Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The five solutions to the equation \[(z-1)(z^2+2z+4)(z^2+4z+6)=0\] may be written in the form $x_k+y_ki$ for $1\le k\le 5,$ where $x_k$ and $y_k$ are real. Let $\mathcal E$ be the unique ellipse that passes through the points $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4),$ and $(x_5,y_5)$ . The eccentricity of $\mathcal E$ can be written in the form $\sqrt{\frac mn}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ? (Recall that the eccentricity of an ellipse $\mathcal E$ is the ratio $\frac ca$ , where $2a$ is the length of the major axis of $\mathcal E$ and $2c$ is the is the distance between its two foci.)
$\textbf{(A) }7 \qquad \textbf{(B) }9 \qquad \textbf{(C) }11 \qquad \textbf{(D) }13\qquad \textbf{(E) }15$

Solution: The five roots are $1,-1+i\sqrt{3},-1-i\sqrt{3},-2+i\sqrt{2},-2-i\sqrt{2}.$
So, we express this conic in the form $ax^2+by^2+cx+z=0.$ Note that this conic cannot have the $ky$ term since the roots are symmetric about the $x$ -axis.
Now we have equations \begin{align*} a+c+z&=0, \\ a+3b-c+z&=0, \\ 4a+2b-2c+z&=0, \end{align*} from which $a:b:c=5:6:9.$
So, the conic can be written in the form $5x^2+6y^2+9x=14.$ If it is written in the form of $\frac{(x-m)^2}{r^2}+\frac{y^2}{s^2}=1,$ then $r^2:s^2=6:5.$
Therefore, the desired eccentricity is $\sqrt{\frac{\sqrt{6-5}}{6}}=\sqrt{\frac{1}{6}},$ and the answer is $1+6=\boxed{7}.$

Answer: A 7
Problem: The five solutions to the equation \[(z-1)(z^2+2z+4)(z^2+4z+6)=0\] may be written in the form $x_k+y_ki$ for $1\le k\le 5,$ where $x_k$ and $y_k$ are real. Let $\mathcal E$ be the unique ellipse that passes through the points $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4),$ and $(x_5,y_5)$ . The eccentricity of $\mathcal E$ can be written in the form $\sqrt{\frac mn}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ? (Recall that the eccentricity of an ellipse $\mathcal E$ is the ratio $\frac ca$ , where $2a$ is the length of the major axis of $\mathcal E$ and $2c$ is the is the distance between its two foci.)
$\textbf{(A) }7 \qquad \textbf{(B) }9 \qquad \textbf{(C) }11 \qquad \textbf{(D) }13\qquad \textbf{(E) }15$

Solution: After calculating the $5$ points that lie on $\mathcal E$ , we try to find a transformation that sends $\mathcal E$ to the unit circle. Scaling about $(1, 0)$ works, since $(1, 0)$ is already on the unit circle and such a transformation will preserve the ellipse's symmetry about the $x$ -axis. If $2a$ and $2b$ are the lengths of the major and minor axes, respectively, then the ellipse will be scaled by a factor of $r := \frac1a$ in the $x$ -dimension and $s := \frac1b$ in the $y$ -dimension.
The transformation then sends the points $\left(-1,\pm\sqrt 3\right)$ and $\left(-2,\pm\sqrt 2\right)$ to the points $\left(1-2r, \pm s\sqrt 3\right)$ and $\left(1-3r, \pm s\sqrt 2\right)$ , respectively. These points are on the unit circle, so \[(1-2r)^2 + 3s^2 = 1 \quad \text{and} \quad (1-3r)^2 + 2s^2 = 1.\] This yields \[4r^2 + 3s^2 = 4r \quad \text{and} \quad 9r^2 + 2s^2 = 6r,\] from which \begin{align*} 12r^2 + 9s^2 &= 18r^2 + 4s^2 \\ \frac{r^2}{s^2} &= \frac56. \end{align*} Recalling that $r = \frac1a$ and $s = \frac1b$ , this implies $\frac{b^2}{a^2} = \frac56$ . From this, we get \[\frac{c^2}{a^2} = \frac{a^2-b^2}{a^2} = 1 - \frac{b^2}{a^2} = \frac{1}{6},\] so $\frac ca = \sqrt{\frac16}$ , giving an answer of $1 + 6 = \boxed{7}$

Answer: A 7
Problem: Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$ . Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$ . If $QR=3\sqrt3$ and $\angle QPR=60^\circ$ , then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$

Solution: Let $O=\Gamma$ be the center of the semicircle and $X=\Omega$ be the center of the circle.
Applying the Extended Law of Sines to $\triangle PQR,$ we find the radius of $\odot X:$ \[XP=\frac{QR}{2\cdot\sin \angle QPR}=\frac{3\sqrt3}{2\cdot\frac{\sqrt3}{2}}=3.\] Alternatively, by the Inscribed Angle Theorem, $\triangle QRX$ is a $30^\circ\text{-}30^\circ\text{-}120^\circ$ triangle with base $QR=3\sqrt3.$ Dividing $\triangle QRX$ into two congruent $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles, we get that the radius of $\odot X$ is $XQ=XR=3$ by the side-length ratios.
Let $M$ be the midpoint of $\overline{QR}.$ By the Perpendicular Chord Bisector Converse, we have $\overline{OM}\perp\overline{QR}$ and $\overline{XM}\perp\overline{QR}.$ Together, points $O, X,$ and $M$ must be collinear.
By the SAS Congruence, we have $\triangle QXM\cong\triangle RXM,$ both of which are $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles. By the side-length ratios, we obtain $RM=\frac{3\sqrt3}{2}, RX=3,$ and $XM=\frac{3}{2}.$ By the Pythagorean Theorem on right $\triangle ORM,$ we get $OM=\frac{13}{2}$ and $OX=OM-XM=5.$ By the Pythagorean Theorem on right $\triangle OXP,$ we get $OP=4.$
Let $C$ be the foot of the perpendicular from $P$ to $\overline{QR},$ and $D$ be the foot of the perpendicular from $X$ to $\overline{PC},$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(300); pair O, X, A, B, P, Q, R, M, C, D; O = (0,0); X = (4,3); A = (-7,0); B = (7,0); P = (4,0); Q = intersectionpoints(Circle(O,7),Circle(X,3))[0]; R = intersectionpoints(Circle(O,7),Circle(X,3))[1]; M = midpoint(Q--R); C = foot(P,Q,R); D = foot(X,P,C); fill(P--Q--R--cycle,yellow); dot("$O$",O,S); dot("$X$",X,N); dot("$A$",A,SW); dot("$B$",B,SE); dot("$P$",P,S); dot("$Q$",Q,E); dot("$R$",R,N); dot("$M$",M,dir(M)); dot("$C$",C,NE); dot("$D$",D,SE); markscalefactor=0.0375; draw(rightanglemark(O,M,R),red); draw(rightanglemark(P,C,M),red); draw(rightanglemark(P,D,X),red); draw(rightanglemark(O,P,X),red); draw(P--Q--R--cycle); draw(arc(O, 7, 0, 180)^^A--B^^Circle(X,3)); draw(O--M^^X--P); draw(P--C^^X--D,dashed); [/asy] Clearly, quadrilateral $XDCM$ is a rectangle. Since $\angle XPD=\angle OXP$ by alternate interior angles, we have $\triangle XPD\sim\triangle OXP$ by the AA Similarity, with the ratio of similitude $\frac{XP}{OX}=\frac 35.$ Therefore, we get $PD=\frac 95$ and $PC=PD+DC=PD+XM=\frac 95 + \frac 32 = \frac{33}{10}.$
The area of $\triangle PQR$ is \[\frac12\cdot QR\cdot PC=\frac12\cdot3\sqrt3\cdot\frac{33}{10}=\frac{99\sqrt3}{20},\] from which the answer is $99+3+20=\boxed{122}.$

Answer: D 122
Problem: Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$ . Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$ . If $QR=3\sqrt3$ and $\angle QPR=60^\circ$ , then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$

Solution: [asy] size(300); pair C = (7, 0); draw(arc(C, 7, 0, 180)); pair A = (0, 0), B = (14, 0); draw(A--B); draw(circle((11,3),3)); label("$A$", A, SSE); label("$B$", B, SSW); label("$C$", (A+B)/2, S); label("$O$", (11, 3), E); label("$P$", (11, 0), S); pair  O = (11, 3), P = (11, 0), Q = intersectionpoints(circle(C, 7), circle(O, 3))[1], R = intersectionpoints(circle(C, 7), circle(O, 3))[0], S = (Q+R)/2, N = (121/8, 0), T = (8/11)*N + (3/11)*R, X = (4/7)*T + (3/7)*S; draw(C--O, blue); draw(O--S, red); draw(C--Q); draw(C--R); draw(Q--N--B); draw(O--P); draw(O--Q); draw(O--R); draw(P--Q--R--cycle); draw(B--T); draw(P--X); label("$Q$", Q, NNE); label("$R$", R, E); label("$S$", S, ENE); label("$N$", N, SSE); label("$T$", T, ENE); label("$X$", X, NE); draw(rightanglemark(P, X, Q)); draw(rightanglemark(B, T, R)); draw(rightanglemark(C, S, Q)); [/asy] Define points as shown above, where $N=\overleftrightarrow{PA}\cap\overleftrightarrow{QR}$ . The area of $\triangle PQR$ is simply \[\dfrac{1}{2}PX\cdot QR=\dfrac{3\sqrt{3}}{2}PX;\] it remains to compute the value of $PX$ . Note that $PX$ is simply a weighted average of $BT$ and $CS;$ it is $\dfrac{CP}{BP}$ times closer to $BT$ than it is to $CS$ . Observe that \[CS=\sqrt{CQ^{2}-\left(\dfrac{1}{2}QR\right)^{2}}=\sqrt{7^{2}-\left(\dfrac{3\sqrt{3}}{2}\right)^{2}}=6.5\] since the radius of $\Gamma$ is $7$ as its diameter is $14$ . Note also by the Extended Law of Sines the radius of $\Omega$ is $\dfrac{3\sqrt{3}}{2\sin 60^{\circ}}=3,$ so $OS=3\cos 60^{\circ}=1.5$ . Since $C, O,$ and $S$ are collinear by symmetry we have $CO=CS-OS=5,$ so $CP=\sqrt{5^{2}-3^{2}}=4$ and $BP=7-4=3$ . Therefore, $\triangle OPC$ is a $3\text{-}4\text{-}5$ right triangle; $\triangle OPC\sim\triangle NSC$ since $\angle OPC=\angle CSN=90^{\circ}$ and $\angle OCP=\angle NCS=\sin^{-1}\left(\dfrac{3}{5}\right)$ . Therefore $\dfrac{CN}{CS}=\cfrac{CO}{CP}=\dfrac{5}{4}$ so $CN=\dfrac{5}{4}CS=\dfrac{65}{8}$ . Since $\triangle BTN\sim\triangle CSN,$ we have $\dfrac{BT}{BN}=\dfrac{CS}{CN}=\dfrac{4}{5}$ . Therefore \[BT=\dfrac{4}{5}BN=\dfrac{4}{5}\left(CN-7\right)=\dfrac{4}{5}\cdot\dfrac{9}{8}=\dfrac{36}{40}=0.9;\] so $PX$ is $\dfrac{4}{3}$ times as close to $0.9$ as to $6.5;$ we can compute $PX=\dfrac{4}{7}BT+\dfrac{3}{7}CS=\dfrac{4}{7}\cdot0.9+\dfrac{3}{7}\cdot6.5=3.3$ . The area of $\triangle PQR$ is \[\dfrac{3\sqrt{3}}{2}\cdot 3.3=\dfrac{99\sqrt{3}}{20}\] and $99+3+20=\boxed{122}$

Answer: D 122
Problem: Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$ . Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$ . If $QR=3\sqrt3$ and $\angle QPR=60^\circ$ , then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$

Solution: Let $O_{1}$ be the center of $\odot\Gamma, O_2$ be the center of $\odot\Omega,$ and $M$ be the midpoint of $\overline{QR}.$ We have $O_{1}M=\sqrt{7^2-\left(\frac{3\sqrt3}{2}\right)^2}=\frac{13}{2}$ and by Extended Law of Sines, the radius of $\odot\Omega$ is $\frac{3\sqrt3}{2\sin 60^\circ}=3$ so $O_{2}M=3\cos 60^\circ=\frac{3}{2}.$ Therefore $O_{1}O_{2}=O_{1}M-O_{2}M=5$ and $O_{1}P=\sqrt{5^2 - 3^2}=4.$
Let $X=\overline{AB}\cap\overline{QR}.$ Obviously \[\angle O_{1}PO_{2}=\angle O_{1}MX=90^\circ~\text{and}~\angle PO_{1}O_{2}=\angle MO_{1}X=\arcsin\left(\frac{3}{5}\right)\] so $\triangle PO_{1}O_{2}\sim\triangle MO_{1}X$ with ratio $\frac{PO_{1}}{MO_{1}}=\frac{4}{\tfrac{13}{2}}=\frac{8}{13}.$ Therefore $O_1X=\frac{13}{8}\cdot O_{1}O_{2}=\frac{13}{8}\cdot5=\frac{65}{8}$ and $MX=\frac{13}{8}\cdot PO_{2}=\frac{13}{8}\cdot3=\frac{39}{8}.$
Let $H$ denote the foot of the altitude from $P$ to $\overline{QR}.$ Because $\overline{PH}\parallel\overline{O_{1}M},$ it follows that $\triangle PHX\sim\triangle O_{1}MX.$ This similarity has ratio \[\frac{PX}{O_{1}X}=1-\frac{O_{1}P}{O_{1}X}=1-\frac{4}{\tfrac{65}{8}}=1-\frac{32}{65}=\frac{33}{65}.\] We therefore have $PH=\frac{33}{65}\cdot O_{1}M=\frac{33}{65}\cdot\frac{13}{2}=\frac{33}{10}.$
Finally, the area of $\triangle PQR$ is \[\frac{1}{2}\cdot QR\cdot PH=\frac{1}{2}\cdot3\sqrt3\cdot\frac{33}{10}=\frac{99\sqrt3}{20},\] so the answer is $99+3+20=\boxed{122}.$

Answer: D 122
Problem: Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$ . Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$ . If $QR=3\sqrt3$ and $\angle QPR=60^\circ$ , then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$

Solution: By the Law of Sine in $\triangle PQR$ and its circumcircle $\odot \Omega$ $2r_{\Omega} = \frac{QR}{ \sin 60^{\circ} } = \frac{ 3\sqrt{3} }{ \frac{ \sqrt{3} }{2} } = 6$ $r_{\Omega} = 3$
\[\Gamma \Omega = \sqrt{r_{\Gamma}^2 - \left( \frac{ PQ }{2}\right)^2} - \sqrt{r_{\Omega} - \left( \frac{PQ}{2}\right)^2} = \sqrt{7^2 - \left( \frac{ 3 \sqrt{3} }{2}\right)^2} - \sqrt{3^2 - \left( \frac{ 3 \sqrt{3} }{2}\right)^2} = \frac{13}{2} - \frac32 = 5, \quad \Gamma P = \sqrt{5^2 - 3^2} = 4\]
By Power of a Point in $\odot \Gamma$ $PQ \cdot PS = PA \cdot PB = (7+4)(7-4) = 33$
By the Law of Sine in $\triangle PRS$ $\frac{PR}{PS} = \frac{ \sin \angle PSR }{ \sin \angle PRS }$
By the Law of Sine in $\triangle QRS$ and its circumcircle $\odot \Gamma$ $\frac{QR}{ \sin \angle PSR } = 14$ $\frac{ 3\sqrt{3} }{ \sin \angle PSR } = 14$ $\sin \angle PSR = \frac{ 3\sqrt{3} }{14}$ $\cos \angle PSR = \frac{ 13 }{14}$
\[\sin \angle PRS = \sin ( 60^{\circ} - \angle PSR ) = \sin 60^{\circ} \cos \angle PSR - \sin \angle PSR \cos 60^{\circ} = \frac{ \sqrt{3} }{2} \cdot \frac{ 13 }{14} - \frac{ 3\sqrt{3} }{14} \cdot \frac12 = \frac{ 5\sqrt{3} }{14}\]
\[\frac{PR}{PS} = \frac{ \frac{ 3\sqrt{3} }{14} }{ \frac{ 5\sqrt{3} }{14} } = \frac35, \quad PQ \cdot PR = PQ \cdot PS \cdot \frac{PR}{PS} = 33 \cdot \frac35 = \frac{99}{5}\]
\[[PQR] = \frac12 \cdot \sin 60^{\circ} \cdot PQ \cdot PR = \frac12 \cdot \frac12 \cdot \frac{99}{5} = \frac{ 99\sqrt{3} }{20}, \quad 99 + 3 + 20 = \boxed{122}\]

Answer: D 122
Problem: Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$ . Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$ . If $QR=3\sqrt3$ and $\angle QPR=60^\circ$ , then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$

Solution: Following Solution 4, We have $O_{1}$ (0,0) , $O_{2}$ (4,3). 
We can write the equation of the two circles as: \[\odot\Gamma  :  x^{2}+y^{2}=7^{2}...(1)\] \[\odot\Omega :  (x-4)^{2}+(y-3)^{2}=3^{2}...(2)\] By substituting (1) into (2), we get \[8x+6y-65=0...(3)\] Notice (3) is the relationship between $x$ value and $y$ value, in other words, (3) is the linear equation that go through $R$ and $Q$ . 
Let the height drops from $P$ to $QR$ at $H$ . Therefore, we have \[Area \triangle PQR =\frac{1}{2}\cdot{QR}\cdot{PH}\] So \[QR=3\sqrt{3}\] And by distance formula, $PH$ is the distance from $P$ (4,0) to $\overline{QR}$ \[PH={\frac{|8\cdot4+6\cdot0-65|}{\sqrt{8^{2}+6^{2}}}=\frac{33}{10}}\] Thus, We get \[Area \triangle PQR =\frac{1}{2}\cdot3\sqrt{3}\cdot\frac{33}{10}=\frac{99\sqrt{3}}{20}\] So the answer is $99+3+20=\boxed{122}.$

Answer: D 122
Problem: Let $d(n)$ denote the number of positive integers that divide $n$ , including $1$ and $n$ . For example, $d(1)=1,d(2)=2,$ and $d(12)=6$ . (This function is known as the divisor function.) Let \[f(n)=\frac{d(n)}{\sqrt [3]n}.\] There is a unique positive integer $N$ such that $f(N)>f(n)$ for all positive integers $n\ne N$ . What is the sum of the digits of $N?$
$\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9$

Solution: We consider the prime factorization of $n:$ \[n=\prod_{i=1}^{k}p_i^{e_i}.\] By the Multiplication Principle, we have \[d(n)=\prod_{i=1}^{k}(e_i+1).\] Now, we rewrite $f(n)$ as \[f(n)=\frac{d(n)}{\sqrt [3]n}=\frac{\prod_{i=1}^{k}(e_i+1)}{\prod_{i=1}^{k}p_i^{e_i/3}}=\prod_{i=1}^{k}\frac{e_i+1}{p_i^{{e_i}/3}}.\] As $f(n)>0$ for all positive integers $n,$ note that $f(a)>f(b)$ if and only if $f(a)^3>f(b)^3$ for all positive integers $a$ and $b.$ So, $f(n)$ is maximized if and only if \[f(n)^3=\prod_{i=1}^{k}\frac{(e_i+1)^3}{p_i^{{e_i}}}\] is maximized.
For each independent factor $\frac{(e_i+1)^3}{p_i^{e_i}}$ with a fixed prime $p_i,$ where $1\leq i\leq k,$ the denominator grows faster than the numerator, as exponential functions always grow faster than polynomial functions. Therefore, for each prime $p_i$ with $\left(p_1,p_2,p_3,p_4,\ldots\right)=\left(2,3,5,7,\ldots\right),$ we look for the nonnegative integer $e_i$ such that $\frac{(e_i+1)^3}{p_i^{e_i}}$ is a relative maximum: \[\begin{array}{c|c|c|c|c}  & & & & \\ [-2.25ex] \boldsymbol{i} & \boldsymbol{p_i} & \boldsymbol{e_i} & \boldsymbol{\dfrac{(e_i+1)^3}{p_i^{e_i}}} & \textbf{Max?} \\ [2.5ex] \hline\hline  & & & & \\ [-2ex] 1 & 2 & 0 & 1 & \\      & & 1 & 4 & \\     & & 2 & 27/4 &\\     & & 3 & 8 & \checkmark\\     & & 4 & 125/16 & \\ [0.5ex] \hline   & & & & \\ [-2ex] 2 & 3 & 0 & 1 &\\     & & 1 & 8/3 & \\     & & 2 & 3 &  \checkmark\\     & & 3 & 64/27 &  \\ [0.5ex] \hline   & & & & \\ [-2ex] 3 & 5 & 0 & 1 &  \\     & & 1 & 8/5 &  \checkmark\\     & & 2 & 27/25 & \\ [0.5ex] \hline   & & & & \\ [-2ex] 4 & 7 & 0 & 1 &  \\     & & 1 & 8/7 &  \checkmark\\     & & 2 & 27/49 & \\ [0.5ex] \hline   & & & & \\ [-2ex] \geq5 & \geq11 & 0 & 1 & \checkmark \\     & & \geq1 & \leq8/11 &   \\ [0.5ex] \end{array}\] Finally, the positive integer we seek is $N=2^3\cdot3^2\cdot5^1\cdot7^1=2520.$ The sum of its digits is $2+5+2+0=\boxed{9}.$

Answer: E 9
Problem: Let $d(n)$ denote the number of positive integers that divide $n$ , including $1$ and $n$ . For example, $d(1)=1,d(2)=2,$ and $d(12)=6$ . (This function is known as the divisor function.) Let \[f(n)=\frac{d(n)}{\sqrt [3]n}.\] There is a unique positive integer $N$ such that $f(N)>f(n)$ for all positive integers $n\ne N$ . What is the sum of the digits of $N?$
$\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9$

Solution: The question statement asks for the value of $N$ that maximizes $f(N)$ . Let $N$ start out at $1$ ; we will find what factors to multiply $N$ by, in order for $N$ to maximize the function.
First, we will find what power of $2$ to multiply $N$ by. If we multiply $N$ by $2^{a}$ , the numerator of $f$ $d(N)$ , will multiply by a factor of $a+1$ ; this is because the number $2^{a}$ has $a+1$ divisors. The denominator, $\sqrt[3]{N}$ , will simply multiply by $\sqrt[3]{2^{a}}$ . Therefore, the entire function multiplies by a factor of $\frac{a+1}{\sqrt[3]{2^{a}}}$ . We want to find the integer value of $a$ that maximizes this value. By inspection, this is $3$ . Therefore, we multiply $N$ by $8$ ; right now, $N$ is $8$
Next, we will find what power of $3$ to multiply $N$ by. Similar to the previous step, we wish to find the integer value of $a$ that maximizes $\frac{a+1}{\sqrt[3]{3^{a}}}$ . This value, also by inspection, is $2$
We can repeat this step on the rest of the primes to get \[N = 2^{3} \cdot 3^{2} \cdot 5 \cdot 7\] but from $11$ on, $a=0$ will maximize the value of the function, so the prime is not a factor in $N$ .
We evaluate $N$ to be $2520$ , so the answer is $\boxed{9}$

Answer: E 9
Problem: Let $d(n)$ denote the number of positive integers that divide $n$ , including $1$ and $n$ . For example, $d(1)=1,d(2)=2,$ and $d(12)=6$ . (This function is known as the divisor function.) Let \[f(n)=\frac{d(n)}{\sqrt [3]n}.\] There is a unique positive integer $N$ such that $f(N)>f(n)$ for all positive integers $n\ne N$ . What is the sum of the digits of $N?$
$\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9$

Solution: Using the answer choices to our advantage, we can show that $N$ must be divisible by 9 without explicitly computing $N$ , by exploiting the following fact:
Claim : If $n$ is not divisible by 3, then $f(9n) > f(3n) > f(n)$
Proof : Since $d(\cdot)$ is a multiplicative function , we have $d(3n) = d(3)d(n) = 2d(n)$ and $d(9n) = 3d(n)$ . Then \begin{align*} f(3n) &= \frac{2d(n)}{\sqrt[3]{3n}} \approx 1.38 f(n)\\ f(9n) &= \frac{3d(n)}{\sqrt[3]{9n}} \approx 1.44 f(n) \end{align*} Note that the values $\frac{2}{\sqrt[3]{3}}$ and $\frac{3}{\sqrt[3]{9}}$ do not have to be explicitly computed; we only need the fact that $\frac{3}{\sqrt[3]{9}} > \frac{2}{\sqrt[3]{3}} > 1$ which is easy to show by hand.
The above claim automatically implies $N$ is a multiple of 9: if $N$ was not divisible by 9, then $f(9N) > f(N)$ which is a contradiction, and if $N$ was divisible by 3 and not 9, then $f(3N) > f(N) > f\left(\frac{N}{3}\right)$ , also a contradiction. Then the sum of digits of $N$ must be a multiple of 9, so only choice $\boxed{9}$ works.

Answer: E 9
Problem: Let $d(n)$ denote the number of positive integers that divide $n$ , including $1$ and $n$ . For example, $d(1)=1,d(2)=2,$ and $d(12)=6$ . (This function is known as the divisor function.) Let \[f(n)=\frac{d(n)}{\sqrt [3]n}.\] There is a unique positive integer $N$ such that $f(N)>f(n)$ for all positive integers $n\ne N$ . What is the sum of the digits of $N?$
$\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9$

Solution: The problem mentions the sum of digits - recall that if a number is divisible by 9, then so is the sum of its digits. Guess that the answer must therefore be $\boxed{9}$

Answer: E 9
Problem: Two distinct numbers are selected from the set $\{1,2,3,4,\dots,36,37\}$ so that the sum of the remaining $35$ numbers is the product of these two numbers. What is the difference of these two numbers?
$\textbf{(A) }5 \qquad \textbf{(B) }7 \qquad \textbf{(C) }8\qquad \textbf{(D) }9 \qquad \textbf{(E) }10$

Solution: The sum of the first $n$ integers is given by $\frac{n(n+1)}{2}$ , so $\frac{37(37+1)}{2}=703$
Therefore, $703-x-y=xy$
Rearranging, $xy+x+y=703$ . We can factor this equation by SFFT to get
$(x+1)(y+1)=704$
Looking at the possible divisors of $704 = 2^6\cdot11$ $22$ and $32$ are within the constraints of $0 < x \leq y \leq 37$ so we try those:
$(x+1)(y+1) = 22\cdot32$
$x+1=22, y+1 = 32$
$x = 21, y = 31$
Therefore, the difference $y-x=31-21=\boxed{10}$

Answer: E 10
Problem: How many values of $\theta$ in the interval $0<\theta\le 2\pi$ satisfy \[1-3\sin\theta+5\cos3\theta = 0?\] $\textbf{(A) }2 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5\qquad \textbf{(D) }6 \qquad \textbf{(E) }8$

Solution: We rearrange to get \[5\cos3\theta = 3\sin\theta-1.\] We can graph two functions in this case: $y=5\cos{3x}$ and $y=3\sin{x} -1$ .
Using transformation of functions, we know that $5\cos{3x}$ is just a cosine function with amplitude $5$ and period $\frac{2\pi}{3}$ . Similarly, $3\sin{x} -1$ is just a sine function with amplitude $3$ and shifted $1$ unit downward: [asy] import graph;  size(400,200,IgnoreAspect);  real Sin(real t) {return 3*sin(t) - 1;} real Cos(real t) {return 5*cos(3*t);}  draw(graph(Sin,0, 2pi),red,"$3\sin{x} -1 $"); draw(graph(Cos,0, 2pi),blue,"$5\cos{3x}$");  xaxis("$x$",BottomTop,LeftTicks); yaxis("$y$",LeftRight,RightTicks(trailingzero));  add(legend(),point(E),20E,UnFill); [/asy] So, we have $\boxed{6}$ solutions.

Answer: D 6
Problem: Let $g(x)$ be a polynomial with leading coefficient $1,$ whose three roots are the reciprocals of the three roots of $f(x)=x^3+ax^2+bx+c,$ where $1<a<b<c.$ What is $g(1)$ in terms of $a,b,$ and $c?$
$\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\frac{a+b+c}{c^2} \qquad \textbf{(E) }\frac{1+a+b+c}{a+b+c}$

Solution: Note that $f(1/x)$ has the same roots as $g(x)$ , if it is multiplied by some monomial so that the leading term is $x^3$ they will be equal. We have \[f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c\] so we can see that \[g(x) = \frac{x^3}{c}f(1/x)\] Therefore \[g(1) = \frac{1}{c}f(1) = \boxed{1}\]

Answer: A 1