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Calculate the car's average miles-per-gallon for the entire trip given that the odometer readings are $34,500, 34,800, 35,250$, and the gas tank was filled with $8, 10, 15$ gallons of gasoline.
|
22.7
|
deepscale
| 30,672
| ||
Given that \(\triangle ABC\) has sides \(a\), \(b\), and \(c\) corresponding to angles \(A\), \(B\), and \(C\) respectively, and knowing that \(a + b + c = 16\), find the value of \(b^2 \cos^2 \frac{C}{2} + c^2 \cos^2 \frac{B}{2} + 2bc \cos \frac{B}{2} \cos \frac{C}{2} \sin \frac{A}{2}\).
|
64
|
deepscale
| 20,288
| ||
Let the sum of the first $n$ terms of an arithmetic sequence $\{a_n\}$ be $S_n$, and it satisfies $S_{2016} > 0, S_{2017} < 0$. For any positive integer $n$, it holds that $|a_n| \geqslant |a_k|$, find the value of $k$.
|
1009
|
deepscale
| 27,448
| ||
Calculate the value of the following expressions:
1. $\sqrt[4]{(3-\pi )^{4}}+(0.008)\;^{- \frac {1}{3}}-(0.25)\;^{ \frac {1}{2}}×( \frac {1}{ \sqrt {2}})^{-4}$
2. $\log _{3} \sqrt {27}-\log _{3} \sqrt {3}-\lg 625-\lg 4+\ln (e^{2})- \frac {4}{3}\lg \sqrt {8}$
|
-1
|
deepscale
| 9,002
| ||
Let \( S_{n} = 1 + \frac{1}{2} + \cdots + \frac{1}{n} \) for \( n = 1, 2, \cdots \). Find the smallest positive integer \( n \) such that \( S_{n} > 10 \).
|
12367
|
deepscale
| 15,392
| ||
Ann and Sue bought identical boxes of stationery. Ann used hers to write $1$-sheet letters and Sue used hers to write $3$-sheet letters.
Ann used all the envelopes and had $50$ sheets of paper left, while Sue used all of the sheets of paper and had $50$ envelopes left.
The number of sheets of paper in each box was
|
Let $S$ represent the number of sheets of paper in each box, and let $E$ represent the number of envelopes in each box. We can set up the following equations based on the problem statement:
1. Ann used all the envelopes and had 50 sheets of paper left. This implies:
\[ S - E = 50 \]
2. Sue used all the sheets of paper and had 50 envelopes left. This implies:
\[ 3E = S \]
Rearranging this equation gives:
\[ E = \frac{S}{3} \]
Now, substituting $E = \frac{S}{3}$ from the second equation into the first equation:
\[ S - \frac{S}{3} = 50 \]
\[ \frac{3S - S}{3} = 50 \]
\[ \frac{2S}{3} = 50 \]
\[ 2S = 150 \]
\[ S = 75 \]
However, this value of $S$ does not satisfy the condition $E - \frac{S}{3} = 50$. Let's recheck the substitution and solve again:
Substitute $E = \frac{S}{3}$ into $S - E = 50$:
\[ S - \frac{S}{3} = 50 \]
\[ \frac{2S}{3} = 50 \]
\[ 2S = 150 \]
\[ S = 75 \]
This seems incorrect as per the initial setup. Let's re-evaluate the equations:
\[ S - E = 50 \]
\[ E = \frac{S}{3} \]
Substitute $E = \frac{S}{3}$ into $S - E = 50$:
\[ S - \frac{S}{3} = 50 \]
\[ \frac{2S}{3} = 50 \]
\[ 2S = 150 \]
\[ S = 75 \]
This still seems incorrect. Let's recheck the second condition:
\[ E - \frac{S}{3} = 50 \]
\[ E = \frac{S}{3} + 50 \]
Now, substitute $E = \frac{S}{3} + 50$ into $S - E = 50$:
\[ S - \left(\frac{S}{3} + 50\right) = 50 \]
\[ S - \frac{S}{3} - 50 = 50 \]
\[ \frac{2S}{3} = 100 \]
\[ 2S = 300 \]
\[ S = 150 \]
Thus, the number of sheets of paper in each box is $\boxed{150}$, which corresponds to choice $\textbf{(A)}$.
|
150
|
deepscale
| 701
| |
Suppose that $PQRS TUVW$ is a regular octagon. There are 70 ways in which four of its sides can be chosen at random. If four of its sides are chosen at random and each of these sides is extended infinitely in both directions, what is the probability that they will meet to form a quadrilateral that contains the octagon?
|
If the four sides that are chosen are adjacent, then when these four sides are extended, they will not form a quadrilateral that encloses the octagon. If the four sides are chosen so that there are exactly three adjacent sides that are not chosen and one other side not chosen, then when these four sides are extended, they will not form a quadrilateral that encloses the octagon. Any other set of four sides that are chosen will form a quadrilateral that encloses the octagon. We are told that there is a total of 70 ways in which four sides can be chosen. We will count the number of ways in which four sides can be chosen that do not result in the desired quadrilateral, and subtract this from 70 to determine the number of ways in which the desired quadrilateral does result. There are 8 ways in which to choose four adjacent sides: choose one side to start (there are 8 ways to choose one side) and then choose the next three adjacent sides in clockwise order (there is 1 way to do this). There are 8 ways to choose adjacent sides to be not chosen (after picking one such set, there are 7 additional positions into which these sides can be rotated). For each of these choices, there are 3 possible choices for the remaining unchosen sides. Therefore, there are $8 imes 3=24$ ways to choose four sides so that there are 3 adjacent sides unchosen. Therefore, of the 70 ways of choosing four sides, exactly $8+24=32$ of them do not give the desired quadrilateral, and so $70-32=38$ ways do. Thus, the probability that four sides are chosen so that the desired quadrilateral is formed is $ rac{38}{70}= rac{19}{35}$.
|
\frac{19}{35}
|
deepscale
| 5,661
| |
Let $n \geq 3$ be an integer. Rowan and Colin play a game on an $n \times n$ grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is [i]orderly[/i] if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of $n$, how many orderly colorings are there?
|
To determine the number of orderly colorings on an \( n \times n \) grid where each square is either red or blue, we must first understand the conditions of the game as described.
An orderly coloring must satisfy two main conditions:
1. No matter how the rows are permuted by Rowan, Colin can permute the columns to revert to the original grid coloring.
2. Conversely, no matter how the columns are permuted by Colin, Rowan can permute the rows to revert to the original coloring.
These conditions imply that the only configurations that are considered orderly are those that exhibit a certain symmetry or uniformity allowing reversibility through permutations.
### Analysis:
1. **Uniform Grids**:
- The simplest orderly colorings are the grids where all squares are the same color. There are exactly two such grids: one entirely red and another entirely blue.
2. **Symmetric Grids**:
- Beyond the two uniform colorings, we need to consider configurations where each permutation of rows or columns allows for reassembling back to the original coloration. This situation is achieved when all rows (and all columns) are identical.
- Specifically, if each row (or column) presents a permutation of a set, and since the grid needs to be reassembled no matter the shuffle, this set must follow the pattern of a multiset with the same number of red and blue cells for all rows and columns.
3. **Calculation of Symmetric Grids**:
- Choose any particular row pattern; since each row must be identical to allow reconstruction via permutation, we are left with \( n! \) permutations of the row that are distinct.
- For each of these patterns, the columns must mirror the same property, allowing \( n! \) permutations as well.
- However, as verified, it's essentially ensuring all are exact transformations under ordering operations \( n! + 1 \) times (including the all one color options).
4. **Total Orderly Colorings**:
- We account for both completely identical colors and identical individual permutations, yielding \( 2(n! + 1) \).
- The additional permutations (1 more than \( n! \)) consider the allowance of identical symmetric setups across \( n \times n \).
### Conclusion:
Thus, the total number of orderly colorings is given by:
\[
\boxed{2(n! + 1)}
\]
This accounts for the resilience of grid patterns to permutations under the constraints of the game.
|
2(n! + 1)
|
deepscale
| 6,107
| |
Three boys and three girls are lined up for a photo. Boy A is next to boy B, and exactly two girls are next to each other. Calculate the total number of different ways they can be arranged.
|
144
|
deepscale
| 31,393
| ||
Given the curve $C:\begin{cases}x=2\cos a \\ y= \sqrt{3}\sin a\end{cases} (a$ is the parameter) and the fixed point $A(0,\sqrt{3})$, ${F}_1,{F}_2$ are the left and right foci of this curve, respectively. With the origin $O$ as the pole and the positive half-axis of $x$ as the polar axis, a polar coordinate system is established.
$(1)$ Find the polar equation of the line $AF_{2}$;
$(2)$ A line passing through point ${F}_1$ and perpendicular to the line $AF_{2}$ intersects this conic curve at points $M$, $N$, find the value of $||MF_{1}|-|NF_{1}||$.
|
\dfrac{12\sqrt{3}}{13}
|
deepscale
| 32,595
| ||
On a sheet of paper, points \( A, B, C, D \) are marked. A recognition device can perform two types of operations with absolute precision: a) measuring the distance in centimeters between two given points; b) comparing two given numbers. What is the minimum number of operations needed for this device to definitively determine whether the quadrilateral \( ABCD \) is a square?
|
10
|
deepscale
| 14,725
| ||
Suppose that $b$ is a positive integer greater than or equal to $2.$ When $197$ is converted to base $b$, the resulting representation has $4$ digits. What is the number of possible values for $b$?
|
2
|
deepscale
| 38,311
| ||
$x, y$ are positive real numbers such that $x+y^{2}=x y$. What is the smallest possible value of $x$?
|
4 Notice that $x=y^{2} /(y-1)=2+(y-1)+1 /(y-1) \geq 2+2=4$. Conversely, $x=4$ is achievable, by taking $y=2$.
|
4
|
deepscale
| 3,572
| |
Let $\mathcal{P}_1$ and $\mathcal{P}_2$ be two parabolas with distinct directrices $\ell_1$ and $\ell_2$ and distinct foci $F_1$ and $F_2$ respectively. It is known that $F_1F_2||\ell_1||\ell_2$ , $F_1$ lies on $\mathcal{P}_2$ , and $F_2$ lies on $\mathcal{P}_1$ . The two parabolas intersect at distinct points $A$ and $B$ . Given that $F_1F_2=1$ , the value of $AB^2$ can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$ . Find $100m+n$ .
[i]Proposed by Yannick Yao
|
1504
|
deepscale
| 30,912
| ||
The distance between the two intersections of $x=y^4$ and $x+y^2=1$ is $\sqrt{u+v\sqrt5}$. Find the ordered pair, $(u,v)$.
|
(-2,2)
|
deepscale
| 34,098
| ||
The product of $\sqrt[3]{4}$ and $\sqrt[4]{8}$ equals
|
1. **Express the radicals in terms of exponents:**
- The cube root of 4 can be written as $\sqrt[3]{4} = 4^{1/3}$. Since $4 = 2^2$, we have $4^{1/3} = (2^2)^{1/3} = 2^{2/3}$.
- The fourth root of 8 can be written as $\sqrt[4]{8} = 8^{1/4}$. Since $8 = 2^3$, we have $8^{1/4} = (2^3)^{1/4} = 2^{3/4}$.
2. **Calculate the product of these expressions:**
- The product of $2^{2/3}$ and $2^{3/4}$ is given by:
\[
2^{2/3} \cdot 2^{3/4} = 2^{(2/3) + (3/4)}.
\]
- To add the exponents, find a common denominator (which is 12):
\[
\frac{2}{3} = \frac{8}{12}, \quad \frac{3}{4} = \frac{9}{12}.
\]
- Therefore, the sum of the exponents is:
\[
\frac{8}{12} + \frac{9}{12} = \frac{17}{12}.
\]
- Thus, the product is $2^{17/12}$.
3. **Express $2^{17/12}$ in radical form:**
- We can write $2^{17/12}$ as $2^{1 + 5/12} = 2 \cdot 2^{5/12}$.
- Recognizing $2^{5/12}$ as the 12th root of $2^5$, we have:
\[
2^{5/12} = \sqrt[12]{2^5} = \sqrt[12]{32}.
\]
- Therefore, $2^{17/12} = 2 \cdot \sqrt[12]{32}$.
4. **Conclude with the final answer:**
- The product of $\sqrt[3]{4}$ and $\sqrt[4]{8}$ is $2 \cdot \sqrt[12]{32}$.
Thus, the correct answer is $\boxed{\textbf{(E) } 2\sqrt[12]{32}}$.
|
2\sqrt[12]{32}
|
deepscale
| 232
| |
Erika, who is 14 years old, flips a fair coin whose sides are labeled 10 and 20, and then she adds the number on the top of the flipped coin to the number she rolls on a standard die. What is the probability that the sum equals her age in years? Express your answer as a common fraction.
|
\frac{1}{12}
|
deepscale
| 34,794
| ||
A triangular box is to be cut from an equilateral triangle of length 30 cm. Find the largest possible volume of the box (in cm³).
|
500
|
deepscale
| 25,369
| ||
Svetlana takes a triplet of numbers and transforms it by the rule: at each step, each number is replaced by the sum of the other two. What is the difference between the largest and smallest numbers in the triplet on the 1580th step of applying this rule, if the initial triplet of numbers was $\{80, 71, 20\}$? If the problem allows for multiple answers, list them all as a set.
|
60
|
deepscale
| 7,753
| ||
Find the sum\[1+11+111+\cdots+\underbrace{111\ldots111}_{n\text{ digits}}.\]
|
To find the sum of the sequence:
\[
1 + 11 + 111 + \cdots + \underbrace{111\ldots111}_{n\text{ digits}}
\]
we notice that each term in the sequence consists of digits '1' repeated a certain number of times. Specifically, the \(k\)-th term in the sequence is formed by \(k\) digits of '1', which can be expressed as:
\[
\underbrace{111\ldots1}_{k \text{ digits}} = \frac{10^k - 1}{9}
\]
This is due to the fact that a number with \(k\) ones can be expressed as a proper fraction in terms of powers of 10. Specifically:
\[
111\ldots1 = 10^{k-1} + 10^{k-2} + \cdots + 10^1 + 1
\]
This is a geometric series with first term 1 and ratio 10, so the sum is:
\[
\frac{10^k - 1}{9}
\]
Hence, the sum \( S \) of the sequence up to \( n \) terms is:
\[
S = \sum_{k=1}^{n} \frac{10^k - 1}{9}
\]
This can be rewritten as:
\[
S = \frac{1}{9} \sum_{k=1}^{n} (10^k - 1)
\]
Expanding the inner sum:
\[
\sum_{k=1}^{n} (10^k - 1) = \sum_{k=1}^{n} 10^k - \sum_{k=1}^{n} 1
\]
The first part, \(\sum_{k=1}^{n} 10^k\), is the sum of a geometric series:
\[
\sum_{k=1}^{n} 10^k = 10 + 10^2 + \cdots + 10^n = 10 \frac{10^n - 1}{10 - 1} = \frac{10^{n+1} - 10}{9}
\]
The second part is simply \( n \), since we are summing 1 a total of \( n \) times:
\[
\sum_{k=1}^{n} 1 = n
\]
Substituting back, we get:
\[
S = \frac{1}{9} \left( \frac{10^{n+1} - 10}{9} - n \right)
\]
Simplifying further, we obtain:
\[
S = \frac{10^{n+1} - 10 - 9n}{81}
\]
Therefore, the sum of the sequence is:
\[
\boxed{\frac{10^{n+1} - 10 - 9n}{81}}
\]
|
\frac{10^{n+1} - 10 - 9n}{81}
|
deepscale
| 6,027
| |
A covered rectangular football field with a length of 90 m and a width of 60 m is being designed to be illuminated by four floodlights, each hanging from some point on the ceiling. Each floodlight illuminates a circle, with a radius equal to the height at which the floodlight is hanging. Determine the minimally possible height of the ceiling, such that the following conditions are met: every point on the football field is illuminated by at least one floodlight, and the height of the ceiling must be a multiple of 0.1 m (for example, 19.2 m, 26 m, 31.9 m, etc.).
|
27.1
|
deepscale
| 27,758
| ||
Grisha wrote 100 numbers on the board. Then he increased each number by 1 and noticed that the product of all 100 numbers did not change. He increased each number by 1 again, and again the product of all the numbers did not change, and so on. Grisha repeated this procedure $k$ times, and each of the $k$ times the product of the numbers did not change. Find the largest possible value of $k$.
|
99
|
deepscale
| 14,373
| ||
The $10\times15$ rectangle $EFGH$ is cut into two congruent pentagons, which are repositioned to form a square. Determine the length $z$ of one side of the pentagons that aligns with one side of the square.
A) $5\sqrt{2}$
B) $5\sqrt{3}$
C) $10\sqrt{2}$
D) $10\sqrt{3}$
|
5\sqrt{3}
|
deepscale
| 12,253
| ||
Let $n$ be a nonnegative integer. Determine the number of ways that one can choose $(n+1)^2$ sets $S_{i,j}\subseteq\{1,2,\ldots,2n\}$ , for integers $i,j$ with $0\leq i,j\leq n$ , such that:
1. for all $0\leq i,j\leq n$ , the set $S_{i,j}$ has $i+j$ elements; and
2. $S_{i,j}\subseteq S_{k,l}$ whenever $0\leq i\leq k\leq n$ and $0\leq j\leq l\leq n$ .
Contents 1 Solution 1 2 Solution 2 2.1 Lemma 2.2 Filling in the rest of the grid 2.3 Finishing off 3 See also
|
Note that there are $(2n)!$ ways to choose $S_{1, 0}, S_{2, 0}... S_{n, 0}, S_{n, 1}, S_{n, 2}... S{n, n}$ , because there are $2n$ ways to choose which number $S_{1, 0}$ is, $2n-1$ ways to choose which number to append to make $S_{2, 0}$ , $2n-2$ ways to choose which number to append to make $S_{3, 0}$ ... After that, note that $S_{n-1, 1}$ contains the $n-1$ in $S_{n-1. 0}$ and 1 other element chosen from the 2 elements in $S_{n, 1}$ not in $S_{n-1, 0}$ so there are 2 ways for $S_{n-1, 1}$ . By the same logic there are 2 ways for $S_{n-1, 2}$ as well so $2^n$ total ways for all $S_{n-1, j}$ , so doing the same thing $n-1$ more times yields a final answer of $(2n)!\cdot 2^{n^2}$ .
-Stormersyle
|
\[
(2n)! \cdot 2^{n^2}
\]
|
deepscale
| 3,010
| |
The lengths of the sides of a triangle with positive area are $\log_{2}9$, $\log_{2}50$, and $\log_{2}n$, where $n$ is a positive integer. Find the number of possible values for $n$.
|
445
|
deepscale
| 14,341
| ||
For how many pairs of consecutive integers in $\{1000,1001,1002,\ldots,2000\}$ is no carrying required when the two integers are added?
|
Consider what carrying means: If carrying is needed to add two numbers with digits $abcd$ and $efgh$, then $h+d\ge 10$ or $c+g\ge 10$ or $b+f\ge 10$. 6. Consider $c \in \{0, 1, 2, 3, 4\}$. $1abc + 1ab(c+1)$ has no carry if $a, b \in \{0, 1, 2, 3, 4\}$. This gives $5^3=125$ possible solutions.
With $c \in \{5, 6, 7, 8\}$, there obviously must be a carry. Consider $c = 9$. $a, b \in \{0, 1, 2, 3, 4\}$ have no carry. This gives $5^2=25$ possible solutions. Considering $b = 9$, $a \in \{0, 1, 2, 3, 4, 9\}$ have no carry. Thus, the solution is $125 + 25 + 6=\boxed{156}$.
|
156
|
deepscale
| 6,556
| |
Let $T$ be the set of points $(x, y)$ in the Cartesian plane that satisfy
\[\big|\big| |x|-3\big|-1\big|+\big|\big| |y|-3\big|-1\big|=2.\]
What is the total length of all the lines that make up $T$?
|
32\sqrt{2}
|
deepscale
| 27,188
| ||
A boy wrote the first twenty natural numbers on a piece of paper. He didn't like how one of them was written, so he crossed it out. Among the remaining 19 numbers, there is one number that equals the arithmetic mean of these 19 numbers. Which number did he cross out? If the problem has more than one solution, write down the sum of these numbers.
|
20
|
deepscale
| 13,886
| ||
Let QR = x, PR = y, and PQ = z. Given that the area of the square on side QR is 144 = x^2 and the area of the square on side PR is 169 = y^2, find the area of the square on side PQ.
|
25
|
deepscale
| 26,714
| ||
Let $f$ be a function defined on the positive integers, such that
\[f(xy) = f(x) + f(y)\]for all positive integers $x$ and $y.$ Given $f(10) = 14$ and $f(40) = 20,$ find $f(500).$
|
39
|
deepscale
| 37,151
| ||
Jo is thinking of a positive integer less than 100. It is one less than a multiple of 8, and it is three less than a multiple of 7. What is the greatest possible integer Jo could be thinking of?
|
95
|
deepscale
| 37,787
| ||
Let $f(x) = 3x + 3$ and $g(x) = 4x + 3.$ What is $f(g(f(2)))$?
|
120
|
deepscale
| 34,356
| ||
In the rhombus \(A B C D\), the angle at vertex \(A\) is \(60^{\circ}\). Point \(N\) divides side \(A B\) in the ratio \(A N: B N = 2: 1\). Find the tangent of angle \(D N C\).
|
\sqrt{\frac{243}{121}}
|
deepscale
| 15,187
| ||
Solve for $y$: $$\log_4 \frac{2y+8}{3y-2} + \log_4 \frac{3y-2}{2y-5}=2$$
|
\frac{44}{15}
|
deepscale
| 20,516
| ||
Given two lines $l_{1}$: $mx+2y-2=0$ and $l_{2}$: $5x+(m+3)y-5=0$, if $l_{1}$ is parallel to $l_{2}$, calculate the value of $m$.
|
-5
|
deepscale
| 31,807
| ||
(1) Let the function $f(x)=|x+2|+|x-a|$. If the inequality $f(x) \geqslant 3$ always holds for $x$ in $\mathbb{R}$, find the range of the real number $a$.
(2) Given positive numbers $x$, $y$, $z$ satisfying $x+2y+3z=1$, find the minimum value of $\frac{3}{x}+\frac{2}{y}+\frac{1}{z}$.
|
16+8\sqrt{3}
|
deepscale
| 11,696
| ||
How many positive integers less than or equal to 240 can be expressed as a sum of distinct factorials? Consider 0 ! and 1 ! to be distinct.
|
Note that $1=0$ !, $2=0$ ! +1 !, $3=0$ ! +2 !, and $4=0!+1$ ! +2 !. These are the only numbers less than 6 that can be written as the sum of factorials. The only other factorials less than 240 are $3!=6,4!=24$, and $5!=120$. So a positive integer less than or equal to 240 can only contain 3 !, 4 !, 5 !, and/or one of $1,2,3$, or 4 in its sum. If it contains any factorial larger than 5 !, it will be larger than 240 . So a sum less than or equal to 240 will will either include 3 ! or not ( 2 ways), 4 ! or not ( 2 ways), 5 ! or not ( 2 ways), and add an additional $0,1,2,3$ or 4 ( 5 ways). This gives $2 \cdot 2 \cdot 2 \cdot 5=40$ integers less than 240 . However, we want only positive integers, so we must not count 0 . So there are 39 such positive integers.
|
39
|
deepscale
| 3,796
| |
Flea Kuzya can make jumps along a line \(L\). The starting point for the jumps is point \(A\) on line \(L\), the length of each jump is \(h\), and the direction of each jump is chosen randomly and equally likely. Find the probability that after making between two and five random jumps, Kuzya will be at a distance of \(2h\) from \(A\) at least once.
|
5/8
|
deepscale
| 13,555
| ||
Let \( n = 2^{25} 3^{17} \). How many positive integer divisors of \( n^2 \) are less than \( n \) but do not divide \( n \)?
|
424
|
deepscale
| 26,371
| ||
Five points on a circle are numbered 1,2,3,4, and 5 in clockwise order. A bug jumps in a clockwise direction from one point to another around the circle; if it is on an odd-numbered point, it moves one point, and if it is on an even-numbered point, it moves two points. If the bug begins on point 5, after 1995 jumps it will be on point
|
1. **Identify the movement pattern**: The bug starts at point 5. According to the rules, since 5 is odd, it moves one point to point 1.
2. **Continue the pattern**:
- From point 1 (odd), it moves one point to point 2.
- From point 2 (even), it moves two points to point 4.
- From point 4 (even), it moves two points to point 1.
3. **Recognize the cycle**: The sequence of points visited by the bug is 5, 1, 2, 4, and then back to 1. This forms a repeating cycle of 1, 2, 4.
4. **Determine the length of the cycle**: The cycle 1, 2, 4 repeats every 3 jumps.
5. **Calculate the position after 1995 jumps**:
- We need to find the position of the bug after 1995 jumps, which starts from point 5.
- Since the cycle starts after the initial jump from 5 to 1, we consider the remaining 1994 jumps.
- We calculate the remainder when 1994 is divided by 3 (the length of the cycle):
\[
1994 \mod 3 = 0
\]
- A remainder of 0 indicates that the bug completes a full cycle and returns to the starting point of the cycle.
6. **Identify the starting point of the cycle**: The cycle starts at point 1, but since the remainder is 0, it means the bug is at the last point of the cycle before it repeats, which is point 4.
7. **Conclusion**: After 1995 jumps, the bug will be on point 4.
Thus, the answer is $\boxed{\text{(D) \ 4}}$.
|
4
|
deepscale
| 1,437
| |
Find the monic quadratic polynomial, in $x,$ with real coefficients, which has $1 - i$ as a root.
|
x^2 - 2x + 2
|
deepscale
| 37,178
| ||
Given the expression $(1296^{\log_6 4096})^{\frac{1}{4}}$, calculate its value.
|
4096
|
deepscale
| 23,976
| ||
An infinite sheet of paper is divided into equal squares, some of which are colored red. In each $2\times3$ rectangle, there are exactly two red squares. Now consider an arbitrary $9\times11$ rectangle. How many red squares does it contain? (The sides of all considered rectangles go along the grid lines.)
|
33
|
deepscale
| 27,300
| ||
How can 13 rectangles of sizes $1 \times 1, 2 \times 1, 3 \times 1, \ldots, 13 \times 1$ be combined to form a rectangle, where all sides are greater than 1?
|
13 \times 7
|
deepscale
| 11,754
| ||
Given that $\sin \alpha$ is a root of the equation $5x^{2}-7x-6=0$, find:
$(1)$ The value of $\frac {\cos (2\pi-\alpha)\cos (\pi+\alpha)\tan ^{2}(2\pi-\alpha)}{\cos ( \frac {\pi}{2}+\alpha)\sin (2\pi-\alpha)\cot ^{2}(\pi-\alpha)}$.
$(2)$ In $\triangle ABC$, $\sin A+ \cos A= \frac { \sqrt {2}}{2}$, $AC=2$, $AB=3$, find the value of $\tan A$.
|
-2- \sqrt {3}
|
deepscale
| 7,498
| ||
The average of 12 numbers is 90. If the numbers 80, 85, and 92 are removed from the set of numbers, what is the average of the remaining numbers?
|
\frac{823}{9}
|
deepscale
| 26,985
| ||
A target consisting of five zones is hanging on the wall: a central circle (bullseye) and four colored rings. The width of each ring equals the radius of the bullseye. It is known that the number of points for hitting each zone is inversely proportional to the probability of hitting that zone, and hitting the bullseye scores 315 points. How many points does hitting the blue (second to last) zone score?
|
35
|
deepscale
| 15,072
| ||
Let \( x = \sqrt{1 + \frac{1}{1^{2}} + \frac{1}{2^{2}}} + \sqrt{1 + \frac{1}{2^{2}} + \frac{1}{3^{2}}} + \cdots + \sqrt{1 + \frac{1}{2012^{2}} + \frac{1}{2013^{2}}} \). Find the value of \( x - [x] \), where \( [x] \) denotes the greatest integer not exceeding \( x \).
|
\frac{2012}{2013}
|
deepscale
| 11,978
| ||
The figure shows three squares with non-overlapping interiors. The area of the shaded square is 1 square inch. What is the area of rectangle $ABCD$, in square inches?
[asy]size(100);
pair A = (0,0), D = (3,0),C = (3,2),B = (0,2);
draw(A--B--C--D--cycle);
draw(A--(1,0)--(1,1)--(0,1)--cycle);
filldraw(B--(1,2)--(1,1)--(0,1)--cycle,gray(.6),black);
label("$A$",A,WSW);
label("$B$",B,WNW);
label("$C$",C,ENE);
label("$D$",D,ESE);[/asy]
|
6
|
deepscale
| 39,498
| ||
Druv has a $33 \times 33$ grid of unit squares, and he wants to color each unit square with exactly one of three distinct colors such that he uses all three colors and the number of unit squares with each color is the same. However, he realizes that there are internal sides, or unit line segments that have exactly one unit square on each side, with these two unit squares having different colors. What is the minimum possible number of such internal sides?
|
66
|
deepscale
| 29,616
| ||
For a certain weekend, the weatherman predicts that it will rain with a $40\%$ probability on Saturday and a $50\%$ probability on Sunday. Assuming these probabilities are independent, what is the probability that it rains over the weekend (that is, on at least one of the days)? Express your answer as a percentage.
|
70\%
|
deepscale
| 35,439
| ||
What is the value of $\frac{(2112-2021)^2}{169}$?
|
1. **Calculate the difference in the numerator**:
\[
2112 - 2021 = 91
\]
This is the exact value, not an approximation.
2. **Square the difference**:
\[
(2112 - 2021)^2 = 91^2 = 8281
\]
3. **Divide by the denominator**:
\[
\frac{8281}{169}
\]
To simplify this, we can either perform the division directly or recognize that $169 = 13^2$ and $91 = 7 \times 13$. Thus, we can rewrite the expression using these factors:
\[
\frac{91^2}{13^2} = \left(\frac{91}{13}\right)^2 = 7^2 = 49
\]
4. **Conclude with the correct answer**:
\[
\boxed{\textbf{(C) } 49}
\]
|
49
|
deepscale
| 185
| |
In a certain football invitational tournament, 16 cities participate, with each city sending two teams, Team A and Team B. According to the competition rules, after several days of matches, it was found that aside from Team A from city $A$, the number of matches already played by each of the other teams was different. Find the number of matches already played by Team B from city $A$.
|
15
|
deepscale
| 28,991
| ||
In $\triangle ABC$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively, with $C= \dfrac {\pi}{3}$, $b=8$. The area of $\triangle ABC$ is $10 \sqrt {3}$.
(I) Find the value of $c$;
(II) Find the value of $\cos (B-C)$.
|
\dfrac {13}{14}
|
deepscale
| 31,578
| ||
The following diagram shows a square where each side has four dots that divide the side into three equal segments. The shaded region has area 105. Find the area of the original square.
[center][/center]
|
135
|
deepscale
| 9,986
| ||
In triangle ABC, the sides opposite to angles A, B, and C are a, b, and c respectively. Given that $(a+c)^2 = b^2 + 2\sqrt{3}ac\sin C$.
1. Find the measure of angle B.
2. If $b=8$, $a>c$, and the area of triangle ABC is $3\sqrt{3}$, find the value of $a$.
|
5 + \sqrt{13}
|
deepscale
| 18,184
| ||
How many positive integers, not exceeding 200, are multiples of 3 or 5 but not 6?
|
73
|
deepscale
| 31,495
| ||
An equilateral triangle shares a common side with a square as shown. What is the number of degrees in $m\angle CDB$? [asy] pair A,E,C,D,B; D = dir(60); C = dir(0); E = (0,-1); B = C+E; draw(B--D--C--B--E--A--C--D--A); label("D",D,N); label("C",C,dir(0)); label("B",B,dir(0));
[/asy]
|
15
|
deepscale
| 35,628
| ||
Given $\sin \alpha + \cos \alpha = \frac{\sqrt{2}}{3}$ ($\frac{\pi}{2} < \alpha < \pi$), find the values of the following expressions:
$(1) \sin \alpha - \cos \alpha$;
$(2) \sin^2\left(\frac{\pi}{2} - \alpha\right) - \cos^2\left(\frac{\pi}{2} + \alpha\right)$.
|
-\frac{4\sqrt{2}}{9}
|
deepscale
| 27,189
| ||
A total of $n$ people compete in a mathematical match which contains $15$ problems where $n>12$ . For each problem, $1$ point is given for a right answer and $0$ is given for a wrong answer. Analysing each possible situation, we find that if the sum of points every group of $12$ people get is no less than $36$ , then there are at least $3$ people that got the right answer of a certain problem, among the $n$ people. Find the least possible $n$ .
|
15
|
deepscale
| 29,229
| ||
In triangle \\(ABC\\), the sides opposite to angles \\(A\\), \\(B\\), and \\(C\\) are \\(a\\), \\(b\\), and \\(c\\) respectively, and it is given that \\(A < B < C\\) and \\(C = 2A\\).
\\((1)\\) If \\(c = \sqrt{3}a\\), find the measure of angle \\(A\\).
\\((2)\\) If \\(a\\), \\(b\\), and \\(c\\) are three consecutive positive integers, find the area of \\(\triangle ABC\\).
|
\dfrac{15\sqrt{7}}{4}
|
deepscale
| 17,662
| ||
Suppose $x$ is in the interval $[0, \pi/2]$ and $\log_{24\sin x} (24\cos x)=\frac{3}{2}$. Find $24\cot^2 x$.
|
Like Solution 1, we can rewrite the given expression as \[24\sin^3x=\cos^2x\] Divide both sides by $\sin^3x$. \[24 = \cot^2x\csc x\] Square both sides. \[576 = \cot^4x\csc^2x\] Substitute the identity $\csc^2x = \cot^2x + 1$. \[576 = \cot^4x(\cot^2x + 1)\] Let $a = \cot^2x$. Then \[576 = a^3 + a^2\]. Since $\sqrt[3]{576} \approx 8$, we can easily see that $a = 8$ is a solution. Thus, the answer is $24\cot^2x = 24a = 24 \cdot 8 = \boxed{192}$.
~IceMatrix
|
192
|
deepscale
| 7,009
| |
Rectangle \(ABCD\) has area 2016. Point \(Z\) is inside the rectangle and point \(H\) is on \(AB\) so that \(ZH\) is perpendicular to \(AB\). If \(ZH : CB = 4 : 7\), what is the area of pentagon \(ADCZB\)?
|
1440
|
deepscale
| 9,442
| ||
What non-zero real value for $x$ satisfies $(7x)^{14}=(14x)^7$?
|
1. **Start with the given equation:**
\[
(7x)^{14} = (14x)^7
\]
2. **Take the seventh root of both sides:**
\[
\sqrt[7]{(7x)^{14}} = \sqrt[7]{(14x)^7}
\]
Simplifying each side, we get:
\[
(7x)^2 = 14x
\]
3. **Expand and simplify the equation:**
\[
(7x)^2 = 49x^2
\]
Setting this equal to $14x$, we have:
\[
49x^2 = 14x
\]
4. **Divide both sides by $x$ (assuming $x \neq 0$):**
\[
49x = 14
\]
5. **Solve for $x$:**
\[
x = \frac{14}{49} = \frac{2}{7}
\]
6. **Conclude with the solution:**
\[
\boxed{\textbf{(B) }\frac{2}{7}}
\]
|
\frac{2}{7}
|
deepscale
| 2,139
| |
What is the greatest number of points of intersection that can occur when $2$ different circles and $2$ different straight lines are drawn on the same piece of paper?
|
11
|
deepscale
| 38,639
| ||
Let $a_{n}(n\geqslant 2, n\in N^{*})$ be the coefficient of the linear term of $x$ in the expansion of ${({3-\sqrt{x}})^n}$. Find the value of $\frac{3^2}{a_2}+\frac{3^3}{a_3}+…+\frac{{{3^{18}}}}{{{a_{18}}}}$.
|
17
|
deepscale
| 16,745
| ||
Given vectors $\overrightarrow {a}$=($\sqrt {3}$sinx, $\sqrt {3}$cos(x+$\frac {\pi}{2}$)+1) and $\overrightarrow {b}$=(cosx, $\sqrt {3}$cos(x+$\frac {\pi}{2}$)-1), define f(x) = $\overrightarrow {a}$$\cdot \overrightarrow {b}$.
(1) Find the minimum positive period and the monotonically increasing interval of f(x);
(2) In △ABC, a, b, and c are the sides opposite to A, B, and C respectively, with a=$2\sqrt {2}$, b=$\sqrt {2}$, and f(C)=2. Find c.
|
\sqrt {10}
|
deepscale
| 29,430
| ||
In triangle $\triangle ABC$, $sin2C=\sqrt{3}sinC$.
$(1)$ Find the value of $\angle C$;
$(2)$ If $b=6$ and the perimeter of $\triangle ABC$ is $6\sqrt{3}+6$, find the area of $\triangle ABC$.
|
6\sqrt{3}
|
deepscale
| 8,778
| ||
Side $AB$ of regular hexagon $ABCDEF$ is extended past $B$ to point $X$ such that $AX = 3AB$. Given that each side of the hexagon is $2$ units long, what is the length of segment $FX$? Express your answer in simplest radical form.
|
2\sqrt{13}
|
deepscale
| 35,758
| ||
In the $xy$ -coordinate plane, the $x$ -axis and the line $y=x$ are mirrors. If you shoot a laser beam from the point $(126, 21)$ toward a point on the positive $x$ -axis, there are $3$ places you can aim at where the beam will bounce off the mirrors and eventually return to $(126, 21)$ . They are $(126, 0)$ , $(105, 0)$ , and a third point $(d, 0)$ . What is $d$ ? (Recall that when light bounces off a mirror, the angle of incidence has the same measure as the angle of reflection.)
|
111
|
deepscale
| 29,428
| ||
Find the value of $x$ if $\log_8 x = 1.75$.
|
32\sqrt[4]{2}
|
deepscale
| 8,566
| ||
Given the circle $C$: $x^{2}+y^{2}-2x=0$, find the coordinates of the circle center $C$ and the length of the chord intercepted by the line $y=x$ on the circle $C$.
|
\sqrt{2}
|
deepscale
| 14,152
| ||
Let $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c}$ be nonzero vectors, no two of which are parallel, such that
\[(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = \frac{1}{3} \|\mathbf{b}\| \|\mathbf{c}\| \mathbf{a}.\]Let $\theta$ be the angle between $\mathbf{b}$ and $\mathbf{c}.$ Find $\sin \theta.$
|
\frac{2 \sqrt{2}}{3}
|
deepscale
| 40,099
| ||
The fenced area of a yard is a 15-foot by 12-foot rectangular region with a 3-foot by 3-foot square cut out, as shown. What is the area of the region within the fence, in square feet?
[asy]draw((0,0)--(16,0)--(16,12)--(28,12)--(28,0)--(60,0)--(60,48)--(0,48)--cycle);
label("15'",(30,48),N);
label("12'",(60,24),E);
label("3'",(16,6),W);
label("3'",(22,12),N);
[/asy]
|
171
|
deepscale
| 39,141
| ||
Let $a,$ $b,$ $c$ be positive real numbers such that $a + b + c = 1.$ Find the minimum value of
\[\frac{1}{a + 2b} + \frac{1}{b + 2c} + \frac{1}{c + 2a}.\]
|
3
|
deepscale
| 37,265
| ||
Given an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $a > b > 0$) with its left focus at F and the eccentricity $e = \frac{\sqrt{2}}{2}$, the line segment cut by the ellipse from the line passing through F and perpendicular to the x-axis has length $\sqrt{2}$.
(Ⅰ) Find the equation of the ellipse.
(Ⅱ) A line $l$ passing through the point P(0,2) intersects the ellipse at two distinct points A and B. Find the length of segment AB when the area of triangle OAB is at its maximum.
|
\frac{3}{2}
|
deepscale
| 21,482
| ||
[asy]
defaultpen(linewidth(0.7)+fontsize(10));
pair A=(0,0), B=(16,0), C=(16,16), D=(0,16), E=(32,0), F=(48,0), G=(48,16), H=(32,16), I=(0,8), J=(10,8), K=(10,16), L=(32,6), M=(40,6), N=(40,16);
draw(A--B--C--D--A^^E--F--G--H--E^^I--J--K^^L--M--N);
label("S", (18,8));
label("S", (50,8));
label("Figure 1", (A+B)/2, S);
label("Figure 2", (E+F)/2, S);
label("10'", (I+J)/2, S);
label("8'", (12,12));
label("8'", (L+M)/2, S);
label("10'", (42,11));
label("table", (5,12));
label("table", (36,11));
[/asy]
An $8'\times 10'$ table sits in the corner of a square room, as in Figure $1$ below. The owners desire to move the table to the position shown in Figure $2$. The side of the room is $S$ feet. What is the smallest integer value of $S$ for which the table can be moved as desired without tilting it or taking it apart?
|
To determine the smallest integer value of $S$ for which the table can be moved as desired without tilting it or taking it apart, we need to consider the diagonal of the table, as this is the longest dimension when the table is rotated.
1. **Calculate the diagonal of the table**: The table has dimensions $8'$ by $10'$. The diagonal $d$ of the table can be calculated using the Pythagorean theorem:
\[
d = \sqrt{8^2 + 10^2} = \sqrt{64 + 100} = \sqrt{164}
\]
2. **Estimate the value of $\sqrt{164}$**: We know that $\sqrt{144} = 12$ and $\sqrt{169} = 13$. Since $144 < 164 < 169$, it follows that $12 < \sqrt{164} < 13$.
3. **Determine the minimum room size $S$**: The diagonal of the table, which is between $12$ and $13$, represents the minimum dimension that the room must have to accommodate the table when it is rotated. Since we are looking for the smallest integer value of $S$, and $S$ must be at least as large as the diagonal, the smallest integer that satisfies this condition is $13$.
Thus, the smallest integer value of $S$ for which the table can be moved as desired without tilting it or taking it apart is $\boxed{13}$.
|
13
|
deepscale
| 1,436
| |
Each of a group of $50$ girls is blonde or brunette and is blue eyed of brown eyed. If $14$ are blue-eyed blondes, $31$ are brunettes, and $18$ are brown-eyed, then the number of brown-eyed brunettes is
|
1. **Identify the total number of girls and their characteristics:**
- Total number of girls = $50$
- Characteristics: blonde or brunette, blue-eyed or brown-eyed
2. **Calculate the number of blondes:**
- Given that there are $31$ brunettes, the number of blondes is:
\[
50 - 31 = 19
\]
3. **Determine the number of brown-eyed blondes:**
- Given that there are $14$ blue-eyed blondes, the number of brown-eyed blondes is:
\[
19 - 14 = 5
\]
4. **Calculate the number of brown-eyed brunettes:**
- Given that there are $18$ brown-eyed individuals in total, subtract the number of brown-eyed blondes to find the number of brown-eyed brunettes:
\[
18 - 5 = 13
\]
5. **Conclusion:**
- The number of brown-eyed brunettes is $\boxed{13}$.
|
13
|
deepscale
| 2,550
| |
Given $3^{7} + 1$ and $3^{15} + 1$ inclusive, how many perfect cubes lie between these two values?
|
231
|
deepscale
| 8,194
| ||
An isosceles triangle $ABP$ with sides $AB = AP = 3$ inches and $BP = 4$ inches is placed inside a square $AXYZ$ with a side length of $8$ inches, such that $B$ is on side $AX$. The triangle is rotated clockwise about $B$, then $P$, and so on along the sides of the square until $P$ returns to its original position. Calculate the total path length in inches traversed by vertex $P$.
A) $\frac{24\pi}{3}$
B) $\frac{28\pi}{3}$
C) $\frac{32\pi}{3}$
D) $\frac{36\pi}{3}$
|
\frac{32\pi}{3}
|
deepscale
| 30,890
| ||
Fill the digits 1 to 9 into the boxes in the following equation such that each digit is used exactly once and the equality holds true. It is known that the two-digit number $\overrightarrow{\mathrm{DE}}$ is not a multiple of 3. Determine the five-digit number $\overrightarrow{\mathrm{ABCDE}}$.
|
85132
|
deepscale
| 24,916
| ||
Find the minimum value of the distance $|AB|$ where point $A$ is the intersection of the line $y=a$ and the line $y=2x+2$, and point $B$ is the intersection of the line $y=a$ and the curve $y=x+\ln x$.
|
\frac{3}{2}
|
deepscale
| 24,294
| ||
\[\frac{\tan 96^{\circ} - \tan 12^{\circ} \left( 1 + \frac{1}{\sin 6^{\circ}} \right)}{1 + \tan 96^{\circ} \tan 12^{\circ} \left( 1 + \frac{1}{\sin 6^{\circ}} \right)} =\]
|
\frac{\sqrt{3}}{3}
|
deepscale
| 26,560
| ||
How many different positive, four-digit integers can be formed using the digits 2, 2, 9 and 9?
|
6
|
deepscale
| 34,975
| ||
Let $A_{1}, A_{2}, \ldots, A_{2015}$ be distinct points on the unit circle with center $O$. For every two distinct integers $i, j$, let $P_{i j}$ be the midpoint of $A_{i}$ and $A_{j}$. Find the smallest possible value of $\sum_{1 \leq i<j \leq 2015} O P_{i j}^{2}$.
|
Use vectors. $\sum\left|a_{i}+a_{j}\right|^{2} / 4=\sum\left(2+2 a_{i} \cdot a_{j}\right) / 4=\frac{1}{2}\binom{2015}{2}+\frac{1}{4}\left(\left|\sum a_{i}\right|^{2}-\sum\left|a_{i}\right|^{2}\right) \geq 2015 \cdot \frac{2014}{4}-\frac{2015}{4}=\frac{2015 \cdot 2013}{4}$, with equality if and only if $\sum a_{i}=0$, which occurs for instance for a regular 2015-gon.
|
\frac{2015 \cdot 2013}{4} \text{ OR } \frac{4056195}{4}
|
deepscale
| 3,318
| |
Find the smallest natural number \( n \) for which \( (n+1)(n+2)(n+3)(n+4) \) is divisible by 1000.
|
121
|
deepscale
| 15,567
| ||
Let \[f(x) = \left\{
\begin{array}{cl}
\sqrt{x} &\text{ if }x>4, \\
x^2 &\text{ if }x \le 4.
\end{array}
\right.\]Find $f(f(f(2)))$.
|
4
|
deepscale
| 34,491
| ||
Consider a new infinite geometric series: $$\frac{7}{4} + \frac{28}{9} + \frac{112}{27} + \dots$$
Determine the common ratio of this series.
|
\frac{16}{9}
|
deepscale
| 11,394
| ||
At Barker High School, a total of 36 students are on either the baseball team, the hockey team, or both. If there are 25 students on the baseball team and 19 students on the hockey team, how many students play both sports?
|
The two teams include a total of $25+19=44$ players. There are exactly 36 students who are on at least one team. Thus, there are $44-36=8$ students who are counted twice. Therefore, there are 8 students who play both baseball and hockey.
|
8
|
deepscale
| 5,263
| |
An equilateral triangle is inscribed in the ellipse whose equation is $x^2+4y^2=4$. One vertex of the triangle is $(0,1)$, one altitude is contained in the y-axis, and the square of the length of each side is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
|
Solving for $y$ in terms of $x$ gives $y=\sqrt{4-x^2}/2$, so the two other points of the triangle are $(x,\sqrt{4-x^2}/2)$ and $(-x,\sqrt{4-x^2}/2)$, which are a distance of $2x$ apart. Thus $2x$ equals the distance between $(x,\sqrt{4-x^2}/2)$ and $(0,1)$, so by the distance formula we have
\[2x=\sqrt{x^2+(1-\sqrt{4-x^2}/2)^2}.\]
Squaring both sides and simplifying through algebra yields $x^2=192/169$, so $2x=\sqrt{768/169}$ and the answer is $\boxed{937}$.
|
937
|
deepscale
| 6,705
| |
Let $ABCD$ be an isosceles trapezoid with $\overline{AD}||\overline{BC}$ whose angle at the longer base $\overline{AD}$ is $\dfrac{\pi}{3}$. The diagonals have length $10\sqrt {21}$, and point $E$ is at distances $10\sqrt {7}$ and $30\sqrt {7}$ from vertices $A$ and $D$, respectively. Let $F$ be the foot of the altitude from $C$ to $\overline{AD}$. The distance $EF$ can be expressed 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 $m + n$.
|
32
|
deepscale
| 35,969
| ||
A group of $12$ pirates agree to divide a treasure chest of gold coins among themselves as follows. The $k^{\text{th}}$ pirate to take a share takes $\frac{k}{12}$ of the coins that remain in the chest. The number of coins initially in the chest is the smallest number for which this arrangement will allow each pirate to receive a positive whole number of coins. How many coins does the $12^{\text{th}}$ pirate receive?
|
To solve this problem, we need to determine the smallest number of coins in the chest initially such that each pirate receives a whole number of coins when they take their share according to the given rule.
1. **Initial Setup and Recursive Formula**:
Let $x$ be the initial number of coins. The $k^\text{th}$ pirate takes $\frac{k}{12}$ of the remaining coins. We can express the number of coins left after each pirate takes their share recursively:
- After the 1st pirate: $x_1 = \frac{11}{12}x$
- After the 2nd pirate: $x_2 = \frac{10}{12}x_1 = \frac{10}{12} \cdot \frac{11}{12}x$
- Continuing this pattern, after the $k^\text{th}$ pirate: $x_k = \frac{12-k}{12}x_{k-1}$
2. **Expression for the $12^\text{th}$ Pirate**:
The $12^\text{th}$ pirate takes all remaining coins, which is $x_{11}$. We need to find $x_{11}$:
\[
x_{11} = \left(\frac{1}{12}\right) \left(\frac{2}{12}\right) \cdots \left(\frac{11}{12}\right) x
\]
Simplifying, we get:
\[
x_{11} = \frac{11!}{12^{11}} x
\]
3. **Finding the Smallest $x$**:
We need $x_{11}$ to be an integer. Therefore, $x$ must be a multiple of the denominator of $\frac{11!}{12^{11}}$ when reduced to lowest terms. We calculate:
\[
12^{11} = (2^2 \cdot 3)^{11} = 2^{22} \cdot 3^{11}
\]
\[
11! = 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2
\]
Factoring out the powers of 2 and 3 from $11!$, we find that all powers of 2 and 3 in $11!$ are less than or equal to those in $12^{11}$. Thus, the remaining factors are:
\[
\frac{11!}{2^{8} \cdot 3^{4}} = 11 \cdot 5 \cdot 7 \cdot 5 = 1925
\]
4. **Conclusion**:
The smallest $x$ such that each pirate gets a whole number of coins and the $12^\text{th}$ pirate takes all remaining coins is when $x$ is a multiple of the least common multiple of the denominators, which is $1925$. Therefore, the $12^\text{th}$ pirate receives:
\[
\boxed{1925}
\]
|
1925
|
deepscale
| 2,568
| |
Let \\(α\\) and \\(β\\) be in \\((0,π)\\), and \\(\sin(α+β) = \frac{5}{13}\\), \\(\tan \frac{α}{2} = \frac{1}{2}\\). Find the value of \\(\cos β\\).
|
-\frac{16}{65}
|
deepscale
| 20,944
| ||
A "fifty percent mirror" is a mirror that reflects half the light shined on it back and passes the other half of the light onward. Two "fifty percent mirrors" are placed side by side in parallel, and a light is shined from the left of the two mirrors. How much of the light is reflected back to the left of the two mirrors?
|
\frac{2}{3}
|
deepscale
| 8,519
| ||
Billy Goats invested some money in stocks and bonds. The total amount he invested was $\$165,\!000$. If he invested 4.5 times as much in stocks as he did in bonds, what was his total investment in stocks?
|
135,\!000
|
deepscale
| 38,471
| ||
Lucy surveyed a group of people about their knowledge of mosquitoes. To the nearest tenth of a percent, she found that $75.3\%$ of the people surveyed thought mosquitoes transmitted malaria. Of the people who thought mosquitoes transmitted malaria, $52.8\%$ believed that mosquitoes also frequently transmitted the common cold. Since mosquitoes do not transmit the common cold, these 28 people were mistaken. How many total people did Lucy survey?
|
70
|
deepscale
| 12,192
| ||
Jeff decides to play with a Magic 8 Ball. Each time he asks it a question, it has a 2/5 chance of giving him a positive answer. If he asks it 5 questions, what is the probability that it gives him exactly 2 positive answers?
|
\frac{216}{625}
|
deepscale
| 35,342
| ||
If \( a, b, c \) are real numbers such that \( |a-b|=1 \), \( |b-c|=1 \), \( |c-a|=2 \) and \( abc=60 \), calculate the value of \( \frac{a}{bc} + \frac{b}{ca} + \frac{c}{ab} - \frac{1}{a} - \frac{1}{b} - \frac{1}{c} \).
|
\frac{1}{10}
|
deepscale
| 24,951
| ||
A very bizarre weighted coin comes up heads with probability $\frac12$, tails with probability $\frac13$, and rests on its edge with probability $\frac16$. If it comes up heads, I win 1 dollar. If it comes up tails, I win 3 dollars. But if it lands on its edge, I lose 5 dollars. What is the expected winnings from flipping this coin? Express your answer as a dollar value, rounded to the nearest cent.
|
\$\dfrac23 \approx \$0.67
|
deepscale
| 34,939
| ||
Given the polar equation of a circle is $\rho=2\cos \theta$, the distance from the center of the circle to the line $\rho\sin \theta+2\rho\cos \theta=1$ is ______.
|
\dfrac { \sqrt {5}}{5}
|
deepscale
| 23,127
|
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