Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?
$\textbf{(A)}\; 14 \qquad\textbf{(B)}\; 16 \qquad\textbf{(C)}\; 18 \qquad\textbf{(D)}\; 20 \qquad\textbf{(E)}\; 24$

Solution: Label the people sitting at the table $A, B, C, D, E, F,$ and assume that they are initially seated in the order $ABCDEF$ . The possible new positions for $A, B, C, D, E,$ and $F$ are respectively (a dash indicates a non-allowed position):
\[\begin{tabular}{| c | c | c | c | c | c |} \hline - & - & A & A & A & - \\ \hline - & - & - & B & B & B \\ \hline C & - & - & - & C & C \\ \hline D & D & - & - & - & D \\ \hline E & E & E & - & - & - \\ \hline - & F & F & F & - & - \\ \hline \end{tabular}\]
The permutations we are looking for should use one letter from each column, and there should not be any repeated letters:
$\begin{tabular}{c} CDEFAB \\ CEAFBD \\ CEFABD \\ CEFBAD \\ CFEABD \\ CFEBAD \\ DEAFBC \\ DEAFCB \\ DEFABC \\ DEFACB \\ DEFBAC \\ DFEABC \\ DFEACB \\ DFEBAC \\ EDAFBC \\ EDAFCB \\ EDFABC \\ EDFACB \\ EDFBAC \\ EFABCD \end{tabular}$
There are $\boxed{20}$ such permutations.

Answer: D 20
Problem: Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $\overline{AR}+\overline{BR}+\overline{CR}+\overline{DR}$
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$

Solution: First, note that $PQ$ lies on the radical axis of any of the pairs of circles. Suppose that $O_1$ and $O_2$ are the centers of two circles $C_1$ and $C_2$ that intersect exactly at $P$ and $Q$ , with $O_1$ and $O_2$ lying on the same side of $PQ$ , and $O_1 O_2=39$ . Let $x=O_1 R$ $y=O_2 R$ , and suppose that the radius of circle $C_1$ is $r$ and the radius of circle $C_2$ is $\tfrac{5}{8}r$
Then the power of point $R$ with respect to $C_1$ is
\[(r+x)(r-x) = r^2 - x^2 = 24^2\]
and the power of point $R$ with respect to $C_2$ is
\[\left(\frac{5}{8}r + y\right) \left(\frac{5}{8}r - y\right) = \frac{25}{64}r^2 - y^2 = 24^2.\]
Also, note that $x-y=39$
Subtract the above two equations to find that $\tfrac{39}{64}r^2 - x^2 + y^2 = 0$ or $39 r^2 = 64(x^2-y^2)$ . As $x-y=39$ , we find that $r^2=64(x+y) = 64(2y+39)$ . Plug this into an earlier equation to find that $25(2y+39)-y^2=24^2$ . This is a quadratic equation with solutions $y=\tfrac{50 \pm 64}{2}$ , and as $y$ is a length, it is positive, hence $y=57$ , and $x=y+39=96$ . This is the only possibility if the two centers lie on the same side of their radical axis.
On the other hand, if they lie on opposite sides, then it is clear that there is only one possibility, and then it is clear that $O_1 R + O_2 R = O_1 O_2 = 39$ . Therefore, we obtain exactly four possible centers, and the sum of the desired lengths is $57+96+39 = \boxed{192}$

Answer: D 192
Problem: Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $\overline{AR}+\overline{BR}+\overline{CR}+\overline{DR}$
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$

Solution: Note that if a circle passes through a pair of points, the center of the circle is on the perpendicular bisector of the line segment between the pair of points. This means that $A$ $B$ $C$ , and $D$ are all on the perpendicular bisector of $PQ$ . Let us say the distance from $A$ to the line $PQ$ is some $a$ . Therefore, the distance from $B$ to the line $PQ$ is $\sqrt{(\frac{8}{5}\sqrt{a^2 + 24^2})^2 - 24^2}$ , which comes out to be $\sqrt{\frac{64}{25}a^2 + \frac{39}{25}\cdot 576}$ . Since $AB = 39$ , we have one of $\sqrt{\frac{64}{25}a^2 + \frac{39}{25}\cdot 576} \pm a$ to be equal to $39$ . We can solve both equations to get that out of the four possible solutions, and only two are positive: $7$ and $57$ . Note that since no two circles can be congruent, we need the radius of one of $A$ or $C$ to be $7$ and the other to be $57$ . Plugging in to find the corresponding radii of $B$ and $D$ gives $32$ and $96$ , and adding everything up gives $\boxed{192}$

Answer: D 192
Problem: Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $\overline{AR}+\overline{BR}+\overline{CR}+\overline{DR}$
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$

Solution: Let's start by drawing $PQ$ . Because all circles contain $P$ and $Q$ , all the centers lie on the perpendicular bisector of $PQ$ , and point $R$ is on this bisector.
For all the circle radii to be different (there can't be two congruent circles), two centers are on the same side of $PQ$ , and two are on the opposite side of $PQ$ . For the latter two circles--call them $A$ and $B$ -- $AR+BR=39$
Let's consider the next case, where $C$ and $D$ lie on the same side. Construct right triangles from the picture, and use the Pythagorean Theorem (divide by 3 to negate big numbers). You will get that the distance from $R$ to the closest circle center is $57$ . Therefore, the answer is $39+2\cdot57+39=\boxed{192}$

Answer: D 192
Problem: Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $\overline{AR}+\overline{BR}+\overline{CR}+\overline{DR}$
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$

Solution: Since the radical axis $PQ$ is perpendicular to the line connecting the center of the circles, we have that $A$ $B$ $C$ $D$ , and $R$ are collinear. WLOG, assume that $A$ and $B$ are on the same side of $R$ and let $AR=y$ and let $BP=x$ so that $AP=\frac{5}{8}x$
Then, using the Pythagorean Theorem on right triangles $PBR$ and $PAR$ \[(39+y)^2+24^2=x^2\qquad(1)\] \[y^2+24^2=\frac{25}{64}x^2\qquad(2)\] Subtracting the $(2)$ from $(1)$ gives \[x^2=64(2y+39)\qquad(3)\] Substituting $(3)$ into $(2)$ gives $y^2-50y-399=0=(y-57)(y+7).$ Taking the positive solution ( $y>0$ ), $y=AR=57$ and $BR=(57)+39=96$
Since none of the circles are congruent, $C$ and $D$ must be on the opposite side of $R$ so $CR+DR=CD=39$ . Hence, $AR+BR+CR+DR=57+96+39=\boxed{192}$

Answer: D 192
Problem: Four circles, no two of which are congruent, have centers at $A$ $B$ $C$ , and $D$ , and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\tfrac{5}{8}$ times the radius of circle $B$ , and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$ . Furthermore, $AB = CD = 39$ and $PQ = 48$ . Let $R$ be the midpoint of $\overline{PQ}$ . What is $\overline{AR}+\overline{BR}+\overline{CR}+\overline{DR}$
$\textbf{(A)}\; 180 \qquad\textbf{(B)}\; 184 \qquad\textbf{(C)}\; 188 \qquad\textbf{(D)}\; 192\qquad\textbf{(E)}\; 196$

Solution: Note that the four circles are coaxial, meaning $A,B,C,D,R$ are all collinear. Let $AR=x$ . By Pythagorean Theorem , the radius of circle $A$ squared would be $r_A^2 = x^2+24^2$ and the radius of circle $B$ squared would be $r_B^2 = (x+39)^2+24^2.$ Since $r_A^2 = \dfrac{25}{64}r_B^2$ \[x^2+24^2 = \dfrac{25}{64}((x+39)^2+24^2)\] Solving this will give $x^2-50x-399 = 0$ , or $x=57, -7$ $AR = 57$ . The same equation will apply to $CR$ $CR$ would be the other root: $CR=7$ (the negative $7$ signals that $C$ is the negative, or opposite, direction of $D$ , about $R$ ). Thus, $AR = 57$ $BR = 57+39 = 96$ $CR = 7$ $DR = 39-7 = 32$ , implying $AR+BR+CR+DR = \boxed{192}$

Answer: D 192
Problem: A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$

Solution: Let $x=e^{i\pi/6}$ , a $30^\circ$ counterclockwise rotation centered at the origin. Notice that $P_k$ on the complex plane is:
\[1+2x+3x^2+\cdots+(k+1)x^k\]
We need to find the magnitude of $P_{2015}$ on the complex plane. This is an arithmetico-geometric series
\begin{align*} S &=1+2x+3x^2+\cdots+2015x^{2014} \\ xS &=x+2x^2+3x^3+\cdots+2015x^{2015} \\ (1-x)S &=1+x+x^2+\cdots+x^{2014}-2015x^{2015} \\ S &= \frac{1-x^{2015}}{(1-x)^2}-\frac{2015x^{2015}}{1-x} \end{align*}
We want to find $|S|$ . First, note that $x^{2015}=x^{11}=x^{-1}$ because $x^{12}=1$ . Therefore
\[S =\frac{1-\frac{1}{x}}{(1-x)^2}-\frac{2015}{x(1-x)}=-\frac{1}{x(1-x)}-\frac{2015}{x(1-x)}=-\frac{2016}{x(1-x)}.\]
Hence, since $|x|=1$ , we have $|S| = \frac{2016}{|1-x|}.$
Now we just have to find $|1-x|$ . This can just be computed directly:
\[1-x=1-\frac{\sqrt{3}}{2}-\frac{1}{2}i\]
\[|1-x|^2=\left(1-\sqrt{3}+\frac{3}{4}\right)+\frac{1}{4}=2-\sqrt{3}={\left(\frac{\sqrt{6}-\sqrt{2}}{2}\right)}^2\]
\[|1-x|=\frac{\sqrt{6}-\sqrt{2}}{2}\]
Therefore $|S|=2016\cdot\frac{2}{\sqrt{6}-\sqrt{2}}=2016\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)=1008\sqrt{2}+1008 \sqrt{6}$
Thus the answer is $1008+2+1008+6=\boxed{2024}$

Answer: B 2024
Problem: A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$

Solution: Here is an alternate solution that does not use complex numbers:
The distance from $P_{2015}$ to $P_0$ can be evaluated using the Pythagorean theorem . Assuming $P_0$ lies at the origin, we can calculate the distance the bee traveled to $P_{2015}$ by evaluating the distance the bee traveled in the x-direction and the y-direction. Let's start by summing each movement:
$x=1\cos{0}+2\cos{30}+3\cos{60}+\cdots+2014\cos{270}+2015\cos{300}$
A movement of $p$ units at $q$ degrees is the same thing as a movement of $-p$ units at $q-180$ degrees, so we can adjust all the cosines with arguments greater than $180$ as follows:
$x=1\cos{0}+2\cos{30}+3\cos{60}+4\cos{90}+5\cos{120}+6\cos{150}-7\cos{0}-8\cos{30}-\cdots-2015\cos{120}$
Grouping terms with like-cosines and factoring out the cosines:
$x=(1-7+13-\cdots+2005-2011)\cos{0}+\cdots+(6-12+18-\cdots-2004+2010)\cos{150}$
Each sum in the parentheses has $336$ terms (except the very last one, which has $335$ ). By pairing each term, we see there are $\frac{336}{2}$ pairs of $-6$ . Therefore, each sum equals $168\cdot-6=-1008$ except the very last sum, which has $167$ pairs of $-6$ plus an extra 2010 and equals $167\cdot-6+2010=1008$ . Plugging in these values:
$x=-1008\cos{0}-1008\cos{30}-1008\cos{60}-1008\cos{90}-1008\cos{120}+1008\cos{150}$ $x=1008(-1-\frac{\sqrt{3}}{2}-\frac{1}{2}-0+\frac{1}{2}-\frac{\sqrt{3}}{2})=-1008(1+\sqrt{3})$
We can find how far the bee traveled in the y-direction using the same logic as above, we arrive at the sum:
$y=-1008\sin{0}-1008\sin{30}-1008\sin{60}-1008\sin{90}-1008\sin{120}+1008\sin{150}$
$y=1008(0-\frac{1}{2}-\frac{\sqrt{3}}{2}-1-\frac{\sqrt{3}}{2}+\frac{1}{2})=-1008(1+\sqrt{3})$
Finally, we use the Pythagorean to find the distance from $P_0$ . This distance is given by:
$\sqrt{x^2+y^2}=\sqrt{(-1008(1+\sqrt{3}))^2+(-1008(1+\sqrt{3}))^2}=\sqrt{2\cdot1008^2\cdot(1+\sqrt{3})^2}=1008(1+\sqrt{3})\sqrt{2}=1008\sqrt{2}+1008\sqrt{6}$ , so the answer is $1008+2+1008+6=\boxed{2024}$

Answer: B 2024
Problem: A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$

Solution: Suppose that the bee makes a move of distance $i$ . After $6$ turns it will be facing the opposite direction and move $i+6$ units. Combining these opposite movements gives a total movement of $-6$ units in the original direction. This means that every $12$ moves, the bee will move $-6$ units in each direction of $0^\circ, 30^\circ, 60^\circ, 90^\circ, 120^\circ, 150^\circ$
We want to find the displacement vector for every $12$ moves. Factoring out the $-6$ for now (which flips the direction), we draw a quick diagram of one unit in each direction. [asy] draw((0,0)--(1,0)--(1+sqrt(3)/2, 1/2)--(3/2+sqrt(3)/2, 1/2+sqrt(3)/2)--(3/2+sqrt(3)/2, 3/2+sqrt(3)/2)--(1+sqrt(3)/2, 3/2+sqrt(3))--(1, 2+sqrt(3)), EndArrow); draw((1,0)--(3/2+sqrt(3)/2, 0), dashed); draw((1,2+sqrt(3))--(3/2+sqrt(3)/2, 2+sqrt(3)), dashed); draw((3/2+sqrt(3)/2, 2+sqrt(3))--(3/2+sqrt(3)/2, 0), dashed); draw((1+sqrt(3)/2,0)--(1+sqrt(3)/2, 1/2)--(3/2+sqrt(3)/2, 1/2), dashed); draw((1+sqrt(3)/2,2+sqrt(3))--(1+sqrt(3)/2, 3/2+sqrt(3))--(3/2+sqrt(3)/2, 3/2+sqrt(3)), dashed); [/asy] Using the 30-60-90 triangles, it is clear that the displacement vector (factoring back in the $-6$ ) is $-6\left\langle 1, 2+\sqrt{3}\right\rangle$
To compute the distance to $P_{2015}$ , we can compute the position of $P_{2016}$ (a multiple of $12$ moves) and then subtract the vector from $P_{2015}$ to $P_{2016}$
The bee reaches $P_{2016}$ after $\frac{2016}{12} = 168$ sets of $12$ moves, so the total displacement vector to $P_{2016}$ is $168(-6)\left\langle 1, 2+\sqrt{3}\right\rangle = \left\langle -1008, -2006-1008\sqrt{3}\right\rangle$
The bee moves at an angle of $-30^\circ$ from $P_{2015}$ to $P_{2016}$ , so subtracting it means moving an angle of $150^\circ$ . Since the vector is $2016$ units long, by a 30-60-90 triangle, it is $\left\langle -1008\sqrt{3}, 1008\right\rangle$
Therefore the total displacement vector to $P_{2015}$ is $\left\langle -1008, -2006-1008\sqrt{3}\right\rangle + \left\langle -1008\sqrt{3}, 1008\right\rangle = \left\langle -1008-1008\sqrt{3}, -1008-1008\sqrt{3}\right\rangle$ . The displacement is thus $\sqrt{2}\left|-1008-1008\sqrt{3}\right| = 1008\sqrt{2}+1008\sqrt{6} \implies 1008+2+1008+6 = \boxed{2024}$

Answer: B 2024
Problem: A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$

Solution: Let $P_0$ be the origin. East would be the real axis in the positive direction. Then we can assign each $P_n$ a complex value. The displacement would then be the magnitude of the complex number.
Notice that after the $n$ th move the value of $P_n$ is $P_{n-1}+ne^{\frac{i(n-1)\pi}{6}}$ . Also notice that after six moves the bee is facing in the opposite direction. And because we have found a recursion, we can add these up.
Then we have \[P_n=e^{0}+2e^{\frac{i\pi}{6}}+\cdots+2015e^{\frac{2014i\pi}{6}},\] and this becomes \[(1-7+13-\cdots-2011)e^0+(2-8+\cdots-2012)e^{\frac{i\pi}{6}}+\cdots+(6-12+\cdots+2010)e^{\frac{5i\pi}{6}}.\]
Simplifying, we have \[-6\cdot168-6\cdot168e^{\frac{i\pi}{6}}-6\cdot168e^{\frac{2i\pi}{6}}-6\cdot168e^{\frac{3i\pi}{6}}-6\cdot168e^{\frac{4i\pi}{6}}-(2010-6\cdot167)e^{\frac{5i\pi}{6}},\] which eventually simplifies to \[-1008-(2+\sqrt3)1008i,\] and this is a $15-75-90$ triangle which has ratios of $1:(2+\sqrt3):(\sqrt2+\sqrt6)$ so the magnitude is $1008\sqrt2+1008\sqrt6$ and the answer is $1008+2+1008+6=\boxed{2024}$

Answer: B 2024
Problem: A bee starts flying from point $P_0$ . She flies $1$ inch due east to point $P_1$ . For $j \ge 1$ , once the bee reaches point $P_j$ , she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$ . When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$ , where $a$ $b$ $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$

Solution: After each 12 moves, the bee will be facing the same direction as it started. Let $P_0$ be the origin and let $P_n$ (with $n$ divisible by $12$ ) be $(x,y)$ . We now notice that each of the move pairs with lengths $n+1$ $n+7$ $n+2$ $n+8$ $n+3$ $n+9$ $n+4$ $n+10$ $n+5$ $n+11$ $n+6$ $n+12$ will move the bee 6 units in the directions corresponding to the moves with lengths $n+7$ $n+8$ $n+9$ $n+10$ $n+11$ , and $n+12$ . This equates to moving the bee from $(x,y)$ to $(x-6,y-12-6\sqrt{3})$ , a move that repeats every 12 moves. Since $\frac{2016}{12} = 168$ , we have that $P_(2016) = (-1008, -2016-1008\sqrt{3})$ . It follows that $P_(2015) = (-1008-1008\sqrt{3}, -1008-1008\sqrt{3})$ so the distance to $P_0$ is $1008(\sqrt{2} + \sqrt{6})$ so the answer is $1008 + 2 + 1008 + 6 = \boxed{2024}$

Answer: B 2024
Problem: At the theater children get in for half price.  The price for $5$ adult tickets and $4$ child tickets is $$24.50$ .  How much would $8$ adult tickets and $6$ child tickets cost?
$\textbf{(A) }$35\qquad \textbf{(B) }$38.50\qquad \textbf{(C) }$40\qquad \textbf{(D) }$42\qquad \textbf{(E) }$42.50$

Solution: Suppose $x$ is the price of an adult ticket. The price of a child ticket would be $\frac{x}{2}$
\begin{eqnarray*} 5x + 4(x/2) = 7x &=& 24.50\\ x &=& 3.50\\ \end{eqnarray*}
Plug in for 8 adult tickets and 6 child tickets.
\begin{eqnarray*} 8x + 6(x/2) &=& 8(3.50) + 3(3.50)\\  &=&\boxed{38.50}

Answer: B 38.50
Problem: 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, $72 + 27 = \boxed{99}.$

Answer: D 99
Problem: 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: 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: D 99
Problem: A fancy bed and breakfast inn has $5$ rooms, each with a distinctive color-coded decor.  One day $5$ friends arrive to spend the night.  There are no other guests that night.  The friends can room in any combination they wish, but with no more than $2$ friends per room.  In how many ways can the innkeeper assign the guests to the rooms?
$\textbf{(A) }2100\qquad \textbf{(B) }2220\qquad \textbf{(C) }3000\qquad \textbf{(D) }3120\qquad \textbf{(E) }3125\qquad$

Solution: We can work in reverse by first determining the number of combinations in which there are more than $2$ friends in at least one room. There are three cases:
Case 1: Three friends are in one room. Since there are $5$ possible rooms in which this can occur, we are choosing three friends from the five, and the other two friends can each be in any of the four remaining rooms,
there are $5\cdot\binom{5}{3}\cdot4\cdot4 = 800$ possibilities.
Case 2: Four friends are in one room. Again, there are $5$ possible rooms, we are choosing four of the five friends, and the other one can be in any of the other four rooms, so there are $5\cdot\binom{5}{4}\cdot4= 100$ possibilities.
Case 3: Five friends are in one room. There are $5$ possible rooms in which this can occur, so there are $5$ possibilities.
Since there are $5^5 = 3125$ possible combinations of the friends, the number fitting the given criteria is $3125 - (800+100+5) = \boxed{2220}$

Answer:  2220
Problem: A five-digit palindrome is a positive integer with respective digits $abcba$ , where $a$ is non-zero.  Let $S$ be the sum of all five-digit palindromes.  What is the sum of the digits of $S$
$\textbf{(A) }9\qquad \textbf{(B) }18\qquad \textbf{(C) }27\qquad \textbf{(D) }36\qquad \textbf{(E) }45\qquad$

Solution: For each digit $a=1,2,\ldots,9$ there are $10\cdot10$ (ways of choosing $b$ and $c$ ) palindromes. So the $a$ s contribute $(1+2+\cdots+9)(100)(10^4+1)$ to the sum.
For each digit $b=0,1,2,\ldots,9$ there are $9\cdot10$ (since $a \neq 0$ ) palindromes. So the $b$ s contribute $(0+1+2+\cdots+9)(90)(10^3+10)$ to the sum.
Similarly, for each $c=0,1,2,\ldots,9$ there are $9\cdot10$ palindromes, so the $c$ contributes $(0+1+2+\cdots+9)(90)(10^2)$ to the sum.
It just so happens that \[(1+2+\cdots+9)(100)(10^4+1)+(1+2+\cdots+9)(90)(10^3+10)+(1+2+\cdots+9)(90)(10^2)=49500000\] so the sum of the digits of the sum is $\boxed{18}$

Answer: B 18
Problem: A five-digit palindrome is a positive integer with respective digits $abcba$ , where $a$ is non-zero.  Let $S$ be the sum of all five-digit palindromes.  What is the sum of the digits of $S$
$\textbf{(A) }9\qquad \textbf{(B) }18\qquad \textbf{(C) }27\qquad \textbf{(D) }36\qquad \textbf{(E) }45\qquad$

Solution: Notice that $10001+ 99999 = 110000.$ In fact, ordering the palindromes in ascending order, we find that the sum of the nth palindrome and the nth to last palindrome is $110000.$ We have $9\cdot 10\cdot 10$ palindromes, or $450$ pairs of palindromes summing to $110000.$ Performing the multiplication gives $49500000$ , so the sum $\boxed{18}$

Answer: B 18