Answer: 108.0

A right prism with height $h$ has bases that are regular hexagons with sides of length $12$ . A vertex $A$ of the prism and its three adjacent vertices are the vertices of a triangular pyramid. The dihedral angle (the angle between the two planes) formed by the face of the pyramid that lies in a base of the prism and the face of the pyramid that does not contain $A$ measures $60$ degrees. Find $h^2$

Solution: Let $B$ and $C$ be the vertices adjacent to $A$ on the same base as $A$ , and let $D$ be the last vertex of the triangular pyramid.  Then $\angle CAB = 120^\circ$ . Let $X$ be the foot of the altitude from $A$ to $\overline{BC}$ . Then since $\triangle ABX$ is a $30-60-90$ triangle, $AX = 6$ .  Since the dihedral angle between $\triangle ABC$ and $\triangle BCD$ is $60^\circ$ $\triangle AXD$ is a $30-60-90$ triangle and $AD = 6\sqrt{3} = h$ . Thus $h^2 = \boxed{108}$

Solution: Let $B$ and $C$ be the vertices adjacent to $A$ on the same base as $A$ , and let $D$ be the last vertex of the triangular pyramid. Notice that we can already find some lengths. We have $AB=AC=12$ (given) and $BC=BD=\sqrt{144+h^2}$ by the Pythagorean Theorem. Let $M$ be the midpoint of $BC$ . Then, we have $AM=6$ $30-60-90$ ) triangles and $DM=\sqrt{36+h^2}$ by the Pythagorean Theorem. Applying the Law of Cosines, since $\angle AMD=60^{\circ}$ , we get \[h^2=36+h^2+36-\frac12 \cdot 12 \sqrt{36+h^2} \implies h^2=\boxed{108},\] as desired.


Answer: 40.0

A right rectangular prism $P_{}$ (i.e., a rectangular parallelpiped) has sides of integral length $a, b, c,$ with $a\le b\le c.$ A plane parallel to one of the faces of $P_{}$ cuts $P_{}$ into two prisms, one of which is similar to $P_{},$ and both of which have nonzero volume.  Given that $b=1995,$ for how many ordered triples $(a, b, c)$ does such a plane exist?

Solution:
Let $P'$ be the prism similar to $P$ , and let the sides of $P'$ be of length $x,y,z$ , such that $x \le y \le z$ . Then
\[\frac{x}{a} = \frac{y}{b} = \frac zc < 1.\]
Note that if the ratio of similarity was equal to $1$ , we would have a prism with zero volume. As one face of $P'$ is a face of $P$ , it follows that $P$ and $P'$ share at least two side lengths in common. Since $x < a, y < b, z < c$ , it follows that the only possibility is $y=a,z=b=1995$ . Then,
\[\frac{x}{a} = \frac{a}{1995} = \frac{1995}{c} \Longrightarrow ac = 1995^2 = 3^25^27^219^2.\]
The number of factors of $3^25^27^219^2$ is $(2+1)(2+1)(2+1)(2+1) = 81$ . Only in $\left\lfloor \frac {81}2 \right\rfloor = 40$ of these cases is $a < c$ (for $a=c$ , we end with a prism of zero volume). We can easily verify that these will yield nondegenerate prisms, so the answer is $\boxed{040}$


Answer: 576.0

A right rectangular prism whose surface area and volume are numerically equal has edge lengths $\log_{2}x, \log_{3}x,$ and $\log_{4}x.$ What is $x?$
$\textbf{(A)}\ 2\sqrt{6} \qquad\textbf{(B)}\ 6\sqrt{6} \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 48 \qquad\textbf{(E)}\ 576$

Solution:
The surface area of this right rectangular prism is $2(\log_{2}x\log_{3}x+\log_{2}x\log_{4}x+\log_{3}x\log_{4}x).$
The volume of this right rectangular prism is $\log_{2}x\log_{3}x\log_{4}x.$
Equating the numerical values of the surface area and the volume, we have \[2(\log_{2}x\log_{3}x+\log_{2}x\log_{4}x+\log_{3}x\log_{4}x)=\log_{2}x\log_{3}x\log_{4}x.\] Dividing both sides by $\log_{2}x\log_{3}x\log_{4}x,$ we get \[2\left(\frac{1}{\log_{4}x}+\frac{1}{\log_{3}x}+\frac{1}{\log_{2}x}\right)=1. \hspace{15mm} (\bigstar)\] Recall that $\log_{b}a=\frac{1}{\log_{a}b}$ and $\log_{b}\left(a^n\right)=n\log_{b}a,$ so we rewrite $(\bigstar)$ as \begin{align*} 2(\log_{x}4+\log_{x}3+\log_{x}2)&=1 \\ 2\log_{x}24&=1 \\ \log_{x}576&=1 \\ x&=\boxed{576} ~MRENTHUSIASM


Answer: 21.0

A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
Although I can't draw the exact picture of this problem, but it is quite easy to imagine that four vertices of the base of this pyramid is on a circle (Radius $\frac{6}{\sqrt{2}} = 3\sqrt{2}$ ). Since all five vertices are on the sphere, the distances of the spherical center and the vertices are the same: $l$ . Because of the symmetrical property of the pyramid,
we can imagine that the line of the apex and the (sphere's) center will intersect the square at the (base's) center.
Since the volume is $54 = \frac{1}{3} \cdot S \cdot h = \frac{1}{3} \cdot 6^2 \cdot h$ , where $h=\frac{9}{2}$ is the height of this pyramid, we have: $l^2=\left(\frac{9}{2}-l\right)^2+\left(3\sqrt{2}\right)^2$ according to the Pythagorean theorem.
Solve this equation will give us $l = \frac{17}{4},$ therefore $m+n=\boxed{021}.$

Solution: To start, we find the height of the pyramid. By the volume of a pyramid formula, we have \[\frac13 \cdot 6^2 \cdot h=54 \implies h=\frac92.\] Next, let us find the length of the non-base sides of the pyramid. By the Pythagorean Theorem, noting that the distance from one vertex of the base to the center of the base is $\frac12 \cdot 6\sqrt2=3\sqrt2$ , we have \[x=\sqrt{\left(\frac92\right)^2+(3\sqrt2)^2}=\sqrt{\frac{153}4}=\frac{3\sqrt{17}}2.\] Taking the cross section of the pyramid and transforming the problem into $2$ -d, it suffices to find the radius of the circumcircle of a triangle of side lengths $\frac{3\sqrt{17}}2$ $\frac{3\sqrt{17}}2$ $6\sqrt2$ . This turns out to be easy by the formula $R=\frac{abc}{4A}$ , and through computing this value (the work has been left out) we find that $R=\frac{17}4$ , so our answer is $\boxed{021}$

Solution:
By the volume of a pyramid formula, we have that the height of the pyramid is $\frac{9}{2}$ . Since the base is a square with side length 6, the simplest way to place it in the coordinate plane is to put the center of the square at the origin and let the base be on the $xy$ plane. Then, the vertices of the base would be $(3,3,0), (3,-3,0), (-3,3,0), (-3,-3,0)$ in some order. Also, let the vertex be $(0,0,\frac{9}{2})$ . Recall that the formula for a sphere is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ where the center is $(a,b,c)$ and the radius is $r$ . Symmetry gives that $a=b=0$ . Plug in $(3,3,0)$ and $(0,0,\frac{9}{2})$ and you get the system of equation
$18+c^2=r^2$
$(\frac{9}{2}-c)^2=r^2$
Solving gives $c=1/4$ and $r=17/4$ , so our answer is $17+4=\boxed{021}$ .~ Ddk001


Answer: 5.0

A rising number, such as $34689$ , is a positive integer each digit of which is larger than each of the digits to its left. There are $\binom{9}{5} = 126$ five-digit rising numbers. When these numbers are arranged from smallest to largest, the $97^{\text{th}}$ number in the list does not contain the digit
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution:
The list starts with $12345$ .  There are $\binom{8}{4} = 70$ four-digit rising numbers that do not begin with $1$ , and thus also $70$ five digit rising numbers that do begin with $1$ that are formed by simply putting a $1$ before the four digit number.
Thus, the $71^{\text{st}}$ number is $23456$ .  There are $\binom{6}{3} = 20$ three-digit rising numbers that do not begin with a $1,2$ or $3$ , and thus $20$ five digit rising numbers that begin with a $23$
Thus, the $91^{\text{st}}$ number is $24567$ .  Counting up, $24568, 24569, 24578, 24579, 24589, 24678$ is the $97^{\text{th}}$ number, which does not contain the digit $\boxed{5}$


Answer: 600.0

A roll of tape is $4$ inches in diameter and is wrapped around a ring that is $2$ inches in diameter. A cross section of the tape is shown in the figure below. The tape is $0.015$ inches thick. If the tape is completely unrolled, approximately how long would it be? Round your answer to the nearest $100$ inches.
$\textbf{(A) } 300\qquad\textbf{(B) } 600\qquad\textbf{(C) } 1200\qquad\textbf{(D) } 1500\qquad\textbf{(E) } 1800$

Solution: The roll of tape is $1/0.015=$ 66 layers thick. In order to find the total length, we have to find the average of each concentric circle and multiply it by $66$ . Since the diameter of the small circle is $2$ inches and the diameter of the large one is $4$ inches, the "middle value" is $3$ . Therefore, the average circumference is $3\pi$ . Multiplying $3\pi \cdot 66$ gives $(B) \boxed{600}$

Solution: There are about $\dfrac{1}{0.015}=\dfrac{200}{3}$ "full circles" of tape, and with average circumference of $\dfrac{4+2}{2}\pi=3\pi.$ $\dfrac{200}{3} \cdot 3\pi=200\pi,$ which means the answer is $\boxed{600}$

Solution: The volume of the tape is always the same, but we can either calculate it when the tape is unrolled as a really long, thin rectangular prism, or we can calculate it as a cylinder with a hole cut out of it. When we calculate it as a long rectangular prism, we can say that the length is $X$ (this is what the problem wants!) and the width is $Y$ . Then, the volume is, of course, $0.015 \cdot X \cdot Y.$ Now, notice that the "width" of our rectangular prism is also the "height" of our cylinder with a hole cut out of it. Then, we can calculate the volume as base times height, or in this case, $3\pi \cdot Y.$ Now, since the volume always stays the same, we know that $3\pi \cdot Y = 0.015 \cdot X \cdot Y.$ Cancelling the $Y$ 's give us an equation for $X$ , and if we approximate $\pi$ as $3$ , then $X = \boxed{600}$ . Yay!

Solution:
If you cannot notice that the average diameter is $3$ , you can still solve this problem by the following method.
The same with solution 1, we have $\frac{1000}{0.015}$ layers of tape. If we consider every layers with the diameter $2$ , the length should be $\frac{1000}{0.015}2\pi\approx 400$ . If the diameter is seem as $4$ , the length should be $800$ . So, the length is between $400$ and $800$ , the only possible answer is $\boxed{600}$


Answer: 2.0

A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is $1$ foot wide on all four sides. What is the length in feet of the inner rectangle?
[asy] size(6cm); defaultpen(fontsize(9pt)); path rectangle(pair X, pair Y){ return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle; } filldraw(rectangle((0,0),(7,5)),gray(0.5)); filldraw(rectangle((1,1),(6,4)),gray(0.75)); filldraw(rectangle((2,2),(5,3)),white);  label("$1$",(0.5,2.5)); draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead)); draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead));  label("$1$",(1.5,2.5)); draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead)); draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead));  label("$1$",(4.5,2.5)); draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead)); draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead));  label("$1$",(4.1,1.5)); draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead)); draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead));  label("$1$",(3.7,0.5)); draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead)); draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead)); [/asy]
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 6 \qquad \textbf{(E) }8$

Solution:
Let the length of the inner rectangle be $x$
Then the area of that rectangle is $x\cdot1 = x$
The second largest rectangle has dimensions of $x+2$ and $3$ , making its area $3x+6$ . The area of the second shaded area, therefore, is $3x+6-x = 2x+6$
The largest rectangle has dimensions of $x+4$ and $5$ , making its area $5x + 20$ . The area of the largest shaded region is the largest rectangle minus the second largest rectangle, which is $(5x+20) - (3x+6) = 2x + 14$
The problem states that $x, 2x+6, 2x+14$ is an arithmetic progression, meaning that the terms in the sequence increase by the same amount each term.
Therefore, $(2x+6) - (x) = (2x+14) - (2x+6)\implies x+6 = 8\implies x =2\implies \boxed{2}$


Answer: 5.0

A sample consisting of five observations has an arithmetic mean of $10$ and a median of $12$ . The smallest value that the range (largest observation minus smallest) can assume for such a sample is
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 10$

Solution: The minimum range occurs in the set $\{7,7,12,12,12\}$ , so the answer is $\boxed{5}$


Answer: 947.0

A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode (most frequent value). Let $D$ be the difference between the mode and the arithmetic mean of the sample. What is the largest possible value of $\lfloor D\rfloor$ ? (For real $x$ $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$ .)

Solution:
Let the mode be $x$ , which we let appear $n > 1$ times. We let the arithmetic mean be $M$ , and the sum of the numbers $\neq x$ be $S$ . Then \begin{align*} D &= \left|M-x\right| = \left|\frac{S+xn}{121}-x\right| = \left|\frac{S}{121}-\left(\frac{121-n}{121}\right)x\right| \end{align*} As $S$ is essentially independent of $x$ , it follows that we wish to minimize or maximize $x$ (in other words, $x \in [1,1000]$ ). Indeed, $D(x)$ is symmetric about $x = 500.5$ ; consider replacing all of numbers $x_i$ in the sample with $1001-x_i$ , and the value of $D$ remains the same. So, without loss of generality , let $x=1$ . Now, we would like to maximize the quantity
$S$ contains $121-n$ numbers that may appear at most $n-1$ times. Therefore, to maximize $S$ , we would have $1000$ appear $n-1$ times, $999$ appear $n-1$ times, and so forth. We can thereby represent $S$ as the sum of $n-1$ arithmetic series of $1000, 999, \ldots, 1001 - \left\lfloor \frac{121-n}{n-1} \right\rfloor$ . We let $k = \left\lfloor \frac{121-n}{n-1} \right\rfloor$ , so
where $R(n)$ denotes the sum of the remaining $121-(n-1)k$ numbers, namely $R(n) = (121-(n-1)k)(1000-k)$
At this point, we introduce the crude estimate that $k=\frac{121-n}{n-1}$ , so $R(n) = 0$ and \begin{align*}2S+2n &= (121-n)\left(2001-\frac{121-n}{n-1}\right)+2n = (120-(n-1))\left(2002-\frac{120}{n-1}\right)\end{align*} Expanding (ignoring the constants, as these do not affect which $n$ yields a maximum) and scaling, we wish to minimize the expression $5(n-1) + \frac{36}{n-1}$ . By AM-GM , we have $5(n-1) + \frac{36}{n-1} \ge 2\sqrt{5(n-1) \cdot \frac{36}{n-1}}$ , with equality coming when $5(n-1) = \frac{36}{n-1}$ , so $n-1 \approx 3$ . Substituting this result and some arithmetic gives an answer of $\boxed{947}$

Solution:
With the same reasoning as Solution 1, in order to get largest possible value of D, we can construct that our set of numbers as $\underbrace{1,1,1...1,}_\text{n times}\underbrace{2,2,2...2,}_\text{n times}\underbrace{3,3,3...3,}_\text{n times}........\underbrace{1000,1000,1000....}_\text{n+1  times}$ And, we need to find the value of n that makes the sum as low as possible. And, we can create a formula to make it easier. It isn't hard to find the sum. The numbers which are not 1000, average to $\frac{120}{2n}$ or $\frac{60}{n}$ , and there are $120-n$ of them. So, they sum to $\frac{60}{n}(120-n)$ . And, the sum of the numbers that are 1000 is $1000(n+1)$ so, our total sum gets us $1000(n+1)+120/2n(120-n)$ We want to minimize it, since the mode will always be 1000. And, testing the values n = 1, n = 2, n = 3, n = 4, we get these results.
$n = 1: 2000+60*119 = 9140$
$n = 2: 3000+30*118 = 6540$
$n = 3: 4000+20*117 = 6340$
$n = 4: 5000+15*116 = 6740$
And, as n grows larger and larger from 4, the values will start increasing. Thus, the lowest possible sum is 6340. Dividing by 121, the lowest possible mean is 52.396...., and thus, the highest possible value of $D$ is 947.604, and the floor of that is $\boxed{947}$


Answer: 1022.0

A scanning code consists of a $7 \times 7$ grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of $49$ squares. A scanning code is called $\textit{symmetric}$ if its look does not change when the entire square is rotated by a multiple of $90 ^{\circ}$ counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?
$\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}$

Solution: Imagine folding the scanning code along its lines of symmetry. There will be $10$ regions which you have control over coloring. Since we must subtract off $2$ cases for the all-black and all-white cases, the answer is $2^{10}-2=\boxed{1022.}$

Solution:
\[\begin{tabular}{|c|c|c|c|c|c|c|} \hline T & T & T & X & T & T & T \\ \hline T & T & T & Y& T & T & T \\ \hline T & T & T & Z & T & T & T \\ \hline X & Y & Z & W & Z & Y & X \\ \hline T & T & T & Z & T & T & T \\ \hline T & T & T & Y & T & T & T \\ \hline T & T & T & X & T & T & T \\ \hline \end{tabular}\]
There are $3 \times 3$ squares in the corners of this $7 \times 7$ square, and there is a horizontal and vertical stripe through the middle. Because we need to have symmetry when the diagonals and midpoints of the large square is connected, we can create a table like this: (Different letters represent different color choices between black and white)
\[\begin{tabular}{|c|c|c|} \hline A & B & C \\ \hline B & D & E \\ \hline C & E & F \\ \hline \end{tabular}\] (Note that coloring one $3 \times 3$ square will also determine the colorings of the other $3$ because the large $7 \times 7$ square must look the same when it is rotated by $90^\circ$
There are $6$ different letters and $2$ choices of color (black and white) for each letter, so there are $2^6=64$ colorings of a proper $3 \times 3$ square. Now, all that's left are the horizontal and vertical stripes. Using similar logic, we can see that there are $4$ different letters, so there are $2^4=16$ different colorings. Multiplying them together gives $16 \times 64 = 1024$ . Going back to the question, we see that "there must be at least one square of each color in this grid of $49$ squares." We must then eliminate $2$ options: an all-black grid and an all-white grid. $1024-2= \boxed{1022}$ -hansenhe


Answer: 1022.0

Solution: Imagine folding the scanning code along its lines of symmetry. There will be $10$ regions which you have control over coloring. Since we must subtract off $2$ cases for the all-black and all-white cases, the answer is $2^{10}-2=\boxed{1022.}$

Solution:
\[\begin{tabular}{|c|c|c|c|c|c|c|} \hline T & T & T & X & T & T & T \\ \hline T & T & T & Y& T & T & T \\ \hline T & T & T & Z & T & T & T \\ \hline X & Y & Z & W & Z & Y & X \\ \hline T & T & T & Z & T & T & T \\ \hline T & T & T & Y & T & T & T \\ \hline T & T & T & X & T & T & T \\ \hline \end{tabular}\]
There are $3 \times 3$ squares in the corners of this $7 \times 7$ square, and there is a horizontal and vertical stripe through the middle. Because we need to have symmetry when the diagonals and midpoints of the large square is connected, we can create a table like this: (Different letters represent different color choices between black and white)
\[\begin{tabular}{|c|c|c|} \hline A & B & C \\ \hline B & D & E \\ \hline C & E & F \\ \hline \end{tabular}\] (Note that coloring one $3 \times 3$ square will also determine the colorings of the other $3$ because the large $7 \times 7$ square must look the same when it is rotated by $90^\circ$
There are $6$ different letters and $2$ choices of color (black and white) for each letter, so there are $2^6=64$ colorings of a proper $3 \times 3$ square. Now, all that's left are the horizontal and vertical stripes. Using similar logic, we can see that there are $4$ different letters, so there are $2^4=16$ different colorings. Multiplying them together gives $16 \times 64 = 1024$ . Going back to the question, we see that "there must be at least one square of each color in this grid of $49$ squares." We must then eliminate $2$ options: an all-black grid and an all-white grid. $1024-2= \boxed{1022}$ -hansenhe


Answer: 13.5

A school has $100$ students and $5$ teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are $50, 20, 20, 5,$ and $5$ . Let $t$ be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let $s$ be the average value obtained if a student was picked at random and the number of students in their class, including the student, is noted. What is $t-s$
$\textbf{(A)}\ {-}18.5  \qquad\textbf{(B)}\ {-}13.5 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 13.5 \qquad\textbf{(E)}\ 18.5$

Solution: The formula for expected values is \[\text{Expected Value}=\sum(\text{Outcome}\cdot\text{Probability}).\] We have \begin{align*} t &= 50\cdot\frac15 + 20\cdot\frac15 + 20\cdot\frac15 + 5\cdot\frac15 + 5\cdot\frac15 \\ &= (50+20+20+5+5)\cdot\frac15 \\ &= 100\cdot\frac15 \\ &= 20, \\ s &= 50\cdot\frac{50}{100} + 20\cdot\frac{20}{100} + 20\cdot\frac{20}{100} + 5\cdot\frac{5}{100} + 5\cdot\frac{5}{100} \\ &= 25 + 4 + 4 + 0.25 + 0.25 \\ &= 33.5. \end{align*} Therefore, the answer is $t-s=\boxed{13.5}.$


Answer: 13.5

Solution: The formula for expected values is \[\text{Expected Value}=\sum(\text{Outcome}\cdot\text{Probability}).\] We have \begin{align*} t &= 50\cdot\frac15 + 20\cdot\frac15 + 20\cdot\frac15 + 5\cdot\frac15 + 5\cdot\frac15 \\ &= (50+20+20+5+5)\cdot\frac15 \\ &= 100\cdot\frac15 \\ &= 20, \\ s &= 50\cdot\frac{50}{100} + 20\cdot\frac{20}{100} + 20\cdot\frac{20}{100} + 5\cdot\frac{5}{100} + 5\cdot\frac{5}{100} \\ &= 25 + 4 + 4 + 0.25 + 0.25 \\ &= 33.5. \end{align*} Therefore, the answer is $t-s=\boxed{13.5}.$


Answer: 24.2

A scientist walking through a forest recorded as integers the heights of $5$ trees standing in a row. She observed that each tree was either twice as tall or half as tall as the one to its right. Unfortunately some of her data was lost when rain fell on her notebook. Her notes are shown below, with blanks indicating the missing numbers. Based on her observations, the scientist was able to reconstruct the lost data. What was the average height of the trees, in meters?
\[\begingroup \setlength{\tabcolsep}{10pt} \renewcommand{\arraystretch}{1.5} \begin{tabular}{|c|c|} \hline Tree 1 & \rule{0.4cm}{0.15mm} meters \\ Tree 2 & 11 meters \\ Tree 3 & \rule{0.5cm}{0.15mm} meters \\ Tree 4 & \rule{0.5cm}{0.15mm} meters \\ Tree 5 & \rule{0.5cm}{0.15mm} meters \\ \hline Average height & \rule{0.5cm}{0.15mm}\text{ .}2 meters \\ \hline \end{tabular} \endgroup\] $\textbf{(A) }22.2 \qquad \textbf{(B) }24.2 \qquad \textbf{(C) }33.2 \qquad \textbf{(D) }35.2 \qquad \textbf{(E) }37.2$

Solution: We will show that $22$ $11$ $22$ $44$ , and $22$ meters are the heights of the trees from left to right. We are given that all tree heights are integers, so since Tree 2 has height $11$ meters, we can deduce that Trees 1 and 3 both have a height of $22$ meters. There are now three possible cases for the heights of Trees 4 and 5 (in order for them to be integers), namely heights of $11$ and $22$ $44$ and $88$ , or $44$ and $22$ . Checking each of these, in the first case, the average is $17.6$ meters, which doesn't end in $.2$ as the problem requires. Therefore, we consider the other cases. With $44$ and $88$ , the average is $37.4$ meters, which again does not end in $.2$ , but with $44$ and $22$ , the average is $24.2$ meters, which does. Consequently, the answer is $\boxed{24.2}$

Solution:
Notice the average height of the trees ends with $0.2$ ; therefore, the sum of all five heights of the trees must end with $1$ or $6$ . ( $0.2$ $\cdot$ $5$ $1$ )
We already know Tree $2$ is $11$ meters tall. Both Tree $1$ and Tree $3$ must $22$ meters tall - since neither can be $5.5$ .
Once again, apply our observation for solving for the Tree $4$ 's height. Tree $4$ can't be $11$ meters for the sum of the five tree heights to still end with $1$ . Therefore, the Tree $4$ is $44$ meters tall.
Now, Tree $5$ can either be $22$ or $88$ . Find the average height for both cases of Tree $5$ . Doing this, we realize the Tree $5$ must be $22$ for the average height to end with $0.2$ and that the average height is $\boxed{24.2}$

Solution:
As in Solution 1, we shall show that the heights of the trees are $22$ $11$ $22$ $44$ , and $22$ meters. Let $S$ be the sum of the heights, so that the average height will be $\frac{S}{5}$ meters. We note that $0.2 = \frac{1}{5}$ , so in order for $\frac{S}{5}$ to end in $.2$ $S$ must be one more than a multiple of $5$ . Moreover, as all the heights are integers, the heights of Tree 1 and Tree 3 are both $22$ meters. At this point, our table looks as follows: \[\begingroup \setlength{\tabcolsep}{10pt} \renewcommand{\arraystretch}{1.5} \begin{tabular}{|c|c|} \hline Tree 1 & 22 meters \\ Tree 2 & 11 meters \\ Tree 3 & 22 meters \\ Tree 4 & \rule{0.5cm}{0.15mm} meters \\ Tree 5 & \rule{0.5cm}{0.15mm} meters \\ \hline Average height & \rule{0.5cm}{0.15mm}\text{ .}2 meters \\ \hline \end{tabular} \endgroup\]
If Tree 4 now has a height of $11$ , then Tree 5 would need to have height $22$ , but in that case $S$ would equal $88$ , which is not $1$ more than a multiple of $5$ . So we instead take Tree 4 to have height $44$ . Then the sum of the heights of the first 4 trees is $22+11+22+44 = 99$ , so using a height of $22$ for Tree 5 gives $S=121$ , which is $1$ more than a multiple of $5$ (whereas $88$ gives $S = 187$ , which is not). Thus the average height of the trees is $\frac{121}{5} = \boxed{24.2}$ meters.

Solution: Since we know that the tree heights have to be integers, then it is immediate that Tree 1 and 3 have a height of $22$ . Now using the information given by the last column (that the average of the heights of the trees ends in $.2$ ), we can tell that the sum of all the heights of the trees ends in either $1$ or $6$ , because those are the only numbers from $0$ to $9$ that are congruent to $1$ after taking modulo $5$ . The two multiples of eleven (eleven because all of the tree heights have to be a multiple of eleven if they are integers) that come to mind are $66$ and $121$ . Since the sum of the heights of Tree 1, 2, and 3 is already $55$ , we know that $66$ is impossible to obtain. Then, we can decide with relative confidence that the answer should be $\frac{121}{5} = \boxed{24.2}$

Solution:
Since her average ends with 0.2, the sum of her tree heights must be $\equiv 1 \pmod {5}$ .
If Tree 1's height is $a$ , Tree 3's is $b$ , Tree 4's is $c$ , and Tree 5's is $d$ , then we $a$ and $b$ are both 22, since all of the tree heights are integers.
Now we have $22+11+22+c+d \equiv 1 \pmod {5}$ . Simplifying, we get $c+d \equiv 1 \pmod {5}$ .
The only possible combination of $c$ and $d$ that abides the condition are 44 and 22, respectively.
Adding these up, we get 121, and $\frac{121}{5}$ is $\boxed{24.2}$


Answer: 100.0

A scout troop buys $1000$ candy bars at a price of five for $2$ dollars. They sell all the candy bars at the price of two for $1$ dollar.  What was their profit, in dollars?
$\textbf{(A) }\ 100      \qquad \textbf{(B) }\ 200      \qquad \textbf{(C) }\ 300      \qquad \textbf{(D) }\ 400      \qquad \textbf{(E) }\ 500$

Solution: \begin{align*} \mbox{Expenses} &= 1000 \cdot \frac25 = 400 \\ \mbox{Revenue}  &= 1000 \cdot \frac12 = 500 \\ \mbox{Profit}   &= \mbox{Revenue} - \mbox{Expenses} = 500-400 = \boxed{100} Note: Revenue is a gain.

Solution:
Note that the troop buys $10$ candy bars at a price of $4$ dollars and sells $10$ bars at a price of $5$ dollars. So the troop gains $1$ dollar for every $10$ bars. So therefore we divide $1000 \div 10 = 100$ . So our answer is $\boxed{100}$ .
~HyperVoid


Answer: 100.0

Solution: \begin{align*} \mbox{Expenses} &= 1000 \cdot \frac25 = 400 \\ \mbox{Revenue}  &= 1000 \cdot \frac12 = 500 \\ \mbox{Profit}   &= \mbox{Revenue} - \mbox{Expenses} = 500-400 = \boxed{100} Note: Revenue is a gain.

Solution:
Note that the troop buys $10$ candy bars at a price of $4$ dollars and sells $10$ bars at a price of $5$ dollars. So the troop gains $1$ dollar for every $10$ bars. So therefore we divide $1000 \div 10 = 100$ . So our answer is $\boxed{100}$ .
~HyperVoid


Answer: 400000.0

A semipro baseball league has teams with 21 players each. League rules state that a player must be paid at least $15,000 and that the total of all players' salaries for each team cannot exceed $700,000. What is the maximum possible salary, in dollars, for a single player?
$\mathrm{(A)}\ 270,000\qquad\mathrm{(B)}\ 385,000\qquad\mathrm{(C)}\ 400,000\qquad\mathrm{(D)}\ 430,000\qquad\mathrm{(E)}\ 700,000$

Solution: The maximum salary for a single player occurs when the other 20 players receive the minimum salary. The total of all players' salaries is 700000. The answer is $700000-15000*20=400000\Rightarrow \boxed{400,000}$


Answer: 324.0

A sequence $a_1,a_2,\dots$ of non-negative integers is defined by the rule $a_{n+2}=|a_{n+1}-a_n|$ for $n\geq 1$ . If $a_1=999$ $a_2<999$ and $a_{2006}=1$ , how many different values of $a_2$ are possible?
$\mathrm{(A)}\ 165 \qquad \mathrm{(B)}\ 324 \qquad \mathrm{(C)}\ 495 \qquad \mathrm{(D)}\ 499 \qquad \mathrm{(E)}\ 660$

Solution:
We say the sequence $(a_n)$ completes at $i$ if $i$ is the minimal positive integer such that $a_i = a_{i + 1} = 1$ . Otherwise, we say $(a_n)$ does not complete.
Note that if $d = \gcd(999, a_2) \neq 1$ , then $d|a_n$ for all $n \geq 1$ , and $d$ does not divide $1$ , so if $\gcd(999, a_2) \neq 1$ , then $(a_n)$ does not complete. (Also, $a_{2006}$ cannot be 1 in this case since $d$ does not divide $1$ , so we do not care about these $a_2$ at all.)
From now on, suppose $\gcd(999, a_2) = 1$
We will now show that $(a_n)$ completes at $i$ for some $i \leq 2006$ . We will do this with 3 lemmas.
Lemma: If $a_j \neq a_{j + 1}$ , and neither value is $0$ , then $\max(a_j, a_{j + 1}) > \max(a_{j + 2}, a_{j + 3})$
Proof: There are 2 cases to consider.
If $a_j > a_{j + 1}$ , then $a_{j + 2} = a_j - a_{j + 1}$ , and $a_{j + 3} = |a_j - 2a_{j + 1}|$ . So $a_j > a_{j + 2}$ and $a_j > a_{j + 3}$
If $a_j < a_{j + 1}$ , then $a_{j + 2} = a_{j + 1} - a_j$ , and $a_{j + 3} = a_j$ . So $a_{j + 1} > a_{j + 2}$ and $a_{j + 1} > a_{j + 3}$
In both cases, $\max(a_j, a_{j + 1}) > \max(a_{j + 2}, a_{j + 3})$ , as desired.
Lemma: If $a_i = a_{i + 1}$ , then $a_i = 1$ . Moreover, if instead we have $a_i = 0$ for some $i > 2$ , then $a_{i - 1} = a_{i - 2} = 1$
Proof: By the way $(a_n)$ is constructed in the problem statement, having two equal consecutive terms $a_i = a_{i + 1}$ implies that $a_i$ divides every term in the sequence. So $a_i | 999$ and $a_i | a_2$ , so $a_i | \gcd(999, a_2) = 1$ , so $a_i = 1$ . For the proof of the second result, note that if $a_i = 0$ , then $a_{i - 1} = a_{i - 2}$ , so by the first result we just proved, $a_{i - 2} = a_{i - 1} = 1$
Lemma: $(a_n)$ completes at $i$ for some $i \leq 2000$
Proof: Suppose $(a_n)$ completed at some $i > 2000$ or not at all. Then by the second lemma and the fact that neither $999$ nor $a_2$ are $0$ , none of the pairs $(a_1, a_2), ..., (a_{1999}, a_{2000})$ can have a $0$ or be equal to $(1, 1)$ . So the first lemma implies \[\max(a_1, a_2) > \max(a_3, a_4) > \cdots > \max(a_{1999}, a_{2000})  > 0,\] so $999 = \max(a_1, a_2) \geq 1000$ , a contradiction. Hence $(a_n)$ completes at $i$ for some $i \leq 2000$
Now we're ready to find exactly which values of $a_2$ we want to count.
Let's keep in mind that $2006 \equiv 2 \pmod 3$ and that $a_1 = 999$ is odd. We have two cases to consider.
Case 1: If $a_2$ is odd, then $a_3$ is even, so $a_4$ is odd, so $a_5$ is odd, so $a_6$ is even, and this pattern must repeat every three terms because of the recursive definition of $(a_n)$ , so the terms of $(a_n)$ reduced modulo 2 are \[1, 1, 0, 1, 1, 0, ...,\] so $a_{2006}$ is odd and hence $1$ (since if $(a_n)$ completes at $i$ , then $a_k$ must be $0$ or $1$ for all $k \geq i$ ).
Case 2: If $a_2$ is even, then $a_3$ is odd, so $a_4$ is odd, so $a_5$ is even, so $a_6$ is odd, and this pattern must repeat every three terms, so the terms of $(a_n)$ reduced modulo 2 are \[1, 0, 1, 1, 0, 1, ...,\] so $a_{2006}$ is even, and hence $0$
We have found that $a_{2006} = 1$ is true precisely when $\gcd(999, a_2) = 1$ and $a_2$ is odd. This tells us what we need to count.
There are $\phi(999) = 648$ numbers less than $999$ and relatively prime to it ( $\phi$ is the Euler totient function). We want to count how many of these are odd. Note that \[t \mapsto 999 - t\] is a 1-1 correspondence between the odd and even numbers less than and relatively prime to $999$ . So our final answer is $648/2 = 324$ , or $\boxed{324}$


Answer: 996506.0

A sequence of natural numbers is constructed by listing the first $4$ , then skipping one, listing the next $5$ , skipping $2$ , listing $6$ , skipping $3$ , and on the $n$ th iteration, listing $n+3$ and skipping $n$ . The sequence begins $1,2,3,4,6,7,8,9,10,13$ . What is the $500,\!000$ th number in the sequence?
$\textbf{(A)}\ 996,\!506\qquad\textbf{(B)}\ 996,\!507\qquad\textbf{(C)}\ 996,\!508\qquad\textbf{(D)}\ 996,\!509\qquad\textbf{(E)}\ 996,\!510$

Solution:
If we list the rows by iterations, then we get
$1,2,3,4$
$6,7,8,9,10$
$13,14,15,16,17,18$ etc.
so that the $500,000$ th number is the $506$ th number on the $997$ th row because $4+5+6+7......+999 = 499,494$ . The last number of the $996$ th row (when including the numbers skipped) is $499,494 + (1+2+3+4.....+996)= 996,000$ , (we add the $1-996$ because of the numbers we skip) so our answer is $996,000 + 506 = \boxed{996,506}$

Solution: One may also note we simply need to add the number of skipped numbers to $500,000$ to get our answer. The number of skipped numbers is $\frac{996\cdot 997}{2}$ which has a units digit of $6$ . Looking at the answer choices, it becomes apparent that the answer is $\boxed{996,506}$

Solution:
Let's start with natural numbers, with no skips in between.
$1,2,3,4,5,...,500,000$
All we need to do is count how many numbers are skipped, $n$ , and "push" (add on to) $500,000$ according to however many numbers are skipped.
Clearly, $\frac{999(1000)}{2}<500,000<\frac{1000(1001)}{2}$ . This means that there are $999-3=996$ skipped number "blocks" in the sequence because we started counting from 4.
Therefore $n=\frac{996(997)}{2}=496,506$ , and the answer is $496,506+500000=\boxed{996,506}$

Solution: First, we group the numbers together in the following way: ${1, 2, 3, 4, (5)}; {6, 7, 8, 9, 10, (11, 12)}; ...$ We quickly realize that the number of terms in the curly braces follow a pattern: $5, 7, 9, 11, ... , n$ (where $n$ is the $n^\text{th}$ block. Now, we can tell that the last number in a curly brace will be the number of terms in the set added to the number of terms in all the previous sets. Luckily for us, odd numbers are easy to add. If we pretend that there was a $1, 3$ at the beginning, then the sum of all of the numbers before and including $n$ is $(n+2)^2$ . However, we have to subtract $1+3$ which results in $n^2+4n$ . The amount of numbers in the parenthesis are the $n^\text{th}$ triangular number or $\frac{n(n+1)}{2}$ . Next, we want to find the greatest $n$ , where $(n^2+4n) - \frac{n(n+1)}{2}<500000$ . Simplifying, we get $n^2+7n<1000000$ . We realize that $n=1000$ results in a number just $7000$ greater than our target. Next, we square $999$ $(1000 - 1)^2 = 1000000 - 2001$ . As we decrease $n$ by $1$ , we decrease the result of the equation by approximately $2000$ . In order to decrease by at least $7000$ , we have to decrease $4$ times leading to $n=1000-4=996$ . We plug it in to $n^2+4n$ getting $996\cdot1000=996000$ . This is the last number in the $996^\text{th}$ set. The number of terms used is $\frac{n\cdot(n+7)}{2}=\frac{996\cdot1003}{2}=498\cdot1003=498000+1494=499494$ . We need to add $500000-499494=506$ terms to get an answer of $\boxed{996,506}$


Answer: 68.0

A sequence of numbers starts with $1$ $2$ , and $3$ . The fourth number of the sequence is the sum of the previous three numbers in the sequence: $1+2+3=6$ . In the same way, every number after the fourth is the sum of the previous three numbers. What is the eighth number in the sequence?
$\textbf{(A)}\  11  \qquad \textbf{(B)}\   20  \qquad \textbf{(C)}\   37  \qquad \textbf{(D)}\   68  \qquad \textbf{(E)}\    99$

Solution:
List them out, adding the three previous numbers to get the next number,
\[1,2,3,6,11,20,37,\boxed{68}\]


Answer: 13.0

A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?
[asy] path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(3,0)*p); draw(shift(3,1)*p); draw(shift(4,0)*p); draw(shift(4,1)*p); draw(shift(7,0)*p); draw(shift(7,1)*p); draw(shift(7,2)*p); draw(shift(8,0)*p); draw(shift(8,1)*p); draw(shift(8,2)*p); draw(shift(9,0)*p); draw(shift(9,1)*p); draw(shift(9,2)*p);[/asy]
$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

Solution: The first square has a sidelength of $1$ , the second square $2$ , and so on. The seventh square has $7$ and is made of $7^2=49$ unit tiles. The sixth square has $6$ and is made of $6^2=36$ unit tiles. The seventh square has $49-36=\boxed{13}$ more tiles than the sixth square.

Solution: The edge of each square is one tile length longer than the edge of the previous square, which means that each square has $2*$ edge length $- 1$ more tiles than the previous square, because each square is just one edge added on the top and on the right to the previous square, with one overlapping tile. Then the seventh square has $2*7-1=13$ more tiles than the sixth square, which is $\boxed{13}$

Solution: We see a pattern of 1, 4, and 9, all of which are the squares of 1, 2,and 3 respectively. So, the 6th and 7th squares will also follow the same pattern. Via the difference of squares, we see that $7^2 - 6^2$ . Now we can see that the seventh square has $(7-6)(7+6) =$ $\boxed{13}$ more tiles than the sixth square.


Answer: 9.0

A set $S$ is constructed as follows. To begin, $S = \{0,10\}$ . Repeatedly, as long as possible, if $x$ is an integer root of some polynomial $a_{n}x^n + a_{n-1}x^{n-1} + \dots + a_{1}x + a_0$ for some $n\geq{1}$ , all of whose coefficients $a_i$ are elements of $S$ , then $x$ is put into $S$ . When no more elements can be added to $S$ , how many elements does $S$ have?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad\textbf{(C)}\ 7 \qquad\textbf{(D)}\ 9 \qquad\textbf{(E)}\ 11$

Solution:
At first, $S=\{0,10\}$
\[\begin{tabular}{r c l c l} \(10x+10\) & has root & \(x=-1\) & so now & \(S=\{-1,0,10\}\) \\ \(-x^{10}-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x+10\) & has root & \(x=1\) & so now & \(S=\{-1,0,1,10\}\) \\ \(x+10\) & has root & \(x=-10\) & so now & \(S=\{-10,-1,0,1,10\}\) \\ \(x^3+x-10\) & has root & \(x=2\) & so now & \(S=\{-10,-1,0,1,2,10\}\) \\ \(x+2\) & has root & \(x=-2\) & so now & \(S=\{-10,-2,-1,0,1,2,10\}\) \\ \(2x-10\) & has root & \(x=5\) & so now & \(S=\{-10,-2,-1,0,1,2,5,10\}\) \\ \(x+5\) & has root & \(x=-5\) & so now & \(S=\{-10,-5,-2,-1,0,1,2,5,10\}\) \end{tabular}\]
At this point, no more elements can be added to $S$ . To see this, let
\begin{align*} a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + a_0 &= 0 \\ x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) + a_0 &= 0 \\ x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) &= -a_0 \end{align*}
with each $a_i$ in $S$ $x$ is a factor of $a_0$ , and $a_0$ is in $S$ , so $x$ has to be a factor of some element in $S$ . There are no such integers left, so there can be no more additional elements. $\{-10,-5,-2,-1,0,1,2,5,10\}$ has $9$ elements $\to \boxed{9}$


Answer: 8.0

A set $S$ of points in the $xy$ -plane is symmetric about the origin, both coordinate axes, and the line $y=x$ . If $(2,3)$ is in $S$ , what is the smallest number of points in $S$
$\mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 16$

Solution: If $(2,3)$ is in $S$ , then $(3,2)$ is also, and quickly we see that every point of the form $(\pm 2, \pm 3)$ or $(\pm 3, \pm 2)$ must be in $S$ . Now note that these $8$ points satisfy all of the symmetry conditions. Thus the answer is $\boxed{8}$


Answer: 9.0

A set S consists of triangles whose sides have integer lengths less than 5, and no two elements of S are congruent or similar. What is the largest number of elements that S can have?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution:
Define $T$ to be the set of all integral triples $(a, b, c)$ such that $a \ge b \ge c$ $b+c > a$ , and $a, b, c < 5$ . Now we enumerate the elements of $T$
$(4, 4, 4)$
$(4, 4, 3)$
$(4, 4, 2)$
$(4, 4, 1)$
$(4, 3, 3)$
$(4, 3, 2)$
$(3, 3, 3)$
$(3, 3, 2)$
$(3, 3, 1)$
$(3, 2, 2)$
$(2, 2, 2)$
$(2, 2, 1)$
$(1, 1, 1)$
It should be clear that $|S|$ is simply $|T|$ minus the larger "duplicates" (e.g. $(2, 2, 2)$ is a larger duplicate of $(1, 1, 1)$ ). Since $|T|$ is $13$ and the number of higher duplicates is $4$ , the answer is $13 - 4$ or $\boxed{9}$


Answer: 791.0

A set contains four numbers. The six pairwise sums of distinct elements of the set, in no particular order, are $189$ $320$ $287$ $234$ $x$ , and $y$ . Find the greatest possible value of $x+y$

Solution: Let these four numbers be $a$ $b$ $c$ , and $d$ , where $a>b>c>d$ $x+y$ needs to be maximized, so let $x=a+b$ and $y=a+c$ because these are the two largest pairwise sums. Now $x+y=2a+b+c$ needs to be maximized. Notice that $2a+b+c=3(a+b+c+d)-(a+2b+2c+3d)=3((a+c)+(b+d))-((a+d)+(b+c)+(b+d)+(c+d))$ . No matter how the numbers $189$ $320$ $287$ , and $234$ are assigned to the values $a+d$ $b+c$ $b+d$ , and $c+d$ , the sum $(a+d)+(b+c)+(b+d)+(c+d)$ will always be $189+320+287+234$ . Therefore we need to maximize $3((a+c)+(b+d))-(189+320+287+234)$ . The maximum value of $(a+c)+(b+d)$ is achieved when we let $a+c$ and $b+d$ be $320$ and $287$ because these are the two largest pairwise sums besides $x$ and $y$ . Therefore, the maximum possible value of $x+y=3(320+287)-(189+320+287+234)=\boxed{791}$

Solution:
Let the four numbers be $a$ $b$ $c$ , and $d$ , in no particular order. Adding the pairwise sums, we have $3a+3b+3c+3d=1030+x+y$ , so $x+y=3(a+b+c+d)-1030$ . Since we want to maximize $x+y$ , we must maximize $a+b+c+d$
Of the four sums whose values we know, there must be two sums that add to $a+b+c+d$ . To maximize this value, we choose the highest pairwise sums, $320$ and $287$ . Therefore, $a+b+c+d=320+287=607$
We can substitute this value into the earlier equation to find that $x+y=3(607)-1030=1821-1030=\boxed{791}$

Solution:
There are two cases we can consider. Let the elements of our set be denoted $a,b,c,d$ , and say that the largest sums $x$ and $y$ will be consisted of $b+d$ and $c+d$ . Thus, we want to maximize $b+c+2d$ , which means $d$ has to be as large as possible, and $a$ has to be as small as possible to maximize $b$ and $c$ . So, the two cases we look at are:
Case 1:
\[a+d = 287\] \[b+c = 320\] \[a+b = 234\] \[a+c = 189\]
Case 2:
\[a+d = 320\] \[b+c = 189\] \[a+b = 234\] \[a+c = 287\]
Note we have determined these cases by maximizing the value of $a+d$ determined by our previous conditions. So, the answers for each ( after some simple substitution ) will be:
Case 1:
\[(a,b,c,d) = (\frac{103}{2},\frac{365}{2},\frac{275}{2},\frac{471}{2})\]
Case 2:
\[(a,b,c,d) = (166,68,121,154)\]
See the first case has our largest $d$ , so our answer will be $471+\frac{640}{2} = \boxed{791}$


Answer: 600.0

A set of $25$ square blocks is arranged into a $5 \times 5$ square. How many different combinations of $3$ blocks can be selected from that set so that no two are in the same row or column?
$\textbf{(A) } 100 \qquad\textbf{(B) } 125  \qquad\textbf{(C) } 600 \qquad\textbf{(D) } 2300 \qquad\textbf{(E) } 3600$

Solution:
There are $25$ ways to choose the first square. The four remaining squares in its row and column and the square you chose exclude nine squares from being chosen next time.
There are $16$ remaining blocks to be chosen for the second square. The three remaining spaces in its row and column and the square you chose must be excluded from being chosen next time.
Finally, the last square has $9$ remaining choices.
The number of ways to choose $3$ squares is $25 \cdot 16 \cdot 9,$ but the order in which you chose the squares does not matter as the blocks are indistinguishable, so we divide by $3!$
\[\frac{25 \cdot 16 \cdot 9}{3 \cdot 2 \cdot 1} = 25 \cdot 8 \cdot 3 = 100 \cdot 6 = \boxed{600}\]

Solution:
Once we choose our three squares, we will have occupied three separate columns $(A, B, C)$ and three separate rows. There are ${5 \choose 3} \times {5 \choose 3}$ ways to choose these rows and columns.
There are $3$ ways to assign the square in column $A$ to a row, $2$ ways to assign the square in column $B$ to one of the remaining two rows, and poor square in column $C$ doesn't get to choose. $:($
In total, we have \[{5 \choose 3} \times {5 \choose 3} \times 3!\] which totals out to $\boxed{600}$

Solution: We know that there are $\binom{25}{3}=2300$ ways to choose three blocks. However, the restriction clearly limits the number of ways we can choose our blocks. Hence, only $\text{(A)}$ $\text{(B)}$ , or $\text{(C)}$ could be the correct answer. Clearly, there are more than $125$ ways, thus yielding $\boxed{600}$ ways.


Answer: 557.0

A set of $n$ people participate in an online video basketball tournament. Each person may be a member of any number of $5$ -player teams, but no two teams may have exactly the same $5$ members. The site statistics show a curious fact: The average, over all subsets of size $9$ of the set of $n$ participants, of the number of complete teams whose members are among those $9$ people is equal to the reciprocal of the average, over all subsets of size $8$ of the set of $n$ participants, of the number of complete teams whose members are among those $8$ people. How many values $n$ $9\leq n\leq 2017$ , can be the number of participants?
$\textbf{(A) } 477 \qquad \textbf{(B) } 482 \qquad \textbf{(C) } 487 \qquad \textbf{(D) } 557 \qquad \textbf{(E) } 562$

Solution:
Let there be $T$ teams. For each team, there are ${n-5\choose 4}$ different subsets of $9$ players that includes a given full team, so the total number of team-(group of 9) pairs is
\[T{n-5\choose 4}.\]
Thus, the expected value of the number of full teams in a random set of $9$ players is
\[\frac{T{n-5\choose 4}}{{n\choose 9}}.\]
Similarly, the expected value of the number of full teams in a random set of $8$ players is
\[\frac{T{n-5\choose 3}}{{n\choose 8}}.\]
The condition is thus equivalent to the existence of a positive integer $T$ such that
\[\frac{T{n-5\choose 4}}{{n\choose 9}}\frac{T{n-5\choose 3}}{{n\choose 8}} = 1.\]
\[T^2\frac{(n-5)!(n-5)!8!9!(n-8)!(n-9)!}{n!n!(n-8)!(n-9)!3!4!} = 1\]
\[T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{3!4!}{8!9!}\]
\[T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{144}{7!7!8\cdot8\cdot9}\]
\[T^2 = \big((n)(n-1)(n-2)(n-3)(n-4)\big)^2 \frac{1}{4\cdot7!7!}\]
\[T = \frac{(n)(n-1)(n-2)(n-3)(n-4)}{2^5\cdot3^2\cdot5\cdot7}\]
Note that this is always less than ${n\choose 5}$ , so as long as $T$ is integral, $n$ is a possibility. Thus, we have that this is equivalent to
\[2^5\cdot3^2\cdot5\cdot7\big|(n)(n-1)(n-2)(n-3)(n-4).\]
It is obvious that $5$ divides the RHS, and that $7$ does iff $n\equiv 0,1,2,3,4\mod 7$ . Also, $3^2$ divides it iff $n\not\equiv 5,8\mod 9$ . One can also bash out that $2^5$ divides it in $16$ out of the $32$ possible residues $\mod 32$
Note that $2016 = 7*9*32$ so by using all numbers from $2$ to $2017$ , inclusive, it is clear that each possible residue $\mod 7,9,32$ is reached an equal number of times, so the total number of working $n$ in that range is $5\cdot 7\cdot 16 = 560$ . However, we must subtract the number of "working" $2\leq n\leq 8$ , which is $3$ . Thus, the answer is $\boxed{557}$


Answer: 385.0

A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$

Solution:
There are $10 \cdot 2+1=21$ teams. Any of the $\tbinom{21}3=1330$ sets of three teams must either be a fork (in which one team beat both the others) or a cycle:
 But we know that every team beat exactly $10$ other teams, so for each possible $A$ at the head of a fork, there are always exactly $\tbinom{10}2$ choices for $B$ and $C$ as $A$ beat exactly 10 teams and we are choosing 2 of them. Therefore there are $21\cdot\tbinom{10}2=945$ forks, and all the rest must be cycles.
Thus the answer is $1330-945=385$ which is $\boxed{385}$

Solution: Since there are $21$ teams and for each set of three teams there is a cycle, there are a total of $\tbinom{21}3=1330$ cycles of three teams.  Because about $1/4$ of the cycles $\{A, B, C\}$ satisfy the conditions of the problems, our answer is close to $1/4 \cdot 1330=332.5$ . Looking at the answer choices, we find that $332.5$ is closer to $385$ than any other answer choices, and that the next closest is $665$ which is twice of $332.5$ , so our answer is $385$ which is $\boxed{385}$

Solution:
Let's arrange all the teams in a circle and assume that each team won against the first 10 teams clockwise of themselves, and lost against the first 10 teams counter-clockwise of themselves.
Consider a working set of $3$ teams: moving clockwise, we know that the first team beat the team clockwise of itself, that team beat the next team clockwise of itself, and the final team beat the first team, which would be clockwise of itself. However, we also must remember that if the first team beat the second team, moving clockwise, the second team cannot be more than $10$ teams away from the first team; the same applies to the second and third team, and the third and first team.
Let's say, WLOG, that the first team, team $A$ , is at position $0$ out of $20$ on the circle.
If team $A$ is to beat team $B$ , since we are assuming each team beats the 10 members clockwise of themselves, there are $10$ places on the circle that team $B$ could be: positions $1$ through $10$ . Also, if team $C$ is to beat team $A$ , team $C$ must be located from positions $11$ to $20$
If Team $B$ is in position $n$ $C$ must be located from position $n+1$ to position $n+10$ , since $B$ beats $C$ . We also just found that $C$ 's position must be between $11$ and $20$ inclusive. So, when $B$ is in position $1$ $C$ can only be in position $11$ . When $B$ is in position $2$ $C$ can be in position $11$ or $12$ . In general, when $B$ is in position $n$ , there are $n$ choices for where $C$ can be. So, there are $1+2+3+ \dots + 10 = 55$ ways to place $B$ and $C$ . There are $21$ players to choose as player $A$ , and each working set will be counted $3$ times, so our final answer is $55 \cdot 21 \div 3 = \boxed{385,}$

Solution:
This is a Graph Theory problem with directed graph. There are $21$ teams in total. WLOG, pick team $A$ , there are $10$ teams that lost to $A$ and $10$ teams that won over $A$ . Call the group of teams that lost to $A$ group $L$ , and the group of teams that won over $A$ group $W$

Any team from group $L$ that won a team from group $W$ will form a cycle with $A$ . Now we need to count how many teams in group $L$ won over a team from group $W$ . The total number of wins in group $L$ is $10 \cdot 10 =100$ . There are $\tbinom{10}2=45$ wins among the teams inside group $L$ . So group $L$ has $100-45=55$ wins over group $W$
\[\frac{21 \cdot 55}{3}=\boxed{385,}\]


Answer: 385.0

Solution:
We use complementary counting. First, because each team played $20$ other teams, there are $21$ teams total. All sets that do not have $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A$ have one team that beats both the other teams. Thus we must count the number of sets of three teams such that one team beats the two other teams and subtract that number from the total number of ways to choose three teams.
There are $21$ ways to choose the team that beat the two other teams, and $\binom{10}{2} = 45$ to choose two teams that the first team both beat. This is $21 * 45 = 945$ sets. There are $\binom{21}{3} = 1330$ sets of three teams total. Subtracting, we obtain $1330 - 945 = \boxed{385})$ is our answer.

Solution: Note that there are $21$ teams total and $\binom{21}{3}=1330$ ways to pick ${A,B,C}.$ The possible arrangements are one team beats the other two or they each win/lose equally (we want the second case). Approximately $\frac{1}{4}$ of all the arrangements satisfy the second case, and $\frac{1330}{4}=332.5,$ which is by far the closest to $\boxed{385}.$


Answer: 385.0

Solution:
As above, note that there are 21 teams, and call them A, B, C, ... T, U. WLOG, assume that A beat teams B-L and lost to teams M-U. We will count the number of sets satisfying the “cycle-win” condition—e.g. here, A beats a team in X which beats a team in Y which beats A. The first and third part of the condition are already met by our wlog, so we just need to count of number of ways the second condition is true (a team in X beats a team in Y). These are the number of cycle-wins that include A, then multiply by 21 (for each team) and divide by 3 (since every set will be counted by each of the 3 teams that are a part of that set).
To do this, let X $=\{B, ..., L\}$ and Y $=\{M, ..., U\}$ . Since a total of $10*10=100$ losses total were suffered by teams in Y and $\binom{10}{2}=45^{*}$ losses were suffered by teams in Y from teams in Y, we have $100-45=55$ losses suffered by teams in Y from teams in X. Hence, for each of these $55$ losses, there is exactly one set of three teams that includes A that satisfies the problem conditions. Thus, the answer is $\frac{55\cdot 21}{3}=\boxed{385}$


Answer: 18.0

A set of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a perfect square , and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?
$\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20$

Solution:
The pattern is quite simple to see after listing a couple of terms.
\[\begin{tabular}{|r|r|r|} \hline \#&\text{Removed}&\text{Left}\\ \hline 1&10&90\\ 2&9&81\\ 3&9&72\\ 4&8&64\\ 5&8&56\\ 6&7&49\\ 7&7&42\\ 8&6&36\\ 9&6&30\\ 10&5&25\\ 11&5&20\\ 12&4&16\\ 13&4&12\\ 14&3&9\\ 15&3&6\\ 16&2&4\\ 17&2&2\\ 18&1&1\\     \hline \end{tabular}\]
Thus, the answer is $\boxed{18}$

Solution:
Given $n^2$ tiles, a step removes $n$ tiles, leaving $n^2 - n$ tiles behind. Now, $(n-1)^2 = n^2 - n + (1-n) < n^2 - n < n^2$ , so in the next step $n-1$ tiles are removed. This gives $(n^2 - n) - (n-1) = n^2 - 2n + 1 = (n-1)^2$ , another perfect square.
Thus each two steps we cycle down a perfect square, and in $(10-1)\times 2 = 18$ steps, we are left with $1$ tile, hence our answer is $\boxed{18}$


Answer: 18.0

Solution:
We start of with $100 = 10 \cdot 10$ numbers. When we use the certain operation, call if $P(x)$ , have $100 - 10 = 90 = 10 \cdot 9$ .
Then we do $P(x)$ again, to subtract $9$ numbers and get $9 \cdot 9$ . In the end, we will want $1 = 1 \cdot 1$ . We can say we have to use $P(x)$ once to make $n \cdot n$ into $n \cdot (n-1)$ . Thus we must use it twice to get from $n \cdot n$ to $(n-1)(n-1)$ . For example, it takes us $2$ of $P(x)$ to get from $10 \cdot 10$ to $9 \cdot 9$ . Then $2$ of $P(x)$ to get from $9 \cdot 9$ to $8 \cdot 8$ . You should try this with $7$ and $6$ , and see it works. This means we can have $n$ be the number we start with, and $1$ be the number we want. Then we would use $P(x)$ $2(n - 1)$ times to get $1 \cdot 1$ . Substituting $n$ for $10$ we get $2(10-1) = 2 \cdot 9 = \boxed{18}$ .
- Wiselion


Answer: 700800.0

A ship sails $10$ miles in a straight line from $A$ to $B$ , turns through an angle between $45^{\circ}$ and $60^{\circ}$ , and then sails another $20$ miles to $C$ . Let $AC$ be measured in miles. Which of the following intervals contains $AC^2$ [asy] unitsize(2mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4;  pair B=(0,0), A=(-10,0), C=20*dir(50);  draw(A--B--C); draw(A--C,linetype("4 4"));  dot(A); dot(B); dot(C); label("$10$",midpoint(A--B),S); label("$20$",midpoint(B--C),SE); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); [/asy]
$\textbf{(A)}\ [400,500] \qquad \textbf{(B)}\ [500,600] \qquad \textbf{(C)}\ [600,700] \qquad \textbf{(D)}\ [700,800]$ $\textbf{(E)}\ [800,900]$

Solution:
Let $C_1$ be the point the ship would reach if it turned $45^\circ$ , and $C_2$ the point it would reach if it turned $60^\circ$ . Obviously, $C_1$ is the furthest possible point from $A$ , and $C_2$ is the closest possible point to $A$
Hence the interval of possible values for $AC^2$ is $[AC_2^2,AC_1^2]$

We can find $AC_1^2$ and $AC_2^2$ as follows:
Let $D_1$ and $D_2$ be the feet of the heights from $C_1$ and $C_2$ onto $AB$ . The angles in the triangle $BD_1C_1$ are $45^\circ$ $45^\circ$ , and $90^\circ$ , hence $BD_1 = D_1C_1 = BC_1 / \sqrt 2$ . Similarly, the angles in the triangle $BD_2C_2$ are $30^\circ$ $60^\circ$ , and $90^\circ$ , hence $BD_2 = BC_2 / 2$ and $D_2C_2 = BC_2 \sqrt 3 / 2$

Hence we get:
\[AC_2^2 = AD_2^2 + D_2C_2^2 = 20^2 + (10\sqrt 3)^2 = 400 + 300 = 700\]
\[AC_1^2 = AD_1^2 + D_1C_1^2 = (10 + 20/\sqrt 2)^2 + (20\sqrt 2)^2 = 100 + 400/\sqrt 2 + 200 + 200 = 500 + 200\sqrt 2 < 500 + 200\cdot 1.5 = 800\]
Therefore for any valid $C$ the value $AC^2$ is surely in the interval $\boxed{700,800}$


Answer: 60.0

A shop advertises everything is "half price in today's sale." In addition, a coupon gives a 20% discount on sale prices. Using the coupon, the price today represents what percentage off the original price?
$\textbf{(A)}\hspace{.05in}10\qquad\textbf{(B)}\hspace{.05in}33\qquad\textbf{(C)}\hspace{.05in}40\qquad\textbf{(D)}\hspace{.05in}60\qquad\textbf{(E)}\hspace{.05in}70$

Solution:
Let the original price of an item be $x$
First, everything is half-off, so the price is now $\frac{x}{2} = 0.5x$
Next, the extra coupon applies 20% off on the sale price , so the price after this discount will be $100\% - 20\% = 80\%$ of what it was before. (Notice how this is not applied to the original price; if it were, the solution would be applying 50% + 20 % = 70% off the original price.)
$80\% \cdot 0.5 x = \frac{4}{5} \cdot 0.5 x = 0.4x$
The price of the item after all discounts have been applied is $0.4x = 40\% \cdot x$ . However, we need to find the percentage off the original price, not the current percentage of the original price. We then subtract $40\% x$ from $100\% x$ (the original price of the item), to find the answer, $\boxed{60}$

Solution: Since the problem implies that the percentage off the original price will be the same for every item in the store, fakesolving is applicable here. Say we are buying an item worth 10 dollars, a convenient number to work with. First, it is clear that we'll get 50% off, which makes the price then 5 dollars. Taking 20% off of 5 dollars gives us 4 dollars. Therefore, we have saved a total of $\frac{10-4}{10} = \frac{6}{10} = \frac{60}{100} = \boxed{60} \%$

Solution: The price is $50\%$ off, which means the price would be $100\% - 50\% = 50\%$ of what it was before. Then, you get $20\%$ off. Sidenote: A common mistake is finding $20\%$ of the price, but forgetting to subtract it. $20\%$ off the price is $100\% - 20\% = 80\%.$ Now, we can multiply the two discounts as decimals. $0.5 \times 0.8 = 0.4 = 40\%.$ Our final answer is $100\% - 40\% = \boxed{60}$


Answer: 81.0

A shopper buys a $100$ dollar coat on sale for $20\%$ off.  An additional $5$ dollars are taken off the sale price by using a discount coupon.  A sales tax of $8\%$ is paid on the final selling price.  The total amount the shopper pays for the coat is
$\text{(A)}\ \text{81.00 dollars} \qquad \text{(B)}\ \text{81.40 dollars} \qquad \text{(C)}\ \text{82.00 dollars} \qquad \text{(D)}\ \text{82.08 dollars} \qquad \text{(E)}\ \text{82.40 dollars}$

Solution: After the $20\%$ sale, the coat costs $100(0.8)=80$ dollars. Then $5$ dollars are taken off for a cost of $80-5=75$ . Adding on the sales tax, the final amount is $(75)(1.08)=\boxed{81.00}$


Answer: 50.0

A shopper plans to purchase an item that has a listed price greater than $\textdollar 100$ and can use any one of the three coupons. Coupon A gives $15\%$ off the listed price, Coupon B gives $\textdollar 30$ off the listed price, and Coupon C gives $25\%$ off the amount by which the listed price exceeds $\textdollar 100$ Let $x$ and $y$ be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or C. What is $y - x$
$\textbf{(A)}\ 50 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 75 \qquad \textbf{(D)}\ 80  \qquad \textbf{(E)}\ 100$

Solution:
Let the listed price be $(100 + p)$ , where $p > 0$
Coupon A saves us: $0.15(100+p) = (0.15p + 15)$
Coupon B saves us: $30$
Coupon C saves us: $0.25p$
Now, the condition is that A has to be greater than or equal to either B or C which gives us the following inequalities:
$A \geq B \Rightarrow 0.15p + 15 \geq 30 \Rightarrow p \geq 100$
$A \geq C \Rightarrow 0.15p + 15 \geq 0.25p \Rightarrow p \leq 150$
We see here that the greatest possible value for $p$ is $150$ , thus $y = 100 + 150 = 250$ and the smallest value for $p$ is $100$ so $x = 100 + 100 = 200$
The difference between $y$ and $x$ is $y - x = 250 - 200 = \boxed{50}$


Answer: 12.0

A sign at the fish market says, "50 $\%$ off, today only: half-pound packages for just $3 per package." What is the regular price for a full pound of fish, in dollars?
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$

Solution: 50% off the price of half a pound of fish is $3, so 100%, the regular price, of a half pound of fish is $6. If half a pound of fish costs $6, then a whole pound of fish is $\boxed{12}$ dollars.

Solution: Suppose a full pound at normal price costs $x$ dollars. Then, with the 50% off deal, the full pound would cost $x/2$ dollars. A half pound of this would cost $x/4$ dollars (including the 50% off). 50% off half a pound is 3, so we form the equation $x/4=3\Rightarrow x=\boxed{12}$


Answer: 18.0

A single bench section at a school event can hold either $7$ adults or $11$ children. When $N$ bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value of $N?$
$\textbf{(A) } 9 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 27 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 77$

Solution: The least common multiple of $7$ and $11$ is $77$ . Therefore, there must be $77$ adults and $77$ children. The total number of benches is $\frac{77}{7}+\frac{77}{11}=11+7=\boxed{18}$ .~taarunganesh

Solution: Let $x$ denote how many adults there are. Since the number of adults is equal to the number of children we can write $N$ as $\frac{x}{7}+\frac{x}{11}=N$ . Simplifying we get $\frac{18x}{77} = N$ Since both $n$ and $x$ have to be positive integers, $x$ has to equal $77$ . Therefore, $N=\boxed{18}$ is our final answer.


Answer: 2.0

A singles tournament had six players. Each player played every other player only once, with no ties. If Helen won 4 games, Ines won 3 games, Janet won 2 games, Kendra won 2 games and Lara won 2 games, how many games did Monica win?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution: Since there are 6 players, a total of $\frac{6(6-1)}{2}=15$ games are played. So far, $4+3+2+2+2=13$ games finished (one person won from each game), so Monica needs to win $15-13 = \boxed{2}$


Answer: 15.0

A small bottle of shampoo can hold $35$ milliliters of shampoo, whereas a large bottle can hold $500$ milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?
$\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$

Solution: To find how many small bottles we need, we can simply divide $500$ by $35$ . This simplifies to $\frac{100}{7}=14 \frac{2}{7}.$ Since the answer must be an integer greater than $14$ , we have to round up to $15$ bottles, or $\boxed{15}$

Solution:
We double $35$ to get $70.$ We see that $70\cdot7=490,$ which is very close to $500.$ Thus, $2\cdot7+1=\boxed{15}$ bottles are enough.
~Technodoggo


Answer: 15.0

Solution: To find how many small bottles we need, we can simply divide $500$ by $35$ . This simplifies to $\frac{100}{7}=14 \frac{2}{7}.$ Since the answer must be an integer greater than $14$ , we have to round up to $15$ bottles, or $\boxed{15}$

Solution:
We double $35$ to get $70.$ We see that $70\cdot7=490,$ which is very close to $500.$ Thus, $2\cdot7+1=\boxed{15}$ bottles are enough.
~Technodoggo


Answer: 32.0

A small square is constructed inside a square of area 1 by dividing each side of the unit square into $n$ equal parts, and then connecting the vertices to the division points closest to the opposite vertices. Find the value of $n$ if the the area of the small square is exactly $\frac1{1985}$
AIME 1985 Problem 4.png

Solution: The lines passing through $A$ and $C$ divide the square into three parts, two right triangles and a parallelogram .  Using the smaller side of the parallelogram, $1/n$ , as the base, where the height is 1, we find that the area of the parallelogram is $A = \frac{1}{n}$ .  By the Pythagorean Theorem , the longer base of the parallelogram has length $l = \sqrt{1^2 + \left(\frac{n - 1}{n}\right)^2} = \frac{1}{n}\sqrt{2n^2 - 2n + 1}$ , so the parallelogram has height $h = \frac{A}{l} = \frac{1}{\sqrt{2n^2 - 2n + 1}}$ .  But the height of the parallelogram is the side of the little square, so $2n^2 - 2n + 1 = 1985$ .  Solving this quadratic equation gives $n = \boxed{32}$

Solution:
Aime.png
Surrounding the square with area $\frac{1}{1985}$ are $4$ right triangles with hypotenuse $1$ (sides of the large square). Thus, $X + \frac{1}{1985} = 1$ , where $X$ is the area of the of the 4 triangles.
We can thus use proportions to solve this problem. \begin{eqnarray*} \frac{GF}{BE}=\frac{CG}{CB}\implies \frac{\frac{1}{\sqrt{1985}}}{BE}=\frac{\frac{1}{n}}{1}\implies BE=\frac{n\sqrt{1985}}{1985} \end{eqnarray*} Also, \begin{eqnarray*} \frac{BE}{1}=\frac{EC}{\frac{n-1}{n}}\implies EC=\frac{\sqrt{1985}}{1985}(n-1) \end{eqnarray*} Thus, \begin{eqnarray*} 2(BE)(EC)+\frac{1}{1985}=1\\ 2n^{2}-2n+1=1985\\ n(n-1)=992 \end{eqnarray*} Simple factorization and guess and check gives us $\boxed{32}$

Solution:
AIME 1985 Problem 4 Solution 3 Diagram.png
Line Segment $DE = \frac{1}{n}$ , so $EC = 1 - \frac{1}{n} = \frac{n-1}{n}$ . Draw line segment $HE$ parallel to the corresponding sides of the small square, $HE$ has length $\frac{1}{\sqrt{1985}}$ , as it is the same length as the sides of the square. Notice that $\triangle CEL$ is similar to $\triangle HDE$ by $AA$ similarity. Thus, $\frac{LC}{HE} = \frac{EC}{DE} = n-1$ , so $LC = \frac{n-1}{\sqrt{1985}}$ . Notice that $\triangle CEL$ is also similar to $\triangle CDF$ by $AA$ similarity. Thus, $\frac{FC}{EC} = \frac{DC}{LC}$ , and the expression simplifies into a quadratic equation $n^2 - n - 992 = 0$ . Solving this quadratic equation yields $n =\boxed{32}$


Answer: 122.0

A soccer team has $22$ available players. A fixed set of $11$ players starts the game, while the other $11$ are available as substitutes. During the game, the coach may make as many as $3$ substitutions, where any one of the $11$ players in the game is replaced by one of the substitutes. No player removed from the game may reenter the game, although a substitute entering the game may be replaced later. No two substitutions can happen at the same time. The players involved and the order of the substitutions matter. Let $n$ be the number of ways the coach can make substitutions during the game (including the possibility of making no substitutions). Find the remainder when $n$ is divided by $1000$

Solution:
There are $0-3$ substitutions. The number of ways to sub any number of times must be multiplied by the previous number. This is defined recursively. The case for $0$ subs is $1$ , and the ways to reorganize after $n$ subs is the product of the number of new subs ( $12-n$ ) and the players that can be ejected ( $11$ ). The formula for $n$ subs is then $a_n=11(12-n)a_{n-1}$ with $a_0=1$
Summing from $0$ to $3$ gives $1+11^2+11^{3}\cdot 10+11^{4}\cdot 10\cdot 9$ . Notice that $10+9\cdot11\cdot10=10+990=1000$ . Then, rearrange it into $1+11^2+11^3\cdot (10+11\cdot10\cdot9)= 1+11^2+11^3\cdot (1000)$ . When taking modulo $1000$ , the last term goes away. What is left is $1+11^2=\boxed{122}$


Answer: 45.0

A softball team played ten games, scoring $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ , and $10$ runs.  They lost by one run in exactly five games.  In each of their other games, they scored twice as many runs as their opponent. How many total runs did their opponents score?
$\textbf{(A)}\ 35 \qquad\textbf{(B)}\ 40 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 50 \qquad\textbf{(E)}\ 55$

Solution:
We know that, for the games where they scored an odd number of runs, they cannot have scored twice as many runs as their opponents, as odd numbers are not divisible by $2$ .  Thus, from this, we know that the five games where they lost by one run were when they scored $1$ $3$ $5$ $7$ , and $9$ runs, and the others are where they scored twice as many runs.  We can make the following chart:
$\begin{tabular}{|l|l|} \hline Them & Opponent \\ \hline 1 & 2 \\ 2 & 1 \\ 3 & 4 \\ 4 & 2 \\ 5 & 6 \\ 6 & 3 \\ 7 & 8 \\ 8 & 4 \\ 9 & 10 \\ 10 & 5 \\ \hline \end{tabular}$
The sum of their opponent's scores is $2 + 1 + 4 + 2 + 6 + 3 + 8 + 4 + 10 + 5 = \boxed{45}$


Answer: 18.0

A solid box is $15$ cm by $10$ cm by $8$ cm. A new solid is formed by removing a cube $3$ cm on a side from each corner of this box. What percent of the original volume is removed?
$\mathrm{(A) \ } 4.5\%\qquad \mathrm{(B) \ } 9\%\qquad \mathrm{(C) \ } 12\%\qquad \mathrm{(D) \ } 18\%\qquad \mathrm{(E) \ } 24\%$

Solution:
The volume of the original box is $15\cdot10\cdot8=1200.$
The volume of each cube that is removed is $3\cdot3\cdot3=27.$
Since there are $8$ corners on the box, $8$ cubes are removed.
So the total volume removed is $8\cdot27=216$
Therefore, the desired percentage is $\frac{216}{1200}\cdot100 = \boxed{18}.$


Answer: 18.0

Solution:
The volume of the original box is $15\cdot10\cdot8=1200.$
The volume of each cube that is removed is $3\cdot3\cdot3=27.$
Since there are $8$ corners on the box, $8$ cubes are removed.
So the total volume removed is $8\cdot27=216$
Therefore, the desired percentage is $\frac{216}{1200}\cdot100 = \boxed{18}.$


Answer: 7.0

A solid cube has side length 3 inches. A 2-inch by 2-inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 15$

Solution:
Imagine making the cuts one at a time. The first cut removes a box $2\times 2\times 3$ . The second cut removes two boxes, each of dimensions $2\times 2\times 0.5$ , and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is $12 + 4 + 4 = 20$
Therefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \boxed{7}$

Solution:
We can use Principle of Inclusion-Exclusion (PIE) to find the final volume of the cube.
There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has $2 \times 2 \times 3=12$ cubic inches. However, we can not just sum their volumes, as
the central $2\times 2\times 2$ cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice.
Hence the total volume of the cuts is $3(2 \times 2 \times 3) - 2(2\times 2\times 2) = 36 - 16 = 20$
Therefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \boxed{7}$

Solution:
We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.
Each edge can be seen as a $2\times 0.5\times 0.5$ box, and each corner can be seen as a $0.5\times 0.5\times 0.5$ box.
$12\cdot{\frac{1}{2}} + 8\cdot{\frac{1}{8}} = 6+1 = \boxed{7}$


Answer: 7.0

Solution:
Imagine making the cuts one at a time. The first cut removes a box $2\times 2\times 3$ . The second cut removes two boxes, each of dimensions $2\times 2\times 0.5$ , and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is $12 + 4 + 4 = 20$
Therefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \boxed{7}$

Solution:
We can use Principle of Inclusion-Exclusion (PIE) to find the final volume of the cube.
There are 3 "cuts" through the cube that go from one end to the other. Each of these "cuts" has $2 \times 2 \times 3=12$ cubic inches. However, we can not just sum their volumes, as
the central $2\times 2\times 2$ cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the volume of the central cube twice.
Hence the total volume of the cuts is $3(2 \times 2 \times 3) - 2(2\times 2\times 2) = 36 - 16 = 20$
Therefore the volume of the rest of the cube is $3^3 - 20 = 27 - 20 = \boxed{7}$

Solution:
We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners.
Each edge can be seen as a $2\times 0.5\times 0.5$ box, and each corner can be seen as a $0.5\times 0.5\times 0.5$ box.
$12\cdot{\frac{1}{2}} + 8\cdot{\frac{1}{8}} = 6+1 = \boxed{7}$


Answer: 84.0

A solid cube of side length $1$ is removed from each corner of a solid cube of side length $3$ .  How many edges does the remaining solid have?
$\textbf{(A) }36\qquad\textbf{(B) }60\qquad\textbf{(C) }72\qquad\textbf{(D) }84\qquad\textbf{(E) }108\qquad$

Solution: We can use Euler's polyhedron formula that says that $F+V=E+2$ .  We know that there are originally $6$ faces on the cube, and each corner cube creates $3$ more. $6+8(3) = 30$ .  In addition, each cube creates $7$ new vertices while taking away the original $8$ , yielding $8(7) = 56$ vertices.  Thus $E+2=56+30$ , so $E=\boxed{84}$

Solution: The removal of each cube adds nine additional edges to the solid. Since a cube initially has $12$ edges and there are eight vertices, the number of edges will be $12 + 9 \times 8 = \boxed{84}$


Answer: 384.0

A solid rectangular block is formed by gluing together $N$ congruent 1-cm cubes face to face. When the block is viewed so that three of its faces are visible, exactly $231$ of the 1-cm cubes cannot be seen. Find the smallest possible value of $N.$

Solution: The $231$ cubes which are not visible must lie below exactly one layer of cubes.  Thus, they form a rectangular solid which is one unit shorter in each dimension.  If the original block has dimensions $l \times m \times n$ , we must have $(l - 1)\times(m-1) \times(n - 1) = 231$ .  The prime factorization of $231 = 3\cdot7\cdot11$ , so we have a variety of possibilities; for instance, $l - 1 = 1$ and $m - 1 = 11$ and $n - 1 = 3 \cdot 7$ , among others.  However, it should be fairly clear that the way to minimize $l\cdot m\cdot n$ is to make $l$ and $m$ and $n$ as close together as possible, which occurs when the smaller block is $3 \times 7 \times 11$ .  Then the extra layer makes the entire block $4\times8\times12$ , and $N= \boxed{384}$


Answer: 394.0

A solitaire game is played as follows.  Six distinct pairs of matched tiles are placed in a bag.  The player randomly draws tiles one at a time from the bag and retains them, except that matching tiles are put aside as soon as they appear in the player's hand.  The game ends if the player ever holds three tiles, no two of which match; otherwise the drawing continues until the bag is empty.  The probability that the bag will be emptied is $p/q,\,$ where $p\,$ and $q\,$ are relatively prime positive integers.  Find $p+q.\,$

Solution:
Let $P_k$ be the probability of emptying the bag when it has $k$ pairs in it. Let's consider the possible draws for the first three cards:
Therefore, we obtain the recursion $P_k = \frac {3}{2k - 1}P_{k - 1}$ . Iterating this for $k = 6,5,4,3,2$ (obviously $P_1 = 1$ ), we get $\frac {3^5}{11*9*7*5*3} = \frac {9}{385}$ , and $p+q=\boxed{394}$

Solution:
Call the case that we begin with [ABCDEF]. It doesn't matter what letter we choose at first, so WLOG assume we choose A. Now there is BCDEFABCDEF remaining in the bag. We have two cases to consider here.
1. We pick the other A. There's a $\frac{1}{11}$ chance for this to happen. We remain with the case [BCDEF] if this is the case.
2. We pick any other letter that is not an A. There's a $\frac{10}{11}$ chance for this to happen. WLOG, assume we pick the letter B. Now in order for us to continue the game, we must choose either the other A or B. There's a $\frac{2}{10}$ chance for this to happen. WLOG, assume we choose A. Now we have BCDEFCDEF left.
Notice however that in the first case, the probability of emptying the bag with [BCDEF] is the same thing as with BCDEFCDEF, as the only difference is you've removed one of the letters (and it doesn't matter which you chose).
Hence for this case, there is a $\frac{1}{11} + \frac{10}{11}*\frac{2}{10} = \frac{3}{11}$ * [BCDEF] chance to empty the bag.
Continuing this process, we get that:
[BCDEF] = $\frac{3}{9}$ * [CDEF]
[CDEF] = $\frac{3}{7}$ * [DEF]
[DEF] = $\frac{3}{5}$ * [EF]
[EF] = 1 (clearly, since if we are only left with EFEF then we are going to empty the bag).
And hence [ABCDEF] = $1*\frac{3}{5}*\frac{3}{7}*\frac{3}{9}*\frac{3}{11} = \frac{9}{385}$ so our answer is $9+385=\boxed{394}$


Answer: 33.0

A speaker talked for sixty minutes to a full auditorium. Twenty percent of the audience heard the entire talk and ten percent slept through the entire talk. Half of the remainder heard one third of the talk and the other half heard two thirds of the talk. What was the average number of minutes of the talk heard by members of the audience?
$\mathrm{(A) \ } 24 \qquad \mathrm{(B) \ } 27\qquad \mathrm{(C) \ }30 \qquad \mathrm{(D) \ }33 \qquad \mathrm{(E) \ }36$

Solution:
Assume that there are $100$ people in the audience.
$20$ people heard $60$ minutes of the talk, for a total of $20\cdot 60 = 1200$ minutes heard.
$10$ people heard $0$ minutes.
$\frac{70}{2} = 35$ people heard $20$ minutes of the talk, for a total of $35\cdot 20 = 700$ minutes.
$35$ people heard $40$ minutes of the talk, for a total of $1400$ minutes.
Altogether, there were $1200 + 0 + 700 + 1400 = 3300$ minutes heard among $100$ people.
Thus, the average is $\frac{3300}{100} = 33$ minutes, and the answer is $\boxed{33}$


Answer: 13.0

A special deck of cards contains $49$ cards, each labeled with a number from $1$ to $7$ and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her cards and $\textit{still}$ have at least one card of each color and at least one card with each number is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$

Solution:
Without loss of generality, assume that the $8$ numbers on Sharon's cards are $1$ $1$ $2$ $3$ $4$ $5$ $6$ , and $7$ , in that order, and assume the $8$ colors are red, red, and six different arbitrary colors. There are ${8\choose2}-1$ ways of assigning the two red cards to the $8$ numbers; we subtract $1$ because we cannot assign the two reds to the two $1$ 's.
In order for Sharon to be able to remove at least one card and still have at least one card of each color, one of the reds have to be assigned with one of the $1$ s. The number of ways for this to happen is $2 \cdot 6 = 12$ (the first card she draws has to be a $1$ (2 choices), while the second card can be any card but the remaining card with a $1$ (6 choices)). Each of these assignments is equally likely, so desired probability is $\frac{12}{{8\choose2}-1}=\frac{4}{9} \implies 4 + 9 = 13 = \boxed{013}$

Solution:
First note that out of the $8$ selected cards, one pair of cards have to share the same number and another pair of cards have to share the same color. Now, these $2$ pairs of cards can't be the same or else there will be $2$ cards which are completely same. Then, WLOG let the numbers be $1,1,2,3,4,5,6,$ and $7$ and the colors be $a,a,b,c,d,e,f,$ and $g$ . We therefore obtain only $2$ cases:
Case One: $1a,1b,2a,3c,4d,5e,6f,$ and $7g$ In this case, we can discard $1a$ .
There are $2*6=12$ situations in this case.
Case Two: $1b,1c,2a,3a,4d,5e,6f,$ and $7g$ In this case, we can't discard.
There are $\dbinom{6}{2}=15$ situations in this case.
So the probability is $\frac{12}{12+15}=\frac{4}{9}$ , giving us the answer of $4+9=\boxed{013}$

Solution:
There are $7!$ ways to choose a set of 7 cards that have all the numbers from 1-7 and all 7 colors. There are then $42$ cards remaining. Thus, there are $7!(42)$ desired sets.
Now, the next thing to find is the number of ways to choose 8 cards where there is not a set of 7 such cards. In this case, one color must have 2 cards and one number must have 2 cards, and they can't be the same number/color card. The number of ways to pick this is equal to a multiplication of $\binom{7}{2}$ ways to pick 2 numbers, $7$ colors to assign them to, $\binom{6}{2}$ ways to pick 2 nonchosen colors, $5$ ways to pick a number to assign them to, and $4!$ ways to assign the rest.
Thus, the answer is $\frac{7!(42)}{7!(42) + 21(15)(7)(5!)}$ . Dividing out $5!$ yields $\frac{42(42)}{42(42) + 21(15)(7)}$ which is equal to $\frac{2(42)}{2(42) + 15(7)}$ which is equal to $\frac{12}{12 + 15}$ which is equal to $\frac{4}{9}$ giving a final answer of $\boxed{013}$

Solution:
We can rewrite the problem as "What is the probability that, given 8 cards with numbers a, b, c, d, e, f, g, g and some assortment of seven colors A, B, C, D, E, F, G, G where G is repeated, one of the cards is Gg?" Note that this is a valid restatement because Sharon has to have two of one number and two of one color. She needs to be able to take away one of the cards with the duplicate number, but this also has to have the duplicate color. There are two cases.
Case I: One of the cards is Gg. This implies that the other card with color G can be placed in $6$ ways, and the rest of the colors can be paired with cards in $6!$ ways.
Case II: None of the cards are Gg. This implies that the cards with color G can be chosen in $\dbinom{6}{2}=15$ ways, and the rest of the colors can be paired with cards in $\frac{6!}{2}$ ways, with the divide by 2 because of the double-g.
Note that there is no case with Gg, Gg because all 49 cards are unique!
Therefore our answer is $\frac{6\times6!}{\frac{15}{2}\times6! + 6\times6!}=\frac{4}{9}$ , so $\boxed{013}$

Solution:
We count the two entities: The number of sets of eight cards that contain all seven numbers and all seven colors; and The number of sets of eight cards that contain all seven numbers and colors, and one card can be removed and the property still holds.
For the first: It is equivalent to counting:
How many ways can (colors) $A, A, B, C, D, E, F, G$ be matched to (numbers) $1, 1, 2, 3, 4, 5, 6, 7$ such that the two $A's$ cannot both be matched to the two ones. (This would mean we've chosen two identical cards from the deck, both with color $A$ and number one.)
We consider two cases:
1) Neither $A$ is matched with a one. Then there are $\binom{6}{2}$ ways to choose which two numbers are matched with the $A's,$ and $6! / 2!$ ways to order the remaining numbers. (There are two one's, so we divide by two.)
2) One of the $A$ 's is matched with a one. There are $6 \cdot 6!$ ways to arrange the remaining color
Since there are seven ways to choose the extra color and likewise for the extra number,
we have in total $7 \cdot 7 \cdot \left( 6C2 \cdot \frac{6!}{2!} + 6 \cdot 6! \right)$ ways to choose eight cards satisfying the first condition.
For the second: View the cards as rooks on a seven by seven chess board. Seven cards that contain all colors and numbers is can be represented as placing seven rooks on the board so that no two rooks are attacking each other. There are $7!$ ways to do this. For each arrangement, we can choose any of the other 42 positions on the board (cards) to add an extra rook. Thus, there are $7! \cdot 42$ good sets of eight cards. Note that we are not overcounting.
Thus, the probability of getting a good set given the first condition is $\frac{ 7! \cdot 42  }{ 7 \cdot 7 \cdot \left( 6C2 \cdot \frac{6!}{2!} + 6 \cdot 6! \right) } = \frac{4}{9},$ and the answer is $\boxed{013}.$

Solution:
We recast this problem to coloring 8 cells in a 7x7 grid, such that each row and column has at least one colored cell. If a cell in row a and column b is colored, that means we drew a card with color a and number b.
By pigeonhole principle, there exists a row that has 2 colored cells, and intuitively, that row is unique. There also exists a column with 2 colored cells. Call the unique row, "Row A", and Call the unique column, "Column B".
If we are able to remove a card so that we still have a card of each color and number, that means, we can erase a cell, so each row still has at least one colored cell, and so does each column. If we erase a cell not in Row A, then the erased cell's row will be empty. If we erase a cell not in Column B, the erased cell's column will be empty. Thus the cell that we erase must be in both Row A and Column B. There is only one cell in both Row A and Column B, and thus that cell must be colored.
We first count the number of successful outcomes. We must chose which row to be "Row A", and which column to be "Column B". There are 7 ways to chose "Row A", and 7 ways to chose "Column B". Because of the logic used above, the intersection of Row A and Column B must be colored. However, Row A needs to have 2 colored cells. There are 6 other uncolored cells to chose from. Similarly, there are 6 uncolored cells in Column B.
Note that we can ignore all the rows and columns that contain colored cells. This is because Row A already contains 2 colored cells, and thus we can't color any more cells in Row A, similarly for Column B. The other row that contains a colored cell is not Row A, so it can only contain 1 colored cell, which it has, thus we can't color anymore cells in it. Similarly for the other column.
We can actually remove these rows/columns. This leaves a 5x5 grid, full of uncolored cells. Each row can have 1 cell, and so can each column. We count the # of colorings from the perspective of the rows. We look at the first row, it has 5 choices. We can look at the 2nd row, it has 4 choices, as it's selected cell can't occupy the same column. We continue down the line, and see there are $5! = 120$ ways to color.
Thus, the # of successful outcomes is $5! \cdot 6 \cdot 6 \cdot 7 \cdot 7$
We can then count the # of possible outcomes.
We can count the # of unsuccessful outcomes. We select "Row A" and "Column B" as usual, with $7 \cdot 7$ ways. We don't color the intersection, so we have to color 2 cells in each. There are 6 other cells that can be colored in each, so we have ${6 \choose 2} = 15$ ways to color each. Then, we can get rid of rows and columns using the method above and we are left with a 4x4 grid. There are $4! = 24$ ways to color.
Thus, the # of unsuccessful outcomes is $4! \cdot 15 \cdot 15 \cdot 7 \cdot 7$
After some calculation, we get $\frac{successful}{possible} = \frac{successful}{successful+unsuccessful} = \frac{5! \cdot 6^2 \cdot 7^2}{7^2 \cdot(5! \cdot 6^2 + 4! \cdot 15^2} = \frac{5! \cdot 6^2}{5! \cdot 6^2 +4! \cdot 15^2} = \frac{5 \cdot 6^2}{5 \cdot 6^2 + 15^2} = \frac{180}{180+225} = \frac{4}{9} => 4+9 = \boxed{13}$


Answer: 36.0

A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are 6.1 cm, 8.2 cm and 9.7 cm. What is the area of the square in square centimeters?
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 64$

Solution: The perimeter of the triangle is $6.1+8.2+9.7=24$ cm. A square's perimeter is four times its sidelength, since all its sidelengths are equal. If the square's perimeter is $24$ , the sidelength is $24/4=6$ , and the area is $6^2=\boxed{36}$


Answer: 2.0

A square flag has a red cross of uniform width with a blue square in the center on a white background as shown. (The cross is symmetric with respect to each of the diagonals of the square.) If the entire cross (both the red arms and the blue center) takes up 36% of the area of the flag, what percent of the area of the flag is blue?
[asy] unitsize(2.5 cm);  pair[] A, B, C; real t = 0.2;  A[1] = (0,0); A[2] = (1,0); A[3] = (1,1); A[4] = (0,1); B[1] = (t,0); B[2] = (1 - t,0); B[3] = (1,t); B[4] = (1,1 - t); B[5] = (1 - t,1); B[6] = (t,1); B[7] = (0,1 - t); B[8] = (0,t); C[1] = extension(B[1],B[4],B[7],B[2]); C[2] = extension(B[3],B[6],B[1],B[4]); C[3] = extension(B[5],B[8],B[3],B[6]); C[4] = extension(B[7],B[2],B[5],B[8]);  fill(C[1]--C[2]--C[3]--C[4]--cycle,blue); fill(A[1]--B[1]--C[1]--C[4]--B[8]--cycle,red); fill(A[2]--B[3]--C[2]--C[1]--B[2]--cycle,red); fill(A[3]--B[5]--C[3]--C[2]--B[4]--cycle,red); fill(A[4]--B[7]--C[4]--C[3]--B[6]--cycle,red);  draw(A[1]--A[2]--A[3]--A[4]--cycle); draw(B[1]--B[4]); draw(B[2]--B[7]); draw(B[3]--B[6]); draw(B[5]--B[8]); [/asy]
$\text{(A)}\ 0.5\qquad\text{(B)}\ 1\qquad\text{(C)}\ 2\qquad\text{(D)}\ 3\qquad\text{(E)}\ 6$

Solution: The diagram can be quartered as shown:  and reassembled into two smaller squares of side $k$ , each of which looks like this:  The border in this figure is the former cross, which still occupies 36% of the area. Therefore the inner square occupies 64% of the area, from which we deduce that it is $0.8k \times 0.8k$ , and that one blue square must be $0.1k\times 0.1k=0.01k^2$ or 1% each. Thus the blue area is $\boxed{2}$ of the total.


Answer: 17.0

A square in the coordinate plane has vertices whose $y$ -coordinates are $0$ $1$ $4$ , and $5$ . What is the area of the square?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27$

Solution:
Let the points be $A=(x_1,0)$ $B=(x_2,1)$ $C=(x_3,5)$ , and $D=(x_4,4)$
Note that the difference in $y$ value of $B$ and $C$ is $4$ . By rotational symmetry of the square, the difference in $x$ value of $A$ and $B$ is also $4$ . Note that the difference in $y$ value of $A$ and $B$ is $1$ . We now know that $AB$ , the side length of the square, is equal to $\sqrt{1^2+4^2}=\sqrt{17}$ , so the area is $\boxed{17}$

Solution:
By translation, we can move the square with point $A$ at the origin. Then, $A=(0,0), B=(x_1,1), C=(x_2,5), D=(x_3,4)$ . We will use the relationship among the 4 sides of being perpendicular and equal.
The slope of $AB$ is $\frac{1-0}{x_1-0}=\frac{1}{x_1}$
Because $BC$ is perpendicular to $AB$ , the slope of $BC=-x_1$ .
From the information above we could have the equation:
Because $CD$ is perpendicular to $BC$ , the slope of $CD=\frac{1}{x_1}$ .
From the information above we could have the equation:
Because $AD=AB,$
Note that the square with $x_1=-4$ is just the reflection of square with $x_1=4$ over the origin. I will use $x_1=4$ $B=(4,1), AB=\sqrt{17}, [ABCD]=\boxed{17}$

Solution:
In this solution, we will use the fact that the diagonals of a square bisect each other, they are perpendicular to each other, and they are equal in length.
Using the fact that the diagonals bisect each other, we get the equation:
Now we use the fact that the diagonals are perpendicular to each other:
Using the fact that the diagonals are equal in length, we get the equation:
Now we have 3 equations with 3 variables:
$\begin{cases} x_2=x_1+x_3 \\ x_2(x_1-x_3)=15 \\ (x_3-x_1)^2-x_2^2=16 \end{cases}$
We substitute $x_2$ into the 2 other equations:
Now we have 2 equations of $x_1$ and $x_3$
$\begin{cases} x_1^2-x_3^2=15 \\ x_1x_3=-4 \end{cases}$
This is the same equation as solution $2$ . So $x_1= \pm 4, AB=\sqrt{17}, [ABCD]=\boxed{17}$

Solution: Draw it out, by inspection the coordinates are $(-1, 4)$ $(0, 0)$ $(4, 1)$ , and $(3, 5)$ . The side length is $\sqrt{17}, [ABCD]=\boxed{17}$ ~JH. L


Answer: 39.0

A square is drawn in the Cartesian coordinate plane with vertices at $(2, 2)$ $(-2, 2)$ $(-2, -2)$ $(2, -2)$ . A particle starts at $(0,0)$ . Every second it moves with equal probability to one of the eight lattice points (points with integer coordinates) closest to its current position, independently of its previous moves. In other words, the probability is $1/8$ that the particle will move from $(x, y)$ to each of $(x, y + 1)$ $(x + 1, y + 1)$ $(x + 1, y)$ $(x + 1, y - 1)$ $(x, y - 1)$ $(x - 1, y - 1)$ $(x - 1, y)$ , or $(x - 1, y + 1)$ . The particle will eventually hit the square for the first time, either at one of the 4 corners of the square or at one of the 12 lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is $m/n$ , where $m$ and $n$ are relatively prime positive integers. What is $m + n$
$\textbf{(A) } 4 \qquad\textbf{(B) } 5 \qquad\textbf{(C) } 7 \qquad\textbf{(D) } 15 \qquad\textbf{(E) } 39$

Solution:
We let $c, e,$ and $m$ be the probability of reaching a corner before an edge when starting at an "inside corner" (e.g. $(1, 1)$ ), an "inside edge" (e.g. $(1, 0)$ ), and the middle respectively.
Starting in the middle, there is a $\frac{4}{8}$ chance of moving to an inside edge and a $\frac{4}{8}$ chance of moving to an inside corner, so
\[m = \frac{1}{2}e + \frac{1}{2}c.\]
Starting at an inside edge, there is a $\frac{2}{8}$ chance of moving to another inside edge, a $\frac{2}{8}$ chance of moving to an inside corner, a $\frac{1}{8}$ chance of moving into the middle, and a $\frac{3}{8}$ chance of reaching an outside edge and stopping. Therefore,
\[e = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m + \frac{3}{8}\cdot 0 = \frac{1}{4}e + \frac{1}{4}c + \frac{1}{8}m.\]
Starting at an inside corner, there is a $\frac{2}{8}$ chance of moving to an inside edge, a $\frac{1}{8}$ chance of moving into the middle, a $\frac{4}{8}$ chance of moving to an outside edge and stopping, and finally a $\frac{1}{8}$ chance of reaching that elusive outside corner. This gives
\[c = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{2}0 + \frac{1}{8}\cdot 1 = \frac{1}{4}e + \frac{1}{8}m + \frac{1}{8}.\]
Solving this system of equations gives
\[m = \frac{4}{35},\] \[e = \frac{1}{14},\] \[c = \frac{11}{70}.\]
Since the particle starts at $(0, 0),$ it is $m$ we are looking for, so the final answer is
\[4 + 35 = \boxed{39}.\]


Answer: 25.0

A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle?
[asy] defaultpen(linewidth(0.8)); size(100); real r=sqrt(50), s=sqrt(10); draw(Arc(origin, r, 0, 180)); draw((r,0)--(-r,0), dashed); draw((-s,0)--(s,0)--(s,2*s)--(-s,2*s)--cycle); [/asy]
$\textbf{(A) } 20\pi\qquad \textbf{(B) } 25\pi\qquad \textbf{(C) } 30\pi\qquad \textbf{(D) } 40\pi\qquad \textbf{(E) } 50\pi$

Solution:
Since the area of the square is $40$ , the length of a side is $\sqrt{40}=2\sqrt{10}$ . The distance between the center of the semicircle and one of the bottom vertices of the square is half the length of the side, which is $\sqrt{10}$
Using the Pythagorean Theorem to find the radius $r$ of the semicircle, $r^2 = (2\sqrt{10})^2 + (\sqrt{10})^2 = 50$ . So, the area of the semicircle is $\frac{1}{2}\cdot \pi \cdot 50 = \boxed{25}$