Answer: 2.0

A square piece of paper has side length $1$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perimeter of triangle $\bigtriangleup AEC' ?$
$\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\frac{13}{6} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}$ [asy] /* Made by samrocksnature */ pair A=(0,1); pair CC=(0.666666666666,1); pair D=(1,1); pair F=(1,0.440062); pair C=(1,0); pair B=(0,0); pair G=(0,0.22005); pair H=(-0.13,0.41); pair E=(0,0.5); dot(A^^CC^^D^^C^^B^^E); draw(E--A--D--F); draw(G--B--C--F, dashed); fill(E--CC--F--G--H--E--CC--cycle, gray); draw(E--CC--F--G--H--E--CC); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("E",E,NW); label("C'",CC,N); label("F",F,NE); [/asy]

Solution:
We can set the point on $CD$ where the fold occurs as point $F$ . Then, we can set $FD$ as $x$ , and $CF$ as $1-x$ because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for $x$ , we get,
\[x^2 + \left(\frac{1}{3}\right)^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}\]
We know this is a 3-4-5 triangle because the side lengths are $\frac{3}{9}, \frac{4}{9}, \frac{5}{9}$ . We also know that $EAC'$ is similar to $C'DF$ because angle $EC'F$ is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of $C'DF \times \frac{AC'}{DF}$ . That's just $\frac{4}{3} \times \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} \times \frac{3}{2} = 2$ . Therefore, the final answer is $\boxed{2}$

Solution: Let the line we're reflecting over be $\ell$ , and let the points where it hits $AB$ and $CD$ , be $M$ and $N$ , respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line $\ell$ . The segment $CC'$ has slope $\frac{0 - 1}{1 - 2/3} = -3$ , implying line $\ell$ has a slope of $\frac{1}{3}$ . Also, the midpoint of segment $CC'$ is $\left( \frac{5}{6}, \frac{1}{2} \right)$ , so line $\ell$ passes through this point. Then, we get the equation of line $\ell$ is simply $y = \frac{1}{3} x + \frac{2}{9}$ . Then, if the point where $B$ is reflected over line $\ell$ is $B'$ , then we get $BB'$ is the line $y = -3x$ . The intersection of $\ell$ and segment $BB'$ is $\left( - \frac{1}{15}, \frac{1}{5} \right)$ . So, we get $B' = \left(- \frac{2}{15}, \frac{2}{5} \right)$ . Then, line segment $B'C'$ has equation $y = \frac{3}{4} x + \frac{1}{2}$ , so the point $E$ is the $y$ -intercept, or $\left(0, \frac{1}{2} \right)$ . This implies that $AE = \frac{1}{2}, AC' = \frac{2}{3}$ , and by the Pythagorean Theorem, $EC' = \frac{5}{6}$ (or you could notice $\triangle AEC'$ is a $3-4-5$ right triangle). Then, the perimeter is $\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2$ , so our answer is $\boxed{2}$ . ~rocketsri

Solution: Assume that E is the midpoint of $\overline{AB}$ . Then, $\overline{AE}=\frac{1}{2}$ and since $C'D=\frac{1}{3}$ $\overline{AC'}=\frac{2}{3}$ . By the Pythagorean Theorem, $\overline{EC'}=\frac{5}{6}$ . It easily follows that our desired perimeter is $\boxed{2}$ ~samrocksnature

Solution:
As described in Solution 1, we can find that $DF=\frac{4}{9}$ , and $C'F = \frac{5}{9}.$
Then, we can find we can find the length of $\overline{AE}$ by expressing the length of $\overline{EF}$ in two different ways, in terms of $AE$ . If let $AE = a$ , by the Pythagorean Theorem we have that $EC = \sqrt{a^2 + \left(\frac{2}{3}\right)^2} = \sqrt{a^2 + \frac{4}{9}}.$ Therefore, since we know that $\angle EC'F$ is right, by Pythagoras again we have that $EF = \sqrt{\left(\sqrt{a^2+\frac{4}{9}}\right)^2 + \left(\frac{5}{9}\right)^2} = \sqrt{a^2 + \frac{61}{81}}.$
Another way we can express $EF$ is by using Pythagoras on $\triangle XEF$ , where $X$ is the foot of the perpendicular from $F$ to $\overline{AE}.$ We see that $ADFX$ is a rectangle, so we know that $AD = 1 = FX$ . Secondly, since $FD = \frac{4}{9}, EX = a - \frac{4}{9}$ . Therefore, through the Pythagorean Theorem, we find that $EF = \sqrt{\left(a-\frac{4}{9}\right)^2 + 1^2} = \sqrt{a^2 - \frac{8}{9}a +\frac{97}{81}}.$
Since we have found two expressions for the same length, we have the equation $\sqrt{a^2 + \frac{61}{81}} = \sqrt{a^2 - \frac{8}{9}a +\frac{97}{81}}.$ Solving this, we find that $a=\frac{1}{2}$
Finally, we see that the perimeter of $\triangle AEC'$ is $\frac{1}{2} + \frac{2}{3} + \sqrt{\left(\frac{1}{2}\right)^2 + \frac{4}{9}},$ which we can simplify to be $2$ . Thus, the answer is $\boxed{2}.$ ~laffytaffy

Solution: Draw a perpendicular line from $\overline{AB}$ at $E$ , and let it intersect $\overline{DC}$ at $E'$ . The angle between $\overline{AB}$ and $\overline{EE'}$ is $2\theta$ , where $\theta$ is the angle between the fold and a line perpendicular to $\overline{AD}$ . The slope of the fold is $\frac{1}{3}$ because it is perpendicular to $\overline{CC'}$ $\overline{CC'}$ has a slope of $-3$ using points $C'$ and $C$ , and perpendicular lines have slopes negative inverses of each other). Using tangent double angle formula, the slope of $\overline{EC'}$ is $\frac{3}{4}$ , which implies $\overline{AE}$ $\frac{1}{2}$ . By the Pythagorean Theorem, $\overline{EC'}=\frac{5}{6}$ . It easily follows that our desired perimeter is $\boxed{2}$ ~forrestc

Solution:
It is easy to prove that the ratio of the sum of the larger leg and hypotenuse to the smaller leg depends monotonically on the angle of a right triangle, which means: \[C' F + DF = CF + DF = CD = AD = 3 C'D \implies C' D : DF : C' F = 3 : 4 : 5.\]
For a similar triangle, the ratio of the perimeter to the larger leg is $\frac {3 + 4 +5}{4} = 3.$
$\triangle AEC'  \sim \triangle DC'F \implies$ the perimeter of triangle $\bigtriangleup AEC'$ is $3 AC' = \boxed{2}.$


Answer: 871.0

A square piece of paper has sides of length $100$ . From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at a distance $\sqrt{17}$ from the corner, and they meet on the diagonal at an angle of $60^{\circ}$ (see the figure below). The paper is then folded up along the lines joining the vertices of adjacent cuts. When the two edges of a cut meet, they are taped together. The result is a paper tray whose sides are not at right angles to the base. The height of the tray, that is, the perpendicular distance between the plane of the base and the plane formed by the upper edges, can be written in the form $\sqrt[n]{m}$ , where $m$ and $n$ are positive integers, $m<1000$ , and $m$ is not divisible by the $n$ th power of any prime. Find $m+n$

Solution:
In the original picture, let $P$ be the corner, and $M$ and $N$ be the two points whose distance is $\sqrt{17}$ from $P$ . Also, let $R$ be the point where the two cuts intersect.
Using $\triangle{MNP}$ (a 45-45-90 triangle), $MN=MP\sqrt{2}\quad\Longrightarrow\quad MN=\sqrt{34}$ $\triangle{MNR}$ is equilateral , so $MR = NR = \sqrt{34}$ . (Alternatively, we could find this by the Law of Sines .)
The length of the perpendicular from $P$ to $MN$ in $\triangle{MNP}$ is $\frac{\sqrt{17}}{\sqrt{2}}$ , and the length of the perpendicular from $R$ to $MN$ in $\triangle{MNR}$ is $\frac{\sqrt{51}}{\sqrt{2}}$ . Adding those two lengths, $PR=\frac{\sqrt{17}+\sqrt{51}}{\sqrt{2}}$ . (Alternatively, we could have used that $\sin 75^{\circ} = \sin (30+45) = \frac{\sqrt{6}+\sqrt{2}}{4}$ .)
Drop a perpendicular from $R$ to the side of the square containing $M$ and let the intersection be $G$
\begin{align*}PG&=\frac{PR}{\sqrt{2}}=\frac{\sqrt{17}+\sqrt{51}}{2}\\ MG=PG-PM&=\frac{\sqrt{17}+\sqrt{51}}{2}-\sqrt{17}=\frac{\sqrt{51}-\sqrt{17}}{2}\end{align*}
Let $A'B'C'D'$ be the smaller square base of the tray and let $ABCD$ be the larger square, such that $AA'$ , etc, are edges. Let $F$ be the foot of the perpendicular from $A$ to plane $A'B'C'D'$
We know $AA'=MR=\sqrt{34}$ and $A'F=MG\sqrt{2}=\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}$ . Now, use the Pythagorean Theorem on triangle $AFA'$ to find $AF$
\begin{align*}\left(\frac{\sqrt{51}-\sqrt{17}}{\sqrt{2}}\right)^2+AF^2&=\left(\sqrt{34}\right)^2\\ \frac{51-34\sqrt{3}+17}{2}+AF^2&=34\\AF&=\sqrt{34-\frac{68-34\sqrt{3}}{2}}\\AF&=\sqrt{\frac{34\sqrt{3}}{2}}\\AF&=\sqrt[4]{867}\end{align*}
The answer is $867 + 4 = \boxed{871}$

Solution:
In the final pyramid, let $ABCD$ be the smaller square and let $A'B'C'D'$ be the larger square such that $AA'$ , etc. are edges.
It is obvious from the diagram that $\angle A'AB = \angle A'AD = 105^\circ$
Let $AB$ and $AD$ be the positive $x$ and $y$ axes in a 3-d coordinate system such that $A'$ has a positive $z$ coordinate. Let $\alpha$ be the angle made with the positive $x$ axis . Define $\beta$ and $\gamma$ analogously.
It is easy to see that if $P: = (x,y,z)$ , then $x = AA'\cdot \cos\alpha$ . Furthermore, this means that $\frac {x^2}{AA'^2} + \frac {y^2}{AA'^2} + \frac {z^2}{AA'^2} = \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$
We have that $\alpha = \beta = 105^\circ$ , so $\cos^2 105^\circ + \cos^2105^\circ + \cos^2\gamma = 1\implies \cos\gamma = \sqrt [4]{\frac {3}{4}}$
It is easy to see from the Law of Sines that $\frac {AA'}{\sin 45^\circ} = \frac {\sqrt {17}}{\sin 30^\circ}\implies AA' = \sqrt {34}$
Now, $z = AA'\cdot \cos\gamma = \sqrt [4]{34^2\cdot \frac {3}{4}} = \sqrt [4]{867}$
It follows that the answer is $867 + 4 = \boxed{871}$


Answer: 80.0

A square pyramid with base $ABCD$ and vertex $E$ has eight edges of length $4$ .  A plane passes through the midpoints of $AE$ $BC$ , and $CD$ .  The plane's intersection with the pyramid has an area that can be expressed as $\sqrt{p}$ .  Find $p$
AIME I 2007-13.png

Solution:
Note first that the intersection is a pentagon
Use 3D analytical geometry, setting the origin as the center of the square base and the pyramid’s points oriented as shown above. $A(-2,2,0),\ B(2,2,0),\ C(2,-2,0),\ D(-2,-2,0),\ E(0,2\sqrt{2})$ . Using the coordinates of the three points of intersection $(-1,1,\sqrt{2}),\ (2,0,0),\ (0,-2,0)$ , it is possible to determine the equation of the plane. The equation of a plane resembles $ax + by + cz = d$ , and using the points we find that $2a = d \Longrightarrow d = \frac{a}{2}$ $-2b = d \Longrightarrow d = \frac{-b}{2}$ , and $-a + b + \sqrt{2}c = d \Longrightarrow -\frac{d}{2} - \frac{d}{2} + \sqrt{2}c = d \Longrightarrow c = d\sqrt{2}$ . It is then $x - y + 2\sqrt{2}z = 2$
Write the equation of the lines and substitute to find that the other two points of intersection on $\overline{BE}$ $\overline{DE}$ are $\left(\frac{\pm 3}{2},\frac{\pm 3}{2},\frac{\sqrt{2}}{2}\right)$ . To find the area of the pentagon, break it up into pieces (an isosceles triangle on the top, an isosceles trapezoid on the bottom). Using the distance formula $\sqrt{a^2 + b^2 + c^2}$ ), it is possible to find that the area of the triangle is $\frac{1}{2}bh \Longrightarrow \frac{1}{2} 3\sqrt{2} \cdot \sqrt{\frac 52} = \frac{3\sqrt{5}}{2}$ . The trapezoid has area $\frac{1}{2}h(b_1 + b_2) \Longrightarrow \frac 12\sqrt{\frac 52}\left(2\sqrt{2} + 3\sqrt{2}\right) = \frac{5\sqrt{5}}{2}$ . In total, the area is $4\sqrt{5} = \sqrt{80}$ , and the solution is $\boxed{080}$


Answer: 4.0

A square with integer side length is cut into 10 squares, all of which have integer side length and at least 8 of which have area 1. What is the smallest possible value of the length of the side of the original square?
$\text{(A)}\hspace{.05in}3\qquad\text{(B)}\hspace{.05in}4\qquad\text{(C)}\hspace{.05in}5\qquad\text{(D)}\hspace{.05in}6\qquad\text{(E)}\hspace{.05in}7$

Solution: The first answer choice ${\textbf{(A)}\ 3}$ , can be eliminated since there must be $10$ squares with integer side lengths. We then test the next smallest sidelength which is $4$ . The square with area $16$ can be partitioned into $8$ squares with area $1$ and two squares with area $4$ , which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is $\boxed{4}$


Answer: 68.0

A square with side length $8$ is colored white except for $4$ black isosceles right triangular regions with legs of length $2$ in each corner of the square and a black diamond with side length $2\sqrt{2}$ in the center of the square, as shown in the diagram. A circular coin with diameter $1$ is dropped onto the square and lands in a random location where the coin is completely contained within the square. The probability that the coin will cover part of the black region of the square can be written as $\frac{1}{196}\left(a+b\sqrt{2}+\pi\right)$ , where $a$ and $b$ are positive integers. What is $a+b$ [asy] /* Made by samrocksnature */ draw((0,0)--(8,0)--(8,8)--(0,8)--(0,0)); fill((2,0)--(0,2)--(0,0)--cycle, black); fill((6,0)--(8,0)--(8,2)--cycle, black); fill((8,6)--(8,8)--(6,8)--cycle, black); fill((0,6)--(2,8)--(0,8)--cycle, black); fill((4,6)--(2,4)--(4,2)--(6,4)--cycle, black); filldraw(circle((2.6,3.31),0.5),gray); [/asy]
$\textbf{(A)} ~64 \qquad\textbf{(B)} ~66 \qquad\textbf{(C)} ~68 \qquad\textbf{(D)} ~70 \qquad\textbf{(E)} ~72$

Solution:
To find the probability, we look at the $\frac{\text{success region}}{\text{total possible region}}$ . For the coin to be completely contained within the square, we must have the distance from the center of the coin to a side of the square to be at least $\frac{1}{2}$ , as it's the radius of the coin. This implies the $\text{total possible region}$ is a square with side length $8 - \frac{1}{2} - \frac{1}{2} = 7$ , with an area of $49$ . Now, we consider cases where needs to land to partially cover a black region.
Near The Center Square
We can have the center of the coin land within $\frac{1}{2}$ outside of the center square, or inside of the center square.  So, we have a region with $\frac{1}{2}$ emanating from every point on the exterior of the square, forming four quarter circles and four rectangles. The four quarter circles combine to make a full circle of radius $\frac{1}{2}$ , so the area is $\frac{\pi}{4}$ . The area of a rectangle is $2 \sqrt 2 \cdot \frac{1}{2} = \sqrt 2$ , so $4$ of them combine to an area of $4 \sqrt 2$ . The area of the black square is simply $\left(2\sqrt 2\right)^2 = 8$ . So, for this case, we have a combined total of $8 + 4\sqrt 2 + \frac{\pi}{4}$ . Onto the second (and last) case.
Near A Triangle
We can also have the coin land within $\frac{1}{2}$ outside of one of the triangles. By symmetry, we can just find the successful region for one of them, then multiply by $4$ . Consider the above diagram. We can draw an altitude from the bottom corner of the square to hit the hypotenuse of the green triangle. The length of this when passing through the black region is $\sqrt 2$ , and when passing through the white region (while being contained in the green triangle) is $\frac{1}{2}$ . However, we have to subtract off when it doesn't pass through the red square. Then, it's the hypotenuse of a small isosceles right triangle with side lengths of $\dfrac{1}{2}$ which is $\dfrac{\sqrt{2}}{2}.$ So, the altitude of the green triangle is $\sqrt 2 + \frac{1}{2} - \frac{\sqrt 2}{2} = \frac{\sqrt 2 + 1}{2}$ . Then, recall, the area of an isosceles right triangle is $h^2$ , where $h$ is the altitude from the right angle. So, squaring this, we get $\frac{3 + 2\sqrt 2}{4}$ . Now, we have to multiply this by $4$ to account for all of the black triangles, to get $3 + 2\sqrt 2$ as the final area for this case.
Finishing
Then, to have the coin touching a black region, we add up the area of our successful regions, or $8 + 4\sqrt 2 + \frac{\pi}{4} + 3 + 2\sqrt 2 = 11 + 6\sqrt 2 + \frac{\pi}{4} = \frac{44 + 24\sqrt 2 + \pi}{4}$ . The total region is $49$ , so our probability is $\frac{\frac{44 + 24\sqrt 2 + \pi}{4}}{49} = \frac{44 + 24\sqrt 2 + \pi}{196}$ , which implies $a+b = 44+24 = 68$ . This corresponds to answer choice $\boxed{68}$


Answer: 2.0

A square with side length 2 and a circle share the same center. The total area of the regions that are inside the circle and outside the square is equal to the total area of the regions that are outside the circle and inside the square. What is the radius of the circle?
[asy] pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2); draw(a--d--b--c--cycle); draw(circle(o, 2.5)); [/asy]
$\textbf{(A)}\ \frac{2}{\sqrt{\pi}} \qquad \textbf{(B)}\ \frac{1+\sqrt{2}}{2} \qquad \textbf{(C)}\ \frac{3}{2} \qquad \textbf{(D)}\ \sqrt{3} \qquad \textbf{(E)}\ \sqrt{\pi}$

Solution:
Let the region within the circle and square be $a$ . In other words, it is the area inside the circle $\textbf{and}$ the square.  Let $r$ be the radius. We know that the area of the circle minus $a$ is equal to the area of the square, minus $a$
We get:
$\pi r^2 -a=4-a$
$r^2=\frac{4}{\pi}$
$r=\frac{2}{\sqrt{\pi}}$
So the answer is $\boxed{2}$

Solution:
We realize that since the areas of the regions outside of the circle and the square are equal to each other, the area of the circle must be equal to the area of the square.
$\pi r^2=4$
$r^2=\frac{4}{\pi}$
$r=\frac{2}{\sqrt{\pi}}$
So the answer is $\boxed{2}$


Answer: 56.0

A square with sides of length $1$ is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points on three of the sides, as shown. Find $x$ , the length of the longer parallel side of each trapezoid.
$\mathrm{(A) \ } \frac 35 \qquad \mathrm{(B) \ } \frac 23 \qquad \mathrm{(C) \ } \frac 34 \qquad \mathrm{(D) \ } \frac 56 \qquad \mathrm{(E) \ } \frac  78$

Solution: The area of the trapezoid is $\frac{1}{3}$ , and the shorter base and height are both $\frac{1}{2}$ .  Therefore, \[\frac{1}{3}=\frac{1}{2}\cdot \frac{1}{2}\cdot \left(\frac{1}{2}+x\right) \Rightarrow x=\frac{5}{6}\rightarrow \boxed{56}\]

Solution:
Divide the pentagon into 2 small congruent trapezoids by extending the common shorter base of the 2 larger trapezoids.
Since each of the smaller trapezoids has its area half each of the larger trapezoids, and each of them has a base $\frac{1}{2}$ , we have \[b_{large}+\frac{1}{2}=2(b_{small}+ \frac{1}{2})\] \[x+\frac{1}{2}=2((1-x)+\frac{1}{2})\] \[x=\frac{5}{6}\boxed{56}\]


Answer: 361.0

A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?
$\textbf{(A) }148\qquad\textbf{(B) }324\qquad\textbf{(C) }361\qquad\textbf{(D) }1296\qquad\textbf{(E) }1369$

Solution: Since the number of tiles lying on both diagonals is $37$ , counting one tile twice, there are $37=2x-1\implies x=19$ tiles on each side. Therefore, our answer is $19^2=361=\boxed{361}$

Solution: Visualize it as 4 separate diagonals connecting to one square in the middle. Each square on the diagonal corresponds to one square of horizontal/vertical distance (because it's a square). So, we figure out the length of each separate diagonal, multiply by two, and then add 1. (Realize that we can just join two of the separate diagonals on opposite sides together to save some time in calculations.) Therefore, the edge length is: \[\frac{37-1}{4} \cdot 2 + 1 = 19\] Thus, our solution is $19^2 = 361 = \boxed{361}$


Answer: 927.0

A stack of $2000$ cards is labelled with the integers from $1$ to $2000,$ with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$

Solution: We try to work backwards from when there are 2 cards left, since this is when the 1999 card is laid onto the table. When there are 2 cards left, the 1999 card is on the top of the deck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there are over 1000 cards remaining, some cards have not even made one trip through yet, $2(1024 - 1000) = 48$ , to be exact. Once these cards go through, 1999 will be the $512 - 48 = 464^\text{th}$ card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position $464 \times 2 = 928$ , meaning that there were $\boxed{927}$ cards are above the one labeled $1999$

Solution:
To simplify matters, we want a power of $2$ . Hence, we will add $48$ 'fake' cards which we must discard in our actual count. Using similar logic as Solution 1, we find that 1999 has position $1024$ in a $2048$ card stack, where the fake cards towards the front.
Let the fake cards have positions $1, 3, 5, \cdots, 95$ . Then, we know that the real cards filling the gaps between the fake cards must be the cards such that they go to their correct starting positions in the $2000$ card case, where all of them are below $1999$ . From this, we know that the cards from positions $1$ to $96$ alternate in fake-real-fake-real, where we have the correct order of cards once the first $96$ have moved and we can start putting real cards on the table. Hence, $1999$ is in position $1024 - 96 = 928$ , so $\boxed{927}$ cards are above it. - Spacesam

Solution: We work backwards. To reverse the process, we must move the bottom card to the top, and add a new number to the top. Let $d_n$ equal the number of cards below 1999 after $n$ process reversals. Reversing one process, our deck only has $2000$ , so we reverse again to obtain $1999, 2000$ . So, $d_2 = 1$ . When $d_{n-1} > 0$ , after a process reversal, the bottom card is moved to the top (it goes above $1999$ ), so we subtract by one to account for this (the addition of the new card doesn't matter since it goes above $1999$ ), giving us $d_n = d_{n-1} - 1$ . So, $d_3 = 0$ . Then, when $d_{n-1} = 0$ , after a process reversal, $1999$ moves to the top and one more card is added above $1999$ . Since at $d_{n}$ $n$ cards have been added, there must be $n - 2$ cards below $1999$ . So, $d_4 = 2$ . Then, consider all $d_{n-1} = 0$ . Then, $d_n = n-2$ as previously stated. After $n-2$ process reversals, we go back to $d_{2n - 2} = n - 2 - (n - 2) = 0$ . Then, $d_{2n-1} = 2n - 3$ . Next, we can simply calculate $2 \cdot 4 - 1 = 7$ $2 \cdot 7 - 1 = 13$ , and so on (which is a very simple bash, as $2000$ is a small number. If you don't want to do this, define sequence $a_n = 2a_{n-1} - 1$ , and solve for the closed form, which is very easy). Consequently, we derive $d_{1537} = 1537 - 2 = 1535$ $2000 - 1537 = 463$ steps remain. So, $d_{2000} = d_{1537} - 463 = 1072$ . Finally, there are $2000 - 1 - 1072 = \boxed{927}$ numbers above $1999$

Solution:
Let us treat each run through the deck as a separate "round". For example, in round one, you would go through all of the $2000$ cards initially in the deck once, in round two, you would go through all $1000$ cards initially in the deck once, so on and so forth. For each round, let us record what the initial and final actions are ( $r$ for moving the card to the right, $b$ for moving the card to the bottom), the number of cards moved to the right, the number of cards left, and what the position of the cards moved to the right were in the original $2000$ card deck (as $n = a + ck$ where $n$ is the position, $c$ is the spacing of the cards moved, $a$ is an integer such that the correct first card is moved, and $k$ is an integer greater than or equal to $1$ which represents which card out of all the cards moved to the right you are finding the position of). Then,
Round 1: $r$ to $b$ $1000$ to right, $1000$ left in deck, $n = -1 + 2k$
Round 2: $r$ to $b$ $500$ to right, $500$ left in deck, $n = -2 + 4k$
Round 3: $r$ to $b$ $250$ to right, $250$ left in deck, $n = -4 + 8k$
Round 4: $r$ to $b$ $125$ to right, $125$ left in deck, $n = -8 + 16k$
Let us treat the remaining deck of $125$ cards as a totally independent deck, note that the positions of card in this deck are $n = 16k$ . Also note that the first action of a new round is never the same action as the last action of the previous round because actions alternate. Also note that for every new round, the spacing between cards moved doubles, and that the cards remaining in the beginning of a new round have position $n = a + c/2 + ck$ for the values $a, c$ of the previous round. Also note that if there are an odd number of cards in an initial deck, the first and last actions are the same. Then,
Round 5: $r$ to $r$ $63$ to right, $62$ left in deck, $n = -1 + 2k$
Round 6: $b$ to $r$ $31$ to right, $31$ left in deck, $n = 4k$ , because $n = 2k$ positioned cards are left at the beginning of this round and the first card is sent to the bottom, only every second card is sent to the right, and because spacing doubles every round,
Round 7: $b$ to $b$ $15$ to right, $16$ left in deck, $n = -2 + 8k$ , because $n = 2 + 4k$ positioned cards are left at the beginning and only every second card is sent to the right, and because spacing doubles every round,
Round 8: $r$ to $b$ $8$ to right, $8$ left in deck, $n = -14 + 16k$ , by similar reasoning, since the cards $n = 2 + 8k$ are left and spacing doubles every round, from here on things get real easy,
Round 9: $r$ to $b$ $4$ to right, $4$ left in deck, $n = -22 + 32k$
Round 10: $r$ to $b$ $2$ to right, $2$ left in deck, $n = -38 + 64k$
Round 11: $r$ to $b$ $1$ to right, $1$ left in deck, $n = 58$ , since $58 = -38 + 64/2 + 64$
This card must be labeled 1999 since it is second to last. Then, since it is the $58th$ in the deck of $125$ , it must be the $58 * 8 = 928th$ card in the original deck, and because we've been numbering from the top of the deck to the bottom (also because AIME answers are 000-999), there were $928 - 1 = \boxed{927}$ cards above it in the deck - Romulus and minimaxweii

Solution:
Similar to Solution 5, we treat each run-through of the deck from the lowest-indexed card to the highest-indexed card as a separate round. Notice that after each round, approximately half the deck will remain, with the other half having been cast aside to the right in sorted order. Then, we can model each round as if we are running a "Sieve" through the deck with powers of 2s, filtering out the cards to put to the right and the cards to be pushed to the bottom.
At the beginning, when all $2000$ cards are still in the deck, notice that the first run-through of the deck consists of removing all the cards such that their index $I\equiv 1\pmod{2}$ , while we leave behind all the cards such that their index $I\equiv 2\pmod{2}$ . (Since the cards are not $0$ -indexed, using $2$ instead of $0\pmod{2}$ is simpler to keep track of the cards. This will appear again later.) The remaining $1000$ cards, then, will all share the attribute that their index numbers are even. To split this further for round 2, we look more specifically at each index, categorizing them further as $I\equiv 2\pmod 4$ or $I\equiv 4\pmod 4$ in that order (The card with the smaller remainder will always be the card on the top of the new deck). Then, in Round 2, we will remove all the cards with index values that leave a remainder of $2$ when divided by $4$ , while leaving the multiples of $4$ behind. Notice that each time, in each new deck, we are removing all the cards with an odd position in that specific deck, and leaving behind all the cards with an even position in that specific deck.
This process will continue until we reach  Round 5, where the number of cards that remain ( $125$ ) is no longer a multiple of 2. Therefore, when we remove all the cards with indices $I\equiv 16\pmod{32}$ and leave behind all the cards with indices $I\equiv 32\pmod{32}$ , the last card in the deck will be removed instead of being placed to the bottom, with $125-63=62$ cards remaining. Because of this, when Round 6 begins, we cannot follow our normal procedure, since the top card of the Round 6 deck would have been moved to the bottom! Therefore, when we are using our sieve with $I\equiv 32\pmod{64}$ and $I\equiv 64\pmod{64}$ , the order at which the remainders are presented have flipped, so all the cards with indices that are multiples of 64 will be removed, while all the cards with indices that are 32 more than a multiple of 64 will remain. The $62$ cards from Round 6 will be reduced to $31$ , and the last card in the deck will be removed just like in Round 5, which means Round 7 will be affected the same way. Of the choices of $I\equiv 32\pmod{128}$ and $I\equiv 32+64=96\pmod{128}$ , we will remove the latter, leaving behind the cards with indices that are 32 more than a multiple of 128. However, since 31 is an odd number, the last card of the deck will NOT be removed, which means our sieving process will return to normal after Round 7, with $31-15=16$ cards remaining. Since $16$ is a perfect power of $2$ , we will not need to worry about this scenario again in the following rounds.
Round 8: $I\equiv 32\pmod{256}$ removed; $I\equiv 32+128=160\pmod{256}$ remains; $16-8=8$ cards left.
Round 9: $I\equiv 160\pmod{512}$ removed; $I\equiv 160+256=416\pmod{512}$ remains; $8-4=4$ cards left.
Round 10: $I\equiv 416\pmod{1024}$ removed; $I\equiv 416+512=928\pmod{1024}$ remains; $4-2=2$ cards left.
Round 11: $I\equiv 928\pmod{2048}$ removed; $I\equiv 928+1024=1952\pmod{2048}$ remains; $2-1=1$ card left.
Round 12: The last card in position $1952$ is sorted into place (This is card # $2000$ !).
It is obvious that the single card put into place in Round 11 is the second-to-last card in the deck, which is our target # $1999$ . Then, its position in the original deck leaves a remainder of $928$ when divided by $2056$ . It is easy to see that $928$ is the only index that satisfies this, so there will be $928-1=\boxed{927}$ cards above card # $1999$


Answer: 12.0

A standard six-sided die is rolled, and $P$ is the product of the five numbers that are visible. What is the largest number that is certain to divide $P$
$\mathrm{(A) \ } 6 \qquad \mathrm{(B) \ } 12 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 144\qquad \mathrm{(E) \ } 720$

Solution:
The product of all six numbers is $6!=720$ . The products of numbers that can be visible are $720/1$ $720/2$ , ..., $720/6$ .
The answer to this problem is their greatest common divisor -- which is $720/L$ , where $L$ is the least common multiple of $\{1,2,3,4,5,6\}$ .
Clearly $L=60$ and the answer is $720/60=\boxed{12}$

Solution:
The product P can be one of the following six numbers excluding the number that is hidden under, so we have:
\begin{align*}
2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3^2 \cdot 5 \\
1 \cdot 3 \cdot 4 \cdot 5 \cdot 6 = 2^3 \cdot 3^2 \cdot 5 \\
1 \cdot 2 \cdot 4 \cdot 5 \cdot 6 = 2^4 \cdot 3 \cdot 5 \\
1 \cdot 2 \cdot 3 \cdot 5 \cdot 6 = 2^2 \cdot 3^2 \cdot 5 \\
1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 = 2^4 \cdot 3^2 \\
1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 = 2^3 \cdot 3 \cdot 5
\end{align*}
The largest number that is certain to divide product P is basically GCD of all the above 6 products which is $2^2 \cdot 3$
Hence $P=3\cdot2^2=\boxed{12}$


Answer: 187.0

A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are $6^4 = 1296$ outcomes).
Case 1 (easy): Four 5's are rolled. This has probability $\frac{1}{6^4}$ of occurring.
Case 2: Two 5's are rolled.
Case 3: No 5's are rolled.
To find the number of outcomes for the latter two cases, we will use recursion. Consider a 5-sided die with faces numbered 1, 2, 3, 4, 6. For $n \ge 1$ , let $a_n$ equal the number of outcomes after rolling the die $n$ times, with the property that the product is a square. Thus, $a_1 = 2$ as 1 and 4 are the only possibilities.
To find $a_{n+1}$ given $a_n$ (where $n \ge 1$ ), we observe that if the first $n$ rolls multiply to a perfect square, then the last roll must be 1 or 4. This gives $2a_n$ outcomes. Otherwise, the first $n$ rolls do not multiply to a perfect square ( $5^n - a_n$ outcomes). In this case, we claim that the last roll is uniquely determined (either 2, 3, or 6). If the product of the first $n$ rolls is $2^x 3^y$ where $x$ and $y$ are not both even, then we observe that if $x$ and $y$ are both odd, then the last roll must be 6; if only $x$ is odd, the last roll must be 2, and if only $y$ is odd, the last roll must be 3. Thus, we have $5^n - a_n$ outcomes in this case, and $a_{n+1} = 2a_n + (5^n - a_n) = 5^n + a_n$
Computing $a_2$ $a_3$ $a_4$ gives $a_2 = 7$ $a_3 = 32$ , and $a_4 = 157$ . Thus for Case 3, there are 157 outcomes. For case 2, we multiply $a_2$ by $\binom{4}{2} = 6$ to distribute the two 5's among four rolls. Thus the probability is
\[\frac{1 + 6 \cdot 7 + 157}{6^4} = \frac{200}{6^4} = \frac{25}{162} \implies m+n = \boxed{187}\]

Solution:
We can solve this without finding the amount of cases. Notice when we roll a 1 or a 4, it does not affect whether or not the product is a square number. We have a 1/3 chance of rolling either a 1 or 4. We have a 2/3 chance of rolling a 2,3,5 or 6. Let's call rolling 1 or 4 rolling a dud (a perfect square).
Probability of rolling 4 duds: $\left(\frac{1}{3}\right)^4$
Probability of rolling 3 duds: $4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3}$
Probability of rolling 2 duds: $6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2$
Probability of rolling 1 dud: $4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3$
Probability of rolling 0 duds: $\left(\frac{2}{3}\right)^4$
Now we will find the probability of a square product given we have rolled each amount of duds
Probability of getting a square product given 4 duds: 1
Probability of getting a square product given 3 duds: 0 (you will have 1 non-dud and that's never going to be square)
Probability of getting a square product given 2 duds: $\frac{1}{4}$ (as long as our two non-duds are the same, our product will be square)
Probability of getting a square product given 1 dud: $\frac{3!}{4^3}$ $\frac{3}{32}$ (the only way to have a square  product is rolling a 2,3 and 6. There are 3! ways of doing that and a total of $4^3$ ways to roll 3 non-duds).
Probability of getting a square product given 0 duds: $\frac{40}{4^4}$ $\frac{5}{32}$ (We can have any two non-duds twice. For example, 2,2,5,5. There are $\binom{4}{2} = 6$ ways of choosing which two non-duds to use and $\binom{4}{2} = 6$ ways of choosing how to arrange those 4 numbers. That gives us 6*6=36 combinations. We can also have 2,2,2,2 or 3,3,3,3 or 5,5,5,5 or 6,6,6,6. This gives us a total of 40 combinations).
We multiply each probability of rolling k duds with the probability of getting a square product given k duds and then sum all the values. \[\left(\frac{1}{3}\right)^4  * 1 + 4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3} * 0 + 6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2 * \frac{1}{4} + 4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3 * \frac{3}{32} + \left(\frac{2}{3}\right)^4 * \frac{5}{32} = \frac{25}{162}.\]
$25+162$ $\boxed{187}$

Solution:
Note that rolling a 1/4 will not affect whether or not the product is a perfect square. This means that in order for the product to be a perfect square, all non 1/4 numbers rolled must come in pairs, with the only exception being the triplet 2,3, 6. Now we can do casework:
If there are four 1/4's, then there are $2^4=16$ combinations.
If there are three 1/4's, then there are 0 combinations, because the fourth number isn't a square.
If there are two 1/4's, there are $2^2=4$ ways to choose the two 1/4's, 4 ways to choose the remaining pair of numbers, and $\frac{4!}{2!2!}=6$ ways to arrange, so there are $4\cdot 4\cdot 6=96$ combinations for this case.
If there is one 1/4, then there are 2 ways to choose whether it is a 1 or 4, and the remaining three numbers must be 2, 3, and 6, so there are $4!$ ways to order, meaning there are $2\cdot 4!=48$ combinations for this case.
Our final case is if there are no 1/4's, in which case we must have two pairs. If the two pairs are of different numbers, then there $\binom{4}{2}$ to choose the numbers and $\frac{4!}{2!2!}=6$ ways to arrange them, so $6\cdot 6=36$ . If all four numbers are the same there are $4$ combinations, so there are $4+36=40$ combinations for this case.
Hence there are $16+0+96+48+40=200$ combinations where the product of the dice is a perfect square, and there are $6^4=1296$ total combinations, so the desired probability is $\frac{200}{1296}=\frac{25}{162}$ , yielding an answer of $25+162=\boxed{187}$

Solution:
Another way to solve this problem is to do casework on all the perfect squares from $1^2$ to $36^2$ , and how many ways they can be ordered $1^2$ $1,1,1,1$ $1$ way. $2^2$ $4,1,1,1$ or $2,2,1,1$ $\binom{4}{2}+4=10$ ways. $3^2$ $3,3,1,1$ $\binom{4}{2}=6$ ways. $4^2$ $4,4,1,1$ $2,2,2,2$ , or $2,2,4,1$ $\binom{4}{2}+1+12=19$ ways. $5^2$ $5,5,1,1$ $\binom{4}{2}=6$ ways. $6^2$ $6,6,1,1$ $1,2,3,6$ $2,3,2,3$ $3,3,4,1$ $2*\binom{4}{2}+4!+12=48$ ways. $7^2$ - Since there is a prime greater than 6 in its prime factorization there are $0$ ways. $8^2$ $4,4,4,1$ or $2,4,2,4$ $\binom{4}{2}+4=10$ ways. $9^2$ $3,3,3,3$ $1$ way. $10^2$ $2,2,5,5$ or $1,4,5,5$ $6+12=18$ ways. $11^2$ $0$ ways for the same reason as $7^2$ $12^2$ $6,6,2,2$ $4,4,3,3$ $2,3,4,6$ , or $1,4,6,6$ $2*\binom{4}{2}+4!+12=48$ ways. $13^2$ $0$ ways. $14^2$ $0$ ways. $15^2$ $3,3,5,5$ $\binom{4}{2}=6$ ways. $16^2$ $4,4,4,4$ $1$ way. $17^2$ $0$ ways. $18^2$ $3,3,6,6$ $\binom{4}{2}=6$ ways. $19^2$ $0$ ways. $20^2$ $4,4,5,5$ $\binom{4}{2}=6$ ways. $21^2$ $0$ ways. $22^2$ $0$ ways. $23^2$ $0$ ways. $24^2$ $4,4,6,6$ $\binom{4}{2}=6$ ways. $25^2$ $5,5,5,5$ $1$ way. $26^2$ $0$ ways. $27^2$ $0$ ways. $28^2$ $0$ ways. $29^2$ $0$ ways. $30^2$ $5,5,6,6$ $\binom{4}{2}$ ways. $31^2$ $0$ ways. $32^2$ $0$ ways. $33^2$ $0$ ways. $34^2$ $0$ ways. $35^2$ $0$ ways. $36^2$ $6,6,6,6$ $1$ way.
There are $6^4=1296$ ways that the dice can land. Summing up the ways, it is easy to see that there are $200$ ways.
This results in a probability of $\frac{200}{1296}=\frac{25}{162}\implies\boxed{187}$ -superninja2000

Solution:
We can do recursion on the number of rolls to find the number of ways we can get $4$ rolls to multiply to a square.
After $n$ rolls, let us say that the product is $p = 2^a3^b5^c$
Define the following:
$A_{n} =$ the number of ways to have a product after $n$ rolls where $a$ is odd, and $b$ $c$ are even
$B_{n} =$ the number of ways to have a product after $n$ rolls where $b$ is odd, and $a$ $c$ are even
$C_{n} =$ the number of ways to have a product after $n$ rolls where $c$ is odd, and $a$ $b$ are even
$D_{n} =$ the number of ways to have a product after $n$ rolls where $c$ is even, and $a$ $b$ are odd
$E_{n} =$ the number of ways to have a product after $n$ rolls where $b$ is even, and $a$ $c$ are odd
$F_{n} =$ the number of ways to have a product after $n$ rolls where $a$ is even, and $b$ $c$ are odd
$G_{n} =$ the number of ways to have a product after $n$ rolls where $a, b,$ and $c$ are all odd
$S_{n} =$ the number of ways to have a product after $n$ rolls where $a, b,$ and $c$ are all even (square!)
We have the following equations after considering the possible values of the nth roll:
\[A_{n} = S_{n-1}+B_{n-1}+D_{n-1}+E_{n-1}+2A_{n-1}\]
\[B_{n} = A_{n-1}+D_{n-1}+F_{n-1}+S_{n-1}+2B_{n-1}\]
\[C_{n} = S_{n-1}+E_{n-1}+F_{n-1}+G_{n-1}+2C_{n-1}\]
\[D_{n} = S_{n-1}+A_{n-1}+B_{n-1}+G_{n-1}+2D_{n-1}\]
\[E_{n} = A_{n-1}+C_{n-1}+F_{n-1}+G_{n-1}+2E_{n-1}\]
\[F_{n} = B_{n-1}+E_{n-1}+C_{n-1}+G_{n-1}+2F_{n-1}\]
\[G_{n} = C_{n-1}+D_{n-1}+F_{n-1}+E_{n-1}+2G_{n-1}\]
\[S_{n} = A_{n-1}+C_{n-1}+B_{n-1}+D_{n-1}+2S_{n-1}\]
We have the following values after considering the possible values of the 1st roll:
\[A_1 = B_1 = C_1 = D_1 = 1; E_1 = F_1 = G_1 = 0; S_1 = 2\]
After applying recursion twice, we get:
\[A_2 = B_2 = D_2 = 6, C_2 = 4, E_2 = F_2 = G_2 = 2, S_2 = 8\]
\[A_3 = B_3 = D_3 = 34, C_3 = 22, E_3 = F_3 = G_3 = 18, S_3 = 38\]
Finally, we have $S_4 = 200$ $\frac{m}{n} = \frac{200}{1296} = \frac{25}{162}$ meaning our answer is $\boxed{187}$

Solution:
Consider all the distinct "fundamental" groups of integers from $1$ to $6$ whose product is a perfect square. A "fundamental" group is one that cannot be broken into two smaller groups that each have a perfect square product. For example, $\{2,2\}$ is a fundamental group, while $\{3,3,4\}$ is not, because it can be broken up into $\{3,3\}$ and $\{4\}$
$1$ and $4$ are already perfect squares, so they each form a "fundamental" group and cannot belong in any other group. Pairs of the other $4$ numbers ( $\{2,2\}$ $\{3,3\}$ , etc. ) form "fundamental" groups as well. The last "fundamental" group is $\{2,3,6\}$ . It can be easily seen that no more groups exist.
Thus, we have the "fundamental" groups $\{1\}$ $\{4\}$ $\{2,2\}$ $\{3,3\}$ $\{5,5\}$ $\{6,6\}$ , and $\{2,3,6\}$
We now consider the ways to use these groups to form a sequence of $4$ numbers whose product is a perfect square. To form a set, we can simply select zero to two groups of size $2$ or $3$ and fill in any remaining spots with $1$ s and $4$ s. We can do this in one of $5$ ways: Using only $1$ s and $4$ s, using one group of size $2$ , using one group of size $3$ , using two different groups of size $2$ , and using the same group of size $2$ twice.
If we only use $1$ s and $4$ s, each of the $4$ slots can be filled with one of the $2$ numbers, so there are $2^4=16$ possibilities.
If we use one group of size $2$ , there are $4$ options for the group to use, $\binom{4}{2}$ ways to place the two numbers (since they are identical), and $2^2$ ways to fill in the remaining slots with $1$ s and $4$ s, so there are $4\cdot\binom{4}{2}\cdot2^2=96$ possibilities.
If we use one group of size $3$ , there is only $1$ option for the group to use, $4\cdot3\cdot2$ ways to place the three numbers (since they are distinct), and $2$ ways to fill in the remaining slot, so there are $4\cdot3\cdot2\cdot2=48$ possibilities.
If we use two different groups of size $2$ , there are $\binom{4}{2}$ options for the groups to use and $\binom{4}{2}$ ways to place the four numbers (since there are $2$ groups of identical numbers, and one group's placement uniquely determines the other's), so there are $\binom{4}{2}\cdot\binom{4}{2}=36$ possibilities.
If we use the same group of size $2$ twice, there are $4$ options for the group to use and $1$ way to place the four numbers (since they are all identical), so there are $4=4$ possibilities.
This gives us a total of $16+96+48+36+4=200$ possibilities, and since there are $6^4=1296$ total sequences that can be rolled, the probability is equal to $\frac{200}{1296}=\frac{25}{162}$ , so the answer is $25+162=\boxed{187}$ . ~ emerald_block

Solution:
There are a total of $2^4=1296$ possible die rolls.
We use casework:
Case 1 : All 4 numbers are the same.
There are obviously $6$ ways.
Case 2 : Sets of 2 different numbers.
A set of two different numbers is basically $(x,x,y,y)$ . There are a total of $\frac{4!}{2!\cdot 2!}=6$ ways to arrange the numbers.
By listing these cases, we quickly see a pattern:
$(1,1,2,2)$
$(1,1,3,3)$
$...$
$(1,1,6,6)$
$(2,2,3,3)$
$...$
$(2,2,6,6)$
$...$
$(5,5,6,6)$
There are a total of $5+4+3+2+1=15$ cases. Multiplying this by $6$ yields $15\cdot 6=90$ ways.
Case 3 : Sets of numbers in the form of $(x,x,1,4)$
A special case must be made for the number $4$ because $4$ itself is a perfect square.
$(1,1,1,4)$ $4$
$(2,2,1,4)$ $12$
$(3,3,1,4)$ $12$
$(4,4,1,4)$ $4$
$(5,5,1,4)$ $12$
$(6,6,1,4)$ $12$
Summing these up yields a total of $4+12+12+4+12+12=56$ ways.
Case 4 : Sets with all 4 numbers different
Note that the sets
$(1,2,3,6)$
$(2,3,4,6)$
Multiply to perfect square. The total of these cases are $24+24=48$
Adding all these cases together yields $6+90+56+48=200$ ways that the product of the values of the die can be a perfect square.
Therefore the probability is
$\frac{200}{1296}=\frac{25}{162}$
$m+n = 25+162 = \boxed{187}$


Answer: 40.0

A store increased the original price of a shirt by a certain percent and then lowered the new price by the same amount. Given that the resulting price was $84\%$ of the original price, by what percent was the price increased and decreased $?$
$\textbf{(A) }16\qquad\textbf{(B) }20\qquad\textbf{(C) }28\qquad\textbf{(D) }36\qquad\textbf{(E) }40$

Solution: Suppose the fraction of discount is $x$ . That means $(1-x)(1+x)=0.84$ ; so, $1-x^{2}=0.84$ , and $(x^{2})=0.16$ , obtaining $x=0.4$ . Therefore, the price was increased and decreased by $40$ %, or $\boxed{40}$

Solution:
After the first increase by $p$ percent, the shirt price became $(1+p)$ times greater than the original. Upon the decrease in p percent on this price, the shirt price became $(1-p)$ times less than $(1+p)$ , or $(1-p)(1+p)$ . We know that this price is $84$ percent of the original, so $(1-p)(1+p) = 0.84$
From here, we can list the factors of $0.84$ and see which are equidistant from $1$ . We see that $0.6$ and $1.4$ are both $0.4$ from $1$ , so $p = 0.4 = 40 \%$ , or choice $\boxed{40}$

Solution: We can try out every option and see which one works. By this method, we get $\boxed{40}$

Solution:
Let our original cost be $$ 100$ , so we are looking for a whole number of $$ 84$ . Also, we can see that (A), (C), and (D) give us answers with decimals while we know that (B) and (E) give us whole numbers. Therefore, we only need to try these two: (B) $$100$ increased by 20% = $$120$ , and $$120$ decreased by 20% = $$96$ , a whole number, and (E) $$100$ increased by 40% = $$140$ , and $$140$ decreased by 40% = $$84$ , a whole number.
Thus, $40$ % or $\boxed{40}$ is the answer.


Answer: 40.0

Solution:
Let our original cost be $$ 100.$ We are looking for a result of $$ 84,$ then. We try 16% and see it gets us higher than 84. We try 20% and see it gets us lower than 16 but still higher than 84. We know that the higher the percent, the less the value. We try 36, as we are not progressing much, and we are close! We try $\boxed{40}$ , and we have the answer; it worked.
(OR: try (C) first to eliminate 2 answer choices)


Answer: 100.0

A store normally sells windows at $$100$ each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?
$\textbf{(A) } 100\qquad \textbf{(B) } 200\qquad \textbf{(C) } 300\qquad \textbf{(D) } 400\qquad \textbf{(E) } 500$

Solution:
The store's offer means that every $5$ th window is free.
Dave would get $\left\lfloor\frac{7}{5}\right\rfloor=1$ free window.
Doug would get $\left\lfloor\frac{8}{5}\right\rfloor=1$ free window.
This is a total of $2$ free windows.
Together, they would get $\left\lfloor\frac{8+7}{5}\right\rfloor = \left\lfloor\frac{15}{5}\right\rfloor=3$ free windows.
So they get $3-2=1$ additional window if they purchase the windows together.
Therefore they save $1\cdot100= \boxed{100}$


Answer: 1000.0

A store owner bought $1500$ pencils at $$ 0.10$ each.  If he sells them for $$ 0.25$ each, how many of them must he sell to make a profit of exactly $$ 100.00$
$\text{(A)}\ 400 \qquad \text{(B)}\ 667 \qquad \text{(C)}\ 1000 \qquad \text{(D)}\ 1500 \qquad \text{(E)}\ 1900$

Solution: $1500\times 0.1=150$ , so the store owner is $$150$ below profit. Therefore he needs to sell $150+100= 250$ dollars worth of pencils. Selling them at $$0.25$ each gives $250/0.25= \boxed{1000}$


Answer: 2.0

A straight concrete sidewalk is to be $3$ feet wide, $60$ feet long, and $3$ inches thick.  How many cubic yards of concrete must a contractor order for the sidewalk if concrete must be ordered in a whole number of cubic yards?
$\text{(A)}\ 2 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ \text{more than 20}$

Solution: This is a $1$ yard by $20$ yard by $1/12$ yard sidewalk, so its volume in yards is \[1\times 20\times \frac{1}{12} = 1.\overline{6}.\] Since concrete must be ordered in a whole number of cubic yards, we need $2\rightarrow \boxed{2}$


Answer: 550.0

A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per minute. At the same time, Melanie and Sherry begin swimming in straight lines to a point on the north bank of the river that is equidistant from their starting positions. The two women arrive at this point simultaneously. Find $D$

Solution:
Define $m$ as the number of minutes they swim for.
Let their meeting point be $A$ . Melanie is swimming against the current, so she must aim upstream from point $A$ , to compensate for this; in particular, since she is swimming for $m$ minutes, the current will push her $14m$ meters downstream in that time, so she must aim for a point $B$ that is $14m$ meters upstream from point $A$ . Similarly, Sherry is swimming downstream for $m$ minutes, so she must also aim at point $B$ to compensate for the flow of the current.
If Melanie and Sherry were to both aim at point $B$ in a currentless river with the same dimensions, they would still both meet at that point simultaneously. Since there is no current in this scenario, the distances that Melanie and Sherry travel, respectively, are $80m$ and $60m$ meters. We can draw out this new scenario, with the dimensions that we have:  (While it is indeed true that the triangle above with side lengths $60m$ $80m$ and $D$ is a right triangle, we do not know this yet, so we cannot assume this based on the diagram.)
By the Pythagorean Theorem, we have \begin{align*} 264^{2} + \left( \frac{D}{2} - 14m \right) ^{2} &= 3600m^{2} \\ 264^{2} + \left( \frac{D}{2} + 14m \right) ^{2} &= 6400m^{2}. \end{align*}
Subtracting the first equation from the second gives us $28Dm = 2800m^{2}$ , so $D = 100m$ . Substituting this into our first equation, we have that \begin{align*}264^{2} + 36^{2} m^{2} &= 60^{2}m^{2} \\ 264^{2} &= 96 \cdot 24 \cdot m^{2} \\ 11^{2} &= 4 \cdot m^{2} \\ m &= \frac{11}{2}. \end{align*}
So $D = 100m = \boxed{550}$

Solution:
Claim
Median $AM$ and altitude $AH$ are drawn in triangle $ABC$ $AB = c, AC = b < c, BC = a$ are known. Let's denote $MH = x$
Prove that \begin{align*}2ax = c^{2} - b^{2}\end{align*}
Proof \[BH + CH = a,\] \begin{align*} BH^{2} - CH^{2} = c^{2} - b^{2}\implies BH - CH &= \frac{c^{2} - b^{2}} {a},\end{align*} \[BH =  \frac{c^{2} - b^{2}}{2a} +  \frac{a}{2},\] \begin{align*}MH = BH - BM  &= \frac{c^{2} - b^{2}} {2a}.\end{align*}
Solution
In the coordinate system associated with water, the movement is described by the scheme in the form of a triangle, the side on which Melanie floats is $80t$ , where t is the time of  Melanie's movement, the side along which Sherry floats is $60t$
The meeting point floated away at a distance of $14t$ from the midpoint between the starting points of Melanie and Sherry.
In the notation of the Claim \begin{align*}  c = 80t, b = 60t, x = 14t \implies a = \frac{(80t)^2-(60t)^2}{2 \cdot 14t}=\frac{20^2}{4}\cdot \frac{16-9}{7}t = 100t.\end{align*} Hence, \begin{align*}  AH = \sqrt{BC^2-BH^2}= \sqrt{(80t)^2-(50t+14t)^2}=16t \cdot \sqrt{5^2-4^2}= 48t = 264 \implies t = 5.5.\end{align*} \[D = a = 100t = \boxed{550}\] vladimir.shelomovskii@gmail.com, vvsss

Solution:
We have the following diagram:  Since Melanie and Sherry swim for the same distance and the same amount of time, they swim at the same net speed.
Let $x$ and $y$ be some positive numbers. We have the following table: \[\begin{array}{c||c|c|c}     & \textbf{Net Velocity Vector (m/min)} & \textbf{Natural Velocity Vector (m/min)} & \textbf{Natural Speed (m/min)} \\ \hline \hline &&& \\ [-2.25ex] \textbf{Melanie} & \langle -x,y\rangle & \langle -x-14,y\rangle & 80  \\ \hline   &&& \\ [-2.25ex] \textbf{Sherry} & \langle x,y\rangle & \langle x-14,y\rangle & 60 \end{array}\] Recall that $|\text{velocity}|=\text{speed},$ so \begin{align*} (-x-14)^2 + y^2 &= 80^2, &&(1) \\ (x-14)^2 + y^2 &= 60^2. &&(2) \end{align*} We subtract $(2)$ from $(1)$ to get $56x=2800,$ from which $x=50.$ Substituting this into either equation, we have $y=48.$
It follows that Melanie and Sherry both swim for $264\div y=5.5$ minutes. Therefore, the answer is \[D=2x\cdot5.5=\boxed{550}.\] ~MRENTHUSIASM

Solution:
We can break down movement into two components: the $x$ -component and the $y$ -component. Suppose that Melanie travels a distance of $a$ in the $x$ -direction and a distance of $c$ in the $y$ -direction in one minute when there is no current. Similarly, suppose that Sherry travels a distance of $a$ in the $x$ -direction but a distance of $b$ in the $y$ -direction in one minute when there is no current. The current only affects the $x$ -components because it goes in the $x$ -direction.

Now, note that $a^2 + b^2 = 60^2$ because Sherry travels 60 meters in a minute. Thus, $a^2 + c^2 = 80^2$ because Melanie travels 80 meters in a minute. Also, the distance they travel with the current must be the same in one minute because they reach the point equidistant from them at the same time. That means $b + 14 = c - 14$ or $b = c - 28$ . So now we can plug that into the two equations to get: \begin{align*} a^2 + c^2 &= 80^2, \\ a^2 + (c-28)^2 &= 60^2. \end{align*} We can solve the system of equations to get $a = 48$ and $c = 64$ . From this, we can figure out that it must've taken them $5.5$ minutes to get to the other side, because $264/48 = 5.5$ . This means that there are $5.5$ lengths of $48$ in each person's travel. Also, $D$ must be equal to $11(b+14) = 11(c-14)$ because there are $(5.5)2 = 11$ lengths of $b-14$ between them, $5.5$ on each person's side. Since $c = 64$ , we have $c-14 = 50$ , so the answer is \[D=11\cdot50=\boxed{550}.\] ~Curious_crow


Answer: 12.0

A street has parallel curbs $40$ feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is $15$ feet and each stripe is $50$ feet long. Find the distance, in feet, between the stripes.
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)}\ 25$

Solution:
Drawing the problem out, we see we get a parallelogram with a height of $40$ and a base of $15$ , giving an area of $600$
If we look at it the other way, we see the distance between the stripes is the height and the base is $50$

The area is still the same, so the distance between the stripes is $600/50 = \boxed{12}$

Solution: Alternatively, we could use similar triangles--the $30-40-50$ triangle (created by the length of the bordering stripe and the difference between the two curbs) is similar to the $x-y-15$ triangle, where we are trying to find $y$ (the shortest distance between the two stripes). Therefore, $y$ would have to be $\boxed{12}$


Answer: 504.0

A strictly increasing sequence of positive integers $a_1$ $a_2$ $a_3$ $\cdots$ has the property that for every positive integer $k$ , the subsequence $a_{2k-1}$ $a_{2k}$ $a_{2k+1}$ is geometric and the subsequence $a_{2k}$ $a_{2k+1}$ $a_{2k+2}$ is arithmetic. Suppose that $a_{13} = 2016$ . Find $a_1$

Solution: Instead of setting $a_1$ equal to something and $a_2$ equal to something, note that it is rather easier to set $a_1=x^2$ and $a_3=y^2$ so that $a_2=xy,a_4=y(2y-x),a_5=(2y-x)^2$ and so on until you reached $a_{13}=(6y-5x)^2$ (Or simply notice the pattern), so $6y-5x=\sqrt{2016}=12\sqrt{14}$ . Note that since each of the terms has degree 2 so if you multiply $x$ and $y$ by $\sqrt{14}$ you multiply each term by $14$ so each term is still a integer if the terms are already integers before you multiply $x$ and $y$ by $\sqrt{14}$ , so let $w=\frac{x}{\sqrt{14}}$ and $z=\frac{y}{\sqrt{14}}$ so $6z-5w=12$ . Then, for the sequence to be strictly increasing positive integers we have $(w,z)=(6,7)$ so $x=6\sqrt{14}$ and $a_1=x^2=6^2 \cdot 14=\boxed{504}$ Ddk001

Solution: The thirteenth term of the sequence is $2016$ , which makes that fourteenth term of the sequence $2016+r$ and the $15^{\text{th}}$ term $\frac{(2016+r)^2}{2016}$ . We note that $r$ is an integer so that means $\frac{r^2}{2016}$ is an integer. Thus, we assume the smallest value of $r$ , which is $168$ . We bash all the way back to the first term and get our answer of $\boxed{504}$

Solution:
Let $a_{2k-1}=s$ where $k=1$ . Then, $a_{2k}=sr \Longrightarrow a_{2k+1}=sr^2$ . Continuing on, we get $a_{2k+2}=sr(r-1)+sr^2=sr(2r-1) \Longrightarrow a_{2k+3}=sr^2(\frac{2r-1}{r})^2=s(2r-1)^2$ . Moreover, $a_{2k+4}=s(2r-1)(r-1)+s(2r-1)^2=s(2r-1)(3r-2) \Longrightarrow a_{2k+5}=s(2r-1)^2(\frac{3r-2}{2r-1})^2=s(3r-2)^2$
It is clear now that $a_{2k+2c}=s(cr-(c-1))((c+1)r-c)$ and $a_{2k+2c-1}=s(cr-(c-1))^2$ . Plugging in $c=6$ $a_{13}=s(6r-5)^2=2016$ . The prime factorization of $2016=2^5\cdot3^2\cdot7$ so we look for perfect squares.
$6r-5\equiv (6r-5)^2\equiv 1\pmod{6}$ if $r$ is an integer, and $\frac{\omega+5}{6}=r \Longrightarrow 6\mid{s}$ if $r$ is not an integer and $\omega$ is rational, so $6\mid{s}$ . This forces $s=2\cdot3^2\cdot7\cdot{N}$ . Assuming $(6r-5)$ is an integer, it can only be $2^x$ $x\in{1,2}$
If $6r-5=2^1$ $r=\frac{7}{6}$ . If $6r-5=2^2$ $r=\frac{3}{2}$ . Note that the latter cannot work since $a_{2k+1}=s(\frac{9}{4}) \Longrightarrow 4\mid{s}$ but $N=1 \Longrightarrow s=2\cdot3^2\cdot7$ in this scenario. Therefore, $r=\frac{7}{6} \Longrightarrow s=\frac{2016}{2^2}=504$ . Plugging back $k=1$ $a_1=s=\boxed{504}$


Answer: 10.0

A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly $10$ ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person planning committee be selected?
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 25$

Solution: Let the number of students on the council be $x$ .  To select a two-person committee, we can select a "first person" and a "second person." There are $x$ choices to select a first person; subsequently, there are $x-1$ choices for the second person. This gives a preliminary count of $x(x-1)$ ways to choose a two-person committee. However, this accounts for the order of committees. To understand this, suppose that Alice and Bob are two students in the council. If we choose Alice and then Bob, that is the same as choosing Bob and then Alice and so latter and former arrangements would be considered the same. Therefore, we have to divide by $2$ to account for overcounting. Thus, there are $\dfrac{x(x-1)} 2=10$ ways to choose the two-person committee. Solving this equation, we find that $5$ and $-4$ are integer solutions. $-4$ is a ridiculous situation, so there are $5$ people on the student council. The solution is $\dbinom 5 3=10\implies \boxed{10}$

Solution: To choose $2$ people from $n$ total people and get $10$ as a result, we can establish the equation $\binom{n}{2}=10$ which we can easily see $n=5$ , so there are $5$ people. The question asks how many ways to choose $3$ people from the $5$ , so there are $\binom{5}{3}=\boxed{10}$ ways.


Answer: 9.0

A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 16$

Solution:
Let us split this up into two cases.
Case $1$ : The student chooses both algebra and geometry.
This means that $3$ courses have already been chosen.  We have $3$ more options for the last course, so there are $3$ possibilities here.
Case $2$ : The student chooses one or the other.
Here, we simply count how many ways we can do one, multiply by $2$ , and then add to the previous.
Assume the mathematics course is algebra.  This means that we can choose $2$ of History, Art, and Latin, which is simply $\dbinom{3}{2} = 3$ .  If it is geometry, we have another $3$ options, so we have a total of $6$ options if only one mathematics course is chosen.
Thus, overall, we can choose a program in $6 + 3 = \boxed{9}$ ways

Solution: We can use complementary counting. Since there must be an English class, we will add that to our list of classes leaving $3$ remaining spots for rest of classes. We are also told that there needs to be at least one math class. This calls for complementary counting. The total number of ways of choosing $3$ classes out of the $5$ is $\binom{5}3$ . The total number of ways of choosing only non-mathematical classes is $\binom{3}3$ . Therefore the amount of ways you can pick classes with at least one math class is $\binom{5}3-\binom{3}3=10-1=\boxed{9}$ ways.

Solution: Similar to Solution 1, note that for Case 1 of solution the answer is simply $\binom{4}2$ and for the second case it is $\binom{3}2$ hence $\boxed{9}$ ways


Answer: 8.0

A student recorded the exact percentage frequency distribution for a set of measurements, as shown below.
However, the student neglected to indicate $N$ , the total number of measurements. What is the smallest possible value of $N$
\[\begin{tabular}{c c}\text{measured value}&\text{percent frequency}\\ \hline 0 & 12.5\\ 1 & 0\\ 2 & 50\\ 3 & 25\\ 4 & 12.5\\ \hline\ & 100\\ \end{tabular}\]
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 16 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 50$

Solution: Note that $12.5\% = \frac{1}{8}$ $25\% = \frac{1}{4}$ , and $50\% = \frac{1}{2}$ . Thus, since the frequencies must be integers, $N$ must be divisible by $2$ $4$ , and $8$ (so that $\frac{N}{8}$ etc. are integers), or in other words, $N$ is divisible by $8$ . Thus the smallest possible value of $N$ is the smallest positive multiple of $8$ , which is $8$ itself, or $\boxed{8}$


Answer: 6.0

A supermarket has $128$ crates of apples. Each crate contains at least $120$ apples and at most $144$ apples.
What is the largest integer $n$ such that there must be at least $n$ crates containing the same number of apples?
$\textbf{(A) }4\qquad \textbf{(B) }5\qquad \textbf{(C) }6\qquad \textbf{(D) }24\qquad  \textbf{(E) }25$

Solution:
To find the largest number of "repeated" crates necessary, we must account for all the possibilities of the number of apples in each crate. Since each crate contains a minimum of $120$ apples and a maximum of $144$ apples, there are $144 - 120 + 1 = 25$ different amounts possible for the number of apples per crate.
Now, we have to count for the worst case scenario: the $25$ amounts are repeated as many times as possible.
$25$ can go into $128$ exactly $5$ times because $5 \cdot 25 = 125$ , which is less than $128$ . This leaves a remainder of $3$ crates.
The worst case scenario would be that these $3$ crates have a different number of apples each. It doesn't actually matter how many apples are in these $3$ crates because any of the $25$ values would be repeated again anyway. So, the answer is $5 + 1 = \boxed{6}$ jiang147369


Answer: 2.75

A taxi ride costs $$1.50$ plus $$0.25$ per mile traveled.  How much does a $5$ -mile taxi ride cost?
$\textbf{(A)}\ 2.25 \qquad\textbf{(B)}\ 2.50  \qquad\textbf{(C)}\ 2.75 \qquad\textbf{(D)}\ 3.00 \qquad\textbf{(E)}\ 3.75$

Solution: There are five miles which need to be traveled. The cost of these five miles is $(0.25\cdot5) = 1.25$ .   Adding this to $1.50$ , we get $\boxed{2.75}$


Answer: 93.0

A teacher gave a test to a class in which $10\%$ of the students are juniors and $90\%$ are seniors. The average score on the test was $84.$ The juniors all received the same score, and the average score of the seniors was $83.$ What score did each of the juniors receive on the test?
$\textbf{(A) } 85 \qquad\textbf{(B) } 88 \qquad\textbf{(C) } 93 \qquad\textbf{(D) } 94 \qquad\textbf{(E) } 98$

Solution: We can assume there are $10$ people in the class. Then there will be $1$ junior and $9$ seniors. The sum of everyone's scores is $10 \cdot 84 = 840$ . Since the average score of the seniors was $83$ , the sum of all the senior's scores is $9 \cdot 83 = 747$ . The only score that has not been added to that is the junior's score, which is $840 - 747 = \boxed{93}$

Solution: Let the average score of the juniors be $j$ . The problem states the average score of the seniors is $83$ . The equation for the average score of the class (juniors and seniors combined) is $\frac{j}{10} + \frac{83 \cdot 9}{10} = 84$ . Simplifying this equation yields $j = \boxed{93}$


Answer: 93.0

Solution: We can assume there are $10$ people in the class. Then there will be $1$ junior and $9$ seniors. The sum of everyone's scores is $10 \cdot 84 = 840$ . Since the average score of the seniors was $83$ , the sum of all the senior's scores is $9 \cdot 83 = 747$ . The only score that has not been added to that is the junior's score, which is $840 - 747 = \boxed{93}$

Solution: Let the average score of the juniors be $j$ . The problem states the average score of the seniors is $83$ . The equation for the average score of the class (juniors and seniors combined) is $\frac{j}{10} + \frac{83 \cdot 9}{10} = 84$ . Simplifying this equation yields $j = \boxed{93}$


Answer: 26.0

A teacher tells the class,
Ben thinks of $6$ , and gives his answer to Sue.  What should Sue's answer be?
$\text{(A)}\ 18 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 26 \qquad \text{(D)}\ 27 \qquad \text{(E)}\ 30$

Solution: Ben adds $1$ to $6$ , getting $1 + 6 = 7$ .  Then Ben doubles it, to get $7\cdot 2 = 14$ .  He gives this number to Sue.  Then Sue subtracts $1$ from the number, to get $14 - 1 = 13$ .  Finally, Sue doubles the result to get $13\cdot 2 = 26$ , and the answer is $\boxed{26}$


Answer: 258.0

A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $6$ , and a different number in $S$ was divisible by $7$ . The teacher then asked if any of the students could deduce what $S$ is, but in unison, all of the students replied no.
However, upon hearing that all four students replied no, each student was able to determine the elements of $S$ . Find the sum of all possible values of the greatest element of $S$

Solution:
Note that $\operatorname{lcm}(6,7)=42.$ It is clear that $42\not\in S$ and $84\not\in S,$ otherwise the three other elements in $S$ are divisible by neither $6$ nor $7.$
In the table below, the multiples of $6$ are colored in yellow, and the multiples of $7$ are colored in green. By the least common multiple, we obtain cycles: If $n$ is a possible maximum value of $S,$ then $n+42$ must be another possible maximum value of $S,$ and vice versa. By observations, we circle all possible maximum values of $S.$  From the second row of the table above, we perform casework on the possible maximum value of $S:$ \[\begin{array}{c||c|c|l} & & & \\ [-2.5ex] \textbf{Max Value} & \boldsymbol{S} & \textbf{Valid?} & \hspace{16.25mm}\textbf{Reasoning/Conclusion} \\ [0.5ex] \hline & & & \\ [-2ex] 49 & \{46,47,48,49\} & & \text{The student who gets } 46 \text{ will reply yes.} \\ 50 & \{47,48,49,50\} & \checkmark & \text{Another possibility is } S=\{89,90,91,92\}. \\ 51 & \{48,49,50,51\} & & \text{The student who gets } 51 \text{ will reply yes.} \\ 56 & \{53,54,55,56\} & & \text{The student who gets } 53 \text{ will reply yes.} \\ 57 & \{54,55,56,57\} & & \text{The student who gets } 57 \text{ will reply yes.} \\ 63 & \{60,61,62,63\} & & \text{The students who get } 60,61,62 \text{ will reply yes.} \\ 66 & \{63,64,65,66\} & & \text{The students who get } 64,65,66 \text{ will reply yes.} \\ 72 & \{69,70,71,72\} & & \text{The student who gets } 69 \text{ will reply yes.} \\ 73 & \{70,71,72,73\} & & \text{The student who gets } 73 \text{ will reply yes.} \\ 78 & \{75,76,77,78\} & & \text{The student who gets } 75 \text{ will reply yes.} \\ 79 & \{76,77,78,79\} & \checkmark & \text{Another possibility is } S=\{34,35,36,37\}. \\ 80 & \{77,78,79,80\} & & \text{The student who gets } 80 \text{ will reply yes.} \end{array}\] Finally, all possibilities for $S$ are $\{34,35,36,37\}, \{47,48,49,50\}, \{76,77,78,79\},$ and $\{89,90,91,92\},$ from which the answer is $37+50+79+92=\boxed{258}.$

Solution:
We know right away that $42\not\in S$ and $84\not\in S$ as stated in Solution 1.
To get a feel for the problem, let’s write out some possible values of $S$ based on the teacher’s remarks. The first multiple of 7 that is two-digit is 14. The closest multiple of six from 14 is 12, and therefore there are two possible sets of four consecutive integers containing 12 and 14; $\{11,12,13,14\}$ and $\{12,13,14,15\}$ . Here we get our first crucial idea; that if the multiples of 6 and 7 differ by two, there will be 2 possible sets of $S$ without any student input. Similarly, if they differ by 3, there will be only 1 possible set, and if they differ by 1, 3 possible sets.
Now we read the student input. Each student says they can’t figure out what $S$ is just based on the teacher’s information, which means each student has to have a number that would be in 2 or 3 of the possible sets (This is based off of the first line of student input). However, now that each student knows that all of them have numbers that fit into more than one possible set, this means that S cannot have two possible sets because otherwise, when shifting from one set to the other, one of the end numbers would not be in the shifted set, but we know each number has to fall in two or more possible sets. For example, take $\{11,12,13,14\}$ and $\{12,13,14,15\}$ . The numbers at the end, 11 and 15, only fall in one set, but each number must fall in at least two sets. This means that there must be three possible sets of S, in which case the actual S would be the middle S.
Take for example $\{33,34,35,36\}$ $\{34,35,36,37\}$ , and $\{35,36,37,38\}$ . 37 and 34 fall in two sets while 35 and 36 fall in all three sets, so the condition is met. Now, this means that the multiple of 6 and 7 must differ by 1.
Since 42 means the difference is 0, when you add/subtract 6 and 7, you will obtain the desired difference of 1, and similarly adding/subtracting 6 or 7 from 84 will also obtain the difference of 1. Thus there are four possible sets of $S$ $\{34,35,36,37\}$ $\{47,48,49,50\}$ $\{76,77,78,79\}$ and $\{89,90,91,92.\}$ . Therefore the sum of the greatest elements of the possible sets $S$ is $37+50+79+92=\boxed{258}$

Solution:
In a solution that satisfies these constraints, the multiple of 6 must be adjacent to multiple of 7. The other two numbers must be on either side.
WLOG assume the set is $\{a,6j,7k,b\}$ . The student with numbers $a$ $6j$ , and $7k$ can think the set is $\{a-1, a,6j,7k\}$ or $\{a,6j,7k,b\}$ , and the students with number $6j$ $7k$ , and $b$ can think the set is $\{a,6j,7k,b\}$ or $\{6j,7k,b, b+1\}$ . Therefore, none of the students know the set for sure.
Playing around with the arrangement of the multiple of 6 and multiple of 7 shows that this is the only configuration viewed as ambiguous to all the students. (Therefore when they hear nobody else knows either, they can find out it is this configuration)
Considering $S$ as $\{a,6j,7k,b\}$ ,b is 2 mod 6 and 1 mod 7, so $b$ is 8 mod 42. (since it is all 2-digit, the values are either 50 or 92).
Similarly, considering $S$ as $\{a,7j,6k,b\}$ $b$ is 1 mod 6 and 2 mod 7, so $b$ is 37 mod 42. The values that satisfy that are 37 and 79.
The total sum of all these values is therefore $50+92+37+89=\boxed{258}$

Solution:
Consider the tuple $(a, a+1, a+2, a+3)$ as a possible $S$ . If one of the values in $S$ is $3$ or $4 \pmod{7}$ , observe the student will be able to deduce $S$ with no additional information. This is because, if a value is $b = 3 \pmod{7}$ and $S$ contains a $0 \pmod{7}$ , then the values of $S$ must be $(b-3, b-2, b-1, b)$ . Similarly, if we are given a $b \equiv 4 \pmod{7}$ and we know that $0 \pmod{7}$ is in $S$ $S$ must be $(b, b+1, b+2, b+3).$ Hence, the only possibility for $a$ is $5, 6 \pmod{7}.$
In either case, we are guaranteed there is a $6, 0, 1 \pmod{7}$ value in $7$ . The difference comes down to if there is a $5 \pmod{7}$ value or a $1 \pmod{7}$ value. The person receiving such value will be able to determine all of $S$ but the $6, 0, 1 \pmod{7}$ people will not be able to differentiate the two cases ... yet.
Now consider which value among the consecutive integers is $c \equiv 3 \pmod{6}$ , if any. The person will know that $S$ is either $(c-3, c-2, c-1, c)$ or $(c, c+1, c+2, c+3)$ to have a $0 \pmod{6}$ value in $S$ . Neither the $6, 1 \pmod{7}$ person can be $3 \pmod{7}$ , else they can decipher what $S$ is right off the bat by considering which set has $0 \pmod{7}.$ This translates to the possible $5 \pmod{7}$ or $1 \pmod{7}$ person cannot be $0 \pmod{6}$ . We are given that $0 \pmod{7}$ and $0 \pmod{6}$ cannot be the same person. Hence we conclude one of the $6 \pmod{7}$ or $1 \pmod{7}$ must be the $0 \pmod{6}$ person.
Let the $6 \pmod{7}$ person be $0 \pmod{6}.$ Then--hypothetical-- $c \equiv 2 \pmod{7}$ person is $3 \pmod{6}.$ After the first round, the $6, 0, 1 \pmod{7}$ people realize that $2 \pmod{7}$ is not in $S$ else they would have deduced $S$ by noting $S$ was either $(c-3, c-2, c-1, c)$ or $(c, c+1, c+2, c+3)$ to have a $0 \pmod{6}$ and choosing the former based on where $0 \pmod{7}$ is. Hence they figure out $S$ by knowing $5 \pmod{7}.$ So $a \equiv 5 \pmod{7}$ and $a \equiv 5 \pmod{6}$ (from $6 \pmod{7}$ person being $0 \pmod{6}$ ).
Similarly, if $1 \pmod{7}$ person is $0 \pmod{6}$ we find that $a \equiv 6 \pmod{7}$ (so a $2 \pmod{7}$ is in $S$ ) and $a \equiv 4 \pmod{6}.$
By CRT, the possibilities are $a \equiv 5,  34 \pmod{42}.$ The sum of the greatest values of $S$ are the sum of $a + 3$ and so we get $(34 + 3) + (47 + 3) + (76 + 3) + (89 + 3) = \boxed{258}.$


Answer: 23.0

A team won $40$ of its first $50$ games.  How many of the remaining $40$ games must this team win so it will have won exactly $70 \%$ of its games for the season?
$\text{(A)}\ 20 \qquad \text{(B)}\ 23 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$

Solution: Noting that 70% is the same as $\frac{70}{100}=\frac{7}{10}$ , and that, when x is the amount of wins in the last 40 games, the fraction of games won is $\frac{40+x}{50+40}=\frac{40+x}{90}$ , all we have to do is set them equal: \[\frac{40+x}{90}=\frac{7}{10}\] \[40+x=63\] \[x=\boxed{23}\]

Solution: Alternatively we can note that they will play a total of $40+50=90$ games and must win $0.7(90)=63$ games. Since they won $40$ games already they need $63-40=\boxed{23}$ more games.


Answer: 164.0

A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly $.500$ . During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than $.503$ . What's the largest number of matches she could've won before the weekend began?

Solution:
Let $n$ be the number of matches won, so that $\frac{n}{2n}=\frac{1}{2}$ , and $\frac{n+3}{2n+4}>\frac{503}{1000}$
Cross multiplying $1000n+3000>1006n+2012$ , so $n<\frac{988}{6}=164 \dfrac {4}{6}=164 \dfrac{2}{3}$ . Thus, the answer is $\boxed{164}$

Solution: Let $n$ be the number of matches she won before the weekend began. Since her win ratio started at exactly . $500 = \tfrac{1}{2},$ she must have played exactly $2n$ games total before the weekend began. After the weekend, she would have won $n+3$ games out of $2n+4$ total. Therefore, her win ratio would be $(n+3)/(2n+4).$ This means that \[\frac{n+3}{2n+4} > .503 = \frac{503}{1000}.\] Cross-multiplying, we get $1000(n+3) > 503(2n+4),$ which is equivalent to $n < \frac{988}{6} = 164.\overline{6}.$ Since $n$ must be an integer, the largest possible value for $n$ is $\boxed{164}.$


Answer: 0.2

A tetrahedron with four equilateral triangular faces has a sphere inscribed within it and a sphere circumscribed about it. For each of the four faces, there is a sphere tangent externally to the face at its center and to the circumscribed sphere. A point $P$ is selected at random inside the circumscribed sphere. The probability that $P$ lies inside one of the five small spheres is closest to
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }0.1 \qquad \mathrm{(C) \ }0.2 \qquad \mathrm{(D) \ }0.3 \qquad \mathrm{(E) \ }0.4$

Solution:
Let the side length of the tetrahedron be $1$
The altitude of the equilateral triangle base is $\frac{\sqrt{3}}{2}$ . Thus, the distance from the center of the equilateral triangle to its vertex is $\frac{\sqrt{3}}{2} \cdot \frac 23 = \frac{\sqrt{3}}{3}$ . Therefore, the altitude of the tetrahedron is $\sqrt{1^2-(\frac{\sqrt{3}}{3})^2} = \frac{\sqrt{6}}{3}$
Let the radius of the large sphere be $R$ \[(\frac{\sqrt{3}}{3})^2 + (\frac{\sqrt{6}}{3} - R)^2 = R^2\] \[R = \frac{\sqrt{6}}{4}\]
The distance from the center of the tetrahedron to the center of one of the bases is $\frac{\sqrt{6}}{3} - R = \frac{\sqrt{6}}{3} - \frac{\sqrt{6}}{4} = \frac{\sqrt{6}}{12}$
The distance from the center of the tetrahedron to the center of one of the bases is also the same as the radius of the small sphere in the center of the tetrahedron.
The radius of the smaller spheres tangent to the large sphere and the tetrahedron is $\frac{\frac{\sqrt{6}}{4}-\frac{\sqrt{6}}{12}}{2} = \frac{\sqrt{6}}{12}$
Therefore, the probability that $P$ lies inside one of the five small spheres is $\frac{5 \cdot (\frac{\sqrt{6}}{12})^3 \cdot \frac43 \cdot \pi }{(\frac{\sqrt{6}}{4})^3 \cdot \frac43 \cdot \pi} = \frac{5}{27} \approx \boxed{0.2}$


Answer: 33.3

A thin piece of wood of uniform density in the shape of an equilateral triangle with side length $3$ inches weighs $12$ ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of $5$ inches. Which of the following is closest to the weight, in ounces, of the second piece?
$\textbf{(A)}\ 14.0\qquad\textbf{(B)}\ 16.0\qquad\textbf{(C)}\ 20.0\qquad\textbf{(D)}\ 33.3\qquad\textbf{(E)}\ 55.6$

Solution:
We can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use
$\left(\frac{3}{5}\right)^2=\frac{12}{x}$
We can then solve the equation to get $x=\frac{100}{3}$ which is closest to $\boxed{33.3}$

Solution:
Also recall that the area of an equilateral triangle is $\frac{a^2\sqrt3}{4}$ so we can give a ratio as follows:
$\frac{\frac{9\sqrt3}{4}}{12}$ $=$ $\frac{\frac{25\sqrt3}{4}}{x}$
Cross multiplying and simplifying, we get $12 \cdot \frac{25}{9}$
Which is $33.\overline{3}$ $\approx$ $\boxed{33.3}$

Solution: Note that the ratio of the two triangle's weights is equal to the ratio of their areas, as the height is the same. The ratio of their areas is equal to the square of the ratio of their sides. So if $x$ denotes the  weight of the second triangle, we have \[\frac{x}{12}=\frac{5^2}{3^2}=\frac{25}{9}\] Solving gives us $x \approx 33.33$ so the answer is $\boxed{33.3}$


Answer: 33.3

Solution:
By: dragonfly
We can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use
$\left(\frac{3}{5}\right)^2=\frac{12}{x}$
We can then solve the equation to get $x=\frac{100}{3}$ which is closest to $\boxed{33.3}$

Solution:
Another approach to this problem, very similar to the previous one but perhaps explained more thoroughly, is to use proportions. First, since the thickness and density are the same, we can set up a proportion based on the principle that $d=\frac{m}{V}$ , thus $dV=m$
However, since density and thickness are the same and $A$ is proportional to $s^2$ (recognizing that the area of an equilateral triangle is $\frac{(s)^2\sqrt{3}}{4}$ ), we can say that $m$ is proportional to $s^2$
Then, by increasing s by a factor of $\frac{5}{3}$ $s^2$ is increased by a factor of $\frac{25}{9}$ , thus $m=12*\frac{25}{9}$ or $\boxed{33.3}$


Answer: 13.0

A three-digit integer contains one of each of the digits $1$ $3$ , and $5$ . What is the probability that the integer is divisible by $5$
$\textbf{(A)}\  \frac{1}{6}  \qquad \textbf{(B)}\   \frac{1}{3}  \qquad \textbf{(C)}\   \frac{1}{2}  \qquad \textbf{(D)}\  \frac{2}{3}   \qquad \textbf{(E)}\   \frac{5}{6}$

Solution: The three digit numbers are $135,153,351,315,513,531$ . The numbers that end in $5$ are divisible are $5$ , and the probability of choosing those numbers is $\boxed{13}$

Solution:
The number is divisible by 5 if and only if the number ends in $5$ (also $0$ , but that case can be ignored, as none of the digits are $0$ )
If we randomly arrange the three digits, the probability of the last digit being $5$ is $\boxed{13}$


Answer: 391.0

A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of $|y|=|x|$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
Perform the coordinate transformation $(x, y)\rightarrow (x+y, x-y)$ . Then we can see that a movement up, right, left, or down in the old coordinates adds the vectors $\langle 1, -1 \rangle$ $\langle 1, 1 \rangle$ $\langle -1, -1 \rangle$ $\langle -1, 1 \rangle$ respectively. Moreover, the transformation takes the equation $|y| = |x|$ to the union of the x and y axis. Exactly half of the moves go up in the new coordinates, and half of them go down. In order to end up on the x axis, we need to go up thrice and down thrice. The number of possible sequences of up and down moves is the number of permutations of $UUUDDD$ , which is just $\binom63 = 20$ . The probability of any of these sequences happening is $\left(\frac12\right)^6$ . Thus, the probability of ending on the x axis is $\frac{20}{2^6}$ . Similarly, the probability of ending on the y axis is the same.
However, we overcount exactly one case: ending at $(0, 0)$ . Since ending on the x axis and ending on the y axis are independent events, the probability of both is simply $\left(\frac{20}{2^6}\right)^2 = \frac{25}{256}$ . Using PIE, the total probability is $\frac{20}{64} + \frac{20}{64} - \frac{25}{256} = \frac{135}{256}$ , giving an answer of $\boxed{391}$

Solution:
We have 4 possible moves: U, D, R, and L. The total number of paths that could be taken is $4^6$ , or $4096$ . There are 4 possible cases that land along the line $y = x$ $x,y = \pm 1; x,y = \pm 2; x,y = \pm 3;$ or $x = y = 0$ . We will count the number of ways to end up at $(1,1), (2,2),$ and $(3,3)$ , multiply them by 4 to account for the other quadrants, and add this to the number of ways to end up at $(0,0)$
Thus, the total number of ways to end up at $(1,1)$ is $300$
Thus, the total number of ways to end up at $(0,0)$ is $400$
Adding these cases together, we get that the total number of ways to end up on $y = x$ is $4(20 + 120 + 300) + 400 = 2160$ . Thus, our probability is $\frac{2160}{4096}$ . When this fraction is fully reduced, it is $\frac{135}{256}$ , so our answer is $135 + 256 = \boxed{391}.$

Solution:
We split this into cases by making a chart. In each row, the entries $(\pm1)$ before the dividing line represent the right or left steps (without regard to order), and the entries after the dividing line represent the up or down steps (again, without regard to order). This table only represents the cases where the ending position $(x,y)$ satisfies $x=y$ \[\begin{array}{ccccccccccccl} \multicolumn{5}{c}{R (+)\qquad L (-)}& |&\multicolumn{5}{c}{U (+)\qquad D (-)}\\ +1&&  +1&&  +1&| &  +1&&  +1&&  +1\\ +1&& +1&& -1& | & +1&& +1&& -1\\ +1&& -1&&  -1& | & +1&& -1&& -1\\ -1 && -1&&  -1& |  & -1&&  -1&& -1\\ \\ +1&& +1&& +1&& -1 &|& +1&& +1\\ +1&& +1&& -1 && -1 &|& +1 && -1\\ +1&& -1&&  -1 && -1 &|& -1 &&  -1 &&(\times 2 \text{ for symmetry by swapping }R-L\text{ and }U-D)\\ \\ +1&& +1 &&+1 &&-1&& -1& |& +1\\ +1&& +1 &&-1&&  -1&& -1 &|& -1&& (\times 2\text{ symmetry})\\ \\ +1&& +1 &&+1&& -1&& -1 &&-1&| & (\times2 \text{ symmetry})\\ \end{array}\] Note that to account for the cases when $x=-y$ , we can simply multiply the $U-D$ steps by $-1$ , so for example, the first row would become \[+1 \qquad+1\qquad +1 \ \ \ \ |\ \ \  -1\qquad -1\qquad -1.\] Therefore, we must multiply the number of possibilities in each case by $2$ , except for when $x=y=0$
Now, we compute the number of possibilities for each case. In particular, we must compute the number of $RLUD$ words, where $R$ represents $+1$ to the left of $|$ $L$ represents $-1$ to the left of $|$ $U$ represents $+1$ to the right of $|$ , and $D$ represents $-1$ to the right of $|$ . Using multinomials, we compute the following numbers of possibilities for each case. \[{6\choose 3}\cdot  2+ \frac{6!}{2!2!}\cdot 2 + \frac{6!}{2!2!} \cdot 2 + {6\choose 3} \cdot 2 = 2(20 + 180 + 180 + 20) = 800\] \[\frac{6!}{3!2!}\cdot 2 + \frac{6!}{2!2!} + \frac{6!}{3!2!}\cdot 2 = 120 + 180 + 120 = 420\  (\times2\text{ for symmetry})\] \[\frac{6!}{3!2!} \cdot 2 + \frac{6!}{3!2!} \cdot 2 = 120 + 120 = 240\  (\times2\text{ for symmetry})\] \[{6\choose 3} = 20\ (\times 2\text{ for symmetry})\]
Thus, there are $800 + 840 + 480 + 40 = 2160$ possibilities where $|x|=|y|$ . Because there are $4^6$ total possibilities, the probability is $\frac{2160}{4^6} = \frac{135}{256}$ , so the answer is $\boxed{391}.$

Solution: Denote $(x, y)_n$ the probability that starting from point $(x, y)$ , the token reaches a point on the graph of $|y| = |x|$ in exactly $n$ moves. The problem asks us to find $(0, 0)_6$ . We start by breaking this down: \[(0, 0)_6 = \frac14 \cdot ((0, 1)_5 + (0, -1)_5 + (1, 0)_5 + (-1, 0)_5)\] Notice that by symmetry, $(0, 1)_5 = (0, -1)_5 = (1, 0)_5 = (-1, 0)_5$ , so the equation simplifies to \[(0, 0)_6 = (0, 1)_5\] We now expand $(0, 1)_5$ \[(0, 1)_5 = \frac14 \cdot ((0, 0)_4 + (0, 2)_4 + 2(1, 1)_4)\] First, we find $(0, 0)_4$ \[(0, 0)_4 = (0, 1)_3\] \[(0, 1)_3 = \frac14 \cdot ((0, 0)_2 + (0, 2)_2 + 2(1, 1)_2)\] At this point, we can just count the possibilities to find $(0, 0)_2 = \frac34$ $(0, 2)_2 = \frac{7}{16}$ , and $(1, 1)_2 = \frac58$ . Therefore, \[(0, 1)_3 = \frac14 \cdot (\frac34 + \frac{7}{16} + 2 \cdot \frac58)\] \[(0, 1)_3 = \frac{39}{64}\] Next, we find $(0, 2)_4$ \[(0, 2)_4 = \frac14 \cdot ((0, 1)_3 + (0, 3)_3 + 2(1, 2)_3)\] We already calculated $(0, 1)_3$ , so we just need to find $(0, 3)_3$ and $(1, 2)_3$ \[(0, 3)_3 = \frac14 \cdot ((0, 2)_2 + (0, 4)_2 + 2(1, 3)_2)\] \[(0, 3)_3 = \frac14 \cdot (\frac{7}{16} + 0 + 2 \cdot \frac{1}{4})\] \[(0, 3)_3 = \frac{15}{64}\] \[(1, 2)_3 = \frac14 \cdot ((1, 3)_2 + (1, 1)_2 + (0, 2)_2 + (2, 2)_2)\] \[(1, 2)_3 = \frac14 \cdot (\frac14 + \frac58 + \frac{7}{16} + \frac12)\] \[(1, 2)_3 = \frac{29}{64}\] Therefore, \[(0, 2)_4 = \frac14 \cdot (\frac{39}{64} + \frac{15}{64} + 2 \cdot \frac{29}{64})\] \[(0, 2)_4 = \frac{7}{16}\] Finally, we find $(1, 1)_4$ \[(1, 1)_4 = \frac12 \cdot ((0, 1)_3 + (1, 2)_3)\] \[(1, 1)_4 = \frac12 \cdot (\frac{39}{64} + \frac{29}{64})\] \[(1, 1)_4 = \frac{17}{32}\] Putting it all together, \[(0, 0)_6 = (0, 1)_5 =\frac14 \cdot (\frac{39}{64} + \frac{7}{16} + 2 \cdot \frac{17}{32})\] \[(0, 0)_6 = \frac{135}{256}\] Thus, the answer is $135 + 256 = \boxed{391}$


Answer: 600.0

A train, an hour after starting, meets with an accident which detains it a half hour, after which it proceeds at $\frac{3}{4}$ of its former rate and arrives $3\tfrac{1}{2}$ hours late. Had the accident happened $90$ miles farther along the line, it would have arrived only $3$ hours late. The length of the trip in miles was:
$\textbf{(A)}\ 400 \qquad \textbf{(B)}\ 465 \qquad \textbf{(C)}\ 600 \qquad \textbf{(D)}\ 640 \qquad \textbf{(E)}\ 550$

Solution:
Let the speed of the train be $x$ miles per hour, and let $D$ miles be the total distance of the trip, where $x$ and $D$ are unit-less quantities. Then for the trip that actually occurred, the train travelled 1 hour before the crash, and then travelled $D-x$ miles after the crash. In other words, the train travelled for $\frac{(D-x)\text{mi}}{\frac{3x}{4}\frac{\text{mi}}{\text{hr}}}=\frac{D-x}{3x/4}\text{hr}$ after the crash. So, not including the half hour detention after the crash, the entire trip took \[\left(1+\frac{D-x}{3x/4}\right)\text{hours}.\] Now let us consider the alternative trip, where the accident happened $90$ miles farther along the line. Originally, the accident happened $x\frac{\text{mi}}{\text{hr}}\cdot(1\text{ hr})=x\text{ mi}$ down the line, but in this situation it happens $x+90$ miles from the line. Therefore the accident happens $\frac{(x+90)\text{ mi}}{x\frac{\text{mi}}{\text{hr}}}=\frac{x+90}{x}\text{hr}$ into the trip. The length of the remaining part of the trip is now $D-x-90$ , so the remaining trip takes $\frac{(D-x-90)\text{mi}}{\frac{3x}{4}\frac{\text{mi}}{\text{hr}}}=\frac{D-x-90}{3x/4}\text{hr}$ . So, not including the half hour detention after the crash, the entire trip took \[\left(\frac{x+90}{x}+\frac{D-x-90}{3x/4}\right)\text{hours}.\] We are given that the trip that actually happened resulted in being $3.5$ hours late, while the alternative trip would have resulted in being only $3$ hours late. Therefore the first trip took exactly $\frac{1}{2}$ hour more: \[1+\frac{D-x}{3x/4}=\frac{x+90}{x}+\frac{D-x-90}{3x/4}+\frac{1}{2}.\] After simplification and cancellation, we get the equation \[0=\frac{90}{x}-\frac{90}{3x/4}+\frac{1}{2}.\] Therefore $(90/x)+(1/2)=(120/x)$ , so $1/2=30/x$ . We solve for $x$ and have the original speed of the train: $x=60$
We must now solve for the length of the actual trip. If the train had gone $60\frac{\text{mi}}{\text{hr}}$ for the entire trip, then it would have taken $\frac{D}{60}\text{hr}$ , and the train would have been on time. But the trip actually took $1+\frac{D-60}{45}$ hours, not including the half-hour detention, and the train was $3.5$ hours late. Therefore \[\frac{D}{60}+3.5=1+\frac{D-60}{45}+0.5\Rightarrow 3D+630=180+4D-240+90\Rightarrow D=600.\] The length of the trip in miles was $\boxed{600}$


Answer: 314.0

A transformation of the first quadrant of the coordinate plane maps each point $(x,y)$ to the point $(\sqrt{x},\sqrt{y}).$ The vertices of quadrilateral $ABCD$ are $A=(900,300), B=(1800,600), C=(600,1800),$ and $D=(300,900).$ Let $k_{}$ be the area of the region enclosed by the image of quadrilateral $ABCD.$ Find the greatest integer that does not exceed $k_{}.$

Solution:
\begin{eqnarray*}A' = & (\sqrt {900}, \sqrt {300})\\ B' = & (\sqrt {1800}, \sqrt {600})\\ C' = & (\sqrt {600}, \sqrt {1800})\\ D' = & (\sqrt {300}, \sqrt {900}) \end{eqnarray*}
First we see that lines passing through $AB$ and $CD$ have equations $y = \frac {1}{3}x$ and $y = 3x$ , respectively. Looking at the points above, we see the equations for $A'B'$ and $C'D'$ are $y^2 = \frac {1}{3}x^2$ and $y^2 = 3x^2$ , or, after manipulation $y = \frac {x}{\sqrt {3}}$ and $y = \sqrt {3}x$ , respectively, which are still linear functions. Basically the square of the image points gives back the original points and we could plug them back into the original equation to get the equation of the image lines.
Now take a look at $BC$ and $AD$ , which have the equations $y = - x + 2400$ and $y = - x + 1200$ . The image equations hence are $x^2 + y^2 = 2400$ and $x^2 + y^2 = 1200$ , respectively, which are the equations for circles
1999 AIME-6.png
To find the area between the circles (actually, parts of the circles), we need to figure out the angle of the arc . This could be done by $\arctan \sqrt {3} - \arctan \frac {1}{\sqrt {3}} = 60^\circ - 30^\circ = 30^\circ$ . So the requested areas are the area of the enclosed part of the smaller circle subtracted from the area enclosed by the part of the larger circle = $\frac {30^\circ}{360^\circ}(R^2\pi - r^2\pi) = \frac {1}{12}(2400\pi - 1200\pi) = 100\pi$ . Hence the answer is $\boxed{314}$


Answer: 63.0

A trapezoid has side lengths 3, 5, 7, and 11. The sum of all the possible areas of the trapezoid can be written in the form of $r_1\sqrt{n_1}+r_2\sqrt{n_2}+r_3$ , where $r_1$ $r_2$ , and $r_3$ are rational numbers and $n_1$ and $n_2$ are positive integers not divisible by the square of any prime. What is the greatest integer less than or equal to $r_1+r_2+r_3+n_1+n_2$
$\textbf{(A)}\ 57\qquad\textbf{(B)}\ 59\qquad\textbf{(C)}\ 61\qquad\textbf{(D)}\ 63\qquad\textbf{(E)}\ 65$

Solution:
Name the trapezoid $ABCD$ , where $AB$ is parallel to $CD$ $AB<CD$ , and $AD<BC$ . Draw a line through $B$ parallel to $AD$ , crossing the side $CD$ at $E$ . Then $BE=AD$ $EC=DC-DE=DC-AB$ . One needs to guarantee that $BE+EC>BC$ , so there are only three possible trapezoids:
\[AB=3, BC=7, CD=11, DA=5, CE=8\] \[AB=5, BC=7, CD=11, DA=3, CE=6\] \[AB=7, BC=5, CD=11, DA=3, CE=4\]
In the first case, by Law of Cosines, $\cos(\angle BCD) = (8^2+7^2-5^2)/(2\cdot 7\cdot 8) = 11/14$ , so $\sin (\angle BCD) = \sqrt{1-121/196} = 5\sqrt{3}/14$ . Therefore the area of this trapezoid is $\frac{1}{2} (3+11) \cdot 7 \cdot 5\sqrt{3}/14 = \frac{35}{2}\sqrt{3}$
In the second case, $\cos(\angle BCD) = (6^2+7^2-3^2)/(2\cdot 6\cdot 7) = 19/21$ , so $\sin (\angle BCD) = \sqrt{1-361/441} = 4\sqrt{5}/21$ . Therefore the area of this trapezoid is $\frac{1}{2} (5+11) \cdot 7 \cdot 4\sqrt{5}/21 =\frac{32}{3}\sqrt{5}$
In the third case, $\angle BCD = 90^{\circ}$ , therefore the area of this trapezoid is $\frac{1}{2} (7+11) \cdot 3 = 27$
So $r_1 + r_2 + r_3 + n_1 + n_2 = 17.5 + 10.666... + 27 + 3 + 5$ , which rounds down to $\boxed{63}$

Solution:
Let the area of the trapezoid be $S$ , the area of the triangle be $S_1$ , the area of the parallelogram be $S_2$
By Heron's Formula $S_1 = \sqrt{\frac{b+c+d-a}{2} \cdot \frac{c+d-a-b}{2} \cdot \frac{a+b+d-c}{2} \cdot \frac{b+c-a-d}{2}}$
$S_2 = \frac{S_1 \cdot 2}{c-a} \cdot a = \frac{2aS_1}{c-a}$
$S = S_1 + S_2 = S_1(1+\frac{2a}{c-a}) = S_1 \cdot \frac{c+a}{c-a} = \frac14 \cdot \frac{c+a}{c-a} \cdot \sqrt{(b+c+d-a)(c+d-a-b)(a+b+d-c)(b+c-a-d)}$
If $a = 3$ $b = 7$ $c = 11$ $d = 5$
$S = \frac14 \cdot \frac{14}{8} \cdot \sqrt{(7+11+5-3)(11+5-3-7)(3+7+5-11)(7+11-3-5)} = \frac{35\sqrt{3}}{2}$
If $a = 3$ $b = 11$ $c = 5$ $d = 7$
$S = \frac14 \cdot \frac{8}{2} \cdot \sqrt{(11+5+7-3)(5+7-3-11)(3+5+7-11)(11+5-3-7)}$ , which is impossible as $5+7-3-11 = -2$
If $a = 3$ $b = 5$ $c = 7$ $d = 11$
$S = \frac14 \cdot \frac{10}{4} \cdot \sqrt{(5+7+11-3)(7+11-3-5)(3+5+11-7)(5+7-11-3)}$ , which is impossible as $5+7-11-3 = -2$
If $a = 5$ $b = 11$ $c = 7$ $d = 3$
$S = \frac14 \cdot \frac{12}{2} \cdot \sqrt{(11+7+3-5)(7+3-5-11)(5+11+3-7)(11+7-5-3)}$ , which is impossible as $7+3-5-11 = -6$
If $a = 5$ $b = 3$ $c = 11$ $d = 7$
$S = \frac14 \cdot \frac{16}{6} \cdot \sqrt{(3+11+7-5)(11+7-5-3)(5+3+7-11)(3+11-5-7)} = \frac{32\sqrt{5}}{3}$
If $a = 7$ $b = 3$ $c = 11$ $d = 5$
$S = \frac14 \cdot \frac{18}{4} \cdot \sqrt{(3+11+5-7)(11+5-7-)(7+5+5-11)(3+11-7-5)} = 27$
Thus the answer is $\frac{35}{2} + \frac{32}{3} + 27 + 3 + 5$ , which rounds down to $\boxed{63}$


Answer: 23.0

A triangle has area $30$ , one side of length $10$ , and the median to that side of length $9$ .  Let $\theta$ be the acute angle formed by that side and the median.  What is $\sin{\theta}$
$\textbf{(A)}\ \frac{3}{10}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{9}{20}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{9}{10}$

Solution: $AB$ is the side of length $10$ , and $CD$ is the median of length $9$ . The altitude of $C$ to $AB$ is $6$ because the 0.5(altitude)(base)=Area of the triangle. $\theta$ is $\angle CDE$ . To find $\sin{\theta}$ , just use opposite over hypotenuse with the right triangle $\triangle DCE$ . This is equal to $\frac69=\boxed{23}$


Answer: 32.0

A triangle has side lengths $10$ $10$ , and $12$ . A rectangle has width $4$ and area equal to the
area of the triangle. What is the perimeter of this rectangle?
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 36$

Solution:
The triangle is isosceles. The height of the triangle is therefore given by $h = \sqrt{10^2 - \left( \dfrac{12}{2} \right)^2} =  \sqrt{64} = 8$
Now, the area of the triangle is $\dfrac{bh}{2} = \dfrac{12*8}{2} = \dfrac{96}{2} = 48$
We have that the area of the rectangle is the same as the area of the triangle, namely $48$ . We also have the width of the rectangle: $4$
The length of the rectangle therefore is: $l = \dfrac{48}{4} = 12$
The perimeter of the rectangle then becomes: $2l + 2w = 2*12 + 2*4 = 32$
The answer is:
$\boxed{32}$

Solution: An alternative way to find the area of the triangle is by using Heron's formula, $A=\sqrt{(s)(s-a)(s-b)(s-c)}$ where $s$ is the semi-perimeter of the triangle (meaning half the perimeter). Here, the semi-perimeter is $(10+10+12)/2 = 16$ . Thus the area equals $\sqrt{(16)(16-10)(16-10)(16-12)} = \sqrt{16*6*6*4} = 48.$ We know that the width of the rectangle is $4$ , so $48/4 = 12$ , which is the length. The perimeter of the rectangle is $2(4+12) =$ $\boxed{32}$

Solution:
Note that a triangle with side lengths $10,10,12$ is essentially $2$ “6,8,10” right triangles stuck together. Hence, the height is $8$ , and our area is $48$
So, the length of the rectangle is $\frac{48}{4}=12$ , and our perimeter $P=2(4+12)=\boxed{32}$