Answer: 4.0

The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?
[asy]size(170);defaultpen(linewidth(0.9)+fontsize(13pt));draw(unitcircle^^circle((0,1.5),0.5)); path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle; for(int i=1;i<=12;i=i+1){draw(0.9*dir(90-30*i)--dir(90-30*i));label("$"+(string) i+"$",0.78*dir(90-30*i));} dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]
$\textbf{(A) }\text{2 o' clock} \qquad\textbf{(B) }\text{3 o' clock} \qquad\textbf{(C) }\text{4 o' clock} \qquad\textbf{(D) }\text{6 o' clock} \qquad\textbf{(E) }\text{8 o' clock}$

Solution: The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is $\boxed{4}$ . Similarly, the arrow would be pointing downward at 6:00. It would already have completed three 180 degree turns. Therefore, two 180 degree turns would be completed at 4:00.

Solution: The rotation factor of the arrow is the sum of the rates of the regular rotation of the arrow (360° every 360° rotation = 1) and the rotation of the disk around the clock with twice the circumference (360° every 180° = 2). Thus, the rotation factor of the arrow is 3, and so our answer corresponds to 360°/3 = 120°, which is 4 o' clock. $\boxed{4}$

Solution: The arrow travels a path of radius 30 (20 from the interior clock and 10 from the radius of the disk itself). We note that 1 complete rotation of 360 degrees is needed for the arrow to appear up again, so, therefore, the disk must travel its circumference before the arrow goes up. Its circumference is $20\pi$ , so that is $20\pi$ traveled on a $60\pi$ arrow path. This is a ratio of 1/3, so the angle it carves is 120 degrees, which leads us to the correct answer of  4 o' clock. $\boxed{4}$

Solution: Suppose that the small disk also had a clock face on it, and that both disks were toothed wheels, free to rotate around their centers. The part of the picture where they engage would look like this:  The small cog has half the radius, and therefore half the circumference. If the large cog turns $30^\circ$ anticlockwise (i.e. 1 hour), the small cog turns $60^\circ$ clockwise (i.e. 2 hours).  However, in the original problem the large cog does not rotate; it stays where it is. Therefore we must turn the whole diagram above $30^\circ$ clockwise to see what happens when the small cog rolls around it.  It turns out that, when the point of tangency moves $30^\circ$ clockwise (one hour), from our point of view the small disk rotates $90^\circ$ clockwise (three hours) around its center. Thus, for the small disk to perform a complete rotation of $360^\circ$ (twelve hours) around its center from our point of view, the point of tangency must move round four hours. So the answer is $\boxed{4}$

Solution: We can approach this problem with angle measures. As the circumference of the disk is $10\pi,$ and the clock $20 \pi,$ we have that 30 degrees, or the angular measure between hours, of the disk is only 15 degrees of the clock. This yields that every two hour ticks that the clock rotates, on the third one, the ticks will meet. However, the disk must rotate 360 degrees in order to come back to its original position, so the angular measure that the disk has covered relative to the clock is simply $12 \cdot 15 \cdot \frac{2}{3},$ or $120^\circ$ from the 12 starting point, so $\boxed{4}$

Solution:
If the big clock were a flat plane, then the smaller clock could travel $\dfrac{40\pi}{20\pi}=2$ full revolutions.
But we also need to account for rotation. If we mark a red dot on the bottom of the small clock/bottom of the arrow, and then drag it around the clock, the direction of the arrow would still change. After traveling around the big clock, the small clock would travel $1$ full rotation.
Considering these two movements, the small clock travels 3 full rotations around the big clock so the arrow is next pointing upwards at $\dfrac{12 \text{ o'clock}}{3}=\boxed{4}$

Solution: The center of rotation is in the center of the smaller circle, but extends to the center of the larger circle. That means the circumference of the circle in relation to the arrow is $60 \pi$ . The other circle is $20 \pi$ and so that is $\frac{1}{3}$ . So $\frac{12}{3} = 4$ which is $\boxed{4}$

Solution: Let the center of the disk be Planet X with orbit eccentricity $0$ and the center of the clock be the Sun. Note that the question would then be asking for the solar day of Planet X, rather than the sidereal day. Because planet X is rotating around its axis in the same direction as it is revolving around the Sun, the solar day can be calculated as $d_{solar}=\frac{1}{\frac{1}{y}+\frac{1}{d_{siderial}}}$ , where $y$ is the length of the year of Planet X and $d_{siderial}$ is one sidereal day. It is easy to see that $y=12$ and $d_{siderial}=6$ , therefore the answer is $\frac{1}{\frac{1}{12}+\frac{1}{6}}=\frac{1}{\frac{1}{4}}=\boxed{4}$ ~sigmapie


Answer: 64.0

Solution: In a $1$ -foot-by- $1$ -foot floor, there is $1$ white tile. In a $3$ -by- $3$ , there are $4$ . Continuing on, you can deduce the $n^{th}$ positive odd integer floor has $n^2$ white tiles. $15$ is the $8^{th}$ odd integer, so there are $\boxed{64}$ white tiles.


Answer: 260.0

Solution:
Call the squares' side lengths from smallest to largest $a_1,\ldots,a_9$ , and let $l,w$ represent the dimensions of the rectangle.
The picture shows that \begin{align*} a_1+a_2 &= a_3\\ a_1 + a_3 &= a_4\\ a_3 + a_4 &= a_5\\ a_4 + a_5 &= a_6\\ a_2 + a_3 + a_5 &= a_7\\ a_2 + a_7 &= a_8\\ a_1 + a_4 + a_6 &= a_9\\ a_6 + a_9 &= a_7 + a_8.\end{align*}
Expressing all terms 3 to 9 in terms of $a_1$ and $a_2$ and substituting their expanded forms into the previous equation will give the expression $5a_1 = 2a_2$
We can guess that $a_1 = 2$ . (If we started with $a_1$ odd, the resulting sides would not be integers and we would need to scale up by a factor of $2$ to make them integers; if we started with $a_1 > 2$ even, the resulting dimensions would not be relatively prime and we would need to scale down.) Then solving gives $a_9 = 36$ $a_6=25$ $a_8 = 33$ , which gives us $l=61,w=69$ .  These numbers are relatively prime, as desired. The perimeter is $2(61)+2(69)=\boxed{260}$

Solution: We can just list the equations: \begin{align*} s_3 &= s_1 + s_2 \\ s_4 &= s_3 + s_1 \\ s_5 &= s_4 + s_3 \\ s_6 &= s_5 + s_4 \\ s_7 &= s_5 + s_3 + s_2 \\ s_8 &= s_7 + s_2 \\ s_9 &= s_8 + s_2 - s_1 \\ s_9 + s_8 &= s_7 + s_6 + s_5 \end{align*} We can then write each $s_i$ in terms of $s_1$ and $s_2$ as follows \begin{align*} s_4 &= 2s_1 + s_2 \\ s_5 &= 3s_1 +2s_2 \\ s_6 &= 5s_1 + 3s_2 \\ s_7 &= 4s_1 + 4s_2 \\ s_8 &= 4s_1 + 5s_2 \\ s_9 &= 3s_1 + 6s_2 \\ \end{align*} Since $s_9 + s_8 = s_7 + s_6 + s_5 \implies (3s_1 + 6s_2) + (4s_1 + 5s_2) = (4s_1 + 4s_2) + (5s_1 + 3s_2) + (3s_1 + 2s_2),$ \[2s_2 = 5s_1 \implies \frac{2}{5}s_2 = s_1.\] Since the side lengths of the rectangle are relatively prime, we can see that $s_1 = 2$ and $s_2 = 5.$ Therefore, $2(2s_9 + s_6 + s_8) = 30s_1 + 40s_2 = \boxed{260}.$ ~peelybonehead

Solution: We set the side length of the smallest square to 1, and set the side length of square $a_4$ in the previous question to a. We do some "side length chasing" and get $4a - 4 = 2a + 5$ . Solving, we get $a = 4.5$ and the side lengths are $61$ and $69$ . Thus, the perimeter of the rectangle is $2(61 + 69) = \boxed{260}.$


Answer: 10.3

Solution:
Clearly the arrow marks a value between $10.25$ and $10.5$ , so only $\text{C}$ and $\text{D}$ are possible.
Looking, we see that the arrow is closer to $10.3$ , so $\boxed{10.3}$


Answer: 15.0

Solution: After 4 hours, we see that Bjorn biked 45 miles, and Alberto biked 60.  Thus the answer is $60-45=15$ $\boxed{15}$

Solution: We see that each dot is $15$ units away from the nearest one above it. So the answer is $\boxed{15}$


Answer: 154.0

The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as $\dfrac{1}{2}(\sqrt{p}-q)$ where $p$ and $q$ are positive integers. Find $p+q$

Solution:
Let the radius of the circles be $r$ . The longer dimension of the rectangle can be written as $14r$ , and by the Pythagorean Theorem , we find that the shorter dimension is $2r\left(\sqrt{3}+1\right)$
Therefore, $\frac{14r}{2r\left(\sqrt{3}+1\right)}= \frac{7}{\sqrt{3} + 1} \cdot \left[\frac{\sqrt{3}-1}{\sqrt{3}-1}\right] = \frac{1}{2}\left(7\sqrt{3} - 7\right) = \frac{1}{2}\left(\sqrt{p}-q\right)$ . Thus we have $p=147$ and $q=7$ , so $p+q=\boxed{154}$

Solution: Since we only care about the ratio between the longer side and shorter side, we can set the longer side to $14$ . So, this means that each of the radii is $1$ . Now, we connect the radii of three circles such that they form an equilateral triangle with side length 4. Obviously, the height this triangle is $2\sqrt{3}$ , and the shorter side if the triangle is therefore $2\sqrt{3}+2$ and we use simplification similar to as showed above, and we reach the result $\frac{1}{2} \cdot (\sqrt{147}-7)$ and the final answer is $147+7 = \boxed{154}$


Answer: 99.0

The difference between a two-digit number and the number obtained by reversing its digits is $5$ times the sum of the digits of either number.  What is the sum of the two digit number and its reverse?
$\textbf{(A) }44\qquad \textbf{(B) }55\qquad \textbf{(C) }77\qquad \textbf{(D) }99\qquad \textbf{(E) }110$

Solution: Let the two digits be $a$ and $b$ . Then, $5a + 5b = 10a + b - 10b - a = 9a - 9b$ , or $2a = 7b$ . This yields $a = 7$ and $b = 2$ because $a, b < 10$ . Then, $72 + 27 = \boxed{99}.$

Solution: We start like above. Let the two digits be $a$ and $b$ . Therefore, $5(a+b) = 10a+b-10b-a=9(a-b)$ . Since we are looking for $10a+b+10b+a=11(a+b)$ and we know that $a+b$ must be a multiple of $9$ , the only answer choice that works is $\boxed{99}.$


Answer: 85.0

The difference of the roots of $x^2-7x-9=0$ is:
$\textbf{(A) \ }+7  \qquad \textbf{(B) \ }+\frac{7}{2} \qquad \textbf{(C) \ }+9 \qquad \textbf{(D) \ }2\sqrt{85} \qquad \textbf{(E) \ }\sqrt{85}$

Solution: Denote the $2$ roots of this quadratic as $r_1$ and $r_2$ . Note that $(r_1-r_2)^2=(r_1+r_2)^2-4r_1r_2$ . By Vieta's Formula's $r_1+r_2=7$ , and $r_1r_2=-9$ . Thus, $r_1-r_2=\sqrt{49+4\cdot 9}=\boxed{85}$


Answer: 1.0

The digit-sum of $998$ is $9+9+8=26$ .  How many 3-digit whole numbers, whose digit-sum is $26$ , are even?
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$

Solution: The highest digit sum  for three-digit numbers is $9+9+9=27$ . Therefore, the only possible digit combination is $9, 9, 8$ . Of course, of the three possible numbers, only $998$ works. Thus, the answer is $\boxed{1}$


Answer: 1.0

The digits $1$ $2$ $3$ $4$ , and $5$ are each used once to write a five-digit number $PQRST$ . The three-digit number $PQR$ is divisible by $4$ , the three-digit number $QRS$ is divisible by $5$ , and the three-digit number $RST$ is divisible by $3$ . What is $P$
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution: We see that since $QRS$ is divisible by $5$ $S$ must equal either $0$ or $5$ , but it cannot equal $0$ , so $S=5$ . We notice that since $PQR$ must be even, $R$ must be either $2$ or $4$ . However, when $R=2$ , we see that $T \equiv 2 \pmod{3}$ , which cannot happen because $2$ and $5$ are already used up; so $R=4$ . This gives $T \equiv 3 \pmod{4}$ , meaning $T=3$ . Now, we see that $Q$ could be either $1$ or $2$ , but $14$ is not divisible by $4$ , but $24$ is. This means that $Q=2$ and $P=\boxed{1}$

Solution: We know that out of $PQRST,$ $QRS$ is divisible by $5$ . Therefore $S$ is obviously 5 because $QRS$ is divisible by 5. So we now have $PQR5T$ as our number. Next, let's move on to the second piece of information that was given to us. $RST$ is divisible by 3. So, according to the divisibility by 3 rule, the sum of $RST$ has to be a multiple of 3. The only 2 big enough are 9 and 12 and since 5 is already given. The possible sums of $RT$ are 4 and 7. So, the possible values for $R$ are 1,3,4,3 and the possible values of $T$ are 3,1,3,4. So, using this we can move on to the fact that $PQR$ is divisible by 4. So, using that we know that $R$ has to be even so 4 is the only possible value for $R$ . Using that we also know that 3 is the only possible value for 3. So, we have $PQRST$ $PQ453$ so the possible values are 1 and 2 for $P$ and $Q$ . Using the divisibility rule of 4 we know that $QR$ has to be divisible by 4. So, either 14 or 24 are the possibilities, and 24 is divisible by 4. So the only value left for $P$ is 1. $P=\boxed{1}$

Solution: We know that $QRS$ is divisible by $5$ , so $S$ would be either $5$ or $0$ . However, $0$ is not a choice, so $S=5$ . Also, $PQR$ is divisible by $4$ , so this means that $QR$ is $12$ $32$ $24$ , or $52$ . If $R=2$ , then $T$ has to be $2$ or $5$ $RST$ is divisible by $3$ ), but both are taken. So, $R=4 \Rightarrow QR=24$ $R+S+T$ must equal $9$ or $12$ , but because $4+5=9$ $R+S+T=12 \Rightarrow T=3$ . This leaves $P=\boxed{1}$

Solution: We can simply try each of the answer choice, and we will see which one works. Trying $P=\boxed{1}$


Answer: 9.0

The digits $2,0,2,$ and $3$ are placed in the expression below, one digit per box. What is the maximum possible value of the expression?
[asy] // Diagram by TheMathGuyd. I can compress this later size(5cm); real w=2.2; pair O,I,J; O=(0,0);I=(1,0);J=(0,1); path bsqb = O--I; path bsqr = I--I+J; path bsqt = I+J--J; path bsql = J--O; path lsqb = shift((1.2,0.75))*scale(0.5)*bsqb; path lsqr = shift((1.2,0.75))*scale(0.5)*bsqr; path lsqt = shift((1.2,0.75))*scale(0.5)*bsqt; path lsql = shift((1.2,0.75))*scale(0.5)*bsql; draw(bsqb,dashed); draw(bsqr,dashed); draw(bsqt,dashed); draw(bsql,dashed); draw(lsqb,dashed); draw(lsqr,dashed); draw(lsqt,dashed); draw(lsql,dashed); label(scale(3)*"$\times$",(w,1/3)); draw(shift(1.3w,0)*bsqb,dashed); draw(shift(1.3w,0)*bsqr,dashed); draw(shift(1.3w,0)*bsqt,dashed); draw(shift(1.3w,0)*bsql,dashed); draw(shift(1.3w,0)*lsqb,dashed); draw(shift(1.3w,0)*lsqr,dashed); draw(shift(1.3w,0)*lsqt,dashed); draw(shift(1.3w,0)*lsql,dashed); [/asy]
$\textbf{(A) }0 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }16 \qquad \textbf{(E) }18$

Solution: First, let us consider the case where $0$ is a base: This would result in the entire expression being $0.$ Contrastingly, if $0$ is an exponent, we will get a value greater than $0.$ $3^2\times2^0=9$ is greater than $2^3\times2^0=8$ and $2^2\times3^0=4.$ Therefore, the answer is $\boxed{9}.$

Solution: The maximum possible value of using the digits $2,0,2,$ and $3$ : We can maximize our value by keeping the $3$ and $2$ together in one power (the biggest with the biggest and the smallest with the smallest). This shows $3^{2}\times2^{0}=9\times1=9.$ (We don't want $0^{2}$ because that is $0$ .) It is going to be $\boxed{9}.$

Solution: Trying all $12$ distinct orderings, we see that the only possible values are $0,4,8,$ and $9,$ the greatest of which is $\boxed{9}.$


Answer: 9.0

The digits 1, 2, 3, 4 and 9 are each used once to form the smallest possible even five-digit number. The digit in the tens place is
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 9$

Solution: Since the number is even, the last digit must be $2$ or $4$ . To make the smallest possible number, the ten-thousands digit must be as small as possible, so the ten-thousands digit is $1$ . Simillarly, the thousands digit has second priority, so it must also be as small as possible once the ten-thousands digit is decided, so the thousands digit is $2$ . Similarly, the hundreds digit needs to be the next smallest number, so it is $3$ . However, for the tens digit, we can't use $4$ , since we already used $2$ and the number must be even, so the units digit must be $4$ and the tens digit is $9, \boxed{9}$ (The number is $12394$ .)


Answer: 217.0

The digits of a positive integer $n$ are four consecutive integers in decreasing order when read from left to right. What is the sum of the possible remainders when $n$ is divided by $37$

Solution:
A brute-force solution to this question is fairly quick, but we'll try something slightly more clever:  our numbers have the form ${\underline{(n+3)}}\,{\underline{(n+2)}}\,{\underline{( n+1)}}\,{\underline {(n)}}$ $= 1000(n + 3) + 100(n + 2) + 10(n + 1) + n = 3210 + 1111n$ , for $n \in \lbrace0, 1, 2, 3, 4, 5, 6\rbrace$
Now, note that $3\cdot 37 = 111$ so $30 \cdot 37 = 1110$ , and $90 \cdot 37 = 3330$ so $87 \cdot 37 = 3219$ . So the remainders are all congruent to $n - 9 \pmod{37}$ . However, these numbers are negative for our choices of $n$ , so in fact the remainders must equal $n + 28$
Adding these numbers up, we get $(0 + 1 + 2 + 3 + 4 + 5 + 6) + 7\cdot28 = \boxed{217}$


Answer: 0.0

The dimensions of a rectangle $R$ are $a$ and $b$ $a < b$ . It is required to obtain a rectangle with dimensions $x$ and $y$ $x < a, y < a$ ,
so that its perimeter is one-third that of $R$ , and its area is one-third that of $R$ . The number of such (different) rectangles is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ \infty$

Solution: Using the perimeter and area formulas, \[2(x+y) = \frac{2}{3}(a+b)\] \[x+y = \frac{a+b}{3}\] \[xy = \frac{ab}{3}\] Dividing the second equation by the last equation results in \[\frac1y + \frac1x = \frac1b + \frac1a\] Since $x,y < a$ $\tfrac1a < \tfrac1x, \tfrac1y$ .  Since $a < b$ $\tfrac1b < \tfrac1a$ .  That means \[\tfrac1x + \tfrac1y > \tfrac1a + \tfrac1a > \tfrac1a + \tfrac1b\] This is a contradiction, so there are $\boxed{0}$ rectangles that satisfy the conditions.


Answer: 38.0

The dimensions of a rectangular box in inches are all positive integers and the volume of the box is $2002$ in $^3$ . Find the minimum possible sum of the three dimensions.
$\text{(A) }36 \qquad \text{(B) }38  \qquad \text{(C) }42 \qquad \text{(D) }44 \qquad \text{(E) }92$

Solution: Given an arbitrary product and an arbitrary amount of terms to multiply to get that product, to maximize the sum, make all of the terms $1$ with the last one being the number. To minimize the sum, make all of the terms equal to each other. (This is a corollary that follows from the $AM-GM$ proof.) Since $2002$ is not a perfect cube, we have to make the terms as close to each other as possible. A good rule of thumb is the memorize the prime factorization of the AMC year that you are doing, which is $2002= 2 \cdot 7 \cdot 11 \cdot 13$ . The three terms that are closest to each other that multiply to $2002$ are $11, 13,$ and $14$ , so our answer is $11+13+14=\boxed{38}$


Answer: 294.0

The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.

Solution: If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$ .  If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$ .  Thus, if the number of columns is $n$ , the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$ .  In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$ , so this number works and no larger number can.  Thus, the answer is $\boxed{294}$

Solution:
Let there be $m$ members and $n$ members for the square and $c$ for the number of columns of the other formation. We have $n^2 +5 = c(c+7) \implies n^2+5 = \left(c+\frac{7}{2}\right)^2 -\frac{49}{4} \implies n^2 - \left(c+\frac{7}{2}\right)^2 = -\frac{69}{4} \implies \left(n-c-\frac{7}{2}\right)\left(n + c +\frac{7}{2}\right) \implies (2n-2c-7)(2n+2n+7) = -69.$
To maximize this we let $2n+2c+7 = 68$ and $2n-2c-7 = 1.$ Solving we find $n = 17$ so the desired number of members is $17^2 + 5 = \boxed{294}.$

Solution: Think of the process of moving people from the last column to new rows. Since there are less columns than rows, for each column removed, there are people discarded to the "extra" pile to be placed at the end. To maximize the number of "extra" people to fill in the last few rows. We remove 3 columns and add 4 rows. For the first new row, one more person will be discarded. For the second, 1 extra person are added since there is one more row now, and there are 2 less columns. Thus, there are 3 extra people discarded. Similarly, 5 extra people are discarded for the third column. Now there are $5+1+3+5=14$ people in the extra pile to put as the last row, so there are $14(14+7)=\boxed{294}$ people.

Solution:
Note: Only do this if you have a LOT of time (and you've memorized all your perfect squares up to 1000).
We can see that the number of members in the band must be of the form $n(n + 7)$ for some positive integer $n$ . When $n = 28$ , this product is $980$ , and since AIME answers are nonnegative integers less than $1000$ , we don't have to check any higher $n$ . Also, we know that this product must be 5 more than a perfect square, so we can make a table as shown and bash: \[\begin{tabular}{|c|c|c|} n & n(n+7) & 5 more than a perfect square?\\ \hline 1 & 8 & no\\ \hline 2 & 18 & no\\ \hline 3 & 30 & yes\\ \hline 4 & 44 & no\\ \hline 5 & 60 & no\\ \hline 6 & 78 & no\\ \hline 7 & 98 & no\\ \hline 8 & 120 & no\\ \hline 9 & 144 & no\\ \hline 10 & 170 & no\\ \hline 11 & 198 & no\\ \hline 12 & 228 & no\\ \hline 13 & 260 & no\\ \hline 14 & 294 & yes\\ \hline 15 & 330 & no\\ \hline 16 & 368 & no\\ \hline 17 & 408 & no\\ \hline 18 & 450 & no\\ \hline 19 & 494 & no\\ \hline 20 & 540 & no\\ \hline 21 & 588 & no\\ \hline 22 & 638 & no\\ \hline 23 & 690 & no\\ \hline 24 & 744 & no\\ \hline 25 & 800 & no\\ \hline 26 & 858 & no\\ \hline 27 & 918 & no\\ \hline 28 & 980 & no\\ \hline \end{tabular}\] Thus, we can see that our largest valid $n(n+7)$ is $\boxed{294}$


Answer: 18.0

The distance between the $5^\text{th}$ and $26^\text{th}$ exits on an interstate highway is $118$ miles.  If any two consecutive exits are at least $5$ miles apart, then what is the largest number of miles there can be between two consecutive exits that are between the $5^\text{th}$ and $26^\text{th}$ exits?
$\text{(A)}\ 8 \qquad \text{(B)}\ 13 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 47 \qquad \text{(E)}\ 98$

Solution: There are $21$ pairs of consecutive exits. To find the maximum number of miles of one of these, the other $20$ must be equal to the minimum number yielding a total of $(5)(20)=100$ miles. The longest distance must be $118-100=\boxed{18}$


Answer: 371.0

The domain of the function $f(x) = \arcsin(\log_{m}(nx))$ is a closed interval of length $\frac{1}{2013}$ , where $m$ and $n$ are positive integers and $m>1$ . Find the remainder when the smallest possible sum $m+n$ is divided by 1000.

Solution:
We start with the same method as above. The domain of the arcsin function is $[-1, 1]$ , so $-1 \le \log_{m}(nx) \le 1$
\[\frac{1}{m} \le nx \le m\] \[\frac{1}{mn} \le x \le \frac{m}{n}\] \[\frac{m}{n} - \frac{1}{mn} = \frac{1}{2013}\] \[n = 2013m - \frac{2013}{m}\]
For $n$ to be an integer, $m$ must divide $2013$ , and $m > 1$ . To minimize $n$ $m$ should be as small as possible because increasing $m$ will decrease $\frac{2013}{m}$ , the amount you are subtracting, and increase $2013m$ , the amount you are adding; this also leads to a small $n$ which clearly minimizes $m+n$
We let $m$ equal $3$ , the smallest factor of $2013$ that isn't $1$ . Then we have $n = 2013*3 - \frac{2013}{3} = 6039 - 671 = 5368$
$m + n = 5371$ , so the answer is $\boxed{371}$

Solution:
Note that we need $-1\le f(x)\le 1$ , and this eventually gets to $\frac{m^2-1}{mn}=\frac{1}{2013}$ . From there, break out the quadratic formula and note that \[m= \frac{n+\sqrt{n^2+4026^2}}{2013\times 2}.\] Then we realize that the square root, call it $a$ , must be an integer. Then $(a-n)(a+n)=4026^2.$
Observe carefully that $4026^2 = 2\times 2\times 3\times 3\times 11\times 11\times 61\times 61$ ! It is not difficult to see that to minimize the sum, we want to minimize $n$ as much as possible. Seeing that $2a$ is even, we note that a $2$ belongs in each factor. Now, since we want to minimize $a$ to minimize $n$ , we want to distribute the factors so that their ratio is as small as possible (sum is thus minimum). The smallest allocation of $2, 61, 61$ and $2, 11, 3, 3, 11$ fails; the next best is $2, 61, 11, 3, 3$ and $2, 61, 11$ , in which $a=6710$ and $n=5368$ . That is our best solution, upon which we see that $m=3$ , thus $\boxed{371}$


Answer: 271.0

The domain of the function $f(x)=\log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))))$ is an interval of length $\tfrac mn$ , where $m$ and $n$ are relatively prime positive integers.  What is $m+n$
$\textbf{(A) }19\qquad \textbf{(B) }31\qquad \textbf{(C) }271\qquad \textbf{(D) }319\qquad \textbf{(E) }511\qquad$

Solution:
For all real numbers $a,b,$ and $c$ such that $b>0$ and $b\neq1,$ note that:
Therefore, we have \begin{align*} \log_{\frac12}(\log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))) \text{ is defined} &\implies \log_4(\log_{\frac14}(\log_{16}(\log_{\frac1{16}}x)))>0 \\ &\implies \log_{\frac14}(\log_{16}(\log_{\frac1{16}}x))>1 \\ &\implies 0<\log_{16}(\log_{\frac1{16}}x)<\frac14 \\ &\implies 1<\log_{\frac1{16}}x<2 \\ &\implies \frac{1}{256}<x<\frac{1}{16}. \end{align*} The domain of $f(x)$ is an interval of length $\frac{1}{16}-\frac{1}{256}=\frac{15}{256},$ from which the answer is $15+256=\boxed{271}.$

Solution:
For simplicity, let $a=\log_{\frac{1}{16}}{x},b=\log_{16}a,c=\log_{\frac{1}{4}}b$ , and $d=\log_4c$
The domain of $\log_{\frac{1}{2}}x$ is $x \in (0, \infty)$ , so $d \in (0, \infty)$ .
Thus, $\log_4{c} \in (0, \infty) \Rightarrow c \in (1, \infty)$ .
Since $c=\log_{\frac{1}{4}}b$ we have $b \in \left(0, \left(\frac{1}{4}\right)^1\right)=\left(0, \frac{1}{4}\right)$ .
Since $b=\log_{16}{a}$ , we have $a \in (16^0,16^{1/4})=(1,2)$ .
Finally, since $a=\log_{\frac{1}{16}}{x}$ $x \in \left(\left(\frac{1}{16}\right)^2,\left(\frac{1}{16}\right)^1\right)=\left(\frac{1}{256},\frac{1}{16}\right)$
The length of the $x$ interval is $\frac{1}{16}-\frac{1}{256}=\frac{15}{256}$ and the answer is $\boxed{271}$

Solution:
The domain of $f(x)$ is the range of the inverse function $f^{-1}(x)=\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}$ . Now $f^{-1}(x)$ can be seen to be strictly decreasing, since $\left(\frac12\right)^x$ is decreasing, so $4^{\left(\frac12\right)^x}$ is decreasing, so $\left(\frac14\right)^{4^{\left(\frac12\right)^x}}$ is increasing, so $16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}$ is increasing, therefore $\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}$ is decreasing.
Therefore, the range of $f^{-1}(x)$ is the open interval $\left(\lim_{x\to\infty}f^{-1}(x), \lim_{x\to-\infty}f^{-1}(x)\right)$ . We find: \begin{align*}    \lim_{x\to-\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}&= \lim_{a\to\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^a}}}\\ &= \lim_{b\to\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^b}}\\ &= \left(\frac1{16}\right)^{16^0}\\ &= \frac{1}{16}. \end{align*} Similarly, \begin{align*} \lim_{x\to\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}&=\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^0}}}\\ &= \left(\frac1{16}\right)^{16^{\frac14}}\\ &= \left(\frac1{16}\right)^2\\ &= \frac{1}{256}. \end{align*} Hence the range of $f^{-1}(x)$ (which is then the domain of $f(x)$ ) is $\left(\frac{1}{256},\frac{1}{16}\right)$ and the answer is $\boxed{271}$


Answer: 39.0

The eighth grade class at Lincoln Middle School has $93$ students. Each student takes a math class or a foreign language class or both. There are $70$ eighth graders taking a math class, and there are $54$ eighth graders taking a foreign language class. How many eighth graders take only a math class and not a foreign language class?
$\textbf{(A) }16\qquad\textbf{(B) }53\qquad\textbf{(C) }31\qquad\textbf{(D) }39\qquad\textbf{(E) }70$

Solution:
Let $x$ be the number of students taking both a math and a foreign language class.
By P-I-E, we get $70 + 54 - x$ $93$
Solving gives us $x = 31$
But we want the number of students taking only a math class,
which is $70 - 31 = 39$
$\boxed{39}$

Solution: We have $70 + 54 = 124$ people taking classes. However, we over-counted the number of people who take both classes. If we subtract the original amount of people who take classes we get that $31$ people took the two classes. To find the amount of people who took only math class, we subtract the people who didn't take only one math class, so we get $70 - 31 = \boxed{39}$

Solution:
We are looking for students in math only, which is the complement (exactly the rest of the students) compared to those taking a language class. Since $54$ students take a language (with or without math), we subtract that from the total number of students. Since $93-54= 39,$ our answer is $\boxed{39}.$ (It's not necessary to know that $70$ students take math.)
~hailstone


Answer: 39.0

Solution:
We know that the sum of all three areas is $93$ So, we have: \[93 = 70-x+x+54-x\] \[93 = 70+54-x\] \[93 = 124 - x\] \[-31=-x\] \[x=31\]
We are looking for the number of students in only math. This is $70-x$ . Substituting $x$ with $31$ , our answer is $\boxed{39}$


Answer: 42.0

The entries in a $3 \times 3$ array include all the digits from $1$ through $9$ , arranged so that the entries in every row and column are in increasing order. How many such arrays are there?
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60$

Solution:
Observe that all tables must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each table, there exists a valid table diagonally symmetrical across the diagonal extending from the top left to the bottom right.
\[\begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&\\ \hline &8&9\\ \hline \end{tabular}\]
3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So, there are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases. $2*6=12$
\[\begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&8\\ \hline &&9\\ \hline \end{tabular}\]
Here, no 3s or 7s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above. Remembering that $4<5$ , logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, $2*9=18$
By inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured. $2*6=12$
\[12+18+12=\boxed{42}\]

Solution:
This solution is trivial by the hook length theorem. The hooks look like this:
$\begin{tabular}{|c|c|c|} \hline 5 & 4 & 3 \\ \hline 4 & 3 & 2\\ \hline 3 & 2 & 1\\ \hline \end{tabular}$
So, the answer is $\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}$ $\boxed{42}$


Answer: 42.0

Solution:
Observe that all tables must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each table, there exists a valid table diagonally symmetrical across the diagonal extending from the top left to the bottom right.
\[\begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&\\ \hline &8&9\\ \hline \end{tabular}\]
3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So, there are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases. $2*6=12$
\[\begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&8\\ \hline &&9\\ \hline \end{tabular}\]
Here, no 3s or 7s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above. Remembering that $4<5$ , logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, $2*9=18$
By inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured. $2*6=12$
\[12+18+12=\boxed{42}\]

Solution:
This solution is trivial by the hook length theorem. The hooks look like this:
$\begin{tabular}{|c|c|c|} \hline 5 & 4 & 3 \\ \hline 4 & 3 & 2\\ \hline 3 & 2 & 1\\ \hline \end{tabular}$
So, the answer is $\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}$ $\boxed{42}$


Answer: 850.0

The equation
has 10 complex roots $r_1, \overline{r_1}, r_2, \overline{r_2}, r_3, \overline{r_3}, r_4, \overline{r_4}, r_5, \overline{r_5},\,$ where the bar denotes complex conjugation.  Find the value of

Solution:
Let $t = 1/x$ . After multiplying the equation by $t^{10}$ $1 + (13 - t)^{10} = 0\Rightarrow (13 - t)^{10} = - 1$
Using DeMoivre, $13 - t = \text{cis}\left(\frac {(2k + 1)\pi}{10}\right)$ where $k$ is an integer between $0$ and $9$
$t = 13 - \text{cis}\left(\frac {(2k + 1)\pi}{10}\right) \Rightarrow \bar{t} = 13 - \text{cis}\left(-\frac {(2k + 1)\pi}{10}\right)$
Since $\text{cis}(\theta) + \text{cis}(-\theta) = 2\cos(\theta)$ $t\bar{t} = 170 - 26\cos \left(\frac {(2k + 1)\pi}{10}\right)$ after expanding. Here $k$ ranges from 0 to 4 because two angles which sum to $2\pi$ are involved in the product.
The expression to find is $\sum t\bar{t} = 850 - 26\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}$
But $\cos \frac {\pi}{10} + \cos \frac {9\pi}{10} = \cos \frac {3\pi}{10} + \cos \frac {7\pi}{10} = \cos \frac {\pi}{2} = 0$ so the sum is $\boxed{850}$

Solution:
Divide both sides by $x^{10}$ to get \[1 + \left(13-\dfrac{1}{x}\right)^{10}=0\]
Rearranging: \[\left(13-\dfrac{1}{x}\right)^{10} = -1\]
Thus, $13-\dfrac{1}{x} = \omega$ where $\omega = e^{i(\pi n/5+\pi/10)}$ where $n$ is an integer.
We see that $\dfrac{1}{x}=13-\omega$ . Thus, \[\dfrac{1}{x\overline{x}}=(13\, -\, \omega)(13\, -\, \overline{\omega})=169-13(\omega\, +\, \overline{\omega})\, +\, \omega\overline{\omega}=170\, -\, 13(\omega\, +\, \overline{\omega})\]
Summing over all terms: \[\dfrac{1}{r_1\overline{r_1}}+\cdots + \dfrac{1}{r_5\overline{r_5}} = 5\cdot 170 - 13(e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)})\]
However, note that $e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)}=0$ from drawing the numbers on the complex plane, our answer is just \[5\cdot 170=\boxed{850}\]


Answer: 200.0

The equation $2000x^6+100x^5+10x^3+x-2=0$ has exactly two real roots, one of which is $\frac{m+\sqrt{n}}r$ , where $m$ $n$ and $r$ are integers, $m$ and $r$ are relatively prime, and $r>0$ . Find $m+n+r$

Solution:
We may factor the equation as:
\begin{align*} 2000x^6+100x^5+10x^3+x-2&=0\\ 2(1000x^6-1) + x(100x^4+10x^2+1)&=0\\ 2[(10x^2)^3-1]+x[(10x^2)^2+(10x^2)+1]&=0\\ 2(10x^2-1)[(10x^2)^2+(10x^2)+1]+x[(10x^2)^2+(10x^2)+1]&=0\\ (20x^2+x-2)(100x^4+10x^2+1)&=0\\ \end{align*}
Now $100x^4+10x^2+1\ge 1>0$ for real $x$ . Thus the real roots must be the roots of the equation $20x^2+x-2=0$ . By the quadratic formula the roots of this are:
\[x=\frac{-1\pm\sqrt{1^2-4(-2)(20)}}{40} = \frac{-1\pm\sqrt{1+160}}{40} = \frac{-1\pm\sqrt{161}}{40}.\]
Thus $r=\frac{-1+\sqrt{161}}{40}$ , and so the final answer is $-1+161+40 = \boxed{200}$

Solution:
It would be really nice if the coefficients were symmetrical. What if we make the substitution, $x = -\frac{i}{\sqrt{10}}y$ . The the polynomial becomes
$-2y^6 - (\frac{i}{\sqrt{10}})y^5 + (\frac{i}{\sqrt{10}})y^3 - (\frac{i}{\sqrt{10}})y - 2$
It's symmetric! Dividing by $y^3$ and rearranging, we get
$-2(y^3 + \frac{1}{y^3}) - (\frac{i}{\sqrt{10}})(y^2 + \frac{1}{y^2}) + (\frac{i}{\sqrt{10}})$
Now, if we let $z = y + \frac{1}{y}$ , we can get the equations
$z = y + \frac{1}{y}$
$z^2 - 2 = y^2 + \frac{1}{y^2}$
and
$z^3 - 3z = y^3 + \frac{1}{y^3}$
(These come from squaring $z$ and subtracting $2$ , then multiplying that result by $z$ and subtracting $z$ )
Plugging this into our polynomial, expanding, and rearranging, we get
$-2z^3 - (\frac{i}{\sqrt{10}})z^2 + 6z + (\frac{3i}{\sqrt{10}})$
Now, we see that the two $i$ terms must cancel in order to get this polynomial equal to $0$ , so what squared equals 3? Plugging in $z = \sqrt{3}$ into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying $z = -\sqrt{3}$ , we see that it also works! Great, we use long division on the polynomial by
$(z - \sqrt{3})(z + \sqrt{3}) = (z^2 - 3)$ and we get
$2z -(\frac{i}{\sqrt{10}}) = 0$
We know that the other two solutions for z wouldn't result in real solutions for $x$ since we have to solve a quadratic with a negative discriminant, then multiply by $-(\frac{i}{\sqrt{10}})$ . We get that $z = (\frac{i}{-2\sqrt{10}})$ . Solving for $y$ (using $y + \frac{1}{y} = z$ ) we get that $y = \frac{-i \pm \sqrt{161}i}{4\sqrt{10}}$ , and multiplying this by $-(\frac{i}{\sqrt{10}})$ (because $x = -(\frac{i}{\sqrt{10}})y$ ) we get that $x = \frac{-1 \pm \sqrt{161}}{40}$ for a final answer of $-1 + 161 + 40 = \boxed{200}$

Solution:
Observe that the given equation may be rearranged as $2000x^6-2+(100x^5+10x^3+x)=0$ .
The expression in parentheses is a geometric series with common factor $10x^2$ . Using the geometric sum formula, we rewrite as $2000x^6-2+\frac{1000x^7-x}{10x^2-1}=0, 10x^2-1\neq0$ .
Factoring a bit, we get $2(1000x^6-1)+(1000x^6-1)\frac{x}{10x^2-1}=0, 10x^2-1\neq0 \implies$ $(1000x^6-1)(2+\frac{x}{10x^2-1})=0, 10x^2-1\neq0$ .
Note that setting $1000x^6-1=0$ gives $10x^2-1=0$ , which is clearly extraneous.
Hence, we set $2+\frac{x}{10x^2-1}=0$ and use the quadratic formula to get the desired root $x=\frac{-1+\sqrt{161}}{40} \implies -1+161+40=\boxed{200}$


Answer: 113.0

The equation $2^{333x-2} + 2^{111x+2} = 2^{222x+1} + 1$ has three real roots . Given that their sum is $m/n$ where $m$ and $n$ are relatively prime positive integers , find $m+n.$

Solution: Let $y = 2^{111x}$ .  Then our equation reads $\frac{1}{4}y^3 + 4y = 2y^2 + 1$ or $y^3 - 8y^2 + 16y - 4 = 0$ .  Thus, if this equation has roots $r_1, r_2$ and $r_3$ , by Vieta's formulas we have $r_1\cdot r_2\cdot r_3 = 4$ .  Let the corresponding values of $x$ be $x_1, x_2$ and $x_3$ .  Then the previous statement says that $2^{111\cdot(x_1 + x_2 + x_3)} = 4$ so taking a logarithm of that gives $111(x_1 + x_2 + x_3) = 2$ and $x_1 + x_2 + x_3 = \frac{2}{111}$ .  Thus the answer is $111 + 2 = \boxed{113}$


Answer: 1.0

The equation $x + \sqrt{x-2} = 4$ has:
$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$

Solution:
$x + \sqrt{x-2} = 4$ Original Equation
$\sqrt{x-2} = 4 - x$ Subtract x from both sides
$x-2 = 16 - 8x + x^2$ Square both sides
$x^2 - 9x + 18 = 0$ Get all terms on one side
$(x-6)(x-3) = 0$ Factor
$x = \{6, 3\}$
If you put down A as your answer, it's wrong. You need to check for extraneous roots.
$6 + \sqrt{6 - 2} = 6 + \sqrt{4} = 6 + 2 = 8 \ne 4$
$3 + \sqrt{3-2} = 3 + \sqrt{1} = 3 + 1 = 4 \checkmark$
There is $\boxed{1}$

Solution:
We can create symmetry in the equation: \[x+\sqrt{x-2} = 4\] \[x-2+\sqrt{x-2} = 2.\] Let $y = \sqrt{x-2}$ , then we have \[y^2+y-2 = 0\] \[(y+2)(y-1) = 0\] The two roots are $\sqrt{x-2} = -2, 1$
Notice, that the first root is extraneous as the range for the square root function is always the non-negative numbers (remember, negative numbers in square roots give imaginary numbers - imaginary numbers in square roots don't give negative numbers); thus, the only real root for $x$ occurs for the second root; squaring both sides and solving for $x$ gives $x=3 \Rightarrow \boxed{1}$


Answer: 118.0

The equation $z(z+i)(z+3i)=2002i$ has a zero of the form $a+bi$ , where $a$ and $b$ are positive real numbers. Find $a.$
$\text{(A) }\sqrt{118} \qquad \text{(B) }\sqrt{210} \qquad \text{(C) }2 \sqrt{210} \qquad \text{(D) }\sqrt{2002} \qquad \text{(E) }100 \sqrt{2}$

Solution: According to Wolfram-Alpha, the answer is $\boxed{118}$


Answer: 160.0

The equation $z^6+z^3+1=0$ has complex roots with argument $\theta$ between $90^\circ$ and $180^\circ$ in the complex plane . Determine the degree measure of $\theta$

Solution:
We shall introduce another factor to make the equation easier to solve. If $r$ is a root of $z^6+z^3+1$ , then $0=(r^3-1)(r^6+r^3+1)=r^9-1$ . The polynomial $x^9-1$ has all of its roots with absolute value $1$ and argument of the form $40m^\circ$ for integer $m$ (the ninth degree roots of unity ). Now we simply need to find the root within the desired range that satisfies our original equation $x^6 + x^3 + 1 = 0$
This reduces $\theta$ to either $120^{\circ}$ or $160^{\circ}$ . But $\theta$ can't be $120^{\circ}$ because if $r=\cos 120^\circ +i\sin 120^\circ$ , then $r^6+r^3+1=3$ . (When we multiplied by $r^3 - 1$ at the beginning, we introduced some extraneous solutions, and the solution with $120^\circ$ was one of them.) This leaves $\boxed{160}$

Solution:
The substitution $y=z^3$ simplifies the equation to $y^2+y+1 = 0$ . Applying the quadratic formula gives roots $y=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}$ , which have arguments of $120$ and $240,$ respectively.
We can write them as $z^3 = \cos 240^\circ + i\sin 240^\circ$ and $z^3 = \cos 120^\circ + i\sin 120^\circ$ .
So we can use De Moivre's theorem (which I would suggest looking at if you never heard of it before) to find the fractional roots of the expressions above!
For $\cos 240^\circ + i\sin 240$ we have $(\cos 240^\circ + i\sin 240^\circ)^{1/3}$ $\Rightarrow$ $\cos 80^\circ + i\sin 80^\circ, \cos 200^\circ + i\sin200^\circ,$ and $\cos 320^\circ + i\sin320^\circ.$ Similarly for $(\cos 120^\circ + i\sin 120^\circ)^{1/3}$ , we have $\cos 40^\circ + i\sin 40^\circ, \cos 160^\circ + i\sin 160^\circ,$ and $\cos 280^\circ + i\sin 280^\circ.$ The only argument out of all these roots that fits the description is $\theta = \boxed{160}$


Answer: 4.0

The equations $2x + 7 = 3$ and $bx - 10 = - 2$ have the same solution. What is the value of $b$
$\textbf {(A)} -8 \qquad \textbf{(B)} -4 \qquad \textbf {(C) } 2 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 8$

Solution: $2x + 7 = 3 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = \boxed{4}$


Answer: 4.0

Solution: $2x + 7 = 3 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = \boxed{4}$


Answer: 2.0

The equations of $L_1$ and $L_2$ are $y=mx$ and $y=nx$ , respectively. Suppose $L_1$ makes twice as large of an angle with the horizontal (measured counterclockwise from the positive x-axis ) as does $L_2$ , and that $L_1$ has 4 times the slope of $L_2$ . If $L_1$ is not horizontal, then $mn$ is
$\text{(A)} \ \frac{\sqrt{2}}{2} \qquad \text{(B)} \ -\frac{\sqrt{2}}{2} \qquad \text{(C)} \ 2 \qquad \text{(D)} \ -2 \qquad \text{(E)} \ \text{not uniquely determined}$

Solution:
Solution by e_power_pi_times_i
$4n = m$ , as stated in the question. In the line $L_1$ , draw a triangle with the coordinates $(0,0)$ $(1,0)$ , and $(1,m)$ . Then $m = \tan(\theta_1)$ . Similarly, $n = \tan(\theta_2)$ . Since $4n = m$ and $\theta_1 = 2\theta_2$ $\tan(2\theta_2) = 4\tan(\theta_2)$ . Using the angle addition formula for tangents, $\dfrac{2\tan(\theta_2)}{1-\tan^2(\theta_2)} = 4\tan(\theta_2)$ . Solving, we have $\tan(\theta_2) = 0, \dfrac{\sqrt{2}}{2}$ . But line $L_1$ is not horizontal, so therefore $(m,n) = (2\sqrt{2},\dfrac{\sqrt{2}}{2})$ . Looking at the answer choices, it seems the answer is $(2\sqrt{2})(\dfrac{\sqrt{2}}{2}) = \boxed{2}$


Answer: 1008016.0

The expression
\[(x+y+z)^{2006}+(x-y-z)^{2006}\]
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ }  2,015,028$

Solution:
By the Multinomial Theorem , the summands can be written as
\[\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^ay^bz^c}\]
and
\[\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c},\]
respectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are:
\[{2006+2\choose 2} = 2015028\]
terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of $y$ and $z$ must be opposite. Now we find a pattern:
if the exponent of $y$ is $1$ , the exponent of $z$ can be all even integers up to $2004$ , so there are $1003$ terms.
if the exponent of $y$ is $3$ , the exponent of $z$ can go up to $2002$ , so there are $1002$ terms.
$\vdots$
if the exponent of $y$ is $2005$ , then $z$ can only be 0, so there is $1$ term.
If we add them up, we get $\frac{1003\cdot1004}{2}$ terms. However, we can switch the exponents of $y$ and $z$ and these terms will still have a negative sign. So there are a total of $1003\cdot1004$ negative terms.
By subtracting this number from 2015028, we obtain $\boxed{1,008,016}$ or $1,008,016$ as our answer.

Solution:
Define $P$ such that $P=y+z$ . Then the expression in the problem becomes: $(x+P)^{2006}+(x-P)^{2006}$
Expanding this using binomial theorem gives $(x^n+Px^{n-1}+...+P^{n-1}x+P^n)+(x^n-Px^{n-1}+...-P^{n-1}x+P^n)$ , where $n=2006$ (we may omit the coefficients, as we are seeking for the number of terms, not the terms themselves).
Simplifying gives: $2(x^n+x^{n-2}P^2+...+x^2P^{n-2}+P^n)$ . Note that two terms that come out of different powers of $P$ cannot combine and simplify, as their exponent of $x$ will differ. Therefore, we simply add the number of terms produced from each addend. By the Binomial Theorem, $P^k=(y+z)^k$ will have $k+1$ terms, so the answer is $1+3+5+...+2007=1004^2=1,008,016 \implies \boxed{1,008,016}$

Solution:
Using stars and bars we know that $(x+y+z)^{2006}$ has ${2006+2\choose 2}$ or $2015028$ different terms. Now we need to find out how many of these terms are canceled out by $(x-y-z)^{2006}$ . We know that for any term(let's say $x^{a}(-y)^{b}(-z)^{c}$ ) where $a+b+c=2006$ of the expansion of $(x-y-z)^{2006}$ is only going to cancel out with the corresponding term $x^{a}y^{b}z^{c}$ if only $b$ is odd and $c$ is even or $b$ is even and $c$ is odd. Now let's do some casework to see how many terms fit this criteria:
Case 1: $a$ is even
Now we know that $a$ is even and $a+b+c=2006$ . Thus $b+c$ is also even or both $b$ and $c$ are odd or both $b$ and $c$ are even. This case clearly fails the above criteria and there are 0 possible solutions.
Case 2: $a$ is odd
Now we know that $a$ is odd and $a+b+c=2006$ . Thus $b+c$ is odd and $b$ is odd and $c$ is even or $b$ is even and $c$ is odd. All terms that have $a$ being odd work.
We now need to figure out how many of the terms have $a$ as a odd number. We know that $a$ can be equal to any number between 0 and 2006. There are 1003 odd numbers between this range and 2007 total numbers. Thus $\frac{1003}{2007}$ of the $2015028$ terms will cancel out and $\frac{1004}{2007}$ of the terms will work. Thus there are $(\frac{1004}{2007})(2015028)$ terms. This number comes out to be $1,008,016$ $\implies \boxed{1,008,016}$ (Author: David Camacho)

Solution: Noticing how $y$ and $z$ are negative in the second part of the expression, let $x=a$ and $y+z = b$ . Then we get \[(a+b)^{2006} + (a-b)^{2006}\] We know that the terms that don't cancel out have even powers of $a$ and $b$ . Our expansion will be in the form of \[2a^{2006} + x_1a^{2004}b^{2} + x_2a^{2002}b^{4} + \cdots + 2b^{2006}\] Note that $b^n = (x+y)^n$ has $n+1$ terms. Furthermore, the current expression is irreducible as each term has a different $x$ power. Thus, when we write $a$ and $b$ back to their original terms, we will have $1+3+5+ \cdots + 2007 = 1004^2 = 1,008,016 \implies \boxed{1,008,016}$


Answer: 3.0

The expression $(81)^{-2^{-2}}$ has the same value as:
$\textbf{(A)}\ \frac {1}{81} \qquad \textbf{(B) }\ \frac {1}{3} \qquad \textbf{(C) }\ 3 \qquad \textbf{(D) }\ 81\qquad \textbf{(E) }\ 81^4$

Solution: Let us recall $\text{PEMDAS}$ . We  calculate the exponent first. $(-2)^{-2}=\frac{1}{(-2)^2}=\frac{1}{4}$ When we substitute, we get $81^{\frac{1}{4}}=\sqrt[4]{81}=\boxed{3}$


Answer: 4041.0

The expression $\frac{2021}{2020} - \frac{2020}{2021}$ is equal to the fraction $\frac{p}{q}$ in which $p$ and $q$ are positive integers whose greatest common divisor is ${ }1$ . What is $p?$
$(\textbf{A})\: 1\qquad(\textbf{B}) \: 9\qquad(\textbf{C}) \: 2020\qquad(\textbf{D}) \: 2021\qquad(\textbf{E}) \: 4041$

Solution: We write the given expression as a single fraction: \[\frac{2021}{2020} - \frac{2020}{2021} = \frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}\] by cross multiplication. Then by factoring the numerator, we get \[\frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}=\frac{(2021-2020)(2021+2020)}{2020\cdot2021}.\] The question is asking for the numerator, so our answer is $2021+2020=4041,$ giving $\boxed{4041}$

Solution:
Denote $a = 2020$ . Hence, \begin{align*} \frac{2021}{2020} - \frac{2020}{2021} & = \frac{a + 1}{a} - \frac{a}{a + 1} \\ & = \frac{\left( a + 1 \right)^2 - a^2}{a \left( a + 1 \right)} \\ & = \frac{2 a + 1}{a \left( a + 1 \right)} . \end{align*}
We observe that ${\rm gcd} \left( 2a + 1 , a \right) = 1$ and ${\rm gcd} \left( 2a + 1 , a + 1 \right) = 1$
Hence, ${\rm gcd} \left( 2a + 1 , a \left( a + 1 \right) \right) = 1$
Therefore, $p = 2 a + 1 = 4041$
Therefore, the answer is $\boxed{4041}$


Answer: 2.0

The expression $\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}$ cannot be evaluated for $x=-1$ or $x=2$ ,
since division by zero is not allowed. For other values of $x$
$\textbf{(A)}\ \text{The expression takes on many different values.}\\ \textbf{(B)}\ \text{The expression has only the value 2.}\\ \textbf{(C)}\ \text{The expression has only the value 1.}\\ \textbf{(D)}\ \text{The expression always has a value between }-1\text{ and }+2.\\ \textbf{(E)}\ \text{The expression has a value greater than 2 or less than }-1.$

Solution:
$\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)} = \frac{2x^2-2x-4}{(x+1)(x-2)}$
This can be factored as $\frac{(2)(x^2-x-2)}{(x+1)(x-2)} \implies \frac{(2)(x+1)(x-2)}{(x+1)(x-2)}$ , which cancels out to $2 \implies \boxed{2}$


Answer: 1.0

The expressions $a+bc$ and $(a+b)(a+c)$ are:
$\textbf{(A)}\ \text{always equal}\qquad\textbf{(B)}\ \text{never equal}\qquad\textbf{(C)}\ \text{equal whenever }a+b+c=1\\ \textbf{(D)}\ \text{equal when }a+b+c=0\qquad\textbf{(E)}\ \text{equal only when }a=b=c=0$

Solution: Using the FOIL method, we see that $(a+b)(a+c) = a^2 + ab + ac + bc.$ We want to solve \[a + bc = a^2 + ab + ac + bc\] \[a = a^2 + ab + ac\] \[a((a + b + c) - 1) = 0\] These expressions are $\boxed{1.}$


Answer: 39.0

The faces of each of $7$ standard dice are labeled with the integers from $1$ to $6$ . Let $p$ be the probabilities that when all $7$ dice are rolled, the sum of the numbers on the top faces is $10$ . What other sum occurs with the same probability as $p$
$\textbf{(A)} \text{ 13} \qquad \textbf{(B)} \text{ 26} \qquad \textbf{(C)} \text{ 32} \qquad \textbf{(D)} \text{ 39} \qquad \textbf{(E)} \text{ 42}$

Solution:
It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because the number of ways to add a certain number of ones to an assortment of $7$ ones is the same as the number of ways to take away a certain number of ones from an assortment of $7$ $6$ s.
So, we can match up the values to find the sum with the same probability as $10$ . We can start by noticing that $7$ is the smallest possible roll and $42$ is the largest possible roll. The pairs with the same probability are as follows:
$(7, 42), (8, 41), (9, 40), (10, 39), (11, 38)...$
However, we need to find the number that matches up with $10$ . So, we can stop at $(10, 39)$ and deduce that the sum with equal probability as $10$ is $39$ . So, the correct answer is $\boxed{39}$ , and we are done.

Solution:
Let's call the unknown value $x$ . By symmetry, we realize that the difference between 10 and the minimum value of the rolls is equal to the difference between the maximum and $x$ . So,
$10 - 7 = 42- x$
$x = 39$ and our answer is $\boxed{39}$ By: Soccer_JAMS

Solution: For the sums to have equal probability, the average sum of both sets of $7$ dies has to be $(6+1)\cdot 7 = 49$ . Since having $10$ is similar to not having $10$ , you just subtract 10 from the expected total sum. $49 - 10 = 39$ so the answer is $\boxed{39}$

Solution: The expected value of the sums of the die rolls is $3.5\cdot7=24.5$ , and since the probabilities should be distributed symmetrically on both sides of $24.5$ , the answer is $24.5+(24.5-10)=39$ , which is $\boxed{39}$

Solution: Another faster and easier way of doing this, without using almost any math at all, is realizing that the possible sums are ${7,8,9,10,...,39,40,41,42}$ . By symmetry, (and doing a few similar problems in the past), you can realize that the probability of obtaining $7$ is the same as the probability of obtaining $42$ $P(8)=P(41)$ and on and on and on. This means that $P(10)=P(39)$ , and thus the correct answer is $\boxed{39}$


Answer: 315.0

The fifth and eighth terms of a geometric sequence of real numbers are $7!$ and $8!$ respectively.  What is the first term?
$\mathrm{(A)}\ 60\qquad \mathrm{(B)}\ 75\qquad \mathrm{(C)}\ 120\qquad \mathrm{(D)}\ 225\qquad \mathrm{(E)}\ 315$

Solution: Let the $n$ th term of the series be $ar^{n-1}$ .  Because \[\frac {8!}{7!} = \frac {ar^7}{ar^4} = r^3 = 8,\] it follows that $r = 2$ and the first term is $a = \frac {7!}{r^4} = \frac {7!}{16} = \boxed{315}$


Answer: 147.0

The figure below depicts a regular $7$ -gon inscribed in a unit circle. [asy]         import geometry; unitsize(3cm); draw(circle((0,0),1),linewidth(1.5)); for (int i = 0; i < 7; ++i) {   for (int j = 0; j < i; ++j) {     draw(dir(i * 360/7) -- dir(j * 360/7),linewidth(1.5));   } } for(int i = 0; i < 7; ++i) {    dot(dir(i * 360/7),5+black); } [/asy] What is the sum of the $4$ th powers of the lengths of all $21$ of its edges and diagonals?
$\textbf{(A) }49 \qquad \textbf{(B) }98 \qquad \textbf{(C) }147 \qquad \textbf{(D) }168 \qquad \textbf{(E) }196$

Solution:
There are $7$ segments whose lengths are $2 \sin \frac{\pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{2 \pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{3\pi}{7}$
Therefore, the sum of the $4$ th powers of these lengths is \begin{align*} 7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} & = \frac{7 \cdot 2^4}{(2i)^4} \left( e^{i \frac{\pi}{7}} - e^{i \frac{\pi}{7}} \right)^4 + \frac{7 \cdot 2^4}{(2i)^4} \left( e^{i \frac{2 \pi}{7}} - e^{i \frac{2 \pi}{7}} \right)^4 + \frac{7 \cdot 2^4}{(2i)^4} \left( e^{i \frac{3 \pi}{7}} - e^{i \frac{4 \pi}{7}} \right)^4 \\ & = 7 \left( e^{i \frac{4 \pi}{7}} - 4 e^{i \frac{2 \pi}{7}} + 6 - 4 e^{- i \frac{2 \pi}{7}} + e^{- i \frac{4 \pi}{7}} \right) \\ & \quad + 7 \left( e^{i \frac{8 \pi}{7}} - 4 e^{i \frac{4 \pi}{7}} + 6 - 4 e^{- i \frac{4 \pi}{7}} + e^{- i \frac{8 \pi}{7}} \right) \\ & \quad + 7 \left( e^{i \frac{12 \pi}{7}} - 4 e^{i \frac{6 \pi}{7}} + 6 - 4 e^{- i \frac{6 \pi}{7}} + e^{- i \frac{12 \pi}{7}} \right) \\ & = 7 \left( e^{i \frac{4 \pi}{7}} + e^{i \frac{8 \pi}{7}} + e^{i \frac{12 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{8 \pi}{7}} + e^{-i \frac{12 \pi}{7}} \right) \\ & \quad - 7 \cdot 4 \left( e^{i \frac{2 \pi}{7}} + e^{i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} \right) \\ & \quad + 7 \cdot 6 \cdot 3 \\ & = 7 \left( e^{i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{i \frac{2 \pi}{7}} \right) \\ & \quad - 7 \cdot 4 \left( e^{i \frac{2 \pi}{7}} + e^{i \frac{4 \pi}{7}} + e^{i \frac{6 \pi}{7}} + e^{-i \frac{2 \pi}{7}} + e^{-i \frac{4 \pi}{7}} + e^{-i \frac{6 \pi}{7}} \right) \\ & \quad + 7 \cdot 6 \cdot 3 \\ & = -7 + 7 \cdot 4 + 7 \cdot 6 \cdot 3 \\ & = \boxed{147} ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution:
There are $7$ segments whose lengths are $2 \sin \frac{\pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{2 \pi}{7}$ $7$ segments whose lengths are $2 \sin \frac{3\pi}{7}$
Therefore, the sum of the $4$ th powers of these lengths is \begin{align*} & 7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} \\ & = 7 \cdot 2^4 \left( \frac{1 - \cos \frac{2 \pi}{7}}{2} \right)^2 + 7 \cdot 2^4 \left( \frac{1 - \cos \frac{4 \pi}{7}}{2} \right)^2 + 7 \cdot 2^4 \left( \frac{1 - \cos \frac{6 \pi}{7}}{2} \right)^2 \\ & = 7 \cdot 2^2 \left( 1 - 2 \cos \frac{2 \pi}{7} + \cos^2 \frac{2 \pi}{7} \right) + 7 \cdot 2^2 \left( 1 - 2 \cos \frac{4 \pi}{7} + \cos^2 \frac{4 \pi}{7} \right) + 7 \cdot 2^2 \left( 1 - 2 \cos \frac{6 \pi}{7} + \cos^2 \frac{6 \pi}{7} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2^2 \left( \cos^2 \frac{2 \pi}{7} + \cos^2 \frac{4 \pi}{7}  + \cos^2 \frac{6 \pi}{7} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2^2 \left( \frac{1 + \cos \frac{4 \pi}{7} }{2} + \frac{1 + \cos \frac{8 \pi}{7} }{2} + \frac{1 + \cos \frac{12 \pi}{7} }{2} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2 \cdot 3 + 7 \cdot 2 \left( \cos \frac{4 \pi}{7} + \cos \frac{8 \pi}{7} + \cos \frac{12 \pi}{7} \right) \\ & = 7 \cdot 2^2 \cdot 3 - 7 \cdot 2^3 \left( \cos \frac{2 \pi}{7} + \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} \right) + 7 \cdot 2 \cdot 3 + 7 \cdot 2 \left( \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{2 \pi}{7} \right) \\ & = 7 \cdot 2 \cdot 3 \left( 2 + 1 \right) - 7 \cdot 2 \left( 4 - 1 \right) \left( \cos \frac{4 \pi}{7} + \cos \frac{6 \pi}{7} + \cos \frac{2 \pi}{7} \right) \\ & = 7 \cdot 2 \cdot 3 \left( 2 + 1 \right) - 7 \cdot 2 \left( 4 - 1 \right) \cdot \left( - \frac{1}{2} \right) \\ & = \boxed{147} . \]

Solution:
As explained in Solutions 1 and 2, what we are trying to find is $7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7}$ . Using trig we get \begin{align*}  & \sin^4 \frac{\pi}{7} + \sin^4 \frac{2 \pi}{7} + \sin^4 \frac{3 \pi}{7} \\ = & \sin^2 \frac{\pi}{7} \left(1 - \cos^2 \frac{\pi}{7} \right) + \sin^2 \frac{2\pi}{7} \left(1 - \cos^2 \frac{2\pi}{7} \right) + \sin^2 \frac{3\pi}{7} \left(1 - \cos^2 \frac{3\pi}{7} \right) \\ = & \sin^2 \frac{\pi}{7} - \left(\frac{1}{2} \sin \frac{2\pi}{7}\right)^2 + \sin^2 \frac{2\pi}{7} - \left(\frac{1}{2} \sin \frac{4\pi}{7}\right)^2 + \sin^2 \frac{3\pi}{7} - \left(\frac{1}{2} \sin \frac{6\pi}{7}\right)^2\\ = & \sin^2 \frac{\pi}{7} - \frac{1}{4} \sin^2 \frac{2\pi}{7} + \sin^2 \frac{2\pi}{7} - \frac{1}{4} \sin^2 \frac{4\pi}{7} + \sin^2 \frac{3\pi}{7} - \frac{1}{4} \sin^2 \frac{6\pi}{7} \\ = & \frac{3}{4} \left(\sin^2 \frac{\pi}{7} + \sin^2 \frac{2\pi}{7} + \sin^2 \frac{3\pi}{7}\right) \\ = & \frac{3}{4} \cdot \frac{1}{2} \left(1 - \cos \frac{2\pi}{7} + 1 - \cos \frac{4\pi}{7} + 1 - \cos \frac{6\pi}{7} \right)\\ = & \frac{3}{4} \cdot \frac{1}{2} \left(3 - \left(-\frac{1}{2}\right)\right) \\ = & \frac{21}{16}. \end{align*} Like in the second solution, we also use the fact that $\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}$ , which admittedly might need some explanation.
For explanation see supplement
Notice that \begin{align*} \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} & = \frac{1}{2}\left(e^\frac{2i\pi}{7}+ e^{-\frac{2i\pi}{7}} \right) + \frac{1}{2}\left( e^\frac{4i\pi}{7}+ e^{-\frac{4i\pi}{7}} \right) + \frac{1}{2}\left( e^\frac{6i\pi}{7}+ e^{-\frac{6i\pi}{7}}\right) \\ & = \frac{1}{2}\left(e^\frac{2i\pi}{7}+ e^{-\frac{2i\pi}{7}}+ e^\frac{4i\pi}{7}+ e^{-\frac{4i\pi}{7}} +e^\frac{6i\pi}{7}+ e^{-\frac{6i\pi}{7}} + 1\right) - \frac{1}{2} \end{align*} In the brackets we have the sum of the roots of the polynomial $x^7 - 1 = 0$ . These sum to $0$ by Vieta’s formulas , and the desired identity follows. See Roots of unity if you have not seen this technique.
Going back to the question: \[7 \cdot 2^4 \sin^4 \frac{\pi}{7} + 7 \cdot 2^4 \sin^4 \frac{2 \pi}{7} + 7 \cdot 2^4 \sin^4 \frac{3 \pi}{7} = 7 \cdot 2^4 \left(\sin^4 \frac{\pi}{7} + \sin^4 \frac{2 \pi}{7} + \sin^4 \frac{3 \pi}{7}\right) = 7 \cdot 2^4 \cdot \frac{21}{16} = \boxed{147}.\] ~obscene_kangaroo

Solution:
This solution follows the same steps as the trigonometry solutions (Solutions 2 and 3), except it gives an alternate way to prove the statement below true without complex numbers:
\[\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} = -\frac{1}{2}\]
\begin{align*} S &= \cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7}, \\ S^2 &= \cos^2 \frac{2\pi}{7} + \cos^2 \frac{4\pi}{7} + \cos^2 \frac{6\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{4\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{6\pi}{7} + 2\cos \frac{4\pi}{7}\cos \frac{6\pi}{7} \\ &= \left(\frac{1+ \cos \frac{4\pi}{7}}{2}\right) + \left(\frac{1+ \cos \frac{8\pi}{7}}{2}\right) + \left(\frac{1+ \cos \frac{12\pi}{7}}{2}\right) + 2\cos \frac{2\pi}{7}\cos \frac{4\pi}{7} + 2\cos \frac{2\pi}{7}\cos \frac{6\pi}{7} + 2\cos \frac{4\pi}{7}\cos \frac{6\pi}{7} \\ &= \frac{1}{2}(3 + S)  + \left(\cos \frac{6\pi}{7} + \cos \frac{2\pi}{7}\right) + \left(\cos \frac{8\pi}{7} + \cos \frac{4\pi}{7}\right) + \left(\cos \frac{10\pi}{7} + \cos \frac{2\pi}{7}\right) \\ &= \frac{1}{2}(3 + S) + 2\cos \frac{2\pi}{7} + 2\cos \frac{4\pi}{7} + 2\cos \frac{6\pi}{7}\\ &= \frac{1}{2}(3 + S) + 2S. \\ \end{align*} We end up with \[2S^2 - 5S - 3 = 0.\] Using the quadratic formula, we find the solutions for $S$ to be $-\frac{1}{2}$ and $3$ . Because $3$ is impossible, $S = -\frac{1}{2}$ .
With this result, following similar to steps to Solutions 2 and 3 will get $\boxed{147}$

Solution:
Let $x$ $y$ , and $z$ be the lengths of the chords with arcs $\frac{2\pi}{7}$ $\frac{4\pi}{7}$ and $\frac{6\pi}{7}$ respectively.
Then by the law of cosines we get: \begin{align*} x^2 &= 2\left(1-\cos\frac{2\pi}{7}\right), \\ y^2 &= 2\left(1-\cos\frac{4\pi}{7}\right), \\ z^2 &= 2\left(1-\cos\frac{6\pi}{7}\right). \end{align*} The answer is then just $7(x^4+y^4+z^4)$ (since there's $7$ of each diagonal/side), obtained by summing the squares of the above equations and then multiplying by $7$ \begin{align*} & \quad7\cdot2^2\left( \left(1-\cos\frac{2\pi}{7}\right)^2 + \left(1-\cos\frac{4\pi}{7}\right)^2 + \left(1-\cos\frac{6\pi}{7}\right)^2 \right) \\ &= 7\cdot4\left( \left(1-2\cos\frac{2\pi}{7}+\cos^2\frac{2\pi}{7}\right) + \left(1-2\cos\frac{4\pi}{7}+\cos^2\frac{4\pi}{7}\right) + \left(1-2\cos\frac{6\pi}{7}+\cos^2\frac{6\pi}{7}\right) \right) \\ &= 7\cdot4\left(3-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \left(\cos^2\frac{2\pi}{7}+\cos^2\frac{4\pi}{7}+\cos^2\frac{6\pi}{7}\right) \right) \\ &= 7\cdot4\left(3-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \frac12 \left(1+\cos\frac{4\pi}{7}+1+\cos\frac{8\pi}{7}+1+\cos\frac{12\pi}{7}\right) \right) \\ &= 7\cdot4\left(\frac{9}{2}-2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) + \frac12 \left(\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}\right) \right) \\ &= 7\cdot4\left(\frac{9}{2}-\frac{3}{2}\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right) \right) \\ &= 7\cdot4\left(\frac{9}{2}-\frac{3}{2}\left(-\frac{1}{2}\right)\right) \\ &= \boxed{147}

Solution:
Hope you had a ruler handy! This problem can be done with a ruler and basic estimation.
First, measuring the radius of the circle obtains $2.9$ cm (when done on the paper version). Thus, any other measurement we get for the sides/diagonals should be divided by $2.9$
Measuring the sides of the circle gets $2.5$ cm. The shorter diagonals are $4.5$ cm, and the longest diagonals measure $5.6$ cm. Thus, we'd like to estimate \[7\left(\frac{2.5}{2.9}\right)^4 + 7\left(\frac{4.5}{2.9}\right)^4 + 7\left(\frac{5.6}{2.9}\right)^4.\]
We know $\left(\frac{2.5}{2.9}\right)^4$ is slightly less than $1.$ Let's approximate it as 1 for now. Thus, $7\left(\frac{2.5}{2.9}\right)^4 \approx 7.$
Next, $\left(\frac{4.5}{2.9}\right)^4$ is slightly more than $\left(\frac{4.5}{3}\right)^4.$ We know $\left(\frac{4.5}{3}\right)^4 = 1.5^4 = \frac{81}{16},$ slightly more than $5,$ so we can approximate $\left(\frac{4.5}{2.9}\right)^4$ as $5.5.$ Thus, $7\left(\frac{4.5}{2.9}\right)^4 \approx 38.5.$
Finally, $\left(\frac{5.6}{2.9}\right)^4$ is slightly less than $\left(\frac{5.6}{2.8}\right)^4 = 2^4 = 16.$ We say it's around $15,$ so then $7\left(\frac{5.6}{2.9}\right)^4 \approx 105.$
Adding what we have, we get $105 + 38.5 + 1 = 144.5$ as our estimate. We see $\boxed{147}$ is very close to our estimate, so we have successfully finished the problem.

Solution:
Place the figure on the complex plane and let $w = e^{\frac{2\pi}{7}}$ (a $7$ th root of unity). The vertices of the $7$ -gon are $w^0,w^1,w^2,\dots,w^6$ . We wish to find \[\sum_{i=0}^5\sum_{j=i+1}^6\lvert w^i-w^j\rvert^4 = \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6\lvert w^i-w^j\rvert^4.\] The second expression is more convenient to work with. The factor of $\tfrac{1}{2}$ is because it double-counts each edge (and includes when $i=j$ , but the length is $0$ so it doesn't matter).
Recall the identity $|z|^2 = z\overline{z}$ . Since $|w| = 1$ $\overline{w} = w^{-1}$ , and \begin{align*}     \lvert w^i-w^j\rvert^4 &= ((w^i-w^j)(w^{-i}-w^{-j}))^2 \\     &= (2-w^{i-j}-w^{j-i})^2 \\     &= 4+w^{2i-2j}+w^{2j-2i}+2(-2w^{i-j}-2w^{j-i}+1) \\     &= 6-4w^{i-j}-4w^{j-i}+w^{2i-2j}+w^{2j-2i}. \end{align*}
But notice that, for a fixed value of $i$ , over all values of $j$ from $0$ to $6$ , by properties of modular arithmetic with $w^7 = w^0$ (or expanding),
each of the $w^\bullet$ -terms takes on every value in $w^0,w^1,w^2,\dots,w^6$ exactly once. Since $\sum_{j=0}^6 w^j = 0$ \begin{align*}     \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6\lvert w^i-w^j\rvert^4 &= \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6(6-4w^{i-j}-4w^{j-i}+w^{2i-2j}+w^{2j-2i}) \\     &= \frac{1}{2}\sum_{i=0}^6\left(\sum_{j=0}^6 6-4\sum_{j=0}^6 w^{i-j}-4\sum_{j=0}^6 w^{j-i}+\sum_{j=0}^6 w^{2i-2j}+\sum_{j=0}^6 w^{2j-2i}\right) \\     &= \frac{1}{2}\sum_{i=0}^6\left(\sum_{j=0}^6 6-4\sum_{j=0}^6 w^j-4\sum_{j=0}^6 w^j+\sum_{j=0}^6 w^j+\sum_{j=0}^6 w^j\right) \\     &= \frac{1}{2}\sum_{i=0}^6\sum_{j=0}^6 6 \\     &= \frac{1}{2}\cdot7\cdot7\cdot6 \\     &= \boxed{147}

Solution:
This is how I solve this problem:
It's easy to solve for $3$ -gon, $4$ -gon, and $6$ -gon inscribed in a unit circle. (Okay, it's just the weird names for triangle, square, and hexagon)
For $3$ -gon, the sum is equal to $3$ times the $4$ th power of an edge. Thus, \[S_3=3\,\cdot\,\left(\sqrt{3}\right)^4=27.\]
For $4$ -gon, the sum is equal to $4$ times the $4$ th power of an edge, and $2$ times the $4$ th power of the diagonal. Thus, \[S_4=4\,\cdot\,\left(\sqrt{2}\right)^4+2\,\cdot\,\left(2\right)^4=48.\]
For $6$ -gon, the sum is equal to $6$ times the $4$ th power of an edge, $6$ times the $4$ th power of the short diagonal, and $3$ times the $4$ th power of the long diagonal. Thus, \[S_6=6\,\cdot\,\left(1\right)^4+6\,\cdot\,\left(\sqrt{3}\right)^4+3\,\cdot\,\left(2\right)^4=108.\]
Then, I quickly noticed that $27=3\,\cdot\,3^2$ $48=3\,\cdot\,4^2$ , and $108=3\,\cdot\,6^2$ . So reasonably, it will work out this formula, $S_n=3n^2$ . (This step is purely out of guessing, maybe have a look at Solution 8 for more info...)
By inductive reasoning, we got $S_7=3\,\cdot\,7^2=\boxed{147}$


Answer: 147.0

Solution:
First, we put the figure in the coordinate plane with the center of the circle at the origin and a vertex on the positive x-axis. Thus, the coordinates of the vertices will be the terminal points of integer multiples of the angle $\frac{2\pi}{7},$ which are \[\left(\cos\dfrac{2\pi n}{7}, \sin\dfrac{2\pi n}{7}\right)\] for integer $n.$ Then, we notice there are three types of diagonals: the ones with chords of arcs $\dfrac{2\pi}{7}, \dfrac{4\pi}{7},$ and $\dfrac{6\pi}{7}.$ We notive there are here are $7$ of each type of diagonal. Then, we use the pythagorean theorem to find the distance from $\left(\cos\frac{2\pi n}{7}, \sin\frac{2\pi n}{7}\right)$ to $(1, 0)$ \[\left(\sqrt{\left(\cos\frac{2\pi n}{7}-1\right)^2+\sin^2\frac{2\pi n}{7}}\right)^4=\left(\sqrt{\cos^2\frac{2\pi n}{7}+\sin^2\frac{2\pi n}{7}-2\cos\frac{2\pi n}{7}+1}\right)^4\] \[=\left(\sqrt{2-2\cos\frac{2\pi n}{7}}\right)^4\] \[=4\cos^2\frac{2\pi n}{7}-8\cos\frac{2\pi n}{7}+4.\] By the cosine double angle identity, $\cos{2\theta}=2\cos^2\theta-1.$ This means that $2\cos^2\theta=\cos{2\theta}+1.$ Substituting this in, \[4\cos^2\frac{2\pi n}{7}-8\cos\frac{2\pi n}{7}+4=2\left(\cos\frac{4\pi n}{7}+1\right)-8\cos\frac{2\pi n}{7}+4=2\cos\frac{4\pi n}{7}-8\cos\frac{2\pi n}{7}+6.\] Summing this up for $n=1,2,3,$ \[2\cos\frac{4\pi}{7}-8\cos\frac{2\pi}{7}+6+2\cos\frac{8\pi}{7}-8\cos\frac{4\pi}{7}+6+2\cos\frac{12\pi}{7}-8\cos\frac{6\pi}{7}+6\] \[=2\left(\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}+\cos\frac{12\pi}{7}\right)-8\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)+18\] \[=2\left(\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}+\cos\frac{2\pi}{7}\right)-8\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\right)+18\] \[=2(-0.5)-8(-0.5)+18=21.\] (These equalities are based on $\cos\theta=\cos(2\pi-\theta)$ and $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}.$ ) (For explanation of this see supplement .)
Finally, because there are $7$ of each type of diagonal, the answer is $7\cdot 21=\boxed{147}.$


Answer: 4.0

The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L$ , without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.) [asy] import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { pair A = (j,i); dot(A);  } } for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if (j != 3) { draw((j,i)--(j+1,i)); } if (i != 2) { draw((j,i)--(j,i+1)); } } } label("$A$", (0,2), W); label("$L$", (3,0), E); [/asy]
How many different routes can Paula take?
$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$

Solution:
Note that of the $12$ cities, $6$ of them ( $2$ on the top, $2$ on the bottom, and $1$ on each side) have $3$ edges coming into/out of them (i.e., in graph theory terms, they have degree $3$ ). Therefore, at least $1$ edge connecting to each of these cities cannot be used. Additionally, the same applies to the start and end points, since we don't want to return to them. Thus there are $6+2=8$ vertices that we know have $1$ unused edge, and we have $17-13=4$ unused edges to work with (since there are $17$ edges in total, and we must use exactly $13$ of them). It is not hard to find that there is only one configuration satisfying these conditions:
Note: $\text{X}$ s represent unused edges.

Observe that at each of the $2$ cities marked with an $\text{O}$ on a path, there are $2$ possibilities: we can either continue straight and cross back over the path later, or we can make a left turn, then turn right when we approach the junction again. This gives us a total of $2\cdot 2 = \boxed{4}$ valid paths.

Solution:
Let the bottom-left vertex be $(0,0)$ , and let each of the edges have length $1$ , so that all of the vertices are at lattice points.
Firstly, notice that for any vertex $V$ on the graph (other than $A$ or $L$ ), we can visit it at most $M(V) = \left\lfloor \frac{\text{deg}(V)}{2} \right\rfloor$ times, where $\text{deg}(V)$ is, as usual, the degree of $V$ (i.e. the number of edges coming into/out of $V$ ). This is because to visit any of these vertices, we would have to enter and exit it by two different edges, in order to avoid using any portion of a road more than once. (Those who know some graph theory will recognise this a well-known principle.) We will label each vertex with this number.  Additionally, notice that if we visit $n$ vertices (not necessarily distinct, i.e. counting a vertex which is visited twice as two vertices) along our path, we must traverse $n-1$ edges (this is quite easy to prove). Thus, if we want to traverse $13$ edges in total, we have to visit $14$ vertices. We must visit $A$ and $L$ , leaving $12$ vertices from the rest of the graph to visit.
If we sum the maximum numbers of visits to each vertex, we find that this is exactly equal to the $12$ found above. This means that we have to visit each vertex $M(V)$ times, and must traverse $2\cdot M(V)$ edges connected to each vertex. Specifically, we must traverse all $4$ of the edges connected to the two central vertices at $(1,1)$ and $(1,2)$ , as well as both edges connected to the $2$ corner vertices (excluding $A$ and $L$ ), and any $2$ edges connected to the other vertices along the outside edge of the rectangle.
With this information, we can now proceed by dividing into cases.
Case 1 : We first move down from $A$
In this case, we see that we must immediately move right to $(1,1)$ in order to traverse the edge from $(0,1)$ to $(1,1)$ , since we can never revisit $(0,1)$ . However, by the same logic, we must immediately move to $(0,0)$ . This is a contradiction, so there are no possible paths in this case.
Case 2 : We first move right from $A$
Similar to the last case, we see that we must immediately move to $(1,1)$ . By symmetry, we conclude that our last two moves must be $(2,1) \rightarrow (2,0) \rightarrow (3,0)$ . With this information, we have reduced the problem to traveling from $(1,1)$ to $(2,1)$ with the same constraints as before. We redraw the graph, removing the edges we have already used and updating our labels (note that $(1,1)$ and $(2,1)$ are still labeled with $2$ since we will pass through them twice, at the start and the end). Then, we remove the vertices with label $0$ and the edges we know we can never traverse, giving:
 Now, it is clear that we must complete a cycle in the lower left square, return to $(1,1)$ , go to $(2,1)$ , and complete a cycle in the top right square, returning to $(2,1)$ .
There are two ways to traverse each cycle (clockwise and anti-clockwise), giving us a total of $2\cdot 2 = \boxed{4}$ paths of length 13 from $A$ to $L$


Answer: 4.0

Solution: Observe that only the two central vertices can be visited twice. Since the path is of length 13, we need to repeat a vertex. Using casework on each vertex, we can find there are two paths that go through each central vertex twice, for an answer of $\boxed{4}.$

Solution:
Looking at the answer choices we see that it would probably be easy enough to count the different paths. After some experimenting, we find that we must cross one of the middle vertexes twice, making counting easier.
We finally find that there are $\boxed{4}$ paths that satisfy the conditions. Luckily, $4$ is the largest answer choice so we know it must be correct.


Answer: 192.0

The figure below shows a polygon $ABCDEFGH$ , consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that $AH = EF = 8$ and $GH = 14$ . What is the volume of the prism?
[asy] usepackage("mathptmx"); size(275); defaultpen(linewidth(0.8)); real r = 2, s = 2.5, theta = 14; pair G = (0,0), F = (r,0), C = (r,s), B = (0,s), M = (C+F)/2, I = M + s/2 * dir(-theta); pair N = (B+G)/2, J = N + s/2 * dir(180+theta); pair E = F + r * dir(- 45 - theta/2), D = I+E-F; pair H = J + r * dir(135 + theta/2), A = B+H-J; draw(A--B--C--I--D--E--F--G--J--H--cycle^^rightanglemark(F,I,C)^^rightanglemark(G,J,B)); draw(J--B--G^^C--F--I,linetype ("4 4")); dot("$A$",A,N); dot("$B$",B,1.2*N); dot("$C$",C,N); dot("$D$",D,dir(0)); dot("$E$",E,S); dot("$F$",F,1.5*dir(-100)); dot("$G$",G,S); dot("$H$",H,W); dot("$I$",I,NE); dot("$J$",J,1.5*S); [/asy]
$\textbf{(A)} ~112\qquad\textbf{(B)} ~128\qquad\textbf{(C)} ~192\qquad\textbf{(D)} ~240\qquad\textbf{(E)} ~288$

Solution:
While imagining the folding, $\overline{AB}$ goes on $\overline{BC},$ $\overline{AH}$ goes on $\overline{CI},$ and $\overline{EF}$ goes on $\overline{FG}.$ So, $BJ=CI=8$ and $FG=BC=8.$ Also, $\overline{HJ}$ becomes an edge parallel to $\overline{FG},$ so that means $HJ=8.$
Since $GH=14,$ then $JG=14-8=6.$ So, the area of $\triangle BJG$ is $\frac{8\cdot6}{2}=24.$ If we let $\triangle BJG$ be the base, then the height is $FG=8.$ So, the volume is $24\cdot8=\boxed{192}.$


Answer: 732.0

The figure below shows a ring made of six small sections which you are to paint on a wall. You have four paint colors available and you will paint each of the six sections a solid color. Find the number of ways you can choose to paint the sections if no two adjacent sections can be painted with the same color.
[asy] draw(Circle((0,0), 4)); draw(Circle((0,0), 3)); draw((0,4)--(0,3)); draw((0,-4)--(0,-3)); draw((-2.598, 1.5)--(-3.4641, 2)); draw((-2.598, -1.5)--(-3.4641, -2)); draw((2.598, -1.5)--(3.4641, -2)); draw((2.598, 1.5)--(3.4641, 2)); [/asy]

Solution:
Choose a section to start coloring. Assume, WLOG, that this section is color $1$ . We proceed coloring clockwise around the ring. Let $f(n,C)$ be the number of ways to color the first $n$ sections (proceeding clockwise) such that the last section has color $C$ . In general (except for when we complete the coloring), we see that \[f(n,C_i)=\sum_{j\ne i} f(n-1,C_j),\] i.e., $f(n,C_i)$ is equal to the number of colorings of $n-1$ sections that end in any color other than $C_i$ . Using this, we can compute the values of $f(n,C)$ in the following table.
$\begin{tabular}{c|c|c|c|c } \multicolumn{1}{c}{}&\multicolumn{4}{c}{\(C\)}\\ \(n\)&1 & 2 & 3& 4 \\ \hline 1& 1 & 0 & 0 & 0\\ 2 & 0 & 1 & 1 & 1 \\ 3& 3 & 2 & 2 & 2 \\ 4 & 6 & 7 & 7 & 7 \\ 5 & 21 & 20 & 20 & 20\\ 6& 0& 61 & 61 & 61\\ \end{tabular}$
Note that $f(6, 1)=0$ because then $2$ adjacent sections are both color $1$ . We multiply this by $4$ to account for the fact that the initial section can be any color.  Thus the desired answer is $(61+61+61) \cdot 4 = \boxed{732}$

Solution:
We use complementary counting. There are $4^6$ total colorings of the ring without restriction. To count the complement, we wish to count the number of colorings in which at least one set of adjacent sections are the same color. There are six possible sets of adjacent sections that could be the same color (think of these as borders). Call these $B_1,B_2,\dots,B_6$ . Let $\mathcal{A}_1, \mathcal{A}_2,\dots,\mathcal{A}_6$ be the sets of colorings of the ring where the sections on both sides of $B_1,B_2,\dots,B_6$ are the same color. We wish to determine $|\mathcal{A}_1\cup\mathcal{A}_2\cup\cdots\cup\mathcal{A}_6|$ . Note that all of these cases are symmetric, and in general, $|\mathcal{A}_i|=4^5$ . There are $6$ such sets $\mathcal{A}_i$ . Also, $|\mathcal{A}_i\cup\mathcal{A}_j|=4^4$ , because we can only change colors at borders, so if we have two borders along which we cannot change colors, then there are four borders along which we have a choice of color. There are $\binom{6}{2}$ such pairs $\mathcal{A}_i\cup\mathcal{A}_j$ . Similarly, $|\mathcal{A}_i\cup \mathcal{A}_j\cup \mathcal{A}_k|=4^3$ , with $\binom{6}{3}$ such triples, and we see that the pattern will continue for 4-tuples and 5-tuples. For 6-tuples, however, these cases occur when there are no changes of color along the borders, i.e., each section has the same color. Clearly, there are four such possibilities.
Therefore, by PIE, \[|\mathcal{A}_1\cup\mathcal{A}_2\cup\cdots\cup\mathcal{A}_6|=\binom{6}{1}\cdot 4^5-\binom{6}{2}\cdot 4^4+\binom{6}{3}\cdot 4^3-\binom{6}{4}\cdot 4^2+\binom{6}{5}\cdot 4^1-4.\] We wish to find the complement of this, or \[4^6-\left(\binom{6}{1}\cdot 4^5-\binom{6}{2}\cdot 4^4+\binom{6}{3}\cdot 4^3-\binom{6}{4}\cdot 4^2+\binom{6}{5}\cdot 4^1-4\right).\] By the Binomial Theorem, this is $(4-1)^6+3=\boxed{732}$

Solution:
We use generating functions. Suppose that the colors are $0,1,2,3$ .  Then as we proceed around a valid coloring of the ring in the clockwise direction, we know that between two adjacent sections with colors $s_i$ and $s_{i+1}$ , there exists a number $d_i\in\{1,2,3\}$ such that $s_{i+1}\equiv s_i+d_i\pmod{4}$ . Therefore, we can represent each border between sections by the generating function $(x+x^2+x^3)$ , where $x,x^2,x^3$ correspond to increasing the color number by $1,2,3\pmod4$ , respectively. Thus the generating function that represents going through all six borders is $A(x)=(x+x^2+x^3)^6$ , where the coefficient of $x^n$ represents the total number of colorings where the color numbers are increased by a total of $n$ as we proceed around the ring. But if we go through all six borders, we must return to the original section, which is already colored. Therefore, we wish to find the sum of the coefficients of $x^n$ in $A(x)$ with $n\equiv 0\pmod4$
Now we note that if $P(x)=x^n$ , then \[P(x)+P(ix)+P(-x)+P(-ix)=\begin{cases}4x^n&\text{if } n\equiv0\pmod{4}\\0&\text{otherwise}.\end{cases}\] Therefore, the sum of the coefficients of $A(x)$ with powers congruent to $0\pmod 4$ is \[\frac{A(1)+A(i)+A(-1)+A(-i)}{4}=\frac{3^6+(-1)^6+(-1)^6+(-1)^6}{4}=\frac{732}{4}.\] We multiply this by $4$ to account for the initial choice of color, so our answer is $\boxed{732}$

Solution:
Let $f(n)$ be the number of valid ways to color a ring with $n$ sections (which we call an $n$ -ring), so the answer is given by $f(6)$ . For $n=2$ , we compute $f(n)=4\cdot3=12$ . For $n \ge 3$ , we can count the number of valid colorings as follows: choose one of the sections arbitrarily, which we may color in $4$ ways. Moving clockwise around the ring, there are $3$ ways to color each of the $n-1$ other sections. Therefore, we have $4 \cdot 3^{n-1}$ colorings of an $n$ -ring.
However, note that the first and last sections could be the same color under this method. To count these invalid colorings, we see that by "merging" the first and last sections into one, we get a valid coloring of an $(n-1)$ -ring. That is, there are $f(n-1)$ colorings of an $n$ -ring in which the first and last sections have the same color. Thus, $f(n) = 4 \cdot 3^{n-1} - f(n-1)$ for all $n \ge 3$
To compute the requested value $f(6)$ , we repeatedly apply this formula: \begin{align*} f(6)&=4\cdot3^5-f(5)\\&=4\cdot3^5-4\cdot3^4+f(4)\\&=4\cdot3^5-4\cdot3^4+4\cdot3^3-f(3)\\&=4\cdot3^5-4\cdot3^4+4\cdot3^3-4\cdot3^2+f(2)\\&=4(3^5-3^4+3^3-3^2+3)\\&=4\cdot3\cdot\frac{3^5+1}{3+1}\\&=\boxed{732} (Solution by MSTang.)

Solution:
Label the sections 1, 2, 3, 4, 5, 6 clockwise. We do casework on the colors of sections 1, 3, 5.
Case 1: the colors of the three sections are the same.
In this case, each of sections 2, 4, 6 can be one of 3 colors, so this case yields $4 \times 3^3 = 108$ ways.
Case 2: two of sections 1, 3, 5 are the same color.
Note that there are 3 ways for which two of the three sections have the same color, and $4 \times 3 = 12$ ways to determine their colors. After this, the section between the two with the same color can be one of 3 colors and the other two can be one of 2 colors. This case yields $3 \times (4 \times 3) \times (3 \times 2 \times 2) = 432$ ways.
Case 3: all three sections of 1, 3, 5 are of different colors.
Clearly, there are $4 \times 3 \times 2 = 24$ choices for which three colors to use, and there are 2 ways to choose the colors of each of sections 2, 4, 6. Thus, this case gives $4 \times 3 \times 2 \times 2^3 = 192$ ways.
In total, there are $108 + 432 + 192 = \boxed{732}$ valid colorings.

Solution:
We will take a recursive approach to this problem. We can start by writing out the number of colorings for a circle with $1, 2,$ and $3$ compartments, which are $4, 12,$ and $24.$ Now we will try to find a recursive formula, $C(n)$ , for a circle with an arbitrary number of compartments $n.$ We will do this by focusing on the $n-1$ section in the circle. This section can either be the same color as the first compartment, or it can be a different color as the first compartment. We will focus on each case separately.
Case 1:
If they are the same color, we can say there are $C(n-2)$ to fill the first $n-1$ compartments. The $nth$ compartment must be different from the first and second to last compartments, which are the same color. Hence this case adds $3*C(n-2)$ to our recursive formula.
Case 2:
If they are different colors, we can say that there are $C(n-1)$ to fill the first $n-1$ compartments, and for the the $nth$ compartment, there are $2$ ways to color it because the $n-1$ and $1$ compartments are different colors. Hence this case adds $2*C(n-1).$
So our recursive formula, $C(n)$ , is $3*C(n-2) + 2*C(n-1).$ Using the initial values we calculated, we can evaluate this recursive formula up to $n=6.$ When $n=6,$ we get $\boxed{732}$ valid colorings.

Solution:
WLOG, color the top left section $1$ and the top right section $2$ . Then the left section can be colored $2$ $3$ , or $4$ , and the right section can be colored $1$ $3$ , or $4$ . There are $3 \cdot 3 = 9$ ways to color the left and right sections. We split this up into two cases.
Case 1: The left and right sections are of the same color. There are $2$ ways this can happen: either they both are $3$ or they both are $4$ . We have $3$ colors to choose for the bottom left, and $2$ remaining colors to choose for the bottom right, for a total of $2 \cdot 3 \cdot 2$ cases.
Case 2: The left and right sections are of different colors. There are $9 - 2 = 7$ ways this can happen. Assume the left is $3$ and the right is $4$ . Then the bottom left can be $1$ $2$ , or $4$ , and the bottom right can be $1$ $2$ , or $3$ . However the bottom sections cannot both be $1$ or both be $2$ , so there are $3 \cdot 3 - 2 = 7$ ways to color the bottom sections, for a total for $7 \cdot 7 = 49$ colorings.
Since there were $4 \cdot 3 = 12$ ways to color the top sections, the answer is $12 \cdot (12 + 49) = \boxed{732}$

Solution:
Label the four colors $1, 2, 3,$ and $4$ , respectively.
Now let's imagine a circle with the four numbers $1, 2, 3,$ and $4$ written clockwise. We'll say that a bug is standing on number $a$ . It is easy to see that for the bug to move to a different number, it must walk $1, 2,$ or $3$ steps clockwise. (This is since adjacent numbers can't be the same, as stated in the problem). Note that the sixth number in the bug's walking sequence must not equal the first number. Thus, our total number of ways, $N$ , given the bug's starting number $k$ , is simply the number of ordered quintuplets of positive integers $(a_1, a_2, a_3, a_4, a_5)$ that satisfy $a_i \in \{1, 2, 3\}$ for all $1 \leq i \leq 5$ and \[\sum_{i=1}^{5} a_i \neq 8, 12,\] since the bug cannot land on $k$ again on his fifth and last step.
We know that the number of ordered quintuplets of positive integers $(a_1, a_2, a_3, a_4, a_5)$ that satisfy $a_i \in \{1, 2, 3\}$ without the other restriction is just $3^5$ , so we aim to find the number of quintuplets such that \[a_1 + a_2 + a_3 + a_4 + a_5 = 8, 12.\] Note that the number of ordered quintuplets satisfying $\sum_{i=1}^{5} a_i = 8$ is the same as the number of them satisfying $\sum_{i=1}^{5} a_i = 12$ due to symmetry. By stars and bars, there are $\dbinom{7}{4} = 35$ ways to distribute the three "extra" units to the five variables $a_1, a_2, a_3, a_4, a_5$ (since $a_i \geq 1$ ), but ways of distribution such that one variable is equal to $4$ are illegal, so the actual number of ways is $35 - 5 = 30$ . Since there are four possible values of $k$ (or the starting position for the bug), we obtain \[N = 4 \times (3^5 - 2 \times 30) = \boxed{732}.\] -fidgetboss_4000

Solution:
Let's number the regions $1,2,\dots 6$ . Suppose we color regions $1,2,3$ . Then, how many ways are there to color $4,5,6$
Note: the numbers are numbered as shown:

$\textbf{Case 1:}$ The colors of $1,2,3$ are $BAB$ , in that order.
Then the colors of $6,5,4$ can be $AXA$ $AXC$ $CXA$ $CXC$ , or $CXD$ in that order, where $X$ is any color not equal to its surroundings. Then there are $4$ choices for $A$ $3$ choices for $B$ (it cannot be $A$ ), $2$ choices for $C$ , and $1$ for $D$ , the last color. So, summing up, we have \[4*3(1*3+2*2+2*2+2*3+2*1*2)=12*21=252\] colorings.
$\textbf{Case 2:}$ The colors of $1,2,3$ are $BAC$ , in that order.
Again, we list out the possible arrangements of $6,5,4$ $AXA$ $AXB$ $CXA$ $AXD$ $DXA$ $CXB$ $CXD$ $DXB$ , or $DXD$ . (Easily simplified; listed here for clarity.) Then there are $4$ choices for $A$ as usual, $3$ choices for $B$ , and so on. Hence we have \[4*3*2(1*3+1*2+1*2+1*2+1*2+1*2+1*2+1*2+1*3)=24*20=480\] colorings in this case.
Adding up, we have $252+480=\boxed{732}$ as our answer.

Solution:
We quickly notice that this is just the cycle graph with 6 vertices. The chromatic polynomial for a cycle is $(k-1)^n+(-1)^n (k-1)$ where we use $k$ colors on a cycle of $n$ vertices. Plugging in $k=4$ and $n=6$ we arrive at $\boxed{732}$ .
~chrisdiamond10

Solution:
Let's label the regions as $1,2,3,4,5,6$ in that order. We start with region $1$ . There are no restrictions on the color of region $1$ so it can be any of the four colors. We know move on the region $2$ . It can be any color but color used for region $1$ , giving us $3$ choices. Section $3$ is where it gets a bit complicated; we will have to do casework based on whether the color of region $3$ is that of region $1$ or not.
If we have the color of region $3$ being different from that of region $1$ (in which we color do so in $2$ ways), then we have need for another casework at region $4$ . If the color of region $4$ is different from that of region $1$ (which can be achieved in $2$ ways), then we have yet another casework split.
If the color of region $5$ is different from that of region $1$ (which can be done in $2$ ways, then we would have a total of $2$ possible colorings for region $6$ (for it cannot be the color of regions $1$ nor $5$ ). Moving on to the case where the color of section $5$ is the same as that of section $1$ (which can be done in $1$ way), we will have $3$ ways (region $6$ cannot be that color of both region $1$ and $5$ ). Thus, if the color of region $4$ is different of that of region $1$ , then we have $2\cdot 2 + 3 = 7$ ways.
If the color of region $4$ is the same as that of region $1$ (which can be done in one way), then the color of region $5$ have to be different from that of sector $1$ $3$ ways). That means there will be $2$ choices for the color of region $6$ . So if the color of region $4$ is the same as region $1$ , then we will have $3\cdot 2 = 6$ . That means if the color of section $3$ is different from that of region $1$ , then there are $2\cdot 7 + 6 = 20$
Now, moving on to the case where section $3$ has the same color of region $1$ . That means section $4$ will have to be a different color of that of region $1$ $3$ ways). So, that means we have region $5$ to be either the same color or different color as region $1$ . If it is different (which can be done in $2$ ways), then there will be $2$ possibilities on the color of sector $6$ . If it is same (which can be done in $1$ way), then there are $3$ ways to color region $6$ . So, if section $3$ has the same color as section $1$ , then we have $3(2\cdot 2 + 3) = 3(7) = 21$
Now, in overall we will have $4\cdot 3 (2(20)+21) = 12(61) = \boxed{732}$

Solution:
Let the top-right segment be segment $1,$ and remaining segments are numbered $2,3,4,5,6$ in clockwise order.
We have $4$ choices for segment $1,$ and $3$ choices for segments $2,3,4,5.$ For segment $6,$ we wish to find the expected value of the number of choices for segment $6$ 's color, which depends on whether segment $5$ is red. If we let $P(n)$ denote the probability of segment $P$ being the same color as segment $1$ (for simplicity, denote segment $1$ 's color as red), we get the following recurrence relation: \[P(n)=\dfrac{1-P(n-1)}{3}.\] This is because that you cannot have two reds in a row (hence the $1-P(n-1)$ ) and if segment $n-1$ is not red, there are three possible colors, one of which is red (hence the divide by $3$ ). Using the obvious fact that $P(1)=1$ by the definition of $P(n),$ we find that \[P(5)=\dfrac{7}{27}.\] If segment $5$ is red, then there are three possible colors for segment $6.$ If it is not, there are $2$ possible choices for segment $6.$ This means the expected number of color choices for segment $6$ is \[\dfrac{7}{27}\cdot3+\dfrac{20}{27}\cdot2,\] and the total number of colorings of the ring is \[4\cdot3\cdot3\cdot3\cdot3\cdot\left(\dfrac{7}{27}\cdot3+\dfrac{20}{27}\cdot2\right)=\boxed{732}.\]


Answer: 8.0

The figure consists of alternating light and dark squares. The number of dark squares exceeds the number of light squares by
$\text{(A)}\ 7 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 10 \qquad \text{(E)}\ 11$
[asy] unitsize(12); //Force a white background in middle even when transparent fill((3,1)--(12,1)--(12,4)--(3,4)--cycle,white); //Black Squares, Gray Border (blends better than white) for(int a=0; a<7; ++a)  {   filldraw((2a,0)--(2a+1,0)--(2a+1,1)--(2a,1)--cycle,black,gray);  } for(int b=7; b<15; ++b)  {   filldraw((b,14-b)--(b+1,14-b)--(b+1,15-b)--(b,15-b)--cycle,black,gray);  } for(int c=1; c<7; ++c)  {   filldraw((c,c)--(c+1,c)--(c+1,c+1)--(c,c+1)--cycle,black,gray);  } filldraw((6,4)--(7,4)--(7,5)--(6,5)--cycle,black,gray); filldraw((7,5)--(8,5)--(8,6)--(7,6)--cycle,black,gray); filldraw((8,4)--(9,4)--(9,5)--(8,5)--cycle,black,gray); //White Squares, Black Border filldraw((7,4)--(8,4)--(8,5)--(7,5)--cycle,white,black); for(int a=0; a<7; ++a)  {   filldraw((2a+1,0)--(2a+2,0)--(2a+2,1)--(2a+1,1)--cycle,white,black);  } for(int b=9; b<15; ++b)  {   filldraw((b-1,14-b)--(b,14-b)--(b,15-b)--(b-1,15-b)--cycle,white,black);  } for(int c=1; c<7; ++c)  {   filldraw((c+1,c)--(c+2,c)--(c+2,c+1)--(c+1,c+1)--cycle,white,black);  } label("same",(6.3,2.45),N); label("pattern here",(7.5,1.4),N); [/asy]

Solution: It is simple to notice that in each and every row, there is always one more black square than the white squares. Since there are $8$ rows, there are $8$ more black squares than the white squares. $8\rightarrow \boxed{8}$


Answer: 23.0

Solution: Draw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles $ABC$ and $ADE$ , they each have area $2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}$ . For triangle $ACD$ , we can see that $AC=AD=2\sqrt{3}$ and $CD=2$ . Using Pythagorean Theorem, the altitude for this triangle is $\sqrt{11}$ , so the area is $\sqrt{11}$ . Adding each part up, we get $2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{23}$


Answer: 23.0

Solution: Draw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles $ABC$ and $ADE$ , they each have area $2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}$ . For triangle $ACD$ , we can see that $AC=AD=2\sqrt{3}$ and $CD=2$ . Using Pythagorean Theorem, the altitude for this triangle is $\sqrt{11}$ , so the area is $\sqrt{11}$ . Adding each part up, we get $2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{23}$


Answer: 14.0

Solution:
It is clear that $6$ $5$ , and $4$ will not come together to get a sum of $15$
The faces $6$ $5$ , and $3$ come together at a common vertex, making the maximal sum $6+5+3=14\rightarrow \boxed{14}$


Answer: 761.0

The figures $F_1$ $F_2$ $F_3$ , and $F_4$ shown are the first in a sequence of figures. For $n\ge3$ $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $13$ diamonds. How many diamonds are there in figure $F_{20}$
$\textbf{(A)}\ 401 \qquad \textbf{(B)}\ 485 \qquad \textbf{(C)}\ 585 \qquad \textbf{(D)}\ 626 \qquad \textbf{(E)}\ 761$

Solution: The sequence $\{ d_n\}$ goes $1, 5, 13, 25, 41,\dots$ . The first finite differences go $4, 8, 12, 16, \dots$ . The second finite differences go $4, 4, 4, \dots$ , so we see that the second finite difference is constant. Thus, $d_n$ can be represented as a quadratic, $d_n = an^2 + bn + c$ . However, we already know $d_1 = 1$ $d_2 = 3$ , and $d_3 = 13$ . Thus, \[a + b + c = d_1 = 1\] \[4a + 2b + c = d_2 = 3\] \[9a + 3b + c = d_3 = 13\] Solving this system for $a$ $b$ , and $c$ gives $a = 2$ $b = -2$ $c = 1$ . Finally, $d_n = 2n^2 - 2n + 1\implies d_{20} = \boxed{761}$