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|---|---|---|---|---|---|
Let $N$ be the smallest positive integer such that $N+2N+3N+\ldots +9N$ is a number all of whose digits are equal. What is the sum of digits of $N$ ?
|
37
|
deepscale
| 18,369
| ||
The radius of a cylinder is doubled and its height is tripled. If its original volume was 10 cubic feet, what is its volume now, in cubic feet?
|
120
|
deepscale
| 35,688
| ||
Given the function $f(x) = \frac {1}{2}x^{2} + x - 2\ln{x}$ ($x > 0$):
(1) Find the intervals of monotonicity for $f(x)$.
(2) Find the extreme values of the function $f(x)$.
|
\frac {3}{2}
|
deepscale
| 18,064
| ||
The orthocenter of triangle $ABC$ divides altitude $\overline{CF}$ into segments with lengths $HF = 6$ and $HC = 15.$ Calculate $\tan A \tan B.$
[asy]
unitsize (1 cm);
pair A, B, C, D, E, F, H;
A = (0,0);
B = (5,0);
C = (4,4);
D = (A + reflect(B,C)*(A))/2;
E = (B + reflect(C,A)*(B))/2;
F = (C + reflect(A,B)*(C))/2;
H = extension(A,D,B,E);
draw(A--B--C--cycle);
draw(C--F);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$F$", F, S);
dot("$H$", H, W);
[/asy]
|
\frac{7}{2}
|
deepscale
| 39,638
| ||
Vasya is creating a 4-digit password for a combination lock. He dislikes the digit 2 and does not use it. Additionally, he does not like having two identical digits adjacent to each other. Moreover, he wants the first digit to be the same as the last digit. How many possible combinations need to be checked to guarantee guessing Vasya's password correctly?
|
576
|
deepscale
| 7,701
| ||
Find the largest prime divisor of \( 16^2 + 81^2 \).
|
53
|
deepscale
| 25,278
| ||
Jessica has exactly one of each of the first 30 states' new U.S. quarters. The quarters were released in the same order that the states joined the union. The graph below shows the number of states that joined the union in each decade. What fraction of Jessica's 30 coins represents states that joined the union during the decade 1800 through 1809? Express your answer as a common fraction.
[asy]size(200);
label("1780",(6,0),S);
label("1800",(12,0),S);
label("1820",(18,0),S);
label("1840",(24,0),S);
label("1860",(30,0),S);
label("1880",(36,0),S);
label("1900",(42,0),S);
label("1950",(48,0),S);
label("to",(6,-4),S);
label("to",(12,-4),S);
label("to",(18,-4),S);
label("to",(24,-4),S);
label("to",(30,-4),S);
label("to",(36,-4),S);
label("to",(42,-4),S);
label("to",(48,-4),S);
label("1789",(6,-8),S);
label("1809",(12,-8),S);
label("1829",(18,-8),S);
label("1849",(24,-8),S);
label("1869",(30,-8),S);
label("1889",(36,-8),S);
label("1909",(42,-8),S);
label("1959",(48,-8),S);
draw((0,0)--(50,0));
draw((0,2)--(50,2));
draw((0,4)--(50,4));
draw((0,6)--(50,6));
draw((0,8)--(50,8));
draw((0,10)--(50,10));
draw((0,12)--(50,12));
draw((0,14)--(50,14));
draw((0,16)--(50,16));
draw((0,18)--(50,18));
fill((4,0)--(8,0)--(8,12)--(4,12)--cycle,gray(0.8));
fill((10,0)--(14,0)--(14,5)--(10,5)--cycle,gray(0.8));
fill((16,0)--(20,0)--(20,7)--(16,7)--cycle,gray(0.8));
fill((22,0)--(26,0)--(26,6)--(22,6)--cycle,gray(0.8));
fill((28,0)--(32,0)--(32,7)--(28,7)--cycle,gray(0.8));
fill((34,0)--(38,0)--(38,5)--(34,5)--cycle,gray(0.8));
fill((40,0)--(44,0)--(44,4)--(40,4)--cycle,gray(0.8));
[/asy]
|
\frac{1}{6}
|
deepscale
| 24,100
| ||
Two sectors of a circle of radius $12$ are placed side by side, as shown. Determine the $\textit{area}$ of figure $ABCD.$ [asy]
draw((0,0)--(12,0)..(10.3923,6)..(6,10.3923)--(-6,10.3923)..(-4.3923,4.3923)..(0,0),black+linewidth(1));
draw((0,0)--(6,10.3923),black+linewidth(1)+dashed);
label("$A$",(-6,10.3923),NW);
label("$B$",(6,10.3923),NE);
label("$C$",(12,0),SE);
label("$D$",(0,0),SW);
label("$60^\circ$",(2,1));
label("$60^\circ$",(4,9.3923));
[/asy]
|
48\pi
|
deepscale
| 35,605
| ||
Pentagon $S P E A K$ is inscribed in triangle $N O W$ such that $S$ and $P$ lie on segment $N O, K$ and $A$ lie on segment $N W$, and $E$ lies on segment $O W$. Suppose that $N S=S P=P O$ and $N K=K A=A W$. Given that $E P=E K=5$ and $E A=E S=6$, compute $O W$.
|
Note that $[E S K]=[E P A]$, since one has half the base but double the height. Since the sides are the same, we must have $\sin \angle S E K=\sin \angle P E A$, so $\angle S E K+\angle P E A=180^{\circ}$. Let $O W=3 x$, so $S K=x$ and $P A=2 x$. Then by the law of cosines $$\begin{aligned} x^{2} & =61-60 \cos \angle S E K \\ 4 x^{2} & =61-60 \cos \angle P E A \end{aligned}$$ Summing these two gives $5 x^{2}=122$, since $\cos \angle S E K=-\cos \angle P E A$. Then $x=\sqrt{\frac{122}{5}}$, which means $3 x=\frac{3 \sqrt{610}}{5}$.
|
\frac{3 \sqrt{610}}{5}
|
deepscale
| 4,830
| |
A train is scheduled to arrive at a station randomly between 1:00 PM and 3:00 PM, and it waits for 15 minutes before leaving. If Alex arrives at the station randomly between 1:00 PM and 3:00 PM as well, what is the probability that he will find the train still at the station when he arrives?
|
\frac{105}{1920}
|
deepscale
| 25,082
| ||
Two tangents to a circle are drawn from a point $A$. The points of contact $B$ and $C$ divide the circle into arcs with lengths in the ratio $2 : 3$. What is the degree measure of $\angle{BAC}$?
|
1. **Identify the Geometry and Key Properties**:
Let the center of the circle be $O$. The lines $AB$ and $AC$ are tangents to the circle at points $B$ and $C$, respectively. By the property of tangents, the radii $OB$ and $OC$ are perpendicular to $AB$ and $AC$, respectively. Therefore, $\angle ABO = \angle ACO = 90^\circ$.
2. **Recognize the Cyclic Quadrilateral**:
Since $\angle ABO$ and $\angle ACO$ are both right angles, the sum of $\angle ABO$ and $\angle ACO$ is $180^\circ$. This implies that the quadrilateral $ABOC$ is cyclic (a property of cyclic quadrilaterals is that the opposite angles sum to $180^\circ$).
3. **Calculate the Arcs and Central Angle**:
Given that the arcs $BC$ and $CB'$ (where $B'$ is the point diametrically opposite to $B$ on the circle) are in the ratio $2:3$, let's denote the total circumference of the circle as $360^\circ$. If the arc $BC$ is $2x$ degrees and the arc $CB'$ is $3x$ degrees, then $2x + 3x = 360^\circ$. Solving for $x$, we get:
\[
5x = 360^\circ \implies x = 72^\circ
\]
Therefore, the arc $BC$ is $2x = 144^\circ$ and the arc $CB'$ is $3x = 216^\circ$.
4. **Relate Central Angle to Inscribed Angle**:
The central angle $\angle BOC$ subtended by the arc $BC$ is $144^\circ$. Since $ABOC$ is a cyclic quadrilateral, the inscribed angle $\angle BAC$ subtended by the same arc is half the central angle. Therefore:
\[
\angle BAC = \frac{1}{2} \times \angle BOC = \frac{1}{2} \times 144^\circ = 72^\circ
\]
5. **Correct the Calculation Error**:
The previous step contains a calculation error. Since $\angle BOC$ is the central angle and $\angle BAC$ is the inscribed angle subtending the same arc, the relationship should be:
\[
\angle BAC = 180^\circ - \angle BOC = 180^\circ - 144^\circ = 36^\circ
\]
6. **Conclusion**:
The degree measure of $\angle BAC$ is $\boxed{\textbf{(C)}\ 36}$.
|
36
|
deepscale
| 2,112
| |
In the quadrilateral pyramid \( P-ABCD \), it is given that \( AB \parallel CD \), \( AB \perp AD \), \( AB = 4 \), \( AD = 2\sqrt{2} \), \( CD = 2 \), and \( PA \perp \) plane \( ABCD \) with \( PA = 4 \). Let \( Q \) be a point on the segment \( PB \), and the sine of the angle between the line \( QC \) and the plane \( PAC \) is \( \frac{\sqrt{3}}{3} \). Find the value of \( \frac{PQ}{PB} \).
|
1/2
|
deepscale
| 29,186
| ||
Seven cards numbered $1$ through $7$ are to be lined up in a row. Find the number of arrangements of these seven cards where one of the cards can be removed leaving the remaining six cards in either ascending or descending order.
|
10
|
deepscale
| 28,136
| ||
P is a polynomial. When P is divided by $x-1$, the remainder is -4 . When P is divided by $x-2$, the remainder is -1 . When $P$ is divided by $x-3$, the remainder is 4 . Determine the remainder when $P$ is divided by $x^{3}-6 x^{2}+11 x-6$.
|
The remainder polynomial is simply the order two polynomial that goes through the points $(1,-4),(2,-1)$, and $(3,4): x^{2}-5$.
|
x^{2}-5
|
deepscale
| 3,709
| |
The train arrives at a station randomly between 1:00 PM and 3:00 PM and waits for 30 minutes before departing. If John also arrives randomly at the station within the same time period, what is the probability that he will find the train at the station?
|
\frac{7}{32}
|
deepscale
| 8,854
| ||
There is a $40\%$ chance of rain on Saturday and a $30\%$ chance of rain on Sunday. However, it is twice as likely to rain on Sunday if it rains on Saturday than if it does not rain on Saturday. The probability that it rains at least one day this weekend is $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Find $a+b$.
|
Let $x$ be the probability that it rains on Sunday given that it doesn't rain on Saturday. We then have $\dfrac{3}{5}x+\dfrac{2}{5}2x = \dfrac{3}{10} \implies \dfrac{7}{5}x=\dfrac{3}{10}$ $\implies x=\dfrac{3}{14}$. Therefore, the probability that it doesn't rain on either day is $\left(1-\dfrac{3}{14}\right)\left(\dfrac{3}{5}\right)=\dfrac{33}{70}$. Therefore, the probability that rains on at least one of the days is $1-\dfrac{33}{70}=\dfrac{37}{70}$, so adding up the $2$ numbers, we have $37+70=\boxed{107}$.
|
107
|
deepscale
| 7,167
| |
If $a^{2}-4a+3=0$, find the value of $\frac{9-3a}{2a-4} \div (a+2-\frac{5}{a-2})$ .
|
-\frac{3}{8}
|
deepscale
| 20,885
| ||
What is the last digit of the decimal expansion of $\frac{1}{2^{10}}$?
|
5
|
deepscale
| 37,859
| ||
Compute the number of increasing sequences of positive integers $b_1 \le b_2 \le b_3 \le \cdots \le b_{15} \le 3005$ such that $b_i-i$ is odd for $1\le i \le 15$. Express your answer as ${p \choose q}$ for some integers $p > q$ and find the remainder when $p$ is divided by 1000.
|
509
|
deepscale
| 32,639
| ||
The function $f: \mathbb{Z}^{2} \rightarrow \mathbb{Z}$ satisfies - $f(x, 0)=f(0, y)=0$, and - $f(x, y)=f(x-1, y)+f(x, y-1)+x+y$ for all nonnegative integers $x$ and $y$. Find $f(6,12)$.
|
We claim $f(x, y)=\binom{x+y+2}{x+1}-(x+y+2)$. Indeed, the hypothesis holds true for our base cases $f(x, 0)$ and $f(0, y)$, and moreover, $f(x-1, y)+f(x, y-1)+x+y=\binom{x+y+1}{x}+\binom{x+y+1}{x+1}-2(x+y+1)+x+y=\binom{x+y+2}{x+1}-(x+y+2)$. Thus, the final answer is $\binom{20}{7}-20=77500$. Here is a way to derive this formula from scratch. The idea is that the second condition harks back to the Pascal's triangle rule, sans some modifications. Write $f(x, y)=g(x, y)-x-y$, so then $g(0, t)=g(t, 0)=t$ and $g(x, y)=g(x-1, y)+g(x, y-1)+2$. Then, letting $g(x, y)=h(x, y)-2$ gives $h(x, y)=h(x-1, y)+h(x, y-1)$, which is exactly Pascal's rule. We are given the base cases $h(0, t)=h(t, 0)=t+2$, which is starting "inside" of Pascal's triangle, so $h(x, y)=\binom{x+y+2}{x+1}$.
|
77500
|
deepscale
| 5,099
| |
Given that in triangle $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively, and $\sqrt{3}a\cos C=c\sin A$.
$(1)$ Find the measure of angle $C$.
$(2)$ If $a > 2$ and $b-c=1$, find the minimum perimeter of triangle $\triangle ABC$.
|
9 + 6\sqrt{2}
|
deepscale
| 25,748
| ||
Find the number of ordered 17-tuples $(a_1, a_2, a_3, \dots, a_{17})$ of integers, such that the square of any number in the 17-tuple is equal to the sum of the other 16 numbers.
|
12378
|
deepscale
| 36,809
| ||
Suppose \( x \) and \( y \) are integers such that \( xy + 3x + 2y = -4 \). Find the greatest possible value of \( y \).
|
-1
|
deepscale
| 10,646
| ||
Sasha has $\$3.20$ in U.S. coins. She has the same number of quarters and nickels. What is the greatest number of quarters she could have?
|
10
|
deepscale
| 39,069
| ||
The side of rhombus \(ABCD\) is equal to 5. A circle with a radius of 2.4 is inscribed in this rhombus.
Find the distance between the points where this circle touches the sides \(AB\) and \(BC\), if the diagonal \(AC\) is less than the diagonal \(BD\).
|
3.84
|
deepscale
| 14,379
| ||
Let \( p, q, r, s, \) and \( t \) be the roots of the polynomial
\[ x^5 + 10x^4 + 20x^3 + 15x^2 + 6x + 3 = 0. \]
Find the value of
\[ \frac{1}{pq} + \frac{1}{pr} + \frac{1}{ps} + \frac{1}{pt} + \frac{1}{qr} + \frac{1}{qs} + \frac{1}{qt} + \frac{1}{rs} + \frac{1}{rt} + \frac{1}{st}. \]
|
\frac{20}{3}
|
deepscale
| 26,269
| ||
A class of 10 students took a math test. Each problem was solved by exactly 7 of the students. If the first nine students each solved 4 problems, how many problems did the tenth student solve?
|
Suppose the last student solved $n$ problems, and the total number of problems on the test was $p$. Then the total number of correct solutions written was $7 p$ (seven per problem), and also equal to $36+n$ (the sum of the students' scores), so $p=(36+n) / 7$. The smallest $n \geq 0$ for which this is an integer is $n=6$. But we also must have $n \leq p$, so $7 n \leq 36+n$, and solving gives $n \leq 6$. Thus $n=6$ is the answer.
|
6
|
deepscale
| 3,801
| |
Given vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ satisfy $|\overrightarrow{a}| = 1$, $|\overrightarrow{b}| = 4$, and $\vec{a}\cdot \vec{b}=2$, determine the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$.
|
\dfrac{\pi }{3}
|
deepscale
| 32,798
| ||
What is the sum of all real numbers \(x\) for which \(|x^2 - 14x + 45| = 3?\)
A) 12
B) 14
C) 16
D) 18
|
14
|
deepscale
| 19,070
| ||
A man and his faithful dog simultaneously started moving along the perimeter of a block from point \( A \) at time \( t_{0} = 0 \) minutes. The man moved at a constant speed clockwise, while the dog ran at a constant speed counterclockwise. It is known that the first time they met was \( t_{1} = 2 \) minutes after starting, and this meeting occurred at point \( B \). Given that they continued moving after this, each in their original direction and at their original speed, determine the next time they will both be at point \( B \) simultaneously. Note that \( A B = C D = 100 \) meters and \( B C = A D = 200 \) meters.
|
14
|
deepscale
| 25,995
| ||
Let $\mathcal{T}$ be the set of real numbers that can be represented as repeating decimals of the form $0.\overline{ab}$ where $a$ and $b$ are distinct digits. Find the sum of the elements of $\mathcal{T}$.
|
413.5
|
deepscale
| 20,875
| ||
A cinema is showing four animated movies: Toy Story, Ice Age, Shrek, and Monkey King. The ticket prices are 50 yuan, 55 yuan, 60 yuan, and 65 yuan, respectively. Each viewer watches at least one movie and at most two movies. However, due to time constraints, viewers cannot watch both Ice Age and Shrek. Given that there are exactly 200 people who spent the exact same amount of money on movie tickets today, what is the minimum total number of viewers the cinema received today?
|
1792
|
deepscale
| 27,483
| ||
In $\triangle ABC$, where $\angle C=90^{\circ}$, $\angle B=30^{\circ}$, and $AC=1$, let $M$ be the midpoint of $AB$. $\triangle ACM$ is folded along $CM$ such that the distance between points $A$ and $B$ is $\sqrt{2}$. What is the distance from point $A$ to plane $BCM$?
|
\frac{\sqrt{6}}{3}
|
deepscale
| 14,496
| ||
On Monday at work, David produces $w$ widgets per hour, and works for $t$ hours. Exhausted by this work, on Tuesday, he decides to work for $2$ fewer hours, but manages to produce $4$ additional widgets per hour. If $w = 2t$, how many more widgets did David produce on Monday than on Tuesday?
|
8
|
deepscale
| 33,870
| ||
If the product $\dfrac{4}{3}\cdot \dfrac{5}{4}\cdot \dfrac{6}{5}\cdot \dfrac{7}{6}\cdot \ldots\cdot \dfrac{c}{d} = 12$, determine the sum of $c$ and $d$.
|
71
|
deepscale
| 20,558
| ||
For what value of $x$ is the expression $\frac{2x^3+3}{x^2-20x+100}$ not defined?
|
10
|
deepscale
| 33,127
| ||
A car license plate contains three letters and three digits, for example, A123BE. The allowed letters are А, В, Е, К, М, Н, О, Р, С, Т, У, Х (a total of 12 letters) and all digits except the combination 000. Kira considers a license plate lucky if the second letter is a vowel, the second digit is odd, and the third digit is even (other symbols have no restrictions). How many license plates does Kira consider lucky?
|
359999
|
deepscale
| 25,319
| ||
If \(\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}\) for three positive numbers \(x,y\) and \(z\), all different, then \(\frac{x}{y}=\)
|
Given the equation:
\[
\frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y}
\]
We can use the property that if \(\frac{a_1}{b_1} = \frac{a_2}{b_2} = \ldots = \frac{a_n}{b_n} = k\), then:
\[
\frac{a_1 + a_2 + \ldots + a_n}{b_1 + b_2 + \ldots + b_n} = k
\]
1. **Apply the property to the given equation:**
\[
\frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} = k
\]
Adding the numerators and denominators:
\[
\frac{y + (x+y) + x}{(x-z) + z + y} = \frac{2x+y}{x+y} = k
\]
2. **Simplify the expression:**
\[
\frac{2x+y}{x+y} = k
\]
Since \(k = \frac{x}{y}\), we substitute \(k\) in place of \(\frac{x}{y}\):
\[
\frac{2x+y}{x+y} = \frac{x}{y}
\]
3. **Cross-multiply to solve for \(x\) and \(y\):**
\[
(2x+y)y = x(x+y)
\]
Expanding both sides:
\[
2xy + y^2 = x^2 + xy
\]
Rearranging terms:
\[
x^2 - xy - 2xy + y^2 = 0
\]
Simplifying:
\[
x^2 - 3xy + y^2 = 0
\]
4. **Factorize the quadratic equation:**
\[
(x-y)(x-2y) = 0
\]
This gives us two possible solutions:
- \(x = y\), which contradicts the condition that \(x\), \(y\), and \(z\) are all different.
- \(x = 2y\)
5. **Substitute \(x = 2y\) into \(k = \frac{x}{y}\):**
\[
k = \frac{2y}{y} = 2
\]
Thus, the value of \(\frac{x}{y}\) is \(\boxed{2}\), which corresponds to choice \(\boxed{(E)}\).
|
2
|
deepscale
| 1,600
| |
Estimate
99×71≈
25×39≈
124÷3≈
398÷5≈
|
80
|
deepscale
| 17,529
| ||
Today, Ivan the Confessor prefers continuous functions \(f:[0,1] \rightarrow \mathbb{R}\) satisfying \(f(x)+f(y) \geq|x-y|\) for all pairs \(x, y \in[0,1]\). Find the minimum of \(\int_{0}^{1} f\) over all preferred functions.
|
The minimum of \(\int_{0}^{1} f\) is \(\frac{1}{4}\). Applying the condition with \(0 \leq x \leq \frac{1}{2}, y=x+\frac{1}{2}\) we get $$f(x)+f\left(x+\frac{1}{2}\right) \geq \frac{1}{2}.$$ By integrating, $$\int_{0}^{1} f(x) \mathrm{d} x=\int_{0}^{1 / 2}\left(f(x)+f\left(x+\frac{1}{2}\right)\right) \mathrm{d} x \geq \int_{0}^{1 / 2} \frac{1}{2} \mathrm{~d} x=\frac{1}{4}$$ On the other hand, the function \(f(x)=\left|x-\frac{1}{2}\right|\) satisfies the conditions because $$|x-y|=\left|\left(x-\frac{1}{2}\right)+\left(\frac{1}{2}-y\right)\right| \leq\left|x-\frac{1}{2}\right|+\left|\frac{1}{2}-y\right|=f(x)+f(y)$$ and establishes $$\int_{0}^{1} f(x) \mathrm{d} x=\int_{0}^{1 / 2}\left(\frac{1}{2}-x\right) \mathrm{d} x+\int_{1 / 2}^{1}\left(x-\frac{1}{2}\right) \mathrm{d} x=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$$
|
\frac{1}{4}
|
deepscale
| 5,175
| |
Determine the smallest positive integer $m$ with the property that $m^3-3m^2+2m$ is divisible by both $79$ and $83$ .
|
1660
|
deepscale
| 10,105
| ||
Compute the sum of all possible distinct values of \( m+n \) if \( m \) and \( n \) are positive integers such that
$$
\operatorname{lcm}(m, n) + \operatorname{gcd}(m, n) = 2(m+n) + 11
$$
|
32
|
deepscale
| 15,674
| ||
Given that the random variable $X$ follows a normal distribution $N(0,\sigma^{2})$, if $P(X > 2) = 0.023$, determine the probability $P(-2 \leqslant X \leqslant 2)$.
|
0.954
|
deepscale
| 23,768
| ||
For what value of the parameter \( p \) will the sum of the squares of the roots of the equation
\[ p x^{2}+(p^{2}+p) x-3 p^{2}+2 p=0 \]
be the smallest? What is this smallest value?
|
1.10
|
deepscale
| 12,659
| ||
The game of Penta is played with teams of five players each, and there are five roles the players can play. Each of the five players chooses two of five roles they wish to play. If each player chooses their roles randomly, what is the probability that each role will have exactly two players?
|
Consider a graph with five vertices corresponding to the roles, and draw an edge between two vertices if a player picks both roles. Thus there are exactly 5 edges in the graph, and we want to find the probability that each vertex has degree 2. In particular, we want to find the probability that the graph is composed entirely of cycles. Thus there are two cases. The first case is when the graph is itself a 5-cycle. There are 4! ways to choose such a directed cycle (pick an arbitrary vertex $A$ and consider a vertex it connects to, etc.), and thus $\frac{4!}{2}=12$ ways for the undirected graph to be a 5-cycle. Now, there are 5! ways to assign the edges in this cycle to people, giving a total contribution of $12 \cdot 5$!. The second case is when the graph is composed of a 2-cycle and a 3-cycle, which only requires choosing the two vertices to be the 2-cycle, and so there are $\binom{5}{2}=10$ ways. To assign the players to edges, there are $\binom{5}{2}=10$ ways to assign the players to the 2-cycle. For the 3-cycle, any of the $3!=6$ permutations of the remaining 3 players work. The total contribution is $10 \cdot 10 \cdot 6$. Therefore, our answer is $\frac{12 \cdot 120+10 \cdot 10 \cdot 6}{10^{5}}=\frac{51}{2500}$.
|
\frac{51}{2500}
|
deepscale
| 3,367
| |
What is the value of $x$ if $P Q S$ is a straight line and $\angle P Q R=110^{\circ}$?
|
Since $P Q S$ is a straight line and $\angle P Q R=110^{\circ}$, then $\angle R Q S=180^{\circ}-\angle P Q R=70^{\circ}$. Since the sum of the angles in $\triangle Q R S$ is $180^{\circ}$, then $70^{\circ}+(3 x)^{\circ}+(x+14)^{\circ} =180^{\circ}$. Solving, $4 x =96$ gives $x =24$.
|
24
|
deepscale
| 5,626
| |
Let $\omega$ be a circle with radius $1$ . Equilateral triangle $\vartriangle ABC$ is tangent to $\omega$ at the midpoint of side $BC$ and $\omega$ lies outside $\vartriangle ABC$ . If line $AB$ is tangent to $\omega$ , compute the side length of $\vartriangle ABC$ .
|
\frac{2 \sqrt{3}}{3}
|
deepscale
| 26,761
| ||
If the intended number is multiplied by 6, then 382 is added to the product, the result is the largest three-digit number written with two identical even digits and one odd digit. Find the intended number.
|
101
|
deepscale
| 15,335
| ||
What is the slope of the line $2y = -3x + 6$?
|
-\frac32
|
deepscale
| 33,840
| ||
In triangle $ABC$, the three internal angles are $A$, $B$, and $C$. Find the value of $A$ for which $\cos A + 2\cos\frac{B+C}{2}$ attains its maximum value, and determine this maximum value.
|
\frac{3}{2}
|
deepscale
| 20,610
| ||
At what point does the line containing the points $(1, 7)$ and $(3, 11)$ intersect the $y$-axis? Express your answer as an ordered pair.
|
(0,5)
|
deepscale
| 33,019
| ||
Find the smallest exact square with last digit not $0$ , such that after deleting its last two digits we shall obtain another exact square.
|
121
|
deepscale
| 7,463
| ||
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?
|
1. **Identify the invariant**: The center square remains unchanged after a $90^\circ$ rotation. Since the entire grid must end up black, the center square must initially be black. The probability of this happening is $\frac{1}{2}$.
2. **Consider the effect of rotation on the 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. **Calculate the probability for the 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 $\frac{1}{4} + \frac{1}{4} - \frac{1}{4} \times \frac{1}{4} = \frac{7}{16}$ (using the Principle of Inclusion-Exclusion).
4. **Calculate the probability for the corner squares**: Similarly, there are four corner squares, and each pair of diagonally opposite corner squares must be black. The calculation follows the same logic as for the edge squares, so the probability that all corner squares are black is also $\frac{7}{16}$.
5. **Combine the probabilities**: The probability that the center square is black, all edge squares are black, and all corner squares are black is the product of their independent probabilities:
\[
\frac{1}{2} \times \frac{7}{16} \times \frac{7}{16} = \frac{49}{512}
\]
6. **Conclusion**: The probability that the entire grid is black after the described operation is $\boxed{\textbf{(A)}\ \frac{49}{512}}$.
|
\frac{49}{512}
|
deepscale
| 1,902
| |
John ate a total of 120 peanuts over four consecutive nights. Each night he ate 6 more peanuts than the night before. How many peanuts did he eat on the fourth night?
|
Suppose that John ate \( x \) peanuts on the fourth night. Since he ate 6 more peanuts each night than on the previous night, then he ate \( x-6 \) peanuts on the third night, \((x-6)-6=x-12\) peanuts on the second night, and \((x-12)-6=x-18\) peanuts on the first night. Since John ate 120 peanuts in total, then \( x+(x-6)+(x-12)+(x-18)=120 \), and so \( 4x-36=120 \) or \( 4x=156 \) or \( x=39 \). Therefore, John ate 39 peanuts on the fourth night.
|
39
|
deepscale
| 5,247
| |
Let \(\omega\) denote the incircle of triangle \(ABC\). The segments \(BC, CA\), and \(AB\) are tangent to \(\omega\) at \(D, E\), and \(F\), respectively. Point \(P\) lies on \(EF\) such that segment \(PD\) is perpendicular to \(BC\). The line \(AP\) intersects \(BC\) at \(Q\). The circles \(\omega_1\) and \(\omega_2\) pass through \(B\) and \(C\), respectively, and are tangent to \(AQ\) at \(Q\); the former meets \(AB\) again at \(X\), and the latter meets \(AC\) again at \(Y\). The line \(XY\) intersects \(BC\) at \(Z\). Given that \(AB=15\), \(BC=14\), and \(CA=13\), find \(\lfloor XZ \cdot YZ \rfloor\).
|
196
|
deepscale
| 8,269
| ||
What is the greatest four-digit number which is a multiple of 17?
|
9996
|
deepscale
| 19,334
| ||
In the Cartesian coordinate system $xOy$, it is known that the circle $C: x^{2} + y^{2} + 8x - m + 1 = 0$ intersects with the line $x + \sqrt{2}y + 1 = 0$ at points $A$ and $B$. If $\triangle ABC$ is an equilateral triangle, then the value of the real number $m$ is.
|
-11
|
deepscale
| 31,741
| ||
The teacher fills some numbers into the circles in the diagram below (each circle can and must only contain one number). The sum of the three numbers in each of the left and right closed loops is 30, and the sum of the four numbers in each of the top and bottom closed loops is 40. If the number in circle $X$ is 9, then the number in circle $Y$ is $\qquad$
|
11
|
deepscale
| 11,879
| ||
Let \( k_{1} \) and \( k_{2} \) be two circles with the same center, with \( k_{2} \) inside \( k_{1} \). Let \( A \) be a point on \( k_{1} \) and \( B \) a point on \( k_{2} \) such that \( AB \) is tangent to \( k_{2} \). Let \( C \) be the second intersection of \( AB \) and \( k_{1} \), and let \( D \) be the midpoint of \( AB \). A line passing through \( A \) intersects \( k_{2} \) at points \( E \) and \( F \) such that the perpendicular bisectors of \( DE \) and \( CF \) intersect at a point \( M \) which lies on \( AB \). Find the value of \( \frac{AM}{MC} \).
|
5/3
|
deepscale
| 24,970
| ||
In a spelling bee $50\%$ of the students were eliminated after the first round. Only $\frac{1}{3}$ of the remaining students were still in the contest after the second round. If 24 students were still in the contest after the second round, how many students began the contest?
|
144
|
deepscale
| 38,790
| ||
In an opaque bag, there are a total of 50 glass balls in red, black, and white colors. Except for the color, everything else is the same. After several trials of drawing balls, Xiaogang found that the probability of drawing a red or black ball is stable at 15% and 45%, respectively. What could be the possible number of white balls in the bag?
|
20
|
deepscale
| 20,939
| ||
Given that the midpoint of side $BC$ of triangle $\triangle ABC$ is $D$, point $E$ lies in the plane of $\triangle ABC$, and $\overrightarrow{CD}=3\overrightarrow{CE}-2\overrightarrow{CA}$, if $\overrightarrow{AC}=x\overrightarrow{AB}+y\overrightarrow{BE}$, then determine the value of $x+y$.
|
11
|
deepscale
| 28,436
| ||
A mason has bricks with dimensions $2\times5\times8$ and other bricks with dimensions $2\times3\times7$ . She also has a box with dimensions $10\times11\times14$ . The bricks and the box are all rectangular parallelepipeds. The mason wants to pack bricks into the box filling its entire volume and with no bricks sticking out.
Find all possible values of the total number of bricks that she can pack.
|
24
|
deepscale
| 22,469
| ||
A vessel with a capacity of 100 liters is filled with a brine solution containing 10 kg of dissolved salt. Every minute, 3 liters of water flows into it, and the same amount of the resulting mixture is pumped into another vessel of the same capacity, initially filled with water, from which the excess liquid overflows. At what point in time will the amount of salt in both vessels be equal?
|
333.33
|
deepscale
| 12,572
| ||
In a sequence, every (intermediate) term is half of the arithmetic mean of its neighboring terms. What relationship exists between any term, and the terms that are 2 positions before and 2 positions after it? The first term of the sequence is 1, and the 9th term is 40545. What is the 25th term?
|
57424611447841
|
deepscale
| 31,118
| ||
Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
|
Note that we can pick the first and second letters/numbers freely with one choice left for the last letter/number for there to be a palindrome. Thus, the probability of no palindrome is \[\frac{25}{26}\cdot \frac{9}{10}=\frac{45}{52}\] thus we have $1-\frac{45}{52}=\frac{7}{52}$ so our answer is $7+52 = \boxed{059}.$
~Dhillonr25
~ pi_is_3.14
|
59
|
deepscale
| 6,731
| |
Among a group of 120 people, some pairs are friends. A [i]weak quartet[/i] is a set of four people containing exactly one pair of friends. What is the maximum possible number of weak quartets ?
|
Given a group of 120 people, where some pairs are friends, we need to determine the maximum possible number of weak quartets. A weak quartet is defined as a set of four people containing exactly one pair of friends.
To solve this, we need to analyze the structure of weak quartets:
1. **Count the total number of quartets:**
The total number of ways to choose 4 people out of 120 is given by the combination formula:
\[
\binom{120}{4} = \frac{120 \times 119 \times 118 \times 117}{4 \times 3 \times 2 \times 1} = 2550240.
\]
2. **Count the number of quartets that could be considered as weak quartets:**
First, select a pair of friends, and then choose the other two people from the 118 remaining people. If \( f \) is the number of pairs of friends, then:
\[
\text{Number of ways to form a weak quartet involving a specific pair of friends} = f \times \binom{118}{2}.
\]
3. **Maximize the number of weak quartets:**
To maximize the number of weak quartets, assume the maximum possible number of friendship pairs. According to the combinatorial principle, the maximum number of friendship pairs among 120 people occurs when every possible pair of persons is friends:
\[
f = \binom{120}{2} = \frac{120 \times 119}{2} = 7140.
\]
4. **Thus, the maximum possible number of weak quartets is:**
\[
7140 \times \binom{118}{2} = 7140 \times \frac{118 \times 117}{2} = 7140 \times 6903 = 4769280.
\]
Therefore, the maximum possible number of weak quartets is:
\[
\boxed{4769280}.
\]
|
4769280
|
deepscale
| 6,237
| |
What is the smallest number with three different prime factors, none of which can be less than 10?
|
2431
|
deepscale
| 14,409
| ||
In triangle ABC, where AB = 24 and BC = 18, find the largest possible value of $\tan A$.
|
\frac{3\sqrt{7}}{7}
|
deepscale
| 19,733
| ||
Take a standard set of dominoes and remove all duplicates and blanks. Then consider the remaining 15 dominoes as fractions. The dominoes are arranged in such a way that the sum of all fractions in each row equals 10. However, unlike proper fractions, you are allowed to use as many improper fractions (such as \(\frac{4}{3}, \frac{5}{2}, \frac{6}{1}\)) as you wish, as long as the sum in each row equals 10.
|
10
|
deepscale
| 19,749
| ||
Nathaniel and Obediah play a game in which they take turns rolling a fair six-sided die and keep a running tally of the sum of the results of all rolls made. A player wins if, after he rolls, the number on the running tally is a multiple of 7. Play continues until either player wins. If Nathaniel goes first, determine the probability that he ends up winning.
|
\frac{5}{11}
|
deepscale
| 13,808
| ||
From point $A$ outside a circle, a tangent and a secant are drawn to the circle. The distance from point $A$ to the point of tangency is 16, and the distance from point $A$ to one of the intersection points of the secant with the circle is 32. Find the radius of the circle if the distance from its center to the secant is 5.
|
13
|
deepscale
| 9,779
| ||
If $5$ lunks can be traded for $3$ kunks, and $2$ kunks will buy $4$ apples, how many lunks are needed to purchase one dozen apples?
|
10
|
deepscale
| 33,153
| ||
Let $a$, $b$, $c$ be the sides of a triangle, and let $\alpha$, $\beta$, $\gamma$ be the angles opposite them respectively. If $a^2+b^2=2c^2$, find the value of
\[
\frac{\cot \gamma}{\cot \alpha + \cot \beta}.
\]
|
\frac{1}{2}
|
deepscale
| 11,510
| ||
If a computer executes the following program:
1. Initial values \( x = 3 \), \( S = 0 \)
2. \( x = x + 2 \)
3. \( S = S + x \)
4. If \( S \geq 10000 \), go to step 5; otherwise, repeat from step 2
5. Print \( x \)
6. Stop
What value will be printed by step 5?
|
201
|
deepscale
| 21,123
| ||
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 $a_{n+i}=a_i,i=1,2,\ldots,k$
|
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}.
|
n - k
|
deepscale
| 3,000
| |
The store bought a batch of New Year cards at 0.21 yuan each and sold them for a total of 14.57 yuan. If each card was sold at the same price and did not exceed twice the purchase price, how much profit did the store make?
|
4.7
|
deepscale
| 13,370
| ||
The center of the circle with equation $x^2+y^2=8x-6y-20$ is the point $(x,y)$. What is $x+y$?
|
1
|
deepscale
| 33,643
| ||
A rational number written in base eight is $\underline{ab} . \underline{cd}$, where all digits are nonzero. The same number in base twelve is $\underline{bb} . \underline{ba}$. Find the base-ten number $\underline{abc}$.
|
The parts before and after the decimal points must be equal. Therefore $8a + b = 12b + b$ and $c/8 + d/64 = b/12 + a/144$. Simplifying the first equation gives $a = (3/2)b$. Plugging this into the second equation gives $3b/32 = c/8 + d/64$. Multiplying both sides by 64 gives $6b = 8c + d$. $a$ and $b$ are both digits between 1 and 7 (they must be a single non-zero digit in base eight) so using $a = 3/2b$, $(a,b) = (3,2)$ or $(6,4)$. Testing these gives that $(6,4)$ doesn't work, and $(3,2)$ gives $a = 3, b = 2, c = 1$, and $d = 4$. Therefore $abc = \boxed{321}$
~Shreyas S
|
321
|
deepscale
| 7,185
| |
What is the smallest number divisible by integers 1 through 9?
|
2520
|
deepscale
| 37,670
| ||
Compute $\dbinom{9}{2}\times \dbinom{7}{2}$.
|
756
|
deepscale
| 35,170
| ||
The line with equation $y = 3x + 5$ is translated 2 units to the right. What is the equation of the resulting line?
|
The line with equation $y = 3x + 5$ has slope 3 and $y$-intercept 5. Since the line has $y$-intercept 5, it passes through $(0, 5)$. When the line is translated 2 units to the right, its slope does not change and the new line passes through $(2, 5)$. A line with slope $m$ that passes through the point $(x_1, y_1)$ has equation $y - y_1 = m(x - x_1)$. Therefore, the line with slope 3 that passes through $(2, 5)$ has equation $y - 5 = 3(x - 2)$ or $y - 5 = 3x - 6$, which gives $y = 3x - 1$. Alternatively, we could note that when the graph of $y = 3x + 5$ is translated 2 units to the right, the equation of the new graph is $y = 3(x - 2) + 5$ or $y = 3x - 1$.
|
y = 3x - 1
|
deepscale
| 5,229
| |
At the intersection of perpendicular roads, a highway from Moscow to Kazan intersects with a road from Vladimir to Ryazan. Dima and Tolya are traveling at constant speeds from Moscow to Kazan and from Vladimir to Ryazan, respectively.
When Dima crossed the intersection, Tolya was 900 meters away from it. When Tolya crossed the intersection, Dima was 600 meters away from it. How many meters will be between the boys when Tolya travels 900 meters from the moment he crosses the intersection?
|
1500
|
deepscale
| 16,133
| ||
Calculate the sum of the series $1-2-3+4+5-6-7+8+9-10-11+\cdots+1998+1999-2000-2001$.
|
2001
|
deepscale
| 32,152
| ||
Given that the 32-digit integer 64312311692944269609355712372657 is the product of 6 consecutive primes, compute the sum of these 6 primes.
|
Because the product is approximately $64 \cdot 10^{30}$, we know the primes are all around 200000. Say they are $200000+x_{i}$ for $i=1, \ldots, 6$. By expanding $\prod_{i=1}^{6}\left(200000+x_{i}\right)$ as a polynomial in 200000, we see that $$31231 \cdot 10^{25}=200000^{5}\left(x_{1}+\cdots+x_{6}\right)$$ plus the carry from the other terms. Note that $31231=975 \cdot 32+31$, so $x_{1}+\cdots+x_{6} \leq 975$. Thus, $$16\left(x_{1}x_{2}+x_{1}x_{3}+\cdots+x_{5}x_{6}\right) \leq 16 \cdot \frac{5}{12}\left(x_{1}+\cdots+x_{6}\right)^{2}<\frac{20}{3} \cdot 1000^{2}<67 \cdot 10^{5}$$ so the carry term from $200000^{4}\left(x_{1}x_{2}+\cdots+x_{5}x_{6}\right)$ is at most $67 \cdot 10^{25}$. The other terms have negligible carry, so it is pretty clear $x_{1}+\cdots+x_{6}>972$, otherwise the carry term would have to be at least $$31231 \cdot 10^{25}-200000^{5}(972)=127 \cdot 10^{25}$$ It follows that $x_{1}+\cdots+x_{6}$ lies in [973, 975], so the sum of the primes, $6 \cdot 200000+\left(x_{1}+\cdots+x_{6}\right)$, lies in $[1200973,1200975]$. As these primes are all greater than 2, they are all odd, so their sum is even. Thus it must be 1200974.
|
1200974
|
deepscale
| 3,393
| |
In rectangle $ABCD,$ $AB=15$ and $AC=17.$ What is the area of rectangle $ABCD?$ Additionally, find the length of the diagonal $BD.$
|
17
|
deepscale
| 23,265
| ||
Find the maximum value of the expression \( (\sin 2x + \sin 3y + \sin 4z)(\cos 2x + \cos 3y + \cos 4z) \).
|
4.5
|
deepscale
| 15,169
| ||
Let $x, y,$ and $z$ be positive real numbers. Find the minimum value of:
\[
\frac{(x^2 + 4x + 2)(y^2 + 4y + 2)(z^2 + 4z + 2)}{xyz}.
\]
|
216
|
deepscale
| 28,028
| ||
Given the sequence $\left\{a_{n}\right\}$ that satisfies
$$
a_{n-1} = a_{n} + a_{n-2} \quad (n \geqslant 3),
$$
let $S_{n}$ be the sum of the first $n$ terms. If $S_{2018} = 2017$ and $S_{2019} = 2018$, then find $S_{20200}$.
|
1010
|
deepscale
| 7,762
| ||
Let $y=(x-a)^2+(x-b)^2, a, b$ constants. For what value of $x$ is $y$ a minimum?
|
To find the value of $x$ that minimizes $y = (x-a)^2 + (x-b)^2$, we can start by expanding and simplifying the expression for $y$:
1. Expand the squares:
\[
y = (x-a)^2 + (x-b)^2 = (x^2 - 2ax + a^2) + (x^2 - 2bx + b^2)
\]
2. Combine like terms:
\[
y = 2x^2 - 2ax - 2bx + a^2 + b^2 = 2x^2 - 2(a+b)x + (a^2 + b^2)
\]
3. To find the minimum value of $y$, we can complete the square or differentiate and set the derivative to zero. We'll use differentiation here:
\[
\frac{dy}{dx} = 4x - 2(a+b)
\]
Set the derivative equal to zero to find the critical points:
\[
4x - 2(a+b) = 0 \implies x = \frac{a+b}{2}
\]
4. To confirm that this critical point is indeed a minimum, we check the second derivative:
\[
\frac{d^2y}{dx^2} = 4
\]
Since the second derivative is positive, the function has a local minimum at $x = \frac{a+b}{2}$.
5. We can also verify this by substituting $x = \frac{a+b}{2}$ back into the expression for $y$:
\[
y = 2\left(\frac{a+b}{2} - a\right)^2 + 2\left(\frac{a+b}{2} - b\right)^2 = \frac{(b-a)^2}{2} + \frac{(a-b)^2}{2} = (b-a)^2
\]
This expression is clearly minimized when $x = \frac{a+b}{2}$, as it simplifies the terms inside the squares.
Thus, the value of $x$ that minimizes $y$ is $\boxed{\textbf{(A)}\ \frac{a+b}{2}}$.
|
\frac{a+b}{2}
|
deepscale
| 2,399
| |
An experimenter is conducting an experiment that involves implementing five procedures in sequence. Procedure A must only occur either as the first or the last step, and procedures C and D must be implemented consecutively. The number of possible arrangements for the sequence of these procedures is _______.
|
24
|
deepscale
| 22,536
| ||
Lesha's summer cottage has the shape of a nonagon with three pairs of equal and parallel sides. Lesha knows that the area of the triangle with vertices at the midpoints of the remaining sides of the nonagon is 12 sotkas. Help him find the area of the entire summer cottage.
|
48
|
deepscale
| 31,342
| ||
A kite has sides $15$ units and $20$ units meeting at a right angle, and its diagonals are $24$ units and $x$ units. Find the area of the kite.
|
216
|
deepscale
| 18,539
| ||
How many different five-letter words can be formed such that they start and end with the same letter, and the middle letter is always 'A'?
|
17576
|
deepscale
| 19,368
| ||
A natural number undergoes the following operation: the rightmost digit of its decimal representation is discarded, and then the number obtained after discarding is added to twice the discarded digit. For example, $157 \mapsto 15 + 2 \times 7 = 29$, $5 \mapsto 0 + 2 \times 5 = 10$. A natural number is called ‘good’ if after repeatedly applying this operation, the resulting number stops changing. Find the smallest such good number.
|
19
|
deepscale
| 10,547
| ||
The ratio of butter:flour:sugar in a recipe is 1:6:4. When using 8 cups of sugar in this recipe, how many total cups of these three ingredients will be used?
|
22
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deepscale
| 38,461
| ||
Let $A = (0,0)$ and $B = (b,2)$ be points on the coordinate plane. Let $ABCDEF$ be a convex equilateral hexagon such that $\angle FAB = 120^\circ,$ $\overline{AB}\parallel \overline{DE},$ $\overline{BC}\parallel \overline{EF,}$ $\overline{CD}\parallel \overline{FA},$ and the y-coordinates of its vertices are distinct elements of the set $\{0,2,4,6,8,10\}.$ The area of the hexagon can be written in the form $m\sqrt {n},$ where $m$ and $n$ are positive integers and n is not divisible by the square of any prime. Find $m + n.$
|
The y-coordinate of $F$ must be $4$. All other cases yield non-convex and/or degenerate hexagons, which violate the problem statement.
Letting $F = (f,4)$, and knowing that $\angle FAB = 120^\circ$, we can use rewrite $F$ using complex numbers: $f + 4 i = (b + 2 i)\left(e^{i(2 \pi / 3)}\right) = (b + 2 i)\left(-1/2 + \frac{\sqrt{3}}{2} i\right) = -\frac{b}{2}-\sqrt{3}+\left(\frac{b\sqrt{3}}{2}-1\right)i$. We solve for $b$ and $f$ and find that $F = \left(-\frac{8}{\sqrt{3}}, 4\right)$ and that $B = \left(\frac{10}{\sqrt{3}}, 2\right)$.
The area of the hexagon can then be found as the sum of the areas of two congruent triangles ($EFA$ and $BCD$, with height $8$ and base $\frac{8}{\sqrt{3}}$) and a parallelogram ($ABDE$, with height $8$ and base $\frac{10}{\sqrt{3}}$).
$A = 2 \times \frac{1}{2} \times 8 \times \frac{8}{\sqrt{3}} + 8 \times \frac{10}{\sqrt{3}} = \frac{144}{\sqrt{3}} = 48\sqrt{3}$.
Thus, $m+n = \boxed{051}$.
|
51
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deepscale
| 6,789
| |
One of the five faces of the triangular prism shown here will be used as the base of a new pyramid. The numbers of exterior faces, vertices and edges of the resulting shape (the fusion of the prism and pyramid) are added. What is the maximum value of this sum?
[asy]
draw((0,0)--(9,12)--(25,0)--cycle);
draw((9,12)--(12,14)--(28,2)--(25,0));
draw((12,14)--(3,2)--(0,0),dashed);
draw((3,2)--(28,2),dashed);
[/asy]
|
28
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deepscale
| 36,178
| ||
A wholesaler purchased $50$ packs of shirts of size $L$ from a clothing manufacturer. Due to the negligence of the packaging workers, some packs were mixed with shirts of size $M$. The number of $M$ shirts mixed in (pieces) and the corresponding number of packs (packs) are shown in the table below:
|M shirts (pieces)|$0$|$1$|$4$|$5$|$7$|$9$|$10$|$11$|
|---|---|---|---|---|---|---|---|---|
|Packs| $7$| $3$| $10$| $15$| $5$| $4$| $3$| $3$|
A retailer randomly selected one pack from the $50$ packs. Find the probabilities of the following events:
$(1)$ No $M$ shirts were mixed in the pack;
$(2)$ $M$ shirts were mixed in the pack and the number of pieces is less than $7$;
$(3)$ The number of $M$ shirts mixed in the pack is more than $9$.
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\frac{3}{25}
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deepscale
| 16,222
| ||
The Absent-Minded Scientist had a sore knee. The doctor prescribed him 10 pills for his knee: take one pill daily. The pills are effective in $90\%$ of cases, and in $2\%$ of cases, there is a side effect—absent-mindedness disappears, if present.
Another doctor prescribed the Scientist pills for absent-mindedness—also one per day for 10 consecutive days. These pills cure absent-mindedness in $80\%$ of cases, but in $5\%$ of cases, there is a side effect—the knee stops hurting.
The bottles with the pills look similar, and when the Scientist went on a ten-day business trip, he took one bottle with him but didn't pay attention to which one. For ten days, he took one pill per day and returned completely healthy. Both the absent-mindedness and the knee pain were gone. Find the probability that the Scientist took pills for absent-mindedness.
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0.69
|
deepscale
| 13,045
|
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