
67\%
| |
At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At $4{:}00$ the temperature in Minneapolis has fallen by $5$ degrees while the temperature in St. Louis has risen by $3$ degrees, at which time the temperatures in the two cities differ by $2$ degrees. What is the product of all possible values of $N?$
|
60
|
1. **Define the variables:**
Let $M$ represent the temperature in Minneapolis at noon, and $L$ represent the temperature in St. Louis at noon. Given that Minneapolis is $N$ degrees warmer than St. Louis at noon, we can express this relationship as:
\[
M = L + N
\]
2. **Temperature changes by 4:00 PM:**
By 4:00 PM, the temperature in Minneapolis falls by 5 degrees, and the temperature in St. Louis rises by 3 degrees. Therefore, the temperatures at 4:00 PM are:
\[
M_{4:00} = M - 5 = (L + N) - 5 = L + N - 5
\]
\[
L_{4:00} = L + 3
\]
3. **Difference in temperatures at 4:00 PM:**
The problem states that the difference in temperatures at 4:00 PM is 2 degrees. We can set up the equation:
\[
|M_{4:00} - L_{4:00}| = 2
\]
Substituting the expressions for $M_{4:00}$ and $L_{4:00}$, we get:
\[
|(L + N - 5) - (L + 3)| = 2
\]
Simplifying inside the absolute value:
\[
|L + N - 5 - L - 3| = 2
\]
\[
|N - 8| = 2
\]
4. **Solve the absolute value equation:**
The absolute value equation $|N - 8| = 2$ has two solutions:
\[
N - 8 = 2 \quad \text{or} \quad N - 8 = -2
\]
Solving these equations:
\[
N = 10 \quad \text{and} \quad N = 6
\]
5. **Calculate the product of all possible values of $N$:**
The product of the possible values of $N$ is:
\[
10 \times 6 = 60
\]
Thus, the product of all possible values of $N$ is $\boxed{60}$.
|
Ten adults enter a room, remove their shoes, and toss their shoes into a pile. Later, a child randomly pairs each left shoe with a right shoe without regard to which shoes belong together. The probability that for every positive integer $k<5$, no collection of $k$ pairs made by the child contains the shoes from exactly $k$ of the adults is $\frac{m}{n}$, where m and n are relatively prime positive integers. Find $m+n.$
|
28
|
Label the left shoes be $L_1,\dots, L_{10}$ and the right shoes $R_1,\dots, R_{10}$. Notice that there are $10!$ possible pairings.
Let a pairing be "bad" if it violates the stated condition. We would like a better condition to determine if a given pairing is bad.
Note that, in order to have a bad pairing, there must exist a collection of $k<5$ pairs that includes both the left and the right shoes of $k$ adults; in other words, it is bad if it is possible to pick $k$ pairs and properly redistribute all of its shoes to exactly $k$ people.
Thus, if a left shoe is a part of a bad collection, its corresponding right shoe must also be in the bad collection (and vice versa). To search for bad collections, we can start at an arbitrary right shoe (say $R_1$), check the left shoe it is paired with (say $L_i$), and from the previous observation, we know that $R_i$ must also be in the bad collection. Then we may check the left shoe paired with $R_i$, find its counterpart, check its left pair, find its counterpart, etc. until we have found $L_1$. We can imagine each right shoe "sending" us to another right shoe (via its paired left shoe) until we reach the starting right shoe, at which point we know that we have found a bad collection if we have done this less than $5$ times.
Effectively we have just traversed a cycle. (Note: This is the cycle notation of permutations.) The only condition for a bad pairing is that there is a cycle with length less than $5$; thus, we need to count pairings where every cycle has length at least $5$. This is only possible if there is a single cycle of length $10$ or two cycles of length $5$.
The first case yields $9!$ working pairings. The second case yields $\frac{{10\choose 5}}{2}\cdot{4!}^2=\frac{10!}{2 \cdot {5!}^2} \cdot {4!}^2$ pairings. Therefore, taking these cases out of a total of $10!$, the probability is $\frac{1}{10}+\frac{1}{50} = \frac{3}{25}$, for an answer of $\boxed{028}$.
|
If $a$ and $b$ are positive real numbers such that $a \cdot 2^{b}=8$ and $a^{b}=2$, compute $a^{\log _{2} a} 2^{b^{2}}$.
|
128
|
Taking $\log _{2}$ of both equations gives $\log _{2} a+b=3$ and $b \log _{2} a=1$. We wish to find $a^{\log _{2} a} 2^{b^{2}}$; taking $\log _{2}$ of that gives $\left(\log _{2} a\right)^{2}+b^{2}$, which is equal to $\left(\log _{2} a+b\right)^{2}-2 b \log _{2} a=3^{2}-2=7$. Hence, our answer is $2^{7}=128$.
|
Find the sum of the $x$-coordinates of the solutions to the system of equations $y=|x^2-6x+5|$ and $y=\frac{29}{4}-x$.
|
\frac{17}{2}
| |
A standard die is rolled eight times. What is the probability that the product of all eight rolls is odd and that each number rolled is a prime number? Express your answer as a common fraction.
|
\frac{1}{6561}
| |
Solve for \(x\): \(x\lfloor x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\rfloor=122\).
|
\frac{122}{41}
|
This problem can be done without needless casework. (For negative values of \(x\), the left hand side will be negative, so we only need to consider positive values of \(x\).) The key observation is that for \(x \in[2,3), 122\) is an extremely large value for the expression. Indeed, we observe that: \(\lfloor x\rfloor =2 \lfloor x\lfloor x\rfloor\rfloor \leq 2(3)-1 =5 \lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor \leq 3(5)-1 =14 \lfloor x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\rfloor \leq 3(14)-1 =41 x\lfloor x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\rfloor <3(41) =123\). So the expression can only be as large as 122 if ALL of those equalities hold (the the fourth line equaling 40 isn't good enough), and \(x=\frac{122}{41}\). Note that this value is extremely close to 3. We may check that this value of \(x\) indeed works. Note that the expression is strictly increasing in \(x\), so \(x=\frac{122}{41}\) is the only value that works.
|
The integers $1,2,4,5,6,9,10,11,13$ are to be placed in the circles and squares below with one number in each shape. Each integer must be used exactly once and the integer in each circle must be equal to the sum of the integers in the two neighbouring squares. If the integer $x$ is placed in the leftmost square and the integer $y$ is placed in the rightmost square, what is the largest possible value of $x+y$?
|
20
|
From the given information, if $a$ and $b$ are in two consecutive squares, then $a+b$ goes in the circle between them. Since all of the numbers that we can use are positive, then $a+b$ is larger than both $a$ and $b$. This means that the largest integer in the list, which is 13, cannot be either $x$ or $y$ (and in fact cannot be placed in any square). This is because the number in the circle next to it must be smaller than 13 (because 13 is the largest number in the list) and so cannot be the sum of 13 and another positive number from the list. Thus, for $x+y$ to be as large as possible, we would have $x$ and $y$ equal to 10 and 11 in some order. But here we have the same problem: there is only one larger number from the list (namely 13) that can go in the circles next to 10 and 11, and so we could not fill in the circle next to both 10 and 11. Therefore, the next largest possible value for $x+y$ is when $x=9$ and $y=11$. (We could also swap $x$ and $y$.) Here, we could have $13=11+2$ and $10=9+1$, giving the following partial list: The remaining integers $(4,5$ and 6) can be put in the shapes in the following way that satisfies the requirements. This tells us that the largest possible value of $x+y$ is 20.
|
Given sets $M=\{1, 2, a^2 - 3a - 1 \}$ and $N=\{-1, a, 3\}$, and the intersection of $M$ and $N$ is $M \cap N = \{3\}$, find the set of all possible real values for $a$.
|
\{4\}
| |
For any positive integer $n$, let
$f(n) =\begin{cases}\log_{8}{n}, &\text{if }\log_{8}{n}\text{ is rational,}\\ 0, &\text{otherwise.}\end{cases}$
What is $\sum_{n = 1}^{1997}{f(n)}$?
|
\frac{55}{3}
|
1. **Identify when $\log_8 n$ is rational**:
For $\log_8 n$ to be rational, $n$ must be a power of $8$'s base, which is $2$. Thus, $n = 2^k$ for some integer $k$. This is because $\log_8 2^k = k \log_8 2 = \frac{k}{3}$, which is rational.
2. **Determine the range of $k$**:
We need $2^k \leq 1997$. The largest $k$ for which this holds is $k = 10$ because $2^{11} = 2048 > 1997$.
3. **Calculate the sum**:
We need to find $\sum_{k=0}^{10} \log_8 2^k$. Using the property that $\log_8 2^k = \frac{k}{3}$, the sum becomes:
\[
\sum_{k=0}^{10} \frac{k}{3} = \frac{1}{3} \sum_{k=0}^{10} k
\]
The sum of the first $n$ natural numbers is given by $\frac{n(n+1)}{2}$. Applying this formula, we get:
\[
\sum_{k=0}^{10} k = \frac{10 \times 11}{2} = 55
\]
Therefore, the sum $\frac{1}{3} \sum_{k=0}^{10} k$ becomes:
\[
\frac{1}{3} \times 55 = \frac{55}{3}
\]
4. **Conclusion**:
The sum $\sum_{n=1}^{1997} f(n)$ is $\boxed{\frac{55}{3}}$. This corresponds to choice $\textbf{(C)}$.
|
Let $A = \{1, 2, 3, 4, 5, 6, 7\}$, and let $N$ be the number of functions $f$ from set $A$ to set $A$ such that $f(f(x))$ is a constant function. Find the remainder when $N$ is divided by $1000$.
|
399
| |
Let \( a_0 = -3 \), \( b_0 = 2 \), and for \( n \geq 0 \), let:
\[
\begin{align*}
a_{n+1} &= 2a_n + 2b_n + 2\sqrt{a_n^2 + b_n^2}, \\
b_{n+1} &= 2a_n + 2b_n - 2\sqrt{a_n^2 + b_n^2}.
\end{align*}
\]
Find \( \frac{1}{a_{2023}} + \frac{1}{b_{2023}} \).
|
\frac{1}{3}
| |
On each side of a 6 by 8 rectangle, construct an equilateral triangle with that side as one edge such that the interior of the triangle intersects the interior of the rectangle. What is the total area of all regions that are contained in exactly 3 of the 4 equilateral triangles?
|
\frac{96 \sqrt{3}-154}{\sqrt{3}} \text{ OR } \frac{288-154 \sqrt{3}}{3} \text{ OR } 96-\frac{154}{\sqrt{3}} \text{ OR } 96-\frac{154 \sqrt{3}}{3}
|
Let the rectangle be $A B C D$ with $A B=8$ and $B C=6$. Let the four equilateral triangles be $A B P_{1}, B C P_{2}, C D P_{3}$, and $D A P_{4}$ (for convenience, call them the $P_{1^{-}}, P_{2^{-}}, P_{3^{-}}, P_{4^{-}}$triangles). Let $W=A P_{1} \cap D P_{3}, X=A P_{1} \cap D P_{4}$, and $Y=D P_{4} \cap C P_{2}$. Reflect $X, Y$ over the line $P_{2} P_{4}$ (the line halfway between $A B$ and $D C$ ) to points $X^{\prime}, Y^{\prime}$. First we analyze the basic configuration of the diagram. Since $A B=8<2 \cdot 6 \frac{\sqrt{3}}{2}$, the $P_{2^{-}}, P_{4^{-}}$triangles intersect. Furthermore, $A P_{1} \perp B P_{2}$, so if $T=B P_{2} \cap A P_{1}$, then $B P_{2}=6<4 \sqrt{3}=B T$. Therefore $P_{2}$ lies inside triangle $P_{1} B A$, and by symmetry, also triangle $P_{3} D C$. It follows that the area we wish to compute is the union of two (congruent) concave hexagons, one of which is $W X Y P_{2} Y^{\prime} X^{\prime}$. (The other is its reflection over $Y Y^{\prime}$, the mid-line of $A D$ and $B C$.) So we seek $$2\left[W X Y P_{2} Y^{\prime} X^{\prime}\right]=2\left(\left[W X P_{4} X^{\prime}\right]-\left[P_{2} Y P_{4} Y^{\prime}\right]\right)$$ It's easy to see that $\left[W X P_{4} X^{\prime}\right]=\frac{1}{3}\left[A D P_{4}\right]=\frac{1}{3} \frac{6^{2} \sqrt{3}}{4}=3 \sqrt{3}$, since $W X P_{4} X^{\prime}$ and its reflections over lines $D W X^{\prime}$ and $A W X$ partition $\triangle A D P_{4}$. It remains to consider $P_{2} Y P_{4} Y^{\prime}$, a rhombus with (perpendicular) diagonals $P_{2} P_{4}$ and $Y Y^{\prime}$. If $O$ denotes the intersection of these two diagonals (also the center of $A B C D$ ), then $O P_{2}$ is $P_{2} B \frac{\sqrt{3}}{2}-\frac{1}{2} A B=3 \sqrt{3}-4$, the difference between the lengths of the $P_{2}$-altitude in $\triangle C B P_{2}$ and the distance between the parallel lines $Y Y^{\prime}, C B$. Easy angle chasing gives $O Y=\frac{O P_{2}}{\sqrt{3}}$, so $$\left[P_{2} Y P_{4} Y^{\prime}\right]=4 \cdot \frac{O P_{2} \cdot O Y}{2}=\frac{2}{\sqrt{3}} O P_{2}^{2}=\frac{2}{\sqrt{3}}(3 \sqrt{3}-4)^{2}=\frac{86-48 \sqrt{3}}{\sqrt{3}}$$ and our desired area is $$2\left[W X P_{4} X^{\prime}\right]-2\left[P_{2} Y P_{4} Y^{\prime}\right]=6 \sqrt{3}-\frac{172-96 \sqrt{3}}{\sqrt{3}}=\frac{96 \sqrt{3}-154}{\sqrt{3}}$$ or $\frac{288-154 \sqrt{3}}{3}$.
|
A right circular cone has a volume of $12\pi$ cubic centimeters. The height of the cone is 4 cm. How many centimeters is the circumference of the base of the cone, in terms of $\pi$?
|
6\pi
| |
Given a function f(x) defined on ℝ that satisfies f(x-2)=f(-2-x), and when x ≥ -2, f(x)=2^x-3. If the function f(x) has a zero point in the interval (k,k+1) (k ∈ ℤ), determine the value of k.
|
-6
| |
Two swimmers, at opposite ends of a $90$-foot pool, start to swim the length of the pool, one at the rate of $3$ feet per second, the other at $2$ feet per second. They swim back and forth for $12$ minutes. Allowing no loss of times at the turns, find the number of times they pass each other.
|
20
|
1. **Calculate the time taken by each swimmer to swim the length of the pool:**
- The first swimmer swims at $3$ feet per second, so the time to swim $90$ feet is:
\[
\frac{90 \text{ feet}}{3 \text{ feet/second}} = 30 \text{ seconds}
\]
- The second swimmer swims at $2$ feet per second, so the time to swim $90$ feet is:
\[
\frac{90 \text{ feet}}{2 \text{ feet/second}} = 45 \text{ seconds}
\]
2. **Determine the time when both swimmers return to their starting points:**
- The first swimmer's round trip time is $60$ seconds (30 seconds each way).
- The second swimmer's round trip time is $90$ seconds (45 seconds each way).
- The least common multiple (LCM) of $60$ and $90$ seconds gives the time when both swimmers are back at their starting points simultaneously:
\[
\text{LCM}(60, 90) = 180 \text{ seconds} = 3 \text{ minutes}
\]
3. **Calculate the number of times they meet in the first $3$ minutes:**
- The swimmers meet whenever the sum of the distances they have swum equals a multiple of $90$ feet (since the pool is $90$ feet long).
- In $3$ minutes ($180$ seconds), the first swimmer would have completed $3$ round trips (6 lengths), and the second swimmer would have completed $2$ round trips (4 lengths).
- They meet at the start, then again each time one reaches the end of the pool as the other is reaching the opposite end. This happens every time one of their individual distances is a multiple of $90$ feet, considering their relative speeds and directions.
4. **Graphical analysis (not shown here) indicates five meeting points in $3$ minutes.**
5. **Extrapolate to $12$ minutes:**
- Since the pattern repeats every $3$ minutes, and $12$ minutes is four times $3$ minutes, the number of meetings in $12$ minutes is:
\[
4 \times 5 = 20
\]
6. **Conclusion:**
- The swimmers pass each other $20$ times in $12$ minutes.
Thus, the correct answer is $\boxed{\textbf{(C)}\ 20}$.
|
Five glass bottles can be recycled to make a new bottle. Additionally, for every 20 new bottles created, a bonus bottle can be made from residual materials. Starting with 625 glass bottles, how many total new bottles can eventually be made from recycling and bonuses? (Keep counting recycled and bonus bottles until no further bottles can be manufactured. Do not include the original 625 bottles in your count.)
|
163
| |
If \( b \) and \( n \) are positive integers with \( b, n \leq 18 \), what is the greatest number of positive factors \( b^n \) can have?
|
703
| |
A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths 5, 7, and 8. What is the area of the triangle and the radius of the circle?
|
\frac{10}{\pi}
| |
What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$?
|
1120
|
1. **Calculate the numerator**: The numerator of the given expression is the product of the first 8 positive integers:
\[
1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 = 8!
\]
Using the factorial of 8, we find:
\[
8! = 40320
\]
2. **Calculate the denominator**: The denominator is the sum of the first 8 positive integers. Using the formula for the sum of the first $n$ positive integers, $\frac{n(n+1)}{2}$, where $n=8$, we get:
\[
\frac{8 \cdot 9}{2} = 36
\]
3. **Simplify the expression**: The expression now is:
\[
\frac{40320}{36}
\]
To simplify, we can perform the division:
\[
\frac{40320}{36} = 1120
\]
4. **Conclusion**: The value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$ is $\boxed{\textbf{(B)}\ 1120}$.
|
If the integer $k^{}_{}$ is added to each of the numbers $36^{}_{}$, $300^{}_{}$, and $596^{}_{}$, one obtains the squares of three consecutive terms of an arithmetic series. Find $k^{}_{}$.
|
925
| |
If \( e^{i \theta} = \frac{3 + i \sqrt{2}}{4}, \) then find \( \cos 3\theta. \)
|
\frac{9}{64}
| |
Compute
\[
\sin^2 3^\circ + \sin^2 6^\circ + \sin^2 9^\circ + \dots + \sin^2 177^\circ.
\]
|
30
| |
Higher Secondary P4
If the fraction $\dfrac{a}{b}$ is greater than $\dfrac{31}{17}$ in the least amount while $b<17$ , find $\dfrac{a}{b}$ .
|
\frac{11}{6}
| |
How many ordered pairs of integers $(x, y)$ satisfy the equation $x^{2020} + y^2 = 2y$?
|
4
|
We start by analyzing the given equation:
\[ x^{2020} + y^2 = 2y. \]
#### Step 1: Rearrange the equation
We can rearrange the equation to isolate terms involving \( y \):
\[ y^2 - 2y + x^{2020} = 0. \]
#### Step 2: Complete the square
Completing the square for \( y \) gives:
\[ (y-1)^2 - 1 + x^{2020} = 0 \]
\[ (y-1)^2 = 1 - x^{2020}. \]
#### Step 3: Analyze the powers of \( x \)
Since \( x^{2020} \) is a very high even power, the possible values of \( x^{2020} \) for integer \( x \) are limited to the squares of integers. However, since \( (y-1)^2 \geq 0 \), we need \( 1 - x^{2020} \geq 0 \), which implies:
\[ x^{2020} \leq 1. \]
The integers \( x \) that satisfy this are \( x = 0, \pm 1 \) because:
- \( 0^{2020} = 0 \),
- \( 1^{2020} = 1 \),
- \( (-1)^{2020} = 1 \) (since 2020 is even).
#### Step 4: Substitute possible values of \( x \) and solve for \( y \)
- **For \( x = 0 \)**:
\[ (y-1)^2 = 1 \]
\[ y-1 = \pm 1 \]
\[ y = 0 \text{ or } y = 2. \]
This gives pairs \( (0, 0) \) and \( (0, 2) \).
- **For \( x = 1 \) or \( x = -1 \)**:
\[ (y-1)^2 = 0 \]
\[ y-1 = 0 \]
\[ y = 1. \]
This gives pairs \( (1, 1) \) and \( (-1, 1) \).
#### Conclusion
The solutions are \( (0, 0) \), \( (0, 2) \), \( (1, 1) \), and \( (-1, 1) \). Counting these, we find there are 4 solutions.
Thus, the number of ordered pairs \((x, y)\) that satisfy the equation is \(\boxed{\textbf{(D) } 4}\).
|
In the triangle $ABC$, the side lengths are given as $AB=\sqrt{2}$, $BC=\sqrt{5}$, and $AC=3$. Compare the measure of the angle $\angle BOC$ to $112.5^{\circ}$, where $O$ is the center of the circle inscribed in triangle $ABC$.
|
112.5
| |
A truck delivered 4 bags of cement. They are stacked in the truck. A worker can carry one bag at a time either from the truck to the gate or from the gate to the shed. The worker can carry the bags in any order, each time taking the top bag, carrying it to the respective destination, and placing it on top of the existing stack (if there are already bags there). If given a choice to carry a bag from the truck or from the gate, the worker randomly chooses each option with a probability of 0.5. Eventually, all the bags end up in the shed.
a) (7th grade level, 1 point). What is the probability that the bags end up in the shed in the reverse order compared to how they were placed in the truck?
b) (7th grade level, 1 point). What is the probability that the bag that was second from the bottom in the truck ends up as the bottom bag in the shed?
|
\frac{1}{8}
| |
Determine the largest of all integers $n$ with the property that $n$ is divisible by all positive integers that are less than $\sqrt[3]{n}$.
|
420
|
Observation from that $\operatorname{lcm}(2,3,4,5,6,7)=420$ is divisible by every integer less than or equal to $7=[\sqrt[3]{420}]$ and that $\operatorname{lcm}(2,3,4,5,6,7,8)=840$ is not divisible by $9=[\sqrt[3]{840}]$. One may guess 420 is the required integer. Let $N$ be the required integer and suppose $N>420$. Put $t=[\sqrt[3]{N}]$. Then $$t \leq 1(1^{3}+3t+3).$$ Since $t \geq 7, \quad \operatorname{lcm}(2,3,4,5,6,7)=420$ should divide $N$ and hence $N \geq 840$, which implies $t \geq 9$. But then $\operatorname{lcm}(2,3,4,5,6,7,8,9)=2520$ should divide $N$, which implies $t \geq 13=[\sqrt[3]{2520}]$. Observe that any four consecutive integers are divisible by 8 and that any two out of four consecutive integers have gcd either 1, 2, or 3. So, we have $t(t-1)(t-2)(t-3)$ divides $6N$ and in particular, $$t(t-1)(t-2)(t-3) \leq 6N.$$ From this follows $$t(t-1)(t-2)(t-3) \leq 6t(t^{3}+3t+3) \frac{12}{t}+\frac{7}{t^{2}}+\frac{24}{t^{3}} \geq 1.$$ Since $t \geq 13$, $$\frac{12}{t}+\frac{7}{t^{2}}+\frac{24}{t^{3}}<1,$$ which is a contradiction.
|
In a store, there are 21 white shirts and 21 purple shirts hanging in a row. Find the minimum number \( k \) such that, regardless of the initial order of the shirts, it is possible to take down \( k \) white shirts and \( k \) purple shirts so that the remaining white shirts are all hanging consecutively and the remaining purple shirts are also hanging consecutively.
|
10
| |
Rectangle $ABCD$ lies in a plane with $AB = CD = 3$ and $BC = DA = 8$. This rectangle is rotated $90^\circ$ clockwise around $D$, followed by another $90^\circ$ clockwise rotation around the new position of point $C$ after the first rotation. What is the length of the path traveled by point $A$?
A) $\frac{\pi(8 + \sqrt{73})}{2}$
B) $\frac{\pi(8 + \sqrt{65})}{2}$
C) $8\pi$
D) $\frac{\pi(7 + \sqrt{73})}{2}$
E) $\frac{\pi(9 + \sqrt{73})}{2}$
|
\frac{\pi(8 + \sqrt{73})}{2}
| |
The product $8 \cdot 78$ and the increasing sequence of integers ${3, 15, 24, 48, \cdots}$, which consists of numbers that are divisible by 3 and are one less than a perfect square, are given. What is the remainder when the 1994th term in this sequence is divided by 1000?
|
63
| |
Regular octagon $C H I L D R E N$ has area 1. Find the area of pentagon $C H I L D$.
|
\frac{1}{2}
|
The pentagon $C H I L D$ is congruent to the pentagon $N E R D C$, as their corresponding angles and sides are congruent. Moreover, the two pentagons together compose the entire octagon, so each pentagon must have area one-half of the area of the octagon, or $\frac{1}{2}$.
|
Let $a$ and $b$ be five-digit palindromes (without leading zeroes) such that $a<b$ and there are no other five-digit palindromes strictly between $a$ and $b$. What are all possible values of $b-a$?
|
100, 110, 11
|
Let $\overline{x y z y x}$ be the digits of the palindrome $a$. There are three cases. If $z<9$, then the next palindrome greater than $\overline{x y z y x}$ is $\overline{x y(z+1) y x}$, which differs by 100. If $z=9$ but $y<9$, then the next palindrome up is $\overline{x(y+1) 0}(y+1) x$, which differs from $\overline{x y 9 y x}$ by 110. Finally, if $y=z=9$, then the next palindrome after $\overline{x 999 x}$ is $\overline{(x+1) 000(x+1)}$, which gives a difference of 11. Thus, the possible differences are $11,100,110$.
|
To calculate $41^2$, David mentally figures the value $40^2$ and adds 81. David subtracts a number from $40^2$ to calculate $39^2$. What number does he subtract?
|
79
| |
Three of the four endpoints of the axes of an ellipse are, in some order, \[(10, -3), \; (15, 7), \; (25, -3).\] Find the distance between the foci of the ellipse.
|
11.18
| |
How many 3-digit numbers have the property that the units digit is at least twice the tens digit?
|
270
| |
The number $24!$ has many positive integer divisors. What is the probability that a divisor randomly chosen from these is odd?
|
\frac{1}{23}
| |
What is the coefficient of $x^3$ in the product of the polynomials $$x^4 - 2x^3 + 3x^2 - 4x + 5$$ and $$3x^3 - 4x^2 + x + 6$$ after combining like terms?
|
22
| |
In a circle with center $O$, the measure of $\angle SIP$ is $48^\circ$ and $OS=12$ cm. Find the number of centimeters in the length of arc $SP$ and also determine the length of arc $SXP$, where $X$ is a point on the arc $SP$ such that $\angle SXP = 24^\circ$. Express your answer in terms of $\pi$.
|
3.2\pi
| |
Given point O is the circumcenter of ∆ABC, and |→BA|=2, |→BC|=6, calculate →BO⋅→AC.
|
16
| |
Find the smallest positive integer $n$ that has at least $7$ positive divisors $1 = d_1 < d_2 < \ldots < d_k = n$ , $k \geq 7$ , and for which the following equalities hold: $$ d_7 = 2d_5 + 1\text{ and }d_7 = 3d_4 - 1 $$ *Proposed by Mykyta Kharin*
|
2024
| |
The sum of the first 1000 terms of a geometric sequence is 300. The sum of the first 2000 terms is 570. Find the sum of the first 3000 terms.
|
813
| |
How many different $4\times 4$ arrays whose entries are all 1's and -1's have the property that the sum of the entries in each row is 0 and the sum of the entries in each column is 0?
|
90
| |
Let $x_1,$ $x_2,$ $x_3,$ $x_4$ be the roots of the polynomial $f(x) = x^4 - x^3 + x^2 + 1$. Define $g(x) = x^2 - 3$. Find the product:
\[ g(x_1) g(x_2) g(x_3) g(x_4). \]
|
142
| |
If $x - y = 12$ and $x + y = 6$, what is the value of $y$?
|
-3
| |
Given the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ ($a > 0, b > 0$), let F be the right focus of the hyperbola. A perpendicular line from point F to the x-axis intersects the two asymptotes at points A and B, and intersects the hyperbola in the first quadrant at point P. Let O be the origin of the coordinate system. If $\vec{OP} = \lambda \vec{OA} + \mu \vec{OB}$ ($\lambda, \mu \in \mathbb{R}$), and $\lambda^2 + \mu^2 = \frac{5}{8}$, calculate the eccentricity of the hyperbola.
|
\frac{2\sqrt{3}}{3}
| |
Given that the amount of cultural and tourism vouchers issued is $2.51 million yuan, express this amount in scientific notation.
|
2.51 \times 10^{6}
| |
Alex bakes a total of $24$ pies, and each pie is apple, blueberry, or cherry. The ratio of apple to blueberry to cherry pies is $1:4:3$. How many cherry pies did Alex bake?
|
9
| |
Given $\tan\alpha= \frac {1}{2}$ and $\tan(\alpha-\beta)=- \frac {2}{5}$, calculate the value of $\tan(2\alpha-\beta)$.
|
-\frac{1}{12}
| |
**The first term of a sequence is $2089$. Each succeeding term is the sum of the squares of the digits of the previous term. What is the $2089^{\text{th}}$ term of the sequence?**
|
16
| |
How many values of $x,-19<x<98$, satisfy $$\cos ^{2} x+2 \sin ^{2} x=1 ?$$
|
38
|
For any $x, \sin ^{2} x+\cos ^{2} x=1$. Subtracting this from the given equation gives \(\sin ^{2} x=0\), or \(\sin x=0\). Thus $x$ must be a multiple of \(\pi\), so \(-19<k \pi<98\) for some integer $k$, or approximately \(-6.1<k<31.2\). There are 32 values of $k$ that satisfy this, so there are 38 values of $x$ that satisfy \(\cos ^{2} x+2 \sin ^{2} x=1\).
|
In $\triangle ABC$, $a$, $b$, $c$ are the sides opposite to angles $A$, $B$, $C$ respectively, and $B$ is an acute angle. If $\frac{\sin A}{\sin B} = \frac{5c}{2b}$, $\sin B = \frac{\sqrt{7}}{4}$, and $S_{\triangle ABC} = \frac{5\sqrt{7}}{4}$, find the value of $b$.
|
\sqrt{14}
| |
Given real numbers $x$ and $y$ satisfy the equation $x^{2}+y^{2}-4x+1=0$.
$(1)$ Find the maximum and minimum values of $\dfrac {y}{x}$;
$(2)$ Find the maximum and minimum values of $y-x$;
$(3)$ Find the maximum and minimum values of $x^{2}+y^{2}$.
|
7-4 \sqrt {3}
| |
What is the smallest prime number dividing the sum $3^{11}+5^{13}$?
|
2
|
1. **Identify the Parity of Each Term**:
- $3^{11}$ is an odd number because any power of an odd number remains odd.
- $5^{13}$ is also an odd number for the same reason.
2. **Sum of Two Odd Numbers**:
- The sum of two odd numbers is always even. This can be shown by considering that an odd number can be expressed as $2k+1$ for some integer $k$. Thus, the sum of two odd numbers $2k+1$ and $2j+1$ is $2k + 1 + 2j + 1 = 2(k + j + 1)$, which is clearly even.
3. **Check for Divisibility by 2**:
- Since $3^{11} + 5^{13}$ is even, it is divisible by $2$.
4. **Conclusion**:
- The smallest prime number is $2$, and since $2$ divides the sum $3^{11} + 5^{13}$, the smallest prime number dividing the sum is indeed $2$.
$\boxed{\text{A}}$
|
The ferry boat begins transporting tourists to an island every hour starting at 9 AM until its last trip, which starts at 4 PM. On the first trip at 9 AM, there were 120 tourists, and on each successive trip, there were 2 fewer tourists than on the previous trip. Determine the total number of tourists the ferry transported to the island that day.
|
904
| |
Let \\(\triangle ABC\\) have internal angles \\(A\\), \\(B\\), and \\(C\\) opposite to sides of lengths \\(a\\), \\(b\\), and \\(c\\) respectively, and it satisfies \\(a^{2}+c^{2}-b^{2}= \sqrt {3}ac\\).
\\((1)\\) Find the size of angle \\(B\\);
\\((2)\\) If \\(2b\cos A= \sqrt {3}(c\cos A+a\cos C)\\), and the median \\(AM\\) on side \\(BC\\) has a length of \\(\sqrt {7}\\), find the area of \\(\triangle ABC\\).
|
\sqrt {3}
| |
How many rectangles can be formed by the vertices of a cube? (Note: square is also a special rectangle).
|
12
| |
Add $254_{9} + 627_{9} + 503_{9}$. Express your answer in base 9.
|
1485_{9}
| |
In the Cartesian coordinate system, A and B are points moving on the x-axis and y-axis, respectively. If the circle C with AB as its diameter is tangent to the line $3x+y-4=0$, then the minimum area of circle C is \_\_\_\_\_\_.
|
\frac {2}{5}\pi
| |
Let $n$ be a $5$-digit number, and let $q$ and $r$ be the quotient and the remainder, respectively, when $n$ is divided by $100$. For how many values of $n$ is $q+r$ divisible by $11$?
|
9000
|
1. **Understanding the division of $n$ by $100$:**
When a $5$-digit number $n$ is divided by $100$, the quotient $q$ is formed by the first three digits of $n$, and the remainder $r$ is formed by the last two digits. Thus, $n = 100q + r$.
2. **Range of $q$ and $r$:**
- $q$ can range from $100$ to $999$, as $q$ represents a three-digit number.
- $r$ can range from $0$ to $99$, as $r$ represents a two-digit number.
3. **Total possible values of $q$ and $r$:**
- $q$ has $999 - 100 + 1 = 900$ possible values.
- $r$ has $99 - 0 + 1 = 100$ possible values.
4. **Condition for divisibility by $11$:**
We need to find the number of pairs $(q, r)$ such that $q + r$ is divisible by $11$, i.e., $q + r \equiv 0 \pmod{11}$.
5. **Calculating possible values of $r$ for each $q$:**
- For each $q$, we need to find how many values of $r$ satisfy $q + r \equiv 0 \pmod{11}$.
- This is equivalent to finding $r \equiv -q \pmod{11}$.
- Since $r$ ranges from $0$ to $99$, there are $\left\lfloor \frac{100}{11} \right\rfloor + 1 = 9 + 1 = 10$ possible values of $r$ for each residue class modulo $11$.
6. **Counting valid pairs $(q, r)$:**
- Each $q$ has exactly $10$ corresponding values of $r$ that satisfy the condition $q + r \equiv 0 \pmod{11}$.
- Since there are $900$ possible values of $q$, the total number of valid pairs $(q, r)$ is $900 \times 10 = 9000$.
7. **Conclusion:**
The total number of $5$-digit numbers $n$ such that $q + r$ is divisible by $11$ is $\boxed{9000}$.
|
How many two-digit numbers have digits whose sum is a perfect square less than or equal to 25?
|
17
| |
The midpoint of a line segment is located at $(1, -2)$. If one of the endpoints is $(4, 5)$, what is the other endpoint? Express your answer as an ordered pair.
|
(-2,-9)
| |
In $\triangle ABC$, $2\sin 2A\cos A-\sin 3A+\sqrt{3}\cos A=\sqrt{3}$.
(1) Find the measure of angle $A$;
(2) Given that $a$, $b$, and $c$ are the sides opposite to angles $A$, $B$, and $C$ respectively, if $a=1$ and $\sin A+\sin (B-C)=2\sin 2C$, find the area of $\triangle ABC$.
|
\frac{\sqrt{3}}{6}
| |
If we express $3x^2 - 6x - 2$ in the form $a(x - h)^2 + k$, then what is $a + h + k$?
|
-1
| |
What is the inverse of $f(x)=4-5x$?
|
\frac{4-x}{5}
| |
Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that
\[f(f(x) + y) = f(x + y) + xf(y) - xy - x + 1\]for all real numbers $x$ and $y.$
Let $n$ be the number of possible values of $f(1),$ and let $s$ be the sum of all possible values of $f(1).$ Find $n \times s.$
|
2
| |
Calculate the value of the following expressions:
1. $\sqrt[4]{(3-\pi )^{4}}+(0.008)\;^{- \frac {1}{3}}-(0.25)\;^{ \frac {1}{2}}×( \frac {1}{ \sqrt {2}})^{-4}$
2. $\log _{3} \sqrt {27}-\log _{3} \sqrt {3}-\lg 625-\lg 4+\ln (e^{2})- \frac {4}{3}\lg \sqrt {8}$
|
-1
| |
Eleven positive integers from a list of fifteen positive integers are $3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23$. What is the largest possible value of the median of this list of fifteen positive integers?
|
17
| |
Let point $P$ be on the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$. A line through $P$ intersects the asymptotes at $P_{1}$ and $P_{2}$, and $\overrightarrow{P_{1} P} \overrightarrow{P P_{2}}$ $=3$. Let $O$ be the origin. Find the area of $\triangle O P_{1} P_{2}$.
|
16
| |
In convex hexagon $ABCDEF$, all six sides are congruent, $\angle A$ and $\angle D$ are right angles, and $\angle B, \angle C, \angle E,$ and $\angle F$ are congruent. The area of the hexagonal region is $2116(\sqrt{2}+1).$ Find $AB$.
|
46
|
Let the side length be called $x$, so $x=AB=BC=CD=DE=EF=AF$.
The diagonal $BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}$. Then the areas of the triangles AFB and CDE in total are $\frac{x^2}{2}\cdot 2$, and the area of the rectangle BCEF equals $x\cdot x\sqrt{2}=x^2\sqrt{2}$
Then we have to solve the equation
$2116(\sqrt{2}+1)=x^2\sqrt{2}+x^2$.
$2116(\sqrt{2}+1)=x^2(\sqrt{2}+1)$
$2116=x^2$
$x=46$
Therefore, $AB$ is $\boxed{046}$.
|
A semicircle of diameter 1 sits at the top of a semicircle of diameter 2, as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a $\textit{lune}$. Determine the area of this lune. Express your answer in terms of $\pi$ and in simplest radical form.
[asy]
fill((0,2.73)..(1,1.73)--(-1,1.73)..cycle,gray(0.7));
draw((0,2.73)..(1,1.73)--(-1,1.73)..cycle,linewidth(0.7));
fill((0,2)..(2,0)--(-2,0)..cycle,white);
draw((0,2)..(2,0)--(-2,0)..cycle,linewidth(0.7));
draw((-1,1.73)--(1,1.73),dashed);
label("2",(0,0),S);
label("1",(0,1.73),S);
[/asy]
|
\frac{\sqrt{3}}{4} - \frac{1}{24}\pi
| |
Determine the smallest positive real $K$ such that the inequality
\[ K + \frac{a + b + c}{3} \ge (K + 1) \sqrt{\frac{a^2 + b^2 + c^2}{3}} \]holds for any real numbers $0 \le a,b,c \le 1$ .
*Proposed by Fajar Yuliawan, Indonesia*
|
\frac{\sqrt{6}}{3}
| |
How many irreducible fractions with a numerator of 2015 are there that are less than \( \frac{1}{2015} \) and greater than \( \frac{1}{2016} \)?
|
1440
| |
Find the smallest solution to the equation \[\frac{1}{x-2} + \frac{1}{x-4} = \frac{3}{x-3}.\]
|
3 - \sqrt3
| |
Let $x$ and $y$ be real numbers satisfying $3 \leqslant xy^2 \leqslant 8$ and $4 \leqslant \frac{x^2}{y} \leqslant 9$. Find the maximum value of $\frac{x^3}{y^4}$.
|
27
| |
A [i]permutation[/i] of the set of positive integers $[n] = \{1, 2, . . . , n\}$ is a sequence $(a_1 , a_2 , \ldots, a_n ) $ such that each element of $[n]$ appears precisely one time as a term of the sequence. For example, $(3, 5, 1, 2, 4)$ is a permutation of $[5]$. Let $P (n)$ be the number of permutations of $[n]$ for which $ka_k$ is a perfect square for all $1 \leq k \leq n$. Find with proof the smallest $n$ such that $P (n)$ is a multiple of $2010$.
|
4489
|
To solve this problem, we will analyze the given condition involving permutations and perfect squares to determine the smallest \( n \) such that \( P(n) \), the number of permutations of \([n] = \{1, 2, \ldots, n\}\) where \( ka_k \) is a perfect square for all \( 1 \leq k \leq n \), is a multiple of 2010.
### Step-by-Step Analysis
1. **Understanding the Problem:**
Given a permutation \((a_1, a_2, \ldots, a_n)\) of \([n]\), we need each product \( ka_k \) to be a perfect square. This implies \( ka_k = m_k^2 \) for some integer \( m_k \).
Therefore, \( a_k = \frac{m_k^2}{k} \) must be an integer. Hence, \( k \) must divide \( m_k^2 \). Since \( m_k^2 = ka_k \), this implies that \( a_k \) must also divide \( k \).
2. **Condition Analysis:**
The divisibility condition reduces to:
\[
a_k = \frac{m_k^2}{k}
\]
implying \( k \mid m_k^2 \). This is equivalent to saying that \( k \) must be a perfect square itself, because for \( a_k \) to be a positive integer permutation of 1 to \( n \), \( m_k = \sqrt{k} \) is the simplest choice, allowing \( k \) to divide \( m_k^2 = k \).
3. **Valid \( n \) for a Permutation:**
Next, for which values of \( n \) can we construct permutations meeting the conditions? Each \( k \) must be a perfect square, so \( 1, 4, 9, 16, \ldots \) need to be the indices selected for permutation.
4. **Counting the Permutations:**
First, we need to determine how many perfect squares exist within the set \([n]\). Let this count be denoted as \( f(n) \), the floor of the square root of \( n \):
\[
f(n) = \lfloor \sqrt{n} \rfloor
\]
For \( P(n) \) to be non-zero, each \( k \) must be a perfect square up to \( n \). The constraint on \( f(n) \) determining permutations is that it needs to reach a number such that the product of the factorials of the counts of solution possibilities is a multiple of 2010.
5. **Finding the Minimum \( n \):**
We need:
\[
P(n) = f(n)! \equiv 0 \pmod{2010}
\]
**Prime Factorization of 2010:**
\[
2010 = 2 \times 3 \times 5 \times 67
\]
The smallest factorial \( f(n)! \) has at least these factors.
6. **Calculating \( n \):**
Approximate \( f(n)! \) for increasing \( n \) (especially its factorial incremental):
- The smallest \( f(n) \) where \( f(n)! \) is divisible by 67 is when \( f(n) \approx 67 \) because the smallest factorial value divisible by 67 is \( 67! \).
Finding where the number of perfect squares, \( f(n) \), equals 67 should give us the smallest \( n \):
\[
n = (67)^2 = 4489
\]
Thus, the smallest \( n \) such that \( P(n) \) is a multiple of 2010 is:
\[
\boxed{4489}
\]
|
Given that P and Q are points on the graphs of the functions $2x-y+6=0$ and $y=2\ln x+2$ respectively, find the minimum value of the line segment |PQ|.
|
\frac{6\sqrt{5}}{5}
| |
The first term of a given sequence is 1, and each successive term is the sum of all the previous terms of the sequence. What is the value of the first term which exceeds 5000?
|
8192
| |
A subscriber forgot the last digit of a phone number and therefore dials it randomly. What is the probability that they will have to dial the number no more than three times?
|
0.3
| |
Seven dwarfs stood at the corners of their garden, each at one corner, and stretched a rope around the entire garden. Snow White started from Doc and walked along the rope. First, she walked four meters to the east where she met Prof. From there, she continued two meters north before reaching Grumpy. From Grumpy, she walked west and after two meters met Bashful. Continuing three meters north, she reached Happy. She then walked west and after four meters met Sneezy, from where she had three meters south to Sleepy. Finally, she followed the rope by the shortest path back to Doc, thus walking around the entire garden.
How many square meters is the entire garden?
Hint: Draw the shape of the garden, preferably on graph paper.
|
22
| |
Given the singing scores 9.4, 8.4, 9.4, 9.9, 9.6, 9.4, 9.7, calculate the average and variance of the remaining data after removing the highest and lowest scores.
|
0.016
| |
Given $sn(α+ \frac {π}{6})= \frac {1}{3}$, and $\frac {π}{3} < α < \pi$, find $\sin ( \frac {π}{12}-α)$.
|
- \frac {4+ \sqrt {2}}{6}
| |
Given that $\overrightarrow{a}$ and $\overrightarrow{b}$ are both unit vectors, if $|\overrightarrow{a}-2\overrightarrow{b}|=\sqrt{3}$, then the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is ____.
|
\frac{1}{3}\pi
| |
The lines $x=\frac{1}{4}y+a$ and $y=\frac{1}{4}x+b$ intersect at the point $(1,2)$. What is $a+b$?
|
\frac{9}{4}
|
1. **Substitute $(1,2)$ into the first equation:**
Given the equation $x = \frac{1}{4}y + a$, substitute $x = 1$ and $y = 2$:
\[
1 = \frac{1}{4} \cdot 2 + a
\]
Simplify the equation:
\[
1 = \frac{1}{2} + a
\]
Solve for $a$:
\[
a = 1 - \frac{1}{2} = \frac{1}{2}
\]
2. **Substitute $(1,2)$ into the second equation:**
Given the equation $y = \frac{1}{4}x + b$, substitute $x = 1$ and $y = 2$:
\[
2 = \frac{1}{4} \cdot 1 + b
\]
Simplify the equation:
\[
2 = \frac{1}{4} + b
\]
Solve for $b$:
\[
b = 2 - \frac{1}{4} = \frac{8}{4} - \frac{1}{4} = \frac{7}{4}
\]
3. **Calculate $a + b$:**
\[
a + b = \frac{1}{2} + \frac{7}{4}
\]
Convert $\frac{1}{2}$ to quarters for easy addition:
\[
\frac{1}{2} = \frac{2}{4}
\]
Add the fractions:
\[
a + b = \frac{2}{4} + \frac{7}{4} = \frac{9}{4}
\]
4. **Conclusion:**
The value of $a + b$ is $\frac{9}{4}$. Therefore, the correct answer is $\boxed{\textbf{(E) }\frac{9}{4}}$.
|
What is the smallest natural number whose digits in decimal representation are either 0 or 1 and which is divisible by 225? (China Junior High School Mathematics League, 1989)
|
11111111100
| |
Evaluate the expression $8- rac{6}{4-2}$.
|
5
|
Evaluating, $8- rac{6}{4-2}=8- rac{6}{2}=8-3=5$.
|
With all angles measured in degrees, calculate the product $\prod_{k=1}^{30} \csc^2(3k)^\circ$, and express the result as $m^n$, where $m$ and $n$ are integers greater than 1. Find $m+n$.
|
31
| |
Given that $-7$ is a solution to $x^2 + bx -28 = 0$, what is the value of $b$?
|
3
| |
Suppose $b$ is an integer such that $1 \le b \le 30$, and $524123_{81}-b$ is a multiple of $17$. What is $b$?
|
11
| |
Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs $1 more than a pink pill, and Al's pills cost a total of $546 for the two weeks. How much does one green pill cost?
|
$19
|
1. **Identify the total number of days and total cost**: Al takes the pills for two weeks, which is 14 days. The total cost for these two weeks is $\$546$.
2. **Calculate daily expenditure**:
\[
\text{Daily cost} = \frac{\text{Total cost}}{\text{Number of days}} = \frac{546}{14} = 39 \text{ dollars}
\]
3. **Set up the equation for daily costs**:
Let the cost of a green pill be $x$ dollars and the cost of a pink pill be $x-1$ dollars. Since Al takes one green pill and one pink pill each day, the combined daily cost is:
\[
x + (x - 1) = 2x - 1
\]
4. **Solve for $x$**:
Since the daily cost is $39$ dollars, we set up the equation:
\[
2x - 1 = 39
\]
Solving for $x$, we add 1 to both sides:
\[
2x - 1 + 1 = 39 + 1 \implies 2x = 40
\]
Then, divide by 2:
\[
x = \frac{40}{2} = 20
\]
5. **Conclusion**: The cost of one green pill is $\boxed{\textbf{(D) }\textdollar 20}$.
|
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