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Our football team has 10 members, of which only 3 are strong enough to play offensive lineman, while all other positions can be played by anyone. In how many ways can we choose a starting lineup consisting of a quarterback, a running back, an offensive lineman, and a wide receiver?
|
1512
|
deepscale
| 35,298
| ||
What is the smallest positive value of $m$ so that the equation $10x^2 - mx + 630 = 0$ has integral solutions?
|
160
|
deepscale
| 19,376
| ||
Find $d$, given that $\lfloor d\rfloor$ is a solution to \[3x^2 + 19x - 70 = 0\] and $\{d\} = d - \lfloor d\rfloor$ is a solution to \[4x^2 - 12x + 5 = 0.\]
|
-8.5
|
deepscale
| 31,423
| ||
The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares?
|
1. **Define the integers**: Let the three consecutive positive integers be $a-1$, $a$, and $a+1$.
2. **Express the sum and product**:
- The sum of these integers is $(a-1) + a + (a+1) = 3a$.
- The product of these integers is $(a-1)a(a+1)$.
3. **Set up the equation**: According to the problem, the product of the integers is $8$ times their sum. Therefore, we have:
\[
(a-1)a(a+1) = 8 \times 3a
\]
Simplifying the right side, we get:
\[
(a-1)a(a+1) = 24a
\]
4. **Simplify and solve for $a$**:
- Divide both sides by $a$ (assuming $a \neq 0$):
\[
(a-1)(a+1) = 24
\]
- Expanding the left side:
\[
a^2 - 1 = 24
\]
- Solving for $a^2$:
\[
a^2 = 25 \implies a = 5 \quad (\text{since } a \text{ is positive})
\]
5. **Calculate the squares of the integers**:
- The integers are $4$, $5$, and $6$ (since $a = 5$).
- The squares are $4^2 = 16$, $5^2 = 25$, and $6^2 = 36$.
6. **Sum the squares**:
\[
16 + 25 + 36 = 77
\]
7. **Conclusion**: The sum of the squares of the three consecutive integers is $\boxed{77}$.
|
77
|
deepscale
| 2,541
| |
Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each $\frac {1}{3}$. The probability that Club Truncator will finish the season with more wins than losses is $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
|
At first, it wins $6$ games, only one way
Secondly, it wins $5$ games, the other game can be either win or loss, there are $\binom{6}{5}\cdot 2=12$ ways
Thirdly, it wins $4$ games, still the other two games can be either win or loss, there are $\binom{6}{4}\cdot 2^2=60$ ways
Fourthly, it wins $3$ games, this time, it can't lose $3$ games but other arrangements of the three non-winning games are fine, there are $\binom{6}{3}\cdot (2^3-1)=140$ ways
Fifth case, it wins $2$ games, only $0/1$ lose and $4/3$ draw is ok, so there are $\binom{6}{2}(1+\binom{4}{1})=75$ cases
Last case, it only wins $1$ game so the rest games must be all draw, $1$ game
The answer is $\frac{1+12+60+140+75+6}{3^6}=\frac{98}{243}$ leads to $\boxed{341}$
~bluesoul
|
341
|
deepscale
| 6,726
| |
What is the least integer whose square is 36 more than three times its value?
|
-6
|
deepscale
| 29,588
| ||
The number of positive integers less than $1000$ divisible by neither $5$ nor $7$ is:
|
1. **Count the integers divisible by 5:**
The number of positive integers less than $1000$ that are divisible by $5$ can be calculated using the floor function:
\[
\left\lfloor \frac{999}{5} \right\rfloor = 199
\]
This is because $999$ is the largest number less than $1000$, and dividing it by $5$ and taking the floor gives us the count of multiples of $5$.
2. **Count the integers divisible by 7:**
Similarly, the number of positive integers less than $1000$ that are divisible by $7$ is:
\[
\left\lfloor \frac{999}{7} \right\rfloor = 142
\]
Here, we use $999$ again as it is the largest number less than $1000$, and dividing by $7$ and taking the floor gives the count of multiples of $7$.
3. **Count the integers divisible by both 5 and 7 (i.e., 35):**
Numbers divisible by both $5$ and $7$ are divisible by $35$ (since $5 \times 7 = 35$). The count of such numbers is:
\[
\left\lfloor \frac{999}{35} \right\rfloor = 28
\]
4. **Apply the Inclusion-Exclusion Principle:**
To find the number of integers divisible by either $5$ or $7$, we initially add the counts from steps 1 and 2, then subtract the count from step 3 to avoid double-counting:
\[
199 + 142 - 28 = 313
\]
This calculation accounts for the overcounting of numbers divisible by both $5$ and $7$.
5. **Calculate the number of integers divisible by neither 5 nor 7:**
The total number of positive integers less than $1000$ is $999$. Subtracting the number of integers divisible by either $5$ or $7$ gives:
\[
999 - 313 = 686
\]
6. **Conclusion:**
The number of positive integers less than $1000$ that are divisible by neither $5$ nor $7$ is $\boxed{686}$. This corresponds to answer choice $\text{(B)}$.
|
686
|
deepscale
| 363
| |
Given the function $y=\sin 3x$, determine the horizontal shift required to obtain the graph of the function $y=\sin \left(3x+\frac{\pi }{4}\right)$.
|
\frac{\pi}{12}
|
deepscale
| 17,988
| ||
What is 25% of 60?
|
Expressed as a fraction, $25 \%$ is equivalent to $\frac{1}{4}$. Since $\frac{1}{4}$ of 60 is 15, then $25 \%$ of 60 is 15.
|
15
|
deepscale
| 5,771
| |
I live on a very short street with 14 small family houses. The odd-numbered houses from 1 are on one side of the street, and the even-numbered houses from 2 are on the opposite side (e.g., 1 and 2 are opposite each other).
On one side of the street, all families have surnames that are colors, and on the other side, the surnames indicate professions.
Szabó and Fazekas live opposite to Zöld and Fehér, respectively, who are both neighbors of Fekete.
Kovács is the father-in-law of Lakatos.
Lakatos lives in a higher-numbered house than Barna. The sum of the house numbers of Lakatos and Barna is equal to the sum of the house numbers of Fehér and Fazekas. Kádárné's house number is twice the house number of her sister, Kalaposné.
Sárga lives opposite Pék.
If Bordóné's house number is two-digit and she lives opposite her sister, Kádárné, what is the house number of Mr. Szürke?
|
13
|
deepscale
| 27,534
| ||
Roll two dice consecutively. Let the number on the first die be $m$, and the number on the second die be $n$. Calculate the probability that:
(1) $m+n=7$;
(2) $m=n$;
(3) The point $P(m,n)$ is inside the circle $x^2+y^2=16$.
|
\frac{2}{9}
|
deepscale
| 18,774
| ||
What is the maximum number of checkers that can be placed on a $6 \times 6$ board so that no three checkers (more precisely, the centers of the cells they occupy) are collinear (in any direction)?
|
12
|
deepscale
| 14,701
| ||
Lily is riding her bicycle at a constant rate of 15 miles per hour and Leo jogs at a constant rate of 9 miles per hour. If Lily initially sees Leo 0.75 miles in front of her and later sees him 0.75 miles behind her, determine the duration of time, in minutes, that she can see Leo.
|
15
|
deepscale
| 10,430
| ||
A digital watch displays hours and minutes in a 24-hour format. Calculate the largest possible sum of the digits in the display.
|
24
|
deepscale
| 23,467
| ||
Given that $\theta$ is an angle in the third quadrant, if $\sin^{4}{\theta} + \cos^{4}{\theta} = \frac{5}{9}$, find the value of $\sin{2\theta}$.
|
\frac{2\sqrt{2}}{3}
|
deepscale
| 17,590
| ||
If the integers $m,n,k$ hold the equation $221m+247n+323k=2001$ , what is the smallest possible value of $k$ greater than $100$ ?
|
111
|
deepscale
| 24,457
| ||
Five rays $\overrightarrow{OA}$ , $\overrightarrow{OB}$ , $\overrightarrow{OC}$ , $\overrightarrow{OD}$ , and $\overrightarrow{OE}$ radiate in a clockwise order from $O$ forming four non-overlapping angles such that $\angle EOD = 2\angle COB$ , $\angle COB = 2\angle BOA$ , while $\angle DOC = 3\angle BOA$ . If $E$ , $O$ , $A$ are collinear with $O$ between $A$ and $E$ , what is the degree measure of $\angle DOB?$
|
90
|
deepscale
| 19,492
| ||
$\dfrac{10-9+8-7+6-5+4-3+2-1}{1-2+3-4+5-6+7-8+9}=$
|
1. **Simplify the Numerator**:
The numerator of the given fraction is $10-9+8-7+6-5+4-3+2-1$. We can group the terms in pairs:
\[
(10-9) + (8-7) + (6-5) + (4-3) + (2-1)
\]
Each pair simplifies to $1$, so:
\[
1 + 1 + 1 + 1 + 1 = 5
\]
2. **Simplify the Denominator**:
The denominator of the fraction is $1-2+3-4+5-6+7-8+9$. We can rearrange and group the terms to simplify:
\[
(1-2) + (3-4) + (5-6) + (7-8) + 9
\]
Each pair except the last term simplifies to $-1$, and adding $9$ at the end:
\[
-1 -1 -1 -1 + 9 = -4 + 9 = 5
\]
3. **Calculate the Fraction**:
Now, substituting the simplified numerator and denominator back into the fraction:
\[
\dfrac{5}{5} = 1
\]
4. **Conclusion**:
The value of the given expression is $1$. Therefore, the correct answer is $\boxed{\text{B}}$.
|
1
|
deepscale
| 1,686
| |
Let the function \( f(x) \) satisfy the following conditions:
(i) If \( x > y \), and \( f(x) + x \geq w \geq f(y) + y \), then there exists a real number \( z \in [y, x] \), such that \( f(z) = w - z \);
(ii) The equation \( f(x) = 0 \) has at least one solution, and among the solutions of this equation, there exists one solution that is not greater than all other solutions;
(iii) \( f(0) = 1 \);
(iv) \( f(-2003) \leq 2004 \);
(v) \( f(x) \cdot f(y) = f[x \cdot f(y) + y \cdot f(x) + xy] \).
Find the value of \( f(-2003) \).
|
2004
|
deepscale
| 10,587
| ||
Given a cube with its eight vertices labeled with the numbers $1, 2, 3, \cdots, 8$ in any order, define the number on each edge as $|i-j|$, where $i$ and $j$ are the labels of the edge’s endpoints. Let $S$ be the sum of the numbers on all the edges. Find the minimum value of $S$.
|
28
|
deepscale
| 15,685
| ||
Given the function $$f(x)= \begin{cases} 2\cos \frac {\pi }{3}x & x\leq 2000 \\ x-100 & x>2000\end{cases}$$, then $f[f(2010)]$ equals \_\_\_\_\_\_\_\_\_\_\_\_.
|
-1
|
deepscale
| 24,260
| ||
Given an angle $α$ whose terminal side passes through the point $(-1, \sqrt{2})$, find the values of $\tan α$ and $\cos 2α$.
|
-\frac{1}{3}
|
deepscale
| 28,082
| ||
A projectile is launched with an initial velocity of $u$ at an angle of $\phi$ from the horizontal. The trajectory of the projectile is given by the parametric equations:
\[
x = ut \cos \phi,
\]
\[
y = ut \sin \phi - \frac{1}{2} gt^2,
\]
where $t$ is time and $g$ is the acceleration due to gravity. Suppose $u$ is constant but $\phi$ varies from $0^\circ$ to $180^\circ$. As $\phi$ changes, the highest points of the trajectories trace a closed curve. The area enclosed by this curve can be expressed as $d \cdot \frac{u^4}{g^2}$. Find the value of $d$.
|
\frac{\pi}{8}
|
deepscale
| 22,546
| ||
Given a circle $x^2 + (y-1)^2 = 1$ with its tangent line $l$, which intersects the positive x-axis at point A and the positive y-axis at point B. Determine the y-intercept of the tangent line $l$ when the distance AB is minimized.
|
\frac{3+\sqrt{5}}{2}
|
deepscale
| 11,613
| ||
Given that the sum of the first n terms of a geometric sequence {a_n} (where all terms are real numbers) is S_n, if S_10=10 and S_30=70, determine the value of S_40.
|
150
|
deepscale
| 20,107
| ||
Miki's father is saving money in a piggy bank for the family's vacation, adding to it once a week. Miki counts and notes how much money has accumulated every week and looks for patterns in the growth. Let $P_{n}$ denote the amount in the piggy bank in the $n$-th week (in forints). Here are a few observations:
(1) $P_{5} = 2P_{3}$,
(2) $P_{8} = P_{3} + 100$,
(3) $P_{9} = P_{4} + P_{7}$.
"The amount of forints has always been even, but it has never been divisible by 3."
"The number of forints today is a perfect square, and I also noticed that dad increases the deposit each week by the same amount that the third deposit exceeded the second deposit; thus the contents of our piggy bank will never be a perfect square again."
Which week does Miki's last observation refer to, and is Miki's prediction correct?
|
18
|
deepscale
| 31,659
| ||
Given several numbers, one of them, $a$ , is chosen and replaced by the three numbers $\frac{a}{3}, \frac{a}{3}, \frac{a}{3}$ . This process is repeated with the new set of numbers, and so on. Originally, there are $1000$ ones, and we apply the process several times. A number $m$ is called *good* if there are $m$ or more numbers that are the same after each iteration, no matter how many or what operations are performed. Find the largest possible good number.
|
667
|
deepscale
| 28,151
| ||
Determine the sum of all single-digit replacements for $z$ such that the number ${36{,}z72}$ is divisible by both 6 and 4.
|
18
|
deepscale
| 8,010
| ||
In an isosceles trapezoid \(ABCD\), the larger base \(AD = 12\) and \(AB = 6\). Find the distance from point \(O\), the intersection of the diagonals, to point \(K\), the intersection of the extensions of the lateral sides, given that the extensions of the lateral sides intersect at a right angle.
|
\frac{12(3 - \sqrt{2})}{7}
|
deepscale
| 16,064
| ||
The arithmetic mean of several consecutive natural numbers is 5 times greater than the smallest of them. How many times is the arithmetic mean smaller than the largest of these numbers?
|
1.8
|
deepscale
| 15,306
| ||
Compute the sum of all positive integers $n$ for which $9 \sqrt{n}+4 \sqrt{n+2}-3 \sqrt{n+16}$ is an integer.
|
For the expression to be an integer at least one of $n$ and $n+2$ must be a perfect square. We also note that at most one of $n$ and $n+2$ can be a square, so exactly one of them is a square. Case 1: $n$ is a perfect square. By our previous observation, it must be that $4 \sqrt{n+2}=3 \sqrt{n+16} \Rightarrow n=16$. Case 2: $n+2$ is a perfect square. By our previous observation, it must be that $9 \sqrt{n}=3 \sqrt{n+16} \Rightarrow n=2$. Consequently, the answer is $16+2=18$.
|
18
|
deepscale
| 4,435
| |
Given that \(\alpha\) and \(\beta\) are both acute angles, and
$$(1+\tan \alpha)(1+\tan \beta)=2.$$
Find \(\alpha + \beta =\).
|
\frac{\pi}{4}
|
deepscale
| 31,090
| ||
Find the smallest positive integer $n$ such that
\[\begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}^n = \mathbf{I}.\]
|
6
|
deepscale
| 39,943
| ||
Egor wrote a number on the board and encoded it according to the rules of letter puzzles (different letters correspond to different digits, and identical letters correspond to identical digits). The word "GUATEMALA" was the result. How many different numbers could Egor have originally written, if his number was divisible by 25?
|
20160
|
deepscale
| 20,539
| ||
In right triangle $ABC$ with $\angle B = 90^\circ$, sides $AB=1$ and $BC=3$. The bisector of $\angle BAC$ meets $\overline{BC}$ at $D$. Calculate the length of segment $BD$.
A) $\frac{1}{2}$
B) $\frac{3}{4}$
C) $1$
D) $\frac{5}{4}$
E) $2$
|
\frac{3}{4}
|
deepscale
| 22,826
| ||
Given $\lg 2=0.3010$ and $\lg 3=0.4771$, at which decimal place does the first non-zero digit of $\left(\frac{6}{25}\right)^{100}$ occur?
(Shanghai Middle School Mathematics Competition, 1984)
|
62
|
deepscale
| 29,730
| ||
In rectangle $ABCD$, $AB = 6$ cm, $BC = 8$ cm, and $DE = DF$. The area of triangle $DEF$ is one-fourth the area of rectangle $ABCD$. What is the length in centimeters of segment $EF$? Express your answer in simplest radical form.
[asy]
draw((0,0)--(0,24)--(32,24)--(32,0)--cycle);
draw((13,24)--(32,5));
label("$A$",(0,24),W);
label("$B$",(0,0),W);
label("$C$",(32,0),E);
label("$D$",(32,24),E);
label("$E$",(13,24),N);
label("$F$",(32,5),E);
[/asy]
|
4\sqrt{3}
|
deepscale
| 38,588
| ||
Triangle $ABC$ is equilateral with side length $12$ . Point $D$ is the midpoint of side $\overline{BC}$ . Circles $A$ and $D$ intersect at the midpoints of side $AB$ and $AC$ . Point $E$ lies on segment $\overline{AD}$ and circle $E$ is tangent to circles $A$ and $D$ . Compute the radius of circle $E$ .
*2015 CCA Math Bonanza Individual Round #5*
|
3\sqrt{3} - 6
|
deepscale
| 32,433
| ||
Given that point $P$ is on curve $C_1: y^2 = 8x$ and point $Q$ is on curve $C_2: (x-2)^2 + y^2 = 1$. If $O$ is the coordinate origin, find the maximum value of $\frac{|OP|}{|PQ|}$.
|
\frac{4\sqrt{7}}{7}
|
deepscale
| 32,717
| ||
In any finite grid of squares, some shaded and some not, for each unshaded square, record the number of shaded squares horizontally or vertically adjacent to it; this grid's *score* is the sum of all numbers recorded this way. Deyuan shades each square in a blank $n\times n$ grid with probability $k$ ; he notices that the expected value of the score of the resulting grid is equal to $k$ , too! Given that $k > 0.9999$ , find the minimum possible value of $n$ .
*Proposed by Andrew Wu*
|
51
|
deepscale
| 11,308
| ||
Given that the base of the tetrahedron \( S-ABC \) is an isosceles right triangle with \( AB \) as the hypotenuse, and \( SA = SB = SC = AB = 2 \). Assume that the points \( S, A, B, \) and \( C \) lie on a spherical surface with center \( O \). Find the distance from point \( O \) to the plane \( ABC \).
|
\frac{\sqrt{3}}{3}
|
deepscale
| 9,479
| ||
Given the function $f(x) = \sqrt[3]{x}$, calculate the value of $\lim\limits_{\Delta x \to 0} \frac{f(1-\Delta x)-f(1)}{\Delta x}$.
|
-\frac{1}{3}
|
deepscale
| 17,368
| ||
Suppose that $x_1+1=x_2+2=x_3+3=\cdots=x_{2008}+2008=x_1+x_2+x_3+\cdots+x_{2008}+2009$. Find the value of $\left\lfloor|S|\right\rfloor$, where $S=\sum_{n=1}^{2008}x_n$.
|
1005
|
deepscale
| 37,488
| ||
If $\det \mathbf{M} = -2,$ then find $ \det (\mathbf{M}^4).$
|
16
|
deepscale
| 40,299
| ||
If $4^{n}=64^{2}$, what is the value of $n$?
|
We note that $64=4 \times 4 \times 4$. Thus, $64^{2}=64 \times 64=4 \times 4 \times 4 \times 4 \times 4 \times 4$. Since $4^{n}=64^{2}$, then $4^{n}=4 \times 4 \times 4 \times 4 \times 4 \times 4$ and so $n=6$.
|
6
|
deepscale
| 5,525
| |
Let the complex number \( z \) satisfy \( |z|=1 \). Given that the equation \( zx^2 + 2\bar{z}x + 2 = 0 \) in terms of \( x \) has a real root, find the sum of all such complex numbers \( z \).
|
-\frac{3}{2}
|
deepscale
| 14,392
| ||
Given that $x, y > 0$ and $\frac{1}{x} + \frac{1}{y} = 2$, find the minimum value of $x + 2y$.
|
\frac{3 + 2\sqrt{2}}{2}
|
deepscale
| 9,987
| ||
Let $D$ be the circle with equation $x^2 - 10y - 7 = -y^2 - 8x + 4$. Find the center $(a, b)$ and radius $r$ of $D$, and determine the value of $a + b + r$.
|
1 + 2\sqrt{13}
|
deepscale
| 12,210
| ||
Let the complex numbers \( z_1 \) and \( z_2 \) correspond to the points \( A \) and \( B \) on the complex plane respectively, and suppose \( \left|z_1\right| = 4 \) and \( 4z_1^2 - 2z_1z_2 + z_2^2 = 0 \). Let \( O \) be the origin. Calculate the area of triangle \( \triangle OAB \).
|
8\sqrt{3}
|
deepscale
| 18,940
| ||
The expression $10y^2 - 51y + 21$ can be written as $(Cy - D)(Ey - F)$, where $C, D, E, F$ are integers. What is $CE + C$?
|
15
|
deepscale
| 25,558
| ||
A right circular cone is sliced into five equal-height sections by planes parallel to its base. What is the ratio of the volume of the second-largest piece to the volume of the largest piece?
|
\frac{37}{61}
|
deepscale
| 18,421
| ||
A triangular array of numbers has a first row consisting of the odd integers $1,3,5,\ldots,99$ in increasing order. Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry. Each entry in any row after the top row equals the sum of the two entries diagonally above it in the row immediately above it. How many entries in the array are multiples of $67$?
|
The result above is fairly intuitive if we write out several rows and then divide all numbers in row $r$ by $2^{r-1}$ (we can do this because dividing by a power of 2 doesn't affect divisibility by $67$). The second row will be $2, 4, 6, \cdots , 98$, the third row will be $3, 5, \cdots, 97$, and so forth. Clearly, only the odd-numbered rows can have a term divisible by $67$. However, with each row the row will have one less element, and the $99-67+1 = 33$rd row is the last time $67$ will appear. Therefore the number of multiples of 67 in the entire array is $\frac{33+1}{2} = \boxed{017}$.
|
17
|
deepscale
| 6,916
| |
Alice sells an item at $10 less than the list price and receives $10\%$ of her selling price as her commission.
Bob sells the same item at $20 less than the list price and receives $20\%$ of his selling price as his commission.
If they both get the same commission, then the list price is
|
Let $x$ be the list price of the item.
1. **Alice's Selling Price and Commission:**
- Alice sells the item at $x - 10$ dollars.
- Her commission is $10\%$ of her selling price, which is $0.10(x - 10)$.
2. **Bob's Selling Price and Commission:**
- Bob sells the item at $x - 20$ dollars.
- His commission is $20\%$ of his selling price, which is $0.20(x - 20)$.
3. **Equating Commissions:**
- Since both receive the same commission, we set the two expressions for commission equal to each other:
\[
0.10(x - 10) = 0.20(x - 20)
\]
4. **Solving the Equation:**
- Expanding both sides:
\[
0.10x - 1 = 0.20x - 4
\]
- Rearranging the equation to isolate $x$:
\[
0.10x - 0.20x = -4 + 1
\]
\[
-0.10x = -3
\]
- Solving for $x$:
\[
x = \frac{-3}{-0.10} = 30
\]
5. **Conclusion:**
- The list price of the item is $\boxed{\textbf{(B) } \$30}$.
|
$30
|
deepscale
| 903
| |
Triangle $A B C$ has side lengths $A B=15, B C=18, C A=20$. Extend $C A$ and $C B$ to points $D$ and $E$ respectively such that $D A=A B=B E$. Line $A B$ intersects the circumcircle of $C D E$ at $P$ and $Q$. Find the length of $P Q$.
|
WLOG suppose that $P$ is closer to $A$ than to $B$. Let $D A=A B=B E=c=15, B C=a=18, C A=$ $b=20, P A=x$, and $Q B=y$. By Power of a Point on $B$ and $A$, we get $a c=(x+c) y$ and $b c=(y+c) x$, respectively. Subtracting the two equations gives $c y-c x=a c-b c \Rightarrow y-x=a-b$. Substituting $y=x+a-b$ into the first equation gives $a c=(x+c)(x+a-b)=x^{2}+(a-b+c) x+a c-b c$, which is a quadratic with unique positive solution $x=\frac{(b-a-c)+\sqrt{(a-b+c)^{2}+4 b c}}{2}$. Thus, $P Q=x+y+c=(y-x)+2 x+c=(a-b+c)+(b-a-c)+\sqrt{(a-b+c)^{2}+4 b c}=\sqrt{13^{2}+4 \cdot 20 \cdot 15}=37$.
|
37
|
deepscale
| 5,118
| |
Find the smallest value of \(a\) for which the sum of the squares of the roots of the equation \(x^{2}-3ax+a^{2}=0\) is equal to \(0.28\).
|
-0.2
|
deepscale
| 11,894
| ||
Gustave has 15 steel bars of masses $1 \mathrm{~kg}, 2 \mathrm{~kg}, 3 \mathrm{~kg}, \ldots, 14 \mathrm{~kg}, 15 \mathrm{~kg}$. He also has 3 bags labelled $A, B, C$. He places two steel bars in each bag so that the total mass in each bag is equal to $M \mathrm{~kg}$. How many different values of $M$ are possible?
|
The total mass of the six steel bars in the bags is at least $1+2+3+4+5+6=21 \mathrm{~kg}$ and at most $10+11+12+13+14+15=75 \mathrm{~kg}$. This is because the masses of the 15 given bars are $1 \mathrm{~kg}, 2 \mathrm{~kg}, 3 \mathrm{~kg}, \ldots, 14 \mathrm{~kg}$, and 15 kg. Since the six bars are divided between three bags with the same total mass in each bag, then the total mass in each bag is at least $21 \div 3=7 \mathrm{~kg}$ and at most $75 \div 3=25 \mathrm{~kg}$. There are $25-7+1=19$ masses that are an integer number of kilograms in this range (7 kg, $8 \mathrm{~kg}, 9 \mathrm{~kg}, \ldots, 23 \mathrm{~kg}, 24 \mathrm{~kg}, 25 \mathrm{~kg})$. Each of these 19 masses is indeed possible. To see this, we note that $1+6=2+5=3+4=7 \quad 1+7=2+6=3+5=8 \quad 1+8=2+7=3+6=9$, which shows that 7, 8, and 9 are possible values of $M$. Continuing to increase the larger values to $15,14,13$, we eventually obtain $1+15=2+14=3+13=16$ and also that each value of $M$ between 10 and 16, inclusive, will be a possible value of $M$. Now, we increase the smaller values, starting from the last three pairs: $2+15=3+14=4+13=17 \quad 3+15=4+14=5+13=18 \quad \cdots \quad 10+15=11+14=12+13=25$, which shows that $17,18,19,20,21,22,23,24$, and 25 are also possible values of $M$. This shows that every integer value of $M$ with $7 \leq M \leq 25$ is possible. In summary, there are 19 possible values of $M$.
|
19
|
deepscale
| 5,346
| |
The cube of $a$ and the fourth root of $b$ vary inversely. If $a=3$ when $b=256$, then find $b$ when $ab=81$.
|
16
|
deepscale
| 11,231
| ||
The nonzero roots of the equation $x^2 + 6x + k = 0$ are in the ratio $2:1$. What is the value of $k$?
|
8
|
deepscale
| 34,394
| ||
A point $M$ on the parabola $y=4x^{2}$ is at a distance of $1$ from the focus. The ordinate of point $M$ is __________.
|
\frac{15}{16}
|
deepscale
| 23,313
| ||
In a country, there are 110 cities. Between each pair of cities, there is either a road or no road.
A driver starts in a city with exactly one road leading out of it. After traveling along this road, he arrives at a second city, which has exactly two roads leading out of it. After traveling along one of these roads, he arrives at a third city, which has exactly three roads leading out of it, and so on. At some point, after traveling along one of the roads, he arrives in the $N$-th city, which has exactly $N$ roads leading out of it. At this stage, the driver stops his journey. (For each $2 \leqslant k \leqslant N$, the $k$-th city has exactly $k$ roads, including the one by which the driver arrived in this city.)
What is the maximum possible value of $N$?
|
107
|
deepscale
| 29,419
| ||
Given that $a > 0$, $b > 0$, and $a + b = 1$, find the minimum value of $\frac{2}{a} + \frac{3}{b}$.
|
5 + 2\sqrt{6}
|
deepscale
| 28,671
| ||
Roll a die twice in succession, observing the number of points facing up each time, and calculate:
(1) The probability that the sum of the two numbers is 5;
(2) The probability that at least one of the two numbers is odd;
(3) The probability that the point (x, y), with x being the number of points facing up on the first roll and y being the number on the second roll, lies inside the circle $x^2+y^2=15$.
|
\frac{2}{9}
|
deepscale
| 11,113
| ||
Determine the maximal size of a set of positive integers with the following properties: $1.$ The integers consist of digits from the set $\{ 1,2,3,4,5,6\}$ . $2.$ No digit occurs more than once in the same integer. $3.$ The digits in each integer are in increasing order. $4.$ Any two integers have at least one digit in common (possibly at different positions). $5.$ There is no digit which appears in all the integers.
|
32
|
deepscale
| 27,717
| ||
A certain company has two research and development teams, Team A and Team B. The probability of success for developing a new product by Team A is $\frac{4}{5}$, and for Team B is $\frac{3}{4}$. Team A is assigned to develop a new product $A$, and Team B is assigned to develop a new product $B$. It is assumed that the research and development of Teams A and B are independent of each other.
$(1)$ Find the probability that exactly one new product is successfully developed.
$(2)$ If the development of new product $A$ is successful, the company will make a profit of $150$ thousand dollars, otherwise it will incur a loss of $60$ thousand dollars. If the development of new product $B$ is successful, the company will make a profit of $120$ thousand dollars, otherwise it will incur a loss of $40$ thousand dollars. Find the probability distribution and the mathematical expectation $E(\xi)$ of the company's profit (in thousand dollars).
|
188
|
deepscale
| 20,955
| ||
Let $P$ be the point on line segment $\overline{AB}$ such that $AP:PB = 3:2.$ Then
\[\overrightarrow{P} = t \overrightarrow{A} + u \overrightarrow{B}\]for some constants $t$ and $u.$ Enter the ordered pair $(t,u).$
[asy]
unitsize(1 cm);
pair A, B, P;
A = (0,0);
B = (5,1);
P = interp(A,B,3/5);
draw(A--B);
dot("$A$", A, S);
dot("$B$", B, S);
dot("$P$", P, S);
[/asy]
|
\left( \frac{2}{5}, \frac{3}{5} \right)
|
deepscale
| 40,219
| ||
For positive integers $n$, let $f(n)$ return the smallest positive integer $k$ such that $\frac{1}{k}$ has exactly $n$ digits after the decimal point. How many positive integer divisors does $f(2010)$ have?
|
2011
|
deepscale
| 37,976
| ||
If $|x-2|=p$, where $x<2$, then what is $x-p$ in terms of $p$?
|
2-2p
|
deepscale
| 33,625
| ||
Find $x$ such that $\log_{12}3x=2$.
|
48
|
deepscale
| 33,474
| ||
Line segments \( AB \) and \( CD \) are situated between two parallel planes \( \alpha \) and \( \beta \). \( AC \subset \alpha \) and \( BD \subset \beta \). Given \( AB \perp \alpha \), \( AC = BD = 5 \), \( AB = 12 \), and \( CD = 13 \). Points \( E \) and \( F \) divide \( AB \) and \( CD \) in the ratio \( 1:2 \) respectively. Find the length of the line segment \( EF \).
|
\frac{5}{3} \sqrt{7}
|
deepscale
| 14,099
| ||
How many 3-term geometric sequences $a$ , $b$ , $c$ are there where $a$ , $b$ , and $c$ are positive integers with $a < b < c$ and $c = 8000$ ?
|
39
|
deepscale
| 28,797
| ||
There are five positive integers that are divisors of each number in the list $$60, 120, -30, 180, 240$$. Find the sum of these five positive integers.
|
17
|
deepscale
| 21,327
| ||
Compute \[
\left\lfloor \frac{2017! + 2014!}{2016! + 2015!}\right\rfloor.
\] (Note that $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.)
|
2016
|
deepscale
| 20,616
| ||
Each vertex of a regular dodecagon ($12$-gon) is to be colored either red or blue, and thus there are $2^{12}$ possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle.
|
Note that the condition is equivalent to stating that there are no 2 pairs of oppositely spaced vertices with the same color.
Case 1: There are no pairs. This yields $2$ options for each vertices 1-6, and the remaining vertices 7-12 are set, yielding $2^6=64$ cases.
Case 2: There is one pair. Again start with 2 options for each vertex in 1-6, but now multiply by 6 since there are 6 possibilities for which pair can have the same color assigned instead of the opposite. Thus, the cases are: $2^6*6=384$
case 3: There are two pairs, but oppositely colored. Start with $2^6$ for assigning 1-6, then multiply by 6C2=15 for assigning which have repeated colors. Divide by 2 due to half the cases having the same colored opposites. $2^6*15/2=480$
It is apparent that no other cases exist, as more pairs would force there to be 2 pairs of same colored oppositely spaced vertices with the same color. Thus, the answer is: $64+384+480=\boxed{928}$
~SAHANWIJETUNGA
|
928
|
deepscale
| 7,382
| |
In the diagram below, $WXYZ$ is a trapezoid such that $\overline{WX}\parallel \overline{ZY}$ and $\overline{WY}\perp\overline{ZY}$. If $YZ = 12$, $\tan Z = 1.5$, and $\tan X = 2$, then what is $XY$?
[asy]
pair WW,X,Y,Z;
Z = (0,0);
Y = (12,0);
WW = (12,18);
X= (18,18);
draw(WW--X--Y--Z--WW);
label("$W$",WW,N);
label("$X$",X,N);
label("$Y$",Y,S);
label("$Z$",Z,S);
[/asy]
|
9\sqrt{5}
|
deepscale
| 36,261
| ||
Let $S = \{(x,y) | x = 1, 2, \ldots, 1993, y = 1, 2, 3, 4\}$. If $T \subset S$ and there aren't any squares in $T.$ Find the maximum possible value of $|T|.$ The squares in T use points in S as vertices.
|
Let \( S = \{(x,y) \mid x = 1, 2, \ldots, 1993, y = 1, 2, 3, 4\} \). We aim to find the maximum possible value of \( |T| \) for a subset \( T \subset S \) such that there are no squares in \( T \).
To solve this, we need to ensure that no four points in \( T \) form the vertices of a square. The key observation is that for any square in \( S \), we can have at most 3 of its vertices in \( T \). This gives a weak upper bound:
\[
|T| \leq \frac{3}{4} |S|.
\]
We will use a more refined approach to maximize \( |T| \). Consider the columns of \( S \). If a column \( C \) contains all its elements in \( T \), then the adjacent columns can have at most 2 elements in \( T \) to avoid forming squares. Thus, it is more efficient to avoid having all elements of any column in \( T \).
We can choose 3 elements from each column in \( T \) without forming squares. To achieve this, we can use a pattern where each set of 4 adjacent columns has distinct permutations of 3 elements in \( T \) and 1 element not in \( T \). This pattern avoids forming squares and maximizes the number of elements in \( T \).
For example, consider the following arrangement for 4 columns:
\[
\begin{array}{cccc}
\bullet & \circ & \circ & \circ \\
\circ & \circ & \bullet & \circ \\
\circ & \bullet & \circ & \circ \\
\circ & \circ & \circ & \bullet \\
\end{array}
\]
Here, \( \bullet \) represents an element in \( T \) and \( \circ \) represents an element not in \( T \).
This pattern can be repeated, with a separating column containing only 1 element in \( T \) to avoid forming squares. Given that there are 1993 columns, we can divide them into groups of 5 columns (4 columns with 3 elements each and 1 separating column with 1 element).
Thus, we have:
\[
1993 = 5 \cdot 398 + 3.
\]
The maximum number of elements in \( T \) is:
\[
398 \cdot 13 + 3 \cdot 3 = 5183.
\]
Therefore, the maximum possible value of \( |T| \) is:
\[
\boxed{5183}.
\]
|
5183
|
deepscale
| 3,028
| |
Given a cube $ABCD$-$A\_1B\_1C\_1D\_1$ with edge length $1$, point $M$ is the midpoint of $BC\_1$, and $P$ is a moving point on edge $BB\_1$. Determine the minimum value of $AP + MP$.
|
\frac{\sqrt{10}}{2}
|
deepscale
| 14,908
| ||
Let $x<y$ be positive real numbers such that $\sqrt{x}+\sqrt{y}=4$ and $\sqrt{x+2}+\sqrt{y+2}=5$. Compute $x$.
|
Adding and subtracting both equations gives $$\begin{aligned} & \sqrt{x+2}+\sqrt{x}+\sqrt{y+2}+\sqrt{y}=9 \\ & \sqrt{x+2}-\sqrt{x}+\sqrt{y+2}-\sqrt{y}=1 \end{aligned}$$ Substitute $a=\sqrt{x}+\sqrt{x+2}$ and $b=\sqrt{y}+\sqrt{y+2}$. Then since $(\sqrt{x+2}+\sqrt{x})(\sqrt{x+2}-\sqrt{x})=2$, we have $$\begin{gathered} a+b=9 \\ \frac{2}{a}+\frac{2}{b}=1 \end{gathered}$$ Dividing the first equation by the second one gives $$ab=18, a=3, b=6$$ Lastly, $\sqrt{x}=\frac{\sqrt{x+2}+\sqrt{x}-(\sqrt{x+2}-\sqrt{x})}{2}=\frac{3-\frac{2}{3}}{2}=\frac{7}{6}$, so $x=\frac{49}{36}$.
|
\frac{49}{36}
|
deepscale
| 3,509
| |
Compute $35^{1723} - 16^{1723}$ modulo 6.
|
1
|
deepscale
| 37,583
| ||
Given a triangle $ABC$ with sides opposite to angles $A$, $B$, and $C$ being $a$, $b$, and $c$, respectively, and it is given that $(3b-c)\cos A = a\cos C$.
(1) Find the value of $\cos A$;
(2) If the area of $\triangle ABC$ is $S=2\sqrt{2}$, find the minimum value of the perimeter of $\triangle ABC$.
|
2\sqrt{6}+2\sqrt{2}
|
deepscale
| 12,336
| ||
Real numbers \(a, b, c\) and a positive number \(\lambda\) such that \(f(x)=x^{3}+a x^{2}+b x+c\) has three real roots \(x_{1}, x_{2}, x_{3}\), satisfying:
(1) \(x_{2}-x_{1}=\lambda\);
(2) \(x_{3}>\frac{1}{2}\left(x_{1}+x_{2}\right)\).
Find the maximum value of \(\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}\).
|
\frac{3\sqrt{3}}{2}
|
deepscale
| 27,190
| ||
Each vertex of a convex hexagon $ABCDEF$ is to be assigned a color, where there are $7$ colors to choose from. Find the number of possible different colorings such that both the ends of each diagonal and each pair of adjacent vertices have different colors.
|
5040
|
deepscale
| 29,200
| ||
If $\mathbf{v} \times \mathbf{w} = \begin{pmatrix} 5 \\ -2 \\ 4 \end{pmatrix},$ then find $(\mathbf{v} + \mathbf{w}) \times (\mathbf{v} + \mathbf{w}).$
|
\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}
|
deepscale
| 39,655
| ||
Five packages are delivered to five houses, one to each house. If these packages are randomly delivered, what is the probability that exactly three of them are delivered to the correct houses? Express your answer as a common fraction.
|
\frac{1}{12}
|
deepscale
| 22,519
| ||
An engraver makes plates with letters. He engraves the same letters in the same amount of time, but different letters may take different times. On two plates, "ДОМ МОДЫ" (DOM MODY) and "ВХОД" (VKHOD), together he spent 50 minutes, and one plate "В ДЫМОХОД" (V DYMOHOD) took him 35 minutes. How much time will it take him to make the plate "ВЫХОД" (VYKHOD)?
|
20
|
deepscale
| 15,297
| ||
The sides of rectangle $ABCD$ are $AB=3$ and $BC=2$. Point $P$ is on side $AB$ such that line $PD$ touches the circle with diameter $BC$ at point $E$. The line passing through the center of the circle and point $E$ intersects side $AB$ at point $Q$. What is the area of triangle $PQE$?
|
1/24
|
deepscale
| 11,784
| ||
How many positive, three-digit integers contain at least one $3$ as a digit but do not contain a $5$ as a digit?
|
200
|
deepscale
| 34,773
| ||
On each side of a square, a point is taken. It turns out that these points are the vertices of a rectangle whose sides are parallel to the diagonals of the square. Find the perimeter of the rectangle if the diagonal of the square is 6.
|
12
|
deepscale
| 7,907
| ||
In the equation $\frac{1}{(\;\;\;)} + \frac{4}{(\;\;\;)} + \frac{9}{(\;\;\;\;)} = 1$, fill in the three brackets in the denominators with a positive integer, respectively, such that the equation holds true. The minimum value of the sum of these three positive integers is $\_\_\_\_\_\_$.
|
36
|
deepscale
| 28,112
| ||
In February 1983, $789$ millimeters of rain fell in Jorhat, India. What was the average rainfall in millimeters per hour during that particular month?
A) $\frac{789}{672}$
B) $\frac{789 \times 28}{24}$
C) $\frac{789 \times 24}{28}$
D) $\frac{28 \times 24}{789}$
E) $789 \times 28 \times 24$
|
\frac{789}{672}
|
deepscale
| 17,258
| ||
In the polar coordinate system, the polar equation of curve $C$ is $\rho =6\sin \theta$, and the polar coordinates of point $P$ are $(\sqrt{2},\frac{\pi }{4})$. Taking the pole as the origin of coordinates and the positive half-axis of the $x$-axis as the polar axis, a plane rectangular coordinate system is established.
(1) Find the rectangular coordinate equation of curve $C$ and the rectangular coordinates of point $P$;
(2) A line $l$ passing through point $P$ intersects curve $C$ at points $A$ and $B$. If $|PA|=2|PB|$, find the value of $|AB|$.
|
3 \sqrt{2}
|
deepscale
| 18,049
| ||
Find the remainder when $5x^4 - 12x^3 + 3x^2 - 5x + 15$ is divided by $3x - 9$.
|
108
|
deepscale
| 9,571
| ||
Emily cycles at a constant rate of 15 miles per hour, and Leo runs at a constant rate of 10 miles per hour. If Emily overtakes Leo when he is 0.75 miles ahead of her, and she can view him in her mirror until he is 0.6 miles behind her, calculate the time in minutes it takes for her to see him.
|
16.2
|
deepscale
| 21,646
| ||
The game of rock-scissors is played just like rock-paper-scissors, except that neither player is allowed to play paper. You play against a poorly-designed computer program that plays rock with $50 \%$ probability and scissors with $50 \%$ probability. If you play optimally against the computer, find the probability that after 8 games you have won at least 4.
|
Since rock will always win against scissors, the optimum strategy is for you to always play rock; then, you win a game if and only if the computer plays scissors. Let $p_{n}$ be the probability that the computer plays scissors $n$ times; we want $p_{0}+p_{1}+p_{2}+p_{3}+p_{4}$. Note that by symmetry, $p_{n}=p_{8-n}$ for $n=0,1, \ldots, 8$, and because $p_{0}+p_{1}+\cdots+p_{8}=1, p_{0}+\cdots+p_{3}=p_{5}+\cdots+p_{8}=\left(1-p_{4}\right) / 2$. Our answer will thus be $\left(1+p_{4}\right) / 2$. If the computer is to play scissors exactly 4 times, there are $\binom{8}{4}$ ways in which it can do so, compared to $2^{8}$ possible combinations of eight plays. Thus, $p_{4}=\binom{8}{4} / 2^{8}=35 / 128$. Our answer is thus $\frac{1+\frac{35}{128}}{2}=\frac{163}{256}$.
|
\frac{163}{256}
|
deepscale
| 5,077
| |
Calculate the value of $({-\frac{4}{5}})^{2022} \times ({\frac{5}{4}})^{2021}$.
|
\frac{4}{5}
|
deepscale
| 22,703
| ||
In triangle $ABC$, the sides opposite angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. It is given that $a = b\cos C + c\sin B$.
(1) Find angle $B$.
(2) If $b = 4$, find the maximum area of triangle $ABC$.
|
4\sqrt{2} + 4
|
deepscale
| 11,823
| ||
Evaluate the product \[\left(\sqrt{5}+\sqrt{6}+\sqrt{7}\right)\left(\sqrt{5}+\sqrt{6}-\sqrt{7}\right)\left(\sqrt{5}-\sqrt{6}+\sqrt{7}\right)\left(-\sqrt{5}+\sqrt{6}+\sqrt{7}\right).\]
|
We group the first and last factors as well as the two middle factors, then apply the difference of squares repeatedly: \begin{align*} \left(\left(\sqrt{6} + \sqrt{7}\right)^2 - \sqrt{5}^2\right)\left(\sqrt{5}^2 - \left(\sqrt{6} - \sqrt{7}\right)^2\right) &= \left(13 + 2\sqrt{42} - 5\right)\left(5 - \left(13 - 2\sqrt{42}\right)\right) \\ &= \left(2\sqrt{42} + 8\right)\left(2\sqrt{42} - 8\right) \\ &= \left(2\sqrt{42}\right)^2 - 8^2 \\ &= \boxed{104}. \end{align*} ~Azjps (Solution)
~MRENTHUSIASM (Revision)
|
104
|
deepscale
| 6,462
| |
A certain school organized a Chinese traditional culture activity week to promote traditional Chinese culture. During the activity period, a Chinese traditional culture knowledge competition was held, with classes participating in the competition. Each class selected 5 representatives through a Chinese traditional culture knowledge quiz to participate in the grade-level competition. The grade-level competition was divided into two stages: preliminary and final. In the preliminary stage, the questions of Chinese traditional culture were placed in two boxes labeled A and B. Box A contained 5 multiple-choice questions and 3 fill-in-the-blank questions, while box B contained 4 multiple-choice questions and 3 fill-in-the-blank questions. Each class representative team was required to randomly draw two questions to answer from either box A or box B. Each class representative team first drew one question to answer, and after answering, the question was not returned to the box. Then they drew a second question to answer. After answering the two questions, the questions were returned to their original boxes.
$(1)$ If the representative team of Class 1 drew 2 questions from box A, mistakenly placed the questions in box B after answering, and then the representative team of Class 2 answered the questions, with the first question drawn from box B. It is known that the representative team of Class 2 drew a multiple-choice question from box B. Find the probability that the representative team of Class 1 drew 2 multiple-choice questions from box A.
$(2)$ After the preliminary round, the top 6 representative teams from Class 6 and Class 18 entered the final round. The final round was conducted in the form of an idiom solitaire game, using a best-of-five format, meaning the team that won three rounds first won the match and the game ended. It is known that the probability of Class 6 winning the first round is $\frac{3}{5}$, the probability of Class 18 winning is $\frac{2}{5}$, and the probability of the winner of each round winning the next round is $\frac{2}{5}$, with each round having a definite winner. Let the random variable $X$ represent the number of rounds when the game ends. Find the expected value of the random variable $X$, denoted as $E(X)$.
|
\frac{537}{125}
|
deepscale
| 25,430
| ||
A fast train with a weight of $P=150$ tons travels at a maximum speed of $v=72 \frac{\text{km}}{\text{hour}}$ on a horizontal track with a friction coefficient of $\rho=0,005$. What speed can it reach on a track with the same friction conditions but with an incline having $e=0.030$?
(Note: In this problem, $\rho$ is the friction coefficient and $e$ is the sine of the inclination angle.)
|
10.3
|
deepscale
| 15,528
| ||
There are two islands, A and B, that are 20 nautical miles apart. When viewing Island C from Island A, the angle between Island B and Island C is 60°. When viewing Island C from Island B, the angle between Island A and Island C is 75°. Find the distance between Island B and Island C.
|
10\sqrt{6}
|
deepscale
| 11,574
| ||
Let $x_1,$ $x_2,$ $\dots,$ $x_n$ be nonnegative real numbers such that $x_1 + x_2 + \dots + x_n = 1$ and
\[x_1^2 + x_2^2 + \dots + x_n^2 \le \frac{1}{100}.\]Find the smallest possible value of $n.$
|
100
|
deepscale
| 37,345
|
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