Answer: 672.0

Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$ , respectively, intersect at distinct points $A$ and $B$ . A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$ . Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120^{\circ}$ . Find the distance between the centers of $\omega_1$ and $\omega_2$

Solution:
Let $O_i$ and $r_i$ be the center and radius of $\omega_i$ , and let $O$ and $r$ be the center and radius of $\omega$
Since $\overline{AB}$ extends to an arc with arc $120^\circ$ , the distance from $O$ to $\overline{AB}$ is $r/2$ . Let $X=\overline{AB}\cap \overline{O_1O_2}$ . Consider $\triangle OO_1O_2$ . The line $\overline{AB}$ is perpendicular to $\overline{O_1O_2}$ and passes through $X$ . Let $H$ be the foot from $O$ to $\overline{O_1O_2}$ ; so $HX=r/2$ . We have by tangency $OO_1=r+r_1$ and $OO_2=r+r_2$ . Let $O_1O_2=d$  Since $X$ is on the radical axis of $\omega_1$ and $\omega_2$ , it has equal power with respect to both circles, so \[O_1X^2 - r_1^2 = O_2X^2-r_2^2 \implies O_1X-O_2X = \frac{r_1^2-r_2^2}{d}\] since $O_1X+O_2X=d$ . Now we can solve for $O_1X$ and $O_2X$ , and in particular, \begin{align*} O_1H &= O_1X - HX = \frac{d+\frac{r_1^2-r_2^2}{d}}{2} - \frac{r}{2} \\ O_2H &= O_2X + HX = \frac{d-\frac{r_1^2-r_2^2}{d}}{2} + \frac{r}{2}.  \end{align*} We want to solve for $d$ . By the Pythagorean Theorem (twice): \begin{align*} &\qquad -OH^2 = O_2H^2 - (r+r_2)^2 = O_1H^2 - (r+r_1)^2 \\ &\implies \left(d+r-\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_2)^2 = \left(d-r+\tfrac{r_1^2-r_2^2}{d}\right)^2 - 4(r+r_1)^2 \\ &\implies 2dr - 2(r_1^2-r_2^2)-8rr_2-4r_2^2 = -2dr+2(r_1^2-r_2^2)-8rr_1-4r_1^2 \\ &\implies 4dr = 8rr_2-8rr_1 \\ &\implies d=2r_2-2r_1 \end{align*} Therefore, $d=2(r_2-r_1) = 2(961-625)=\boxed{672}$

Solution: Let $O_{1}$ $O_{2}$ , and $O$ be the centers of $\omega_{1}$ $\omega_{2}$ , and $\omega$ with $r_{1}$ $r_{2}$ , and $r$ their radii, respectively. Then, the distance from $O$ to the radical axis $\ell\equiv\overline{AB}$ of $\omega_{1}, \omega_{2}$ is equal to $\frac{1}{2}r$ . Let $x=O_{1}O_{2}$ and $O^{\prime}$ the orthogonal projection of $O$ onto line $\ell$ . Define the function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by \[f(X)=\text{Pow}_{\omega_{1}}(X)-\text{Pow}_{\omega_{2}}(X).\] Then \begin{align*} f(O_{1})=-r_{1}^{2}-(x-r_{2})(x+r_{2})&=-x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O_{2})=(x-r_{1})(x+r_{1})-(-r_{2}^{2})&=x^{2}+r_{2}^{2}-r_{1}^{2}, \\ f(O)=r(r+2r_{1})-r(r+2r_{2})&=2r(r_{1}-r_{2}), \\ f(O^{\prime})&=0. \end{align*} By linearity \[\frac{f(O_{2})-f(O_{1})}{f(O)-f(O^{\prime})}=\frac{O_{1}O_{2}}{OO^{\prime}}=\frac{x}{\tfrac{1}{2}r}=\frac{2x}{r}.\] Notice that $f(O_{2})-f(O_{1})=x^{2}-(-x^{2})=2x^{2}$ and $f(O)-f(O^{\prime})=2r(r_{1}-r_{2})$ , thus \begin{align*}\frac{2x^{2}}{2r(r_{1}-r_{2})}&=\frac{2x}{r}\end{align*} Dividing both sides by $\frac{2x}{r}$ (which is obviously nonzero as $x$ is nonzero) gives us \begin{align*}\frac{x}{2(r_{1}-r_{2})}&=1\end{align*} so $x=2(r_{1}-r_{2})$ . Since $r_{1}=961$ and $r_{2}=625$ , the answer is $x=2\cdot(961-625)=\boxed{672}$

Solution: Denote by $O_1$ $O_2$ , and $O$ the centers of $\omega_1$ $\omega_2$ , and $\omega$ , respectively. Let $R_1 = 961$ and $R_2 = 625$ denote the radii of $\omega_1$ and $\omega_2$ respectively, $r$ be the radius of $\omega$ , and $\ell$ the distance from $O$ to the line $AB$ . We claim that \[\dfrac{\ell}{r} = \dfrac{R_2-R_1}{d},\] where $d = O_1O_2$ . This solves the problem, for then the $\widehat{PQ} = 120^\circ$ condition implies $\tfrac{\ell}r = \cos 60^\circ = \tfrac{1}{2}$ , and then we can solve to get $d = \boxed{672}$

Solution: Suppose we label the points as shown below  By radical axis, the tangents to $\omega$ at $D$ and $E$ intersect on $AB$ . Thus $PDQE$ is harmonic, so the tangents to $\omega$ at $P$ and $Q$ intersect at $X \in DE$ . Moreover, $OX \parallel O_1O_2$ because both $OX$ and $O_1O_2$ are perpendicular to $AB$ , and $OX = 2OP$ because $\angle POQ = 120^{\circ}$ . Thus \[O_1O_2 = O_1Y - O_2Y = 2 \cdot 961 - 2\cdot 625 = \boxed{672}\] by similar triangles.

Solution:
Like in other solutions, let $O$ be the center of $\omega$ with $r$ its radius; also, let $O_{1}$ and $O_{2}$ be the centers of $\omega_{1}$ and $\omega_{2}$ with $R_{1}$ and $R_{2}$ their radii, respectively. Let line $OP$ intersect line $O_{1}O_{2}$ at $T$ , and let $u=TO_{2}$ $v=TO_{1}$ $x=PT$ , where the length $O_{1}O_{2}$ splits as $u+v$ . Because the lines $PQ$ and $O_{1}O_{2}$ are perpendicular, lines $OT$ and $O_{1}O_{2}$ meet at a $60^{\circ}$ angle.
Applying the Law of Cosines to $\triangle O_{2}PT$ $\triangle O_{1}PT$ $\triangle O_{2}OT$ , and $\triangle O_{1}OT$ gives \begin{align*}\triangle O_{2}PT&:O_{2}P^{2}=u^{2}+x^{2}-ux \\ \triangle O_{1}PT&:O_{1}P^{2}=v^{2}+x^{2}+vx \\ \triangle O_{2}OT&:(r+R_{2})^{2}=u^{2}+(r+x)^{2}-u(r+x) \\ \triangle O_{1}OT&:(r+R_{1})^{2}=v^{2}+(r+x)^{2}+v(r+x)\end{align*}
Adding the first and fourth equations, then subtracting the second and third equations gives us \[\left(O_{2}P^{2}-O_{1}P^{2}\right)+\left(R_{1}^{2}-R_{2}^{2}\right)+2r(R_{1}-R_{2})=r(u+v)\]
Since $P$ lies on the radical axis of $\omega_{1}$ and $\omega_{2}$ , the power of point $P$ with respect to either circle is \[O_{2}P^{2}-R_{2}^{2}=O_{1}P^{2}-R_{1}^{2}.\]
Hence $2r(R_{1}-R_{2})=r(u+v)$ which simplifies to \[u+v=2(R_{1}-R_{2}).\]
The requested distance \[O_{1}O_{2}=O_{1}T+O_{2}T=u+v\] is therefore equal to $2\cdot(961-625)=\boxed{672}$

Solution:
Let circle $\omega$ tangent circles $\omega_1$ and $\omega_2,$ respectively at distinct points $C$ and $D$ . Let $O, O_1, O_2 (r, r_1, r_2)$ be the centers (the radii) of $\omega, \omega_1$ and $\omega_2,$ respectively. WLOG $r_1 < r_2.$ Let $F$ be the point of $OO_2$ such, that $OO_1 =OF.$ Let $M$ be the midpoint $FO_1, OE \perp AB, CT$ be the radical axes of $\omega_1$ and $\omega, T \in AB.$
Then $T$ is radical center of $\omega, \omega_1$ and $\omega_2, TD = CT.$
In $\triangle OFO_1 (OF = OO_1) OT$ is bisector of $\angle O, OM$ is median
$\hspace{10mm} \implies O,T,$ and $M$ are collinear.
\[\angle OCT = \angle ODT = \angle OET = 90^\circ \implies\]
$OCTDE$ is cyclic (in $\Omega), OT$ is diameter $\Omega.$ $O_1O_2 \perp AB, OM \perp FO_1 \implies \angle FO_1O_2 = \angle OTE$ $\angle OTE = \angle ODE$ as they intercept the same arc $\overset{\Large\frown}{OE}$ in $\Omega.$ \[OE \perp AB, O_1O_2 \perp AB \implies O_1 O_2 || OE \implies\] \[\angle OO_2O_1 = \angle O_2 OE \implies \triangle ODE \sim \triangle O_2 O_1 F \implies\] \[\frac {OE}{OD} = \frac {O_2F}{O_1O_2} \implies \cos \frac {120^\circ}{2} = \frac{r_2 + r - r_1 -r} {O_1O_2}\implies {O_1O_2}= 2|r_2 – r_1|.\]
Since $r_{1}=625$ and $r_{2}=961$ , the answer is $2\cdot|961-625|=\boxed{672}$

Solution:
We are not given the radius of circle $w$ , but based on the problem statement, that radius isn't important. We can set $w$ to have radius infinity (solution 8), but if you didn't observe that, you could also set the radius to be $2r$ so that the line containing the center of $w$ , call it $W$ , and $w_2$ , call it $W_2$ , is perpendicular to the line containing the center of $w_1$ , call it $W_1$ and $w_2$ . Let $AB = 2h$ and $W_1W_2 = x$ . Also, let the projections of $W$ and $W_1$ onto line $AB$ be $X$ and $Y$ , respectively.
By Pythagorean Theorem on $\triangle WW_1W_2$ , we get \[x^2+(625+2x)^2=(961+2r)^2 \;(1)\] Note that since $\angle PWQ = 120$ $\angle PWX = 60$ . So, $WX = 2r/2 = r = W_1Y$ . We now get two more equations from Pythag: \[h^2+r^2 = 625^2 \; (2)\] \[h^2+(x-r)^2 = 961^2 \; (3)\] From subtracting $(2)$ and $(3)$ $x^2-2rx=961^2-625^2 \; (4)$ . Rearranging $(3)$ yields $x^2-1344r = 961^2+625^2$ . Plugging in our result from $(4)$ $x^2-1344r= x^2-2rx \implies 1344r = 2rx \implies x=\boxed{672}$

Solution:
Let the circle $\omega$ be infinitely big (a line). Then for it to be split into an arc of $120^{\circ}$ $\overline{PQ}$ must intersect at a $60^{\circ}$ with line $\omega$
Notice the 30-60-90 triangle in the image. $O_1R = 961 - 625$
Thus, the distance between the centers of $\omega_1$ and $\omega_2$ is $2(961-625)=\boxed{672}$


Answer: 8.0

Circles of diameter $1$ inch and $3$ inches have the same center. The smaller circle is painted red, and the portion outside the smaller circle and inside the larger circle is painted blue. What is the ratio of the blue-painted area to the red-painted area?
2006amc10b04.gif
$\textbf{(A) } 2\qquad \textbf{(B) } 3\qquad \textbf{(C) } 6\qquad \textbf{(D) } 8\qquad \textbf{(E) } 9$

Solution:
The area painted red is equal to the area of the smaller circle and the area painted blue is equal to the area of the larger circle minus the area of the smaller circle.
So we have:
\begin{align*} A_{red}&=\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4}\\ A_{blue}&=\pi\left(\frac{3}{2}\right)^2-\pi\left(\frac{1}{2}\right)^2=2\pi\\ \end{align*}
So the desired ratio is:
$\frac{A_{blue}}{A_{red}}=\frac{2\pi}{\frac{\pi}{4}}=2\cancel{\pi}\cdot \frac{4}{\cancel{\pi}}=\boxed{8}.$


Answer: 17.0

Circles of radii $5, 5, 8,$ and $\frac mn$ are mutually externally tangent, where $m$ and $n$ are relatively prime positive integers.  Find $m + n.$

Solution:
1997 AIME-4.png
If (in the diagram above) we draw the line going through the centers of the circles with radii $8$ and $\frac mn = r$ , that line is the perpendicular bisector of the segment connecting the centers of the two circles with radii $5$ . Then we form two right triangles , of lengths $5, x, 5+r$ and $5, 8+r+x, 13$ , wher $x$ is the distance between the center of the circle in question and the segment connecting the centers of the two circles of radii $5$ . By the Pythagorean Theorem , we now have two equations with two unknowns:
\begin{eqnarray*} 5^2 + x^2 &=& (5+r)^2 \\ x &=& \sqrt{10r + r^2} \\ && \\ (8 + r + \sqrt{10r+r^2})^2 + 5^2 &=& 13^2\\ 8 + r + \sqrt{10r+r^2} &=& 12\\ \sqrt{10r+r^2}&=& 4-r\\ 10r+r^2 &=& 16 - 8r + r^2\\ r &=& \frac{8}{9} \end{eqnarray*}
So $m+n = \boxed{17}$

Solution: We may also use Descartes' theorem, $k_4=k_1+k_2+k_3\pm 2\sqrt{k_1k_2+k_2k_3+k_3k_1}$ where each of $k_i$ is the curvature of a circle with radius $r_i$ , and the curvature is defined as $k_i=\frac{1}{r_i}$ . The larger solution for $k_4$ will give the curvature of the circle externally tangent to the other circles, while the smaller solution will give the curvature for the circle internally tangent to each of the other circles. Using Descartes' theorem, we get $k_4=\frac15+\frac15+\frac18+2\sqrt{\frac{1}{40}+\frac{1}{40}+\frac{1}{25}}=\frac{21}{40}+2\sqrt{\frac{45}{500}}=\frac{45}{40}$ . Thus, $r_4=\frac{1}{k_4}=\frac{40}{45}=\frac89$ , and the answer is $\boxed{017}$


Answer: 12.0

Circles of radius $2$ and $3$ are externally tangent and are circumscribed by a third circle, as shown in the figure. Find the area of the shaded region.
$\mathrm{(A) \ } 3\pi\qquad \mathrm{(B) \ } 4\pi\qquad \mathrm{(C) \ } 6\pi\qquad \mathrm{(D) \ } 9\pi\qquad \mathrm{(E) \ } 12\pi$

Solution:
A line going through the centers of the two smaller circles also goes through the diameter. The length of this line within the circle is $3+3+2+2=10.$ Because this is the length of the larger circle's diameter, the length of its radius is $5.$
The area of the large circle is $25\pi$ , and the area of the two smaller circles is $9\pi + 4\pi = 13\pi.$ To find the area of the shaded region, subtract the area of the two smaller circles from the area of the large circle. $\longrightarrow 25\pi - 13\pi = \boxed{12}$


Answer: 224.0

Circles of radius $3$ and $6$ are externally tangent to each other and are internally tangent to a circle of radius $9$ . The circle of radius $9$ has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord.

Solution: We label the points as following: the centers of the circles of radii $3,6,9$ are $O_3,O_6,O_9$ respectively, and the endpoints of the chord are $P,Q$ . Let $A_3,A_6,A_9$ be the feet of the perpendiculars from $O_3,O_6,O_9$ to $\overline{PQ}$ (so $A_3,A_6$ are the points of tangency ). Then we note that $\overline{O_3A_3} \parallel \overline{O_6A_6} \parallel \overline{O_9A_9}$ , and $O_6O_9 : O_9O_3 = 3:6 = 1:2$ . Thus, $O_9A_9 = \frac{2 \cdot O_6A_6 + 1 \cdot O_3A_3}{3} = 5$ (consider similar triangles). Applying the Pythagorean Theorem to $\triangle O_9A_9P$ , we find that \[PQ^2 = 4(A_9P)^2 = 4[(O_9P)^2-(O_9A_9)^2] = 4[9^2-5^2] = \boxed{224}\]


Answer: 16.0

Cities $A$ $B$ $C$ $D$ , and $E$ are connected by roads $\widetilde{AB}$ $\widetilde{AD}$ $\widetilde{AE}$ $\widetilde{BC}$ $\widetilde{BD}$ $\widetilde{CD}$ , and $\widetilde{DE}$ . How many different routes are there from $A$ to $B$ that use each road exactly once? (Such a route will necessarily visit some cities more than once.)
[asy] unitsize(10mm); defaultpen(linewidth(1.2pt)+fontsize(10pt)); dotfactor=4; pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08); dot (A); dot (B); dot (C); dot (D); dot (E); label("$A$",A,S); label("$B$",B,SE); label("$C$",C,E); label("$D$",D,N); label("$E$",E,W); guide squiggly(path g, real stepsize, real slope=45) {  real len = arclength(g);  real step = len / round(len / stepsize);  guide squig;  for (real u = 0; u < len; u += step){  real a = arctime(g, u);  real b = arctime(g, u + step / 2);  pair p = point(g, a);  pair q = point(g, b);  pair np = unit( rotate(slope) * dir(g,a));  pair nq = unit( rotate(0 - slope) * dir(g,b));  squig = squig .. p{np} .. q{nq};  }  squig = squig .. point(g, length(g)){unit(rotate(slope)*dir(g,length(g)))};  return squig; } pen pp = defaultpen + 2.718; draw(squiggly(A--B, 4.04, 30), pp); draw(squiggly(A--D, 7.777, 20), pp); draw(squiggly(A--E, 5.050, 15), pp); draw(squiggly(B--C, 5.050, 15), pp); draw(squiggly(B--D, 4.04, 20), pp); draw(squiggly(C--D, 2.718, 20), pp); draw(squiggly(D--E, 2.718, -60), pp); [/asy]
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 18$

Solution:
Note that cities $C$ and $E$ can be removed when counting paths because if a path goes in to $C$ or $E$ , there is only one possible path to take out of cities $C$ or $E$ .
So the diagram is as follows:

Now we proceed with casework. Remember that there are two ways to travel from $A$ to $D$ $D$ to $A$ $B$ to $D$ and $D$ to $B$ .:
Case 1 $A \Rightarrow D$ : From $D$ , if the path returns to $A$ , then the next path must go to $B\Rightarrow D \Rightarrow B$ . There are $2 \cdot 1 \cdot 2 = 4$ possibilities of the path $ADABDB$ . If the path goes to $D$ from $B$ , then the path must continue with either $BDAB$ or $BADB$ . There are $2 \cdot 2 \cdot 2 = 8$ possibilities. So, this case gives $4+8=12$ different possibilities.
Case 2 $A \Rightarrow B$ : The path must continue with $BDADB$ . There are $2 \cdot 2 = 4$ possibilities for this case.
Putting the two cases together gives $12+4 = \boxed{16}$


Answer: 27.0

Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$ . Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? $\textbf{(A) }20\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$

Solution: This is a $d=st$ problem, so let $x$ be the time it takes to meet. We can write the following equation: \[12x+18x=45\] Solving gives us $x=1.5$ . The $18x$ is Alicia so $18\times1.5=\boxed{27}$

Solution: The relative speed of the two is $18+12=30$ , so $\frac{3}{2}$ hours would be required to travel $45$ miles. $d=st$ , so $x=18\cdot\frac{3}{2}=\boxed{27}$

Solution: Since $18$ mph is $\frac{3}{2}$ times $12$ mph, Alicia will travel $\frac{3}{2}$ times as far as Beth. If $x$ is the distance Beth travels, \[\frac{3}{2}x+x=45\] \[\frac{5}{2}x=45\] \[x=18\] Since this is the amount Beth traveled, the amount that Alicia traveled was \[45-18=\boxed{27}\]

Solution: Alice and Barbara close in on each other at 30mph. Since they are 45 miles apart, they will meet in t = d/s = 45miles / 30mph = 3/2 hours. We can either calculate the distance Alice travels at 18mph or the distance Barbara travels at 12mph; since we want the distance from Alice, we go with the former. Alice (and Barbara) will meet in 1 1/2 hours at 18mph x 3/2 hours = 27 miles from A. $\boxed{27}$

Solution:
We know that Beth starts 45 miles away from City A, let’s create two equations:
Alice-> $18t=d$ Beth-> $-12t+45=d$ [-12 is the slope; 45 is the y-intercept]
Solve the system:
$18t=-12t+45 30t=45 t=1.5$
So, $18(1.5)=$ $\boxed{27}$

Solution: Since Alicia and Beth's speeds are constant, they are directly proportional to their distances covered, so the ratio of their speeds is equal to the ratio of their covered distances. Since Alicia travels $\frac{18}{30} = \frac{3}{5}$ of their combined speed, she travels $\frac{3}{5}\cdot 45 = \boxed{27}$ of the total distance.


Answer: 27.0

Solution: This is a $d=st$ problem, so let $x$ be the time it takes to meet. We can write the following equation: \[12x+18x=45\] Solving gives us $x=1.5$ . The $18x$ is Alicia so $18\times1.5=\boxed{27}$

Solution: The relative speed of the two is $18+12=30$ , so $\frac{3}{2}$ hours would be required to travel $45$ miles. $d=st$ , so $x=18\cdot\frac{3}{2}=\boxed{27}$

Solution: Since $18$ mph is $\frac{3}{2}$ times $12$ mph, Alicia will travel $\frac{3}{2}$ times as far as Beth. If $x$ is the distance Beth travels, \[\frac{3}{2}x+x=45\] \[\frac{5}{2}x=45\] \[x=18\] Since this is the amount Beth traveled, the amount that Alicia traveled was \[45-18=\boxed{27}\]

Solution: Alice and Barbara close in on each other at 30mph. Since they are 45 miles apart, they will meet in t = d/s = 45miles / 30mph = 3/2 hours. We can either calculate the distance Alice travels at 18mph or the distance Barbara travels at 12mph; since we want the distance from Alice, we go with the former. Alice (and Barbara) will meet in 1 1/2 hours at 18mph x 3/2 hours = 27 miles from A. $\boxed{27}$

Solution:
We know that Beth starts 45 miles away from City A, let’s create two equations:
Alice-> $18t=d$ Beth-> $-12t+45=d$ [-12 is the slope; 45 is the y-intercept]
Solve the system:
$18t=-12t+45 30t=45 t=1.5$
So, $18(1.5)=$ $\boxed{27}$

Solution: Since Alicia and Beth's speeds are constant, they are directly proportional to their distances covered, so the ratio of their speeds is equal to the ratio of their covered distances. Since Alicia travels $\frac{18}{30} = \frac{3}{5}$ of their combined speed, she travels $\frac{3}{5}\cdot 45 = \boxed{27}$ of the total distance.


Answer: 143.0

Claire adds the degree measures of the interior angles of a convex polygon and arrives at a sum of $2017$ . She then discovers that she forgot to include one angle. What is the degree measure of the forgotten angle?
$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 63\qquad\textbf{(C)}\ 117\qquad\textbf{(D)}\ 143\qquad\textbf{(E)}\ 163$

Solution: We know that the sum of the interior angles of the polygon is a multiple of $180$ . Note that $\left\lceil\frac{2017}{180}\right\rceil = 12$ and $180\cdot 12 = 2160$ , so the angle Claire forgot is $\equiv 2160-2017=143\mod 180$ . Since the polygon is convex, the angle is $\leq 180$ , so the answer is $\boxed{143}$


Answer: 5.0

Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

Solution: Let Claudia have $x$ 5-cent coins and $\left( 12 - x \right)$ 10-cent coins. It is easily observed that any multiple of $5$ between $5$ and $5x + 10(12 - x) = 120 - 5x$ inclusive can be obtained by a combination of coins. Thus, $24 - x = 17$ combinations can be made, so $x = 7$ . But the answer is not $7,$ because we are asked for the number of 10-cent coins, which is $12 - 7 = \boxed{5}$

Solution: Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of $5.$ To have exactly $17$ different multiples of $5,$ we will need to make up to $85$ cents. If all twelve coins were 5-cent coins, we will have $60$ cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain $5$ cents, and as we need to gain $25$ cents, the answer is $\boxed{5}$

Solution: Notice that for every $d$ dimes, any multiple of $5$ less than or equal to $10d + 5(12-d)$ is a valid arrangement. Since there are $17$ in our case, we have $10d + 5(12-d)=17 \cdot 5 \Rightarrow d=5$ . Therefore, the answer is $\boxed{5}$


Answer: 341.0

Club Truncator is in a soccer league with six other teams, each of which it plays once. In any of its 6 matches, the probabilities that Club Truncator will win, lose, or tie are each $\frac {1}{3}$ . The probability that Club Truncator will finish the season with more wins than losses is $\frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution:
Note that the probability that Club Truncator will have more wins than losses is equal to the probability that it will have more losses than wins; the only other possibility is that they have the same number of wins and losses. Thus, by the complement principle , the desired probability is half the probability that Club Truncator does not have the same number of wins and losses.
The possible ways to achieve the same number of wins and losses are $0$ ties, $3$ wins and $3$ losses; $2$ ties, $2$ wins, and $2$ losses; $4$ ties, $1$ win, and $1$ loss; or $6$ ties. Since there are $6$ games, there are $\frac{6!}{3!3!}$ ways for the first, and $\frac{6!}{2!2!2!}$ $\frac{6!}{4!}$ , and $1$ ways for the rest, respectively, out of a total of $3^6$ . This gives a probability of $141/729$ . Then the desired answer is $\frac{1 - \frac{141}{729}}{2} = \frac{98}{243}$ , so the answer is $m+n = \boxed{341}$

Solution:
At first, it wins $6$ games, only one way
Secondly, it wins $5$ games, the other game can be either win or loss, there are $\binom{6}{5}\cdot 2=12$ ways
Thirdly, it wins $4$ games, still the other two games can be either win or loss, there are $\binom{6}{4}\cdot 2^2=60$ ways
Fourthly, it wins $3$ games, this time, it can't lose $3$ games but other arrangements of the three non-winning games are fine, there are $\binom{6}{3}\cdot (2^3-1)=140$ ways
Fifth case, it wins $2$ games, only $0/1$ lose and $4/3$ draw is ok, so there are $\binom{6}{2}(1+\binom{4}{1})=75$ cases
Last case, it only wins $1$ game so the rest games must be all draw, $1$ game
The answer is $\frac{1+12+60+140+75+6}{3^6}=\frac{98}{243}$ leads to $\boxed{341}$


Answer: 375.0

Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $|a|^2 + |b|^2 + |c|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$

Solution: By Vieta's formula, the sum of the roots is equal to 0, or $a+b+c=0$ . Therefore, $\frac{(a+b+c)}{3}=0$ . Because the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin. Let one leg of the right triangle be $x$ and the other leg be $y$ . Without the loss of generality, let $\overline{ac}$ be the hypotenuse. The magnitudes of $a$ $b$ , and $c$ are just $\frac{2}{3}$ of the medians because the origin, or the centroid in this case, cuts the median in a ratio of $2:1$ . So, $|a|^2=\frac{4}{9}\cdot((\frac{x}{2})^2+y^2)=\frac{x^2}{9}+\frac{4y^2}{9}$ because $|a|$ is two thirds of the median from $a$ . Similarly, $|c|^2=\frac{4}{9}\cdot(x^2+(\frac{y}{2})^2)=\frac{4x^2}{9}+\frac{y^2}{9}$ . The median from $b$ is just half the hypotenuse because the median of any right triangle is just half the hypotenuse. So, $|b|^2=\frac{4}{9}\cdot\frac{x^2+y^2}{4}=\frac{x^2}{9}+\frac{y^2}{9}$ . Hence, $|a|^2+|b|^2+|c|^2=\frac{6x^2+6y^2}{9}=\frac{2x^2+2y^2}{3}=250$ . Therefore, $h^2=x^2+y^2=\frac{3}{2}\cdot250=\boxed{375}$

Solution:
Assume $q$ and $r$ are real, so at least one of $a,$ $b,$ and $c$ must be real, with the remaining roots being pairs of complex conjugates. Without loss of generality, we assume $a$ is real and $b$ and $c$ are $x + yi$ and $x - yi$ respectively. By symmetry, the triangle described by $a,$ $b,$ and $c$ must be isosceles and is thus an isosceles right triangle with hypotenuse $\overline{bc}.$ Now since $P(z)$ has no $z^2$ term, we must have $a+b+c = a + (x + yi) + (x - yi) = 0$ and thus $a = -2x.$ Also, since the length of the altitude from the right angle of an isosceles triangle is half the length of the hypotenuse, $a-x=y$ and thus $y=-3x.$ We can then solve for $x$
\begin{align*} |a|^2 + |b|^2 + |c|^2 &= 250\\ |-2x|^2 + |x-3xi|^2 + |x+3xi|^2 &= 250\\ 4x^2 + (x^2 + 9x^2) + (x^2 + 9x^2) &= 250\\ x^2 &= \frac{250}{24} \end{align*}
Now $h$ is the distance between $b$ and $c,$ so $h = 2y = -6x$ and thus $h^2 = 36x^2 = 36 \cdot \frac{250}{24} = \boxed{375.}$

Solution:
Let the roots $a$ $b$ , and $c$ each be represented by complex numbers $m + ni$ $p + qi$ , and $r + ti$ .
By Vieta's formulas, their sum is 0. Breaking into real and imaginary components, we get:
$m + p + r = 0$
$n + q + t = 0$
And, we know that the sum of the squares of the magnitudes of each is 250, so
$m^2 + n^2 + p^2 + q^2 + r^2 + t^2 = 250$
Given the complex plane, we set each of these complex numbers to points: $(m, n)$ $(p, q)$ $(r, t)$ .
WLOG let $(r, t)$ be the vertex opposite the hypotenuse.
If the three points form a right triangle, the vectors from $(r, t)$ to $(m, m)$ and $(p, q)$ 's dot product is 0. $mp + r^2 - r(m + p) + nq + t^2 - t(n + q) = 0$
Substituting $m + p + r = 0$ and likewise, simplifying: $mp + 2r^2 + nq + 2t^2 = 0$
Rearranging we get: $r^2 + t^2 = -\frac{mp + nq}{2}$
The answer is the distance from $(m, n)$ to $(p, q)$ $m^2 + n^2 + p^2 + q^2 - 2(mp + nq)$ .
Substituting the equation equal to 250,
$= 250 - r^2 - t^2 - 2(mp + nq)$ $= 250 + \frac{mp + nq}{2} - 2(mp + nq)$ $= 250 - \frac{3}{2} \cdot (mp + nq)$
Taking our original equations summing to 0, and squaring each we get:
$n + q = -t$ $m + p = -r$
$n^2 + 2nq + q^2 = t^2$ $m^2 + 2mp + p^2 = r^2$
Adding, we get:
$m^2 + n^2 + p^2 + q^2 + 2(mp + nq) = r^2 + t^2$
Substituting again we obtain:
$250 - r^2 - t^2 + 2(mp + nq) = r^2 + t^2$ $2(r^2 + t^2) = 250 + 2(mp + nq)$ $r^2 + t^2 = 125 + (mp + nq)$
Substituting the equivalence of $r^2 + t^2$
$-\frac{mp + nq}{2} = 125 + (mp + nq)$
Solving for $mp + nq$ , we find it equal to $-\frac{250}{3}$
Substituting this value into our answer expression, we get:
$250 - \frac{3}{2} \cdot (-\frac{250}{3})$ ,
Answer = $\boxed{375}$

Solution:
As noted in the previous solutions, $a+b+c = 0$ .
Let $a = a_1+a_2 i$ $b = b_1+b_2 i$ $c = c_1+c_2 i$ and we have $a_1 + b_1 + c_1 = a_2 + b_2 + c_2 = 0$ .
Then the given $|a|^2 + |b|^2 + |c|^2 = 250$ translates to $\sum_{} ( {a_1}^2 + {a_2}^2 ) = 250.$ Note that in a right triangle, the sum of the squares of the three sides is equal to two times the square of the hypotenuse, by the pythagorean theorem.
Thus, we have \[2h^2 = (a_1 - b_1)^2 + (a_2 - b_2)^2 + (b_1 - c_1)^2 + (b_2 - c_2)^2 + (a_1 - c_1)^2 + (a_2 - c_2)^2\] \[= 2 \left( \sum_{} ( {a_1}^2 + {a_2}^2 )  \right) - 2 \left( \sum_{cyc} a_1 b_1 + \sum_{cyc} a_2 b_2 \right)\] \[= 500 - \left( (a_1 + b_1 + c_1)^2 + (a_2 + b_2 + c_2)^2 -  \sum_{cyc} ( {a_1}^2 + {a_2}^2 )     \right)\] \[= 500 - (0^2 + 0^2 - 250)\] so $h^2 = \boxed{375}$ and we may conclude.
~ rzlng

Solution:
First, note that the roots of this cubic will be $a, b$ and $-(a+b)$ due to Vieta's, which means that the sum of the roots are 0.
Now, we use some god level wishful thinking. It would really be nice if one of these roots was on the real axis, and it was a right isosceles triangle, because that would be very convenient and easy to work with. The neat part is that it actually works
Set one of the roots as $r$ , where $r$ is any real number. WLOG, assume that this is the right angle. With symmetry to respect of the x axis (because symmetry makes the imaginary parts of the other 2 roots cancel out, besides the fact that complex conjugate root theorem forces it). This way, we can set the other 2 roots as $\frac{r}{2}+ni$ and $\frac{r}{2}-ni$ , making the roots add up to 0.
Now, as we want the roots to satisfy the original condition (right triangle) we are going to have to set an equation to find $n$ out. We use the fact that it is an isosceles right triangle to find that $\frac{3r}{2}=n$ , which means that the 2 other roots are now $\frac{r}{2}+\frac{3r}{2}i$ and $\frac{r}{2}-\frac{3r}{2}i$
Now we use the fact that $|a|^2+|b|^2+|c|^2=250$ . Clearly one of these is $r$ away from the origin, so that gets $r^2$ , and then we get $2*\frac{r}{2}^2+\frac{3r}{2}^2$ which gets us $5r^2$ , getting $r^2+5r^2=250$ , so $r=\sqrt{\frac{250}{6}}$ . So the final answer comes out to \[(\frac{250}{6}*\frac{9}{4}*2)^2=\boxed{375}\]

Solution:
As shown in the other solutions, $a+b+c = 0$
Without loss of generality, let $b$ be the complex number opposite the hypotenuse.
Note that there is an isomorphism between $\mathbb{C}$ under $+$ and $\mathbb{R}^2$ under $+$
Let $\Vec{a}$ $\Vec{b}$ , and $\Vec{c}$ be the corresponding vectors to $a$ $b$ , and $c$
Thus $\Vec{a} + \Vec{b} + \Vec{c} = \Vec{0}$
$\Rightarrow 0 = \Vec{0}\cdot \Vec{0} = (\Vec{a} + \Vec{b} + \Vec{c})\cdot (\Vec{a} + \Vec{b} + \Vec{c}) = \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} + 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})$
Now $|a|^2 + |b|^2 + |c|^2 = 250$ implies that $\lVert \Vec{a}\rVert^2 + \lVert \Vec{b}\rVert^2 + \lVert \Vec{c}\rVert^2 = 250$
$\Rightarrow \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} = \lVert \Vec{a}\rVert^2 + \lVert \Vec{b}\rVert^2 + \lVert \Vec{c}\rVert^2 = 250$
Also note that because there is a right angle at $b$ $\Vec{a} - \Vec{b}$ and $\Vec{c} - \Vec{b}$ are perpendicular.
$\Rightarrow (\Vec{a} - \Vec{b})\cdot (\Vec{c} - \Vec{b}) = 0$
$\Rightarrow 0 = (\Vec{a} - \Vec{b})\cdot (\Vec{c} - \Vec{b}) = \Vec{a}\cdot \Vec{c} + \Vec{b} \cdot \Vec{b} - \Vec{a} \cdot \Vec{b} - \Vec{b} \cdot \Vec{c}$
Note that $h^2 = |a-c|^2$
$\Rightarrow h^2 = \lVert \Vec{a} - \Vec{c} \rVert^2 = (\Vec{a} - \Vec{c})\cdot (\Vec{a} - \Vec{c}) = \Vec{a} \cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - \Vec{a}\cdot \Vec{c} - \Vec{a}\cdot \Vec{c} = \Vec{a} \cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - 2\Vec{a}\cdot \Vec{c}$
$\Rightarrow 0 = \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} + 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}) = 250 + 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})$
$\Rightarrow -250 = 2(\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c})$
$\Rightarrow -125 = \Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}$
$\Rightarrow -125 = -125 + 0 = (\Vec{a}\cdot \Vec{b} + \Vec{a}\cdot \Vec{c} + \Vec{b}\cdot \Vec{c}) + (\Vec{a}\cdot \Vec{c} + \Vec{b} \cdot \Vec{b} - \Vec{a} \cdot \Vec{b} - \Vec{b} \cdot \Vec{c}) = 2\Vec{a}\cdot \Vec{c} + \Vec{b} \cdot \Vec{b}$
$\Rightarrow 125 = - \Vec{b} \cdot \Vec{b} - 2\Vec{a}\cdot \Vec{c}$
$\Rightarrow 375 = 250 + 125 = \Vec{a}\cdot \Vec{a} + \Vec{b}\cdot \Vec{b} + \Vec{c}\cdot \Vec{c} - \Vec{b} \cdot \Vec{b} - 2\Vec{a}\cdot \Vec{c} = \Vec{a}\cdot \Vec{a} + \Vec{c}\cdot \Vec{c} - 2\Vec{a}\cdot \Vec{c} = h^2$
$\Rightarrow h^2 = \boxed{375}$

Solution:
Note that the roots of the polynomial $(a,b,c)$ must sum to $0$ due to the $z^2$ coefficient equaling $0$ because of Vieta's Formulas. This tells us that $a+b+c = 0$ and $\overline{a} + \overline{b} + \overline{c} = 0$ so $(a+b+c)(\overline{a+b+c}) = |a|^2 + |b|^2 + |c|^2 + a\overline{b} + \overline{a}b + b\overline{c} + \overline{b}c + c\overline{a} + \overline{c}a = 250 + a\overline{b} + \overline{a}b + b\overline{c} + \overline{b}c + c\overline{a} + \overline{c}a =  0$ so $a\overline{b} + \overline{a}b + b\overline{c} + \overline{b}c + c\overline{a} + \overline{c}a = -250.$
Note that terms similar to $\overline{a}b + \overline{b}a$ appear in $|a-b|^2$ so we decide to sum $|a-b|^2 + |b-c|^2 + |c-a|^2$ out of intuition. Note that this corresponds to the sum of the squares of the sidelengths of the right triangle and if WLOG the $|c-a|^2$ side is the hypotenuse then our sum is equal to $2|c-a|^2 = 2h^2.$
\[2|a|^2 + 2|b|^2 + 2|c|^2 -a\overline{b} - \overline{a}b - b\overline{c} - \overline{b}c - c\overline{a} - \overline{c}a = 500 + 250 = 2h^2\]
As a result we know that $h^2 = \boxed{375}$ ~SailS


Answer: 869.0

Solution:
Note that the four numbers to multiply are symmetric with the center at $29.5$ .
Multiply the symmetric pairs to get $31\cdot 28=868$ and $30\cdot 29=870$ $\sqrt{868\cdot 870 + 1} = \sqrt{(869-1)(869+1) + 1} = \sqrt{869^2 - 1^2 + 1} = \sqrt{869^2} = \boxed{869}$

Solution: Notice that $(a+1)^2 = a \cdot (a+2) +1$ . Then we can notice that $30 \cdot 29 =870$ and that $31 \cdot 28 = 868$ . Therefore, $\sqrt{(31)(30)(29)(28) +1} = \sqrt{(870)(868) +1} = \sqrt{(868 +1)^2} = \boxed{869}$ . This is because we have that $a=868$ as per the equation $(a+1)^2 = a \cdot (a+2) +1$

Solution: More generally, we can prove that one more than the product of four consecutive integers must be a perfect square: \begin{align*} (a+3)(a+2)(a+1)(a)+1 &= [(a+3)(a)][(a+2)(a+1)]+1 \\ &= [a^2+3a][a^2+3a+2]+1 \\ &= [a^2+3a]^2+2[a^2+3a]+1 \\ &= [a^2+3a+1]^2. \end{align*} At $a=28,$ we have \[\sqrt{(a+3)(a+2)(a+1)(a)+1}=a^2+3a+1=\boxed{869}.\] ~Novus677 (Fundamental Logic)

Solution: Similar to Solution 1 above, call the consecutive integers $\left(n-\frac{3}{2}\right), \left(n-\frac{1}{2}\right), \left(n+\frac{1}{2}\right), \left(n+\frac{3}{2}\right)$ to make use of symmetry.  Note that $n$ itself is not an integer - in this case, $n = 29.5$ .  The expression becomes $\sqrt{\left(n-\frac{3}{2}\right)\left(n + \frac{3}{2}\right)\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) + 1}$ .  Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives $\sqrt{n^4 - \frac{5}{2}n^2 + \frac{25}{16}}$ .  The inside is a perfect square trinomial, since $b^2 = 4ac$ .  It's equal to $\sqrt{\left(n^2 - \frac{5}{4}\right)^2}$ , which simplifies to $n^2 - \frac{5}{4}$ .  You can plug in the value of $n$ from there, or further simplify to $\left(n - \frac{1}{2}\right)\left(n + \frac{1}{2}\right) - 1$ , which is easier to compute.  Either way, plugging in $n=29.5$ gives $\boxed{869}$

Solution:
We have $(31)(30)(29)(28)+1=755161.$ Since the alternating sum of the digits $7-5+5-1+6-1=11$ is divisible by $11,$ we conclude that $755161$ is divisible by $11.$
We evaluate the original expression by prime factorization: \begin{align*} \sqrt{(31)(30)(29)(28)+1}&=\sqrt{755161} \\ &=\sqrt{11\cdot68651} \\ &=\sqrt{11^2\cdot6241} \\ &=\sqrt{11^2\cdot79^2} \\ &=11\cdot79 \\ &=\boxed{869} ~Vrjmath (Fundamental Logic)

Solution:
The last digit under the radical is $1$ , so the square root must either end in $1$ or $9$ , since $x^2  = 1\pmod {10}$ means $x = \pm 1$ .  Additionally, the number must be near $29 \cdot 30 = 870$ , narrowing the reasonable choices to $869$ and $871$
Continuing the logic, the next-to-last digit under the radical is the same as the last digit of $28 \cdot 29 \cdot 3 \cdot 31$ , which is $6$ .  Quick computation shows that $869^2$ ends in $61$ , while $871^2$ ends in $41$ .  Thus, the answer is $\boxed{869}$


Answer: 373.0

Solution: The Sophie Germain Identity states that $a^4 + 4b^4$ can be factored as $\left(a^2 + 2b^2 - 2ab\right)\left(a^2 + 2b^2 + 2ab\right).$ Each of the terms is in the form of $x^4 + 324.$ Using Sophie Germain, we get that \begin{align*} x^4 + 324 &= x^4 + 4\cdot 3^4 \\ &= \left(x^2 + 2 \cdot 3^2 - 2\cdot 3\cdot x\right)\left(x^2 + 2 \cdot 3^2 + 2\cdot 3\cdot x\right) \\ &= (x(x-6) + 18)(x(x+6)+18), \end{align*} so the original expression becomes \[\frac{[(10(10-6)+18)(10(10+6)+18)][(22(22-6)+18)(22(22+6)+18)]\cdots[(58(58-6)+18)(58(58+6)+18)]}{[(4(4-6)+18)(4(4+6)+18)][(16(16-6)+18)(16(16+6)+18)]\cdots[(52(52-6)+18)(52(52+6)+18)]},\] which simplifies to \[\frac{(10(4)+18)(10(16)+18)(22(16)+18)(22(28)+18)\cdots(58(52)+18)(58(64)+18)}{(4(-2)+18)(4(10)+18)(16(10)+18)(16(22)+18)\cdots(52(46)+18)(52(58)+18)}.\] Almost all of the terms cancel out! We are left with $\frac{58(64)+18}{4(-2)+18} = \frac{3730}{10} = \boxed{373}.$

Solution:
In both the numerator and the denominator, each factor is of the form $N^4+324=N^4+18^2$ for some positive integer $N.$
We factor $N^4+18^2$ by completing the square, then applying the difference of squares: \begin{align*} N^4+18^2&=\left(N^4+36N^2+18^2\right)-36N^2 \\ &=\left(N^2+18\right)^2-(6N)^2 \\ &=\left(N^2-6N+18\right)\left(N^2+6N+18\right) \\ &=\left((N-3)^2+9\right)\left((N+3)^2+9\right). \end{align*} The original expression now becomes \[\frac{\left[(7^2+9)(13^2+9)\right]\left[(19^2+9)(25^2+9)\right]\left[(31^2+9)(37^2+9)\right]\left[(43^2+9)(49^2+9)\right]\left[(55^2+9)(61^2+9)\right]}{\left[(1^2+9)(7^2+9)\right]\left[(13^2+9)(19^2+9)\right]\left[(25^2+9)(31^2+9)\right]\left[(37^2+9)(43^2+9)\right]\left[(49^2+9)(55^2+9)\right]}=\frac{61^2+9}{1^2+9}=\boxed{373}.\] ~MRENTHUSIASM

Solution:
We rewrite $N$ to the polar form \[N=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,\] where $r$ is the magnitude of $N$ such that $r\geq0,$ and $\theta$ is the argument of $N$ such that $0\leq\theta<2\pi.$
By De Moivre's Theorem , we have \[N^4=r^4\operatorname{cis}(4\theta)=18^2(-1),\] from which
By the Factor Theorem , we get \begin{align*} N^4+18^2&=\biggl(N-3\sqrt2\operatorname{cis}\frac{\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{3\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{5\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{7\pi}{4}\biggr) \\ &=\biggl[\biggl(N-3\sqrt2\operatorname{cis}\frac{\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{7\pi}{4}\biggr)\biggr]\biggl[\biggl(N-3\sqrt2\operatorname{cis}\frac{3\pi}{4}\biggr)\biggl(N-3\sqrt2\operatorname{cis}\frac{5\pi}{4}\biggr)\biggr] \\ &=\left[(N-(3+3i))(N-(3-3i))\right]\left[(N-(-3+3i))(N-(-3-3i))\right] \\ &=\left[((N-3)-3i)((N-3)+3i)\right]\left[((N+3)-3i)((N+3)+3i)\right] \\ &=\left[(N-3)^2+9\right]\left[(N+3)^2+9\right]. \end{align*} We continue with the last paragraph of Solution 2 to get the answer $\boxed{373}.$

Solution:
We rewrite $N$ to the rectangular form \[N=a+bi\] for some real numbers $a$ and $b.$
Note that $N^2=\pm18i,$ so there are two cases:
By the Factor Theorem , we get \begin{align*} N^4+18^2&=(N-(3+3i))(N-(-3-3i))(N-(3-3i))(N-(-3+3i)) \\ &=\left[(N-(3+3i))(N-(3-3i))\right]\left[(N-(-3+3i))(N-(-3-3i))\right] \\ &=\left[((N-3)-3i)((N-3)+3i)\right]\left[((N+3)-3i)((N+3)+3i)\right] \\ &=\left[(N-3)^2+9\right]\left[(N+3)^2+9\right]. \end{align*} We continue with the last paragraph of Solution 2 to get the answer $\boxed{373}.$


Answer: 15.0

Connie multiplies a number by $2$ and gets $60$ as her answer. However, she should have divided the number by $2$ to get the correct answer. What is the correct answer?
$\textbf{(A)}\ 7.5\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 240$

Solution: If $x$ is the number, then $2x=60$ and $x=30$ . Dividing the number by $2$ yields $\dfrac{30}{2} = \boxed{15}$


Answer: 4.0

Consider a sequence $x_1,x_2,x_3,\dotsc$ defined by: \begin{align*}&x_1 = \sqrt[3]{3}, \\ &x_2 = \left(\sqrt[3]{3}\right)^{\sqrt[3]{3}},\end{align*} and in general \[x_n = \left(x_{n-1}\right)^{\sqrt[3]{3}} \text{ for } n > 1.\]
What is the smallest value of $n$ for which $x_n$ is an integer?
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \  } 4 \qquad \mathrm{(D) \  } 9 \qquad \mathrm{(E) \  }27$

Solution:
Firstly, we will show by induction that \[x_n = \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)}.\] For the base case, we indeed have \begin{align*}x_1 &= \sqrt[3]{3} \\ &= \left(\sqrt[3]{3}\right)^1 \\ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^0\right)},\end{align*} and for the inductive step, if our claim is true for $x_n$ , then \begin{align*}x_{n+1} &= \left(x_n\right)^{\sqrt[3]{3}} \\ &= \left(\left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)}\right)^{\sqrt[3]{3}} \\ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\cdot\sqrt[3]{3}\right)} \\ &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^n\right)},\end{align*} which completes the proof.
We now rewrite our formula for $x_n$ as follows: \begin{align*}x_n &= \left(\sqrt[3]{3}\right)^{\left(\left(\sqrt[3]{3}\right)^{n-1}\right)} \\ &= \left(\sqrt[3]{3}\right)^{\left(3^{\frac{n-1}{3}}\right)} \\ &=3^{\left(\frac{1}{3} \cdot 3^{\frac{n-1}{3}}\right)} \\ &= 3^{\left(3^{\left(\frac{n-1}{3}-1\right)}\right)} \\ &= 3^{\left(3^{\left(\frac{n-4}{3}\right)}\right)},\end{align*} and as $3$ is not a perfect power, we deduce that $x_n$ is an integer if and only if the exponent, $3^{\left(\frac{n-4}{3}\right)}$ , is itself an integer. By precisely the same argument, this reduces to $\frac{n-4}{3}$ being an integer, so the smallest possible (positive) value of $n$ is $\boxed{4}$


Answer: 1.0

Consider a string of $n$ $7$ 's, $7777\cdots77,$ into which $+$ signs are inserted to produce an arithmetic expression . For example, $7+77+777+7+7=875$ could be obtained from eight $7$ 's in this way. For how many values of $n$ is it possible to insert $+$ signs so that the resulting expression has value $7000$

Solution:
It's obvious that we cannot have any number $\ge 7777$ because $7777 > 7000$ so the max number that an occur is $777$
Let's say we have $a$ 777's , $b$ 77's and $c$ 7's
From here we get our required equation as $777a + 77b + 7c = 7000$
Now comes the main problem , one might think that if we find number of $(a,b,c)$ then we're done , but in reality we are over-counting our number of $n$ 's. This is because $n$ is the total number of 7's and from our equation we'll get $n$ as $3a + 2b + c$ (because there are three 7's , two 7's and one 7)
The reason why we're over-counting is because , say $a_1 , b_1 , c_1$ be a solution of our original equation and $a_2 , b_2 , c_2$ be another solution of our original equation , then there can be a possibility that $3a_1 + 2b_1 + c_1 = 3a_2 + 2b_2 + c_2$ where $a_1 \neq a_2 , b_1 \neq b_2 , c_1 \neq c_2$ (example : $2 + 3 + 4 = 1 + 5 + 3$ but $2 \neq 1 , 3 \neq 5 , 4 \neq 3$
We know that $0 \le a \le 9$ $0 \le b \le 90$ $0 \le c \le 1000$ The bound on $a$ is easier to handle with , so lets start putting values on $a$ and calculate $b , c , n$ by making cases
Reduced equation : $111a + 11b + c = 1000$
Case 1 : $a = 9$
We get $11b + c = 1 \implies b = 0 , c = 1$ is our only solution thus only $\boxed{1}$ value of $n$


Answer: 108.0

Solution:
Suppose we require $a$ $7$ s, $b$ $77$ s, and $c$ $777$ s to sum up to $7000$ $a,b,c \ge 0$ ). Then $7a + 77b + 777c = 7000$ , or dividing by $7$ $a + 11b + 111c = 1000$ . Then the question is asking for the number of values of $n = a + 2b + 3c$
Manipulating our equation, we have $a + 2b + 3c = n = 1000 - 9(b + 12c) \Longrightarrow 0 \le 9(b+12c) < 1000$ . Thus the number of potential values of $n$ is the number of multiples of $9$ from $0$ to $1000$ , or $112$
However, we forgot to consider the condition that $a \ge 0$ . For a solution set $(b,c): n=1000-9(b+12c)$ , it is possible that $a = n-2b-3c < 0$ (for example, suppose we counted the solution set $(b,c) = (1,9) \Longrightarrow n = 19$ , but substituting into our original equation we find that $a = -10$ , so it is invalid). In particular, this invalidates the values of $n$ for which their only expressions in terms of $(b,c)$ fall into the inequality $9b + 108c < 1000 < 11b + 111c$
For $1000 - n = 9k \le 9(7 \cdot 12 + 11) = 855$ , we can express $k$ in terms of $(b,c): n \equiv b \pmod{12}, 0 \le b \le 11$ and $c = \frac{n-b}{12} \le 7$ (in other words, we take the greatest possible value of $c$ , and then "fill in" the remainder by incrementing $b$ ). Then $11b + 111c \le 855 + 2b + 3c \le 855 + 2(11) + 3(7) = 898 < 1000$ , so these values work.
Similarily, for $855 \le 9k \le 9(8 \cdot 12 + 10) = 954$ , we can let $(b,c) = (k-8 \cdot 12,8)$ , and the inequality $11b + 111c \le 954 + 2b + 3c \le 954 + 2(10) + 3(8) = 998 < 1000$ . However, for $9k \ge 963 \Longrightarrow n \le 37$ , we can no longer apply this approach.
So we now have to examine the numbers on an individual basis. For $9k = 972$ $(b,c) = (0,9)$ works. For $9k = 963, 981, 990, 999 \Longrightarrow n = 37, 19, 10, 1$ , we find (using that respectively, $b = 11,9,10,11 + 12p$ for integers $p$ ) that their is no way to satisfy the inequality $11b + 111c < 1000$
Thus, the answer is $112 - 4 = \boxed{108}$

Solution:
To simplify, replace all the $7$ ’s with $1$ ’s.
Because the sum is congruent to $n \pmod 9$ and \[1000 \equiv 1 \pmod 9 \implies n \equiv 1 \pmod 9\] Also, $n \leq 1000$ . There are $\big\lfloor \tfrac{1000}{9} \big\rfloor + 1 = 112$ positive integers that satisfy both conditions i.e. $\{1, 10, 19, 28, 37, 46, . . . , 1000\}.$
For $n = 1, 10, 19$ , the greatest sum that is less than or equal to $1000$ is \[6 \cdot 111 + 1 = 677 \implies 112-3 = 109.\]
Thus $n \geq 28$ and let $S = \{28, 37, 46, . . . , 1000\}$
Note that $n=28$ is possible because $9 \cdot 111+1 \cdot 1 = 1000$
When $n = 37$ , the greatest sum that is at most $1000$ is $8 \cdot 111+6\cdot 11+1 \cdot 1 = 955$
All other elements of $S$ are possible because if any element $n$ of $S$ between $46$ and $991$ is possible, then $(n+ 9)$ must be too.
$\textbf{Case 1:}$ Sum has no $11$ 's
It must have at least one $1$ .
If it has exactly one $1$ , there must be nine $111$ ’s and $n = 28$ .
Thus, for $n \geq 46$ , the sum has more than one $1$ , so it must have at least $1000 - 8 \cdot 111 = 112$ number of $1$ ’s.
For $n \leq 1000$ , at least one $111$ .
To show that if $n$ is possible, then $(n + 9)$ is possible, replace a $111$ with $1 + 1 + 1$ , replace eleven $(1 + 1)$ ’s with eleven $11$ ’s, and include nine new $1$ ’s as $+1$ ’s. The sum remains $1000$
$\textbf{Case 2:}$ Sum has at least one $11$
Replace an $11$ with $1 + 1$ , and include nine new $1$ ’s as $+1$ ’s.
Now note that $46$ is possible because $8 \cdot 111 + 10 \cdot 11 + 2 \cdot 1 = 1000$ .
Thus all elements of $S$ except $37$ are possible.
Thus there are $\boxed{108}$ possible values for $n$


Answer: 431.0

Consider all 1000-element subsets of the set $\{1, 2, 3, ... , 2015\}$ .  From each such subset choose the least element.  The arithmetic mean of all of these least elements is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers.  Find $p + q$

Solution: Let $M$ be the desired mean. Then because $\dbinom{2015}{1000}$ subsets have 1000 elements and $\dbinom{2015 - i}{999}$ have $i$ as their least element, \begin{align*} \binom{2015}{1000} M &= 1 \cdot \binom{2014}{999} + 2 \cdot \binom{2013}{999} + \dots + 1016 \cdot \binom{999}{999} \\      &= \binom{2014}{999} + \binom{2013}{999} + \dots + \binom{999}{999} \\      &                      + \binom{2013}{999} + \binom{2012}{999} + \dots + \binom{999}{999} \\      & \dots \\      & + \binom{999}{999} \\      &= \binom{2015}{1000} + \binom{2014}{1000} + \dots + \binom{1000}{1000} \\      &= \binom{2016}{1001}. \end{align*} Using the definition of binomial coefficient and the identity $n! = n \cdot (n-1)!$ , we deduce that \[M = \frac{2016}{1001} = \frac{288}{143}.\] The answer is $\boxed{431}.$

Solution:
Each 1000-element subset $\left\{ a_1, a_2,a_3,...,a_{1000}\right\}$ of $\left\{1,2,3,...,2015\right\}$ with $a_1<a_2<a_3<...<a_{1000}$ contributes $a_1$ to the sum of the least element of each subset. Now, consider the set $\left\{a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$ . There are $a_1$ ways to choose a positive integer $k$ such that $k<a_1+1<a_2+1,a_3+1<...<a_{1000}+1$ $k$ can be anything from $1$ to $a_1$ inclusive). Thus, the number of ways to choose the set $\left\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$ is equal to the sum. But choosing a set $\left\{k,a_1+1,a_2+1,a_3+1,...,a_{1000}+1\right\}$ is the same as choosing a 1001-element subset from $\left\{1,2,3,...,2016\right\}$
Thus, the average is $\frac{\binom{2016}{1001}}{\binom{2015}{1000}}=\frac{2016}{1001}=\frac{288}{143}$ . Our answer is $p+q=288+143=\boxed{431}$

Solution:
Let $p$ be the size of the large set and $q$ be the size of the subset (i.e. in this problem, $p = 2015$ and $q = 1000$ ). We can easily find the answers for smaller values of $p$ and $q$
For $p = 2$ and $q = 2$ , the answer is $1$
For $p = 3$ and $q = 2$ , the answer is $\frac43$
For $p = 4$ and $q = 2$ , the answer is $\frac53$
For $p = 3$ and $q = 3$ , the answer is $1$
For $p = 4$ and $q = 3$ , the answer is $\frac54$
For $p = 5$ and $q = 3$ , the answer is $\frac32$
At this point, we can see a pattern: our desired answer is always $\frac{p+1}{q+1}$ . Plugging in $p = 2015$ and $q = 1000$ , the answer is $\frac{2016}{1001}=\frac{288}{143}$ , so $288 + 143 = \boxed{431}$


Answer: 360.0

Consider arrangements of the $9$ numbers $1, 2, 3, \dots, 9$ in a $3 \times 3$ array. For each such arrangement, let $a_1$ $a_2$ , and $a_3$ be the medians of the numbers in rows $1$ $2$ , and $3$ respectively, and let $m$ be the median of $\{a_1, a_2, a_3\}$ . Let $Q$ be the number of arrangements for which $m = 5$ . Find the remainder when $Q$ is divided by $1000$

Solution:
Assume that $5 \in \{a_1, a_2, a_3\}$ $m \neq 5$ , and WLOG, $\max{(a_1, a_2, a_3)} = 5$ . Then we know that the other two medians in $\{a_1, a_2, a_3\}$ and the smallest number of rows 1, 2, and 3 are all less than 5. But there are only 4 numbers less than 5 in $1, 2, 3, \dots, 9$ , a Contradiction. Thus, if $5 \in \{a_1, a_2, a_3\}$ , then $m = 5$
WLOG, assume $5$ is in the upper left corner. One of the two other values in the top row needs to be below $5$ , and the other needs to be above $5$ . This can be done in $4\cdot4\cdot2=32$ ways.
The other $6$ can be arranged in $6!=720$ ways.
Finally, accounting for when $5$ is in every other space, our answer is $32\cdot720\cdot9$ , which is $207360$ . But we only need the last $3$ digits, so $\boxed{360}$ is our answer.

Solution:
(Complementary Counting with probability)
Notice that m can only equal 4, 5, or 6, and 4 and 6 are symmetric.
WLOG let $m=4$
1. There is a $\frac{15}{28}$ chance that exactly one of 1, 2, 3 is in the same row with 4.
There are 3 ways to select which of the smaller numbers will get in the row, and then 5
ways to select the number larger than 4.
$\frac{\dbinom{3}{1}\cdot\dbinom{5}{1}}{\dbinom{8}{2}} = \frac{15}{28}$
2. There is a $\frac{2}{5}$ chance that the other two smaller numbers end up in the same row.
There are 2 ways to select the row that the two smaller number are in, and then $\dbinom{3}{2}$ ways
to place the smaller numbers in the row.
$\frac{\dbinom{2}{1}\cdot\dbinom{3}{2}}{\dbinom{6}{2}} = \frac{2}{5}$
$9!(1-2*\frac{15}{28}*\frac{2}{5})=362880*\frac{4}{7}=207\boxed{360}$

Solution:
We will make sure to multiply by $3!$ in the end to account for all the possible permutation of the rows.
WLOG, let $5$ be present in the Row # $1$
Notice that $5$ MUST be placed with a number lower than it and a number higher than it.
This happens in $4\cdot4$ ways. You can permutate Row # $1$ in $3!$ ways.
Now, take a look at Row $2$ and Row $3$
Because there are $6$ numbers to choose from now, you can assign #'s to Row's #2&3 in
$\frac{\binom{6}{3}\cdot\binom{3}{3}}{2}$ ways. There are $3!\cdot3!$ ways to permute the numbers in the individual Rows.
Hence, our answer is $3!(4\cdot4\cdot3!\cdot{10}\cdot{3!}\cdot{3!})=3!(16\cdot60\cdot36)=3!(34560)\implies{207\boxed{360}$

Solution:
We see that if one of the medians is 5, then there are two remaining numbers greater than 5 and two less than 5, so it follows that $m=5$ . There are 3 ways to choose which row to have 5 in, $4\cdot 4=16$ ways to choose the other two numbers in that row, $3!=6$ ways to arrange the numbers in that row, and $6!=720$ ways for the remaining numbers, for our answer is $3\cdot 16\cdot 6\cdot 720=207\boxed{360}$ .
-Stormersyle

Solution:
We take the grid, and we do a bunch of stuff with it. First, we sort each row, smallest on the left, largest on the right, then we arrange these 3 rows such that the middle #s are increasing, from top to bottom. Thus, we get that the cell in the very center of the grid must be 5. (We need to multiply by 1296 at the end)
Let S mean a number that is < 5, L mean a number > 5.
Bobby.jpg
In the 2 blank corners, one can be S, and the other one has to be L.
If the top right corner is S, there are 16 ways, otherwise, there are 144 ways. (This is left as an exercise to the reader).
Thus, there are 160 * 1296 total configurations, which gets us an answer of $207\boxed{360}$

Solution: We note that if $5$ is a median of one of the rows, then $m=5$ . First, focus on the row with $5$ in it. There are $4^2$ ways to choose the other numbers in that row and then $3!$ ways to order it. Now, clearly, there are $6!$ ways to put the other $6$ numbers into the remaining slots so $Q=6!\cdot3!\cdot4^2\cdot3=207360$ . Hence, our answer is $\boxed{360}$


Answer: 19.0

Consider equations of the form $x^2 + bx + c = 0$ . How many such equations have real roots and have coefficients $b$ and $c$ selected
from the set of integers $\{1,2,3, 4, 5,6\}$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 19 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 17 \qquad \textbf{(E)}\ 16$

Solution:
The discriminant of the quadratic is $b^2 - 4c$ .  Since the quadratic has real roots, \[b^2 - 4c \ge 0\] \[b^2 \ge 4c\] If $b = 6$ , then $c$ can be from $1$ to $6$ .  If $b = 5$ , then $c$ can also be from $1$ to $6$ .  If $b=4$ , then $c$ can be from $1$ to $4$ .  If $b=3$ , then $c$ can be $1$ or $2$ .  If $b=2$ , then $c$ can only be $1$ .  If $b = 1$ , no values of $c$ in the set would work.
Thus, there are a total of $19$ equations that work.  The answer is $\boxed{19}$


Answer: 50.0

Consider functions $f$ that satisfy \[|f(x)-f(y)|\leq \frac{1}{2}|x-y|\] for all real numbers $x$ and $y$ . Of all such functions that also satisfy the equation $f(300) = f(900)$ , what is the greatest possible value of \[f(f(800))-f(f(400))?\] $\textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 200$

Solution:
We have \begin{align*} |f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| &&(\bigstar) \\ &\leq \frac12\left|\frac12|800-400|\right| \\ &= 100, \end{align*} from which we eliminate answer choices $\textbf{(D)}$ and $\textbf{(E)}.$
Note that \begin{alignat*}{8} |f(800)-f(300)| &\leq \frac12|800-300| &&= 250, \\ |f(800)-f(900)| &\leq \frac12|800-900| &&= 50, \\ |f(400)-f(300)| &\leq \frac12|400-300| &&= 50, \\ |f(400)-f(900)| &\leq \frac12|400-900| &&= 250. \\ \end{alignat*} Let $a=f(300)=f(900).$ Together, it follows that \begin{align*} |f(800)-a|&\leq 50, \\ |f(400)-a|&\leq 50. \\ \end{align*} We rewrite $(\bigstar)$ as \begin{align*} |f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| \\ &= \frac12|(f(800)-a)-(f(400)-a)| \\ &\leq \frac12|50-(-50)| \\ &=\boxed{50} ~MRENTHUSIASM

Solution:
Denote $f(900)-f(600) = a$ .
Because $f(300) = f(900)$ $f(300) - f(600) = a$
Following from the Lipschitz condition given in this problem, $|a| \leq 150$ and \[ f(800) - f(600) \leq \min \left\{ a + 50 , 100 \right\} \] and \[ f(400) - f(600) \geq \max \left\{ a - 50 , -100 \right\} . \] Thus, \begin{align*} f(800) - f(400) & \leq \min \left\{ a + 50 , 100 \right\} - \max \left\{ a - 50 , -100 \right\}  \\ & = 100 + \min \left\{ a, 50 \right\} - \max \left\{ a , - 50 \right\} \\ & = 100 + \left\{ \begin{array}{ll} a + 50 & \mbox{ if } a \leq -50 \\ 0 & \mbox{ if } -50 < a < 50 \\ -a + 50 & \mbox{ if } a \geq 50 \end{array} \right. . \end{align*} Thus, $f(800) - f(400)$ is maximized at $a = 0$ $f(800)-f(600) = 50$ $f(400)-f(600)=-50$ , with the maximal value 100.
By symmetry, following from an analogous argument, we can show that $f(800) - f(400)$ is minimized at $a = 0$ $f(800)-f(600) = -50$ $f(400)-f(600)=50$ , with the minimal value $-100$
Following from the Lipschitz condition, \begin{align*} f(f(800)) - f(f(400)) & \leq \frac{1}{2} \left| f(800) - f(400) \right| \\ & \leq 50 . \end{align*} We have already construct instances in which the second inequality above is augmented to an equality.
Now, we construct an instance in which the first inequality above is augmented to an equality.
Consider the following piecewise-linear function: \[ f(x) = \left\{ \begin{array}{ll} \frac{1}{2} \left( x - 300 \right) & \mbox{ if } x \leq 300 \\ -\frac{1}{2} \left( x - 300 \right) & \mbox{ if } 300 < x \leq 400 \\ \frac{1}{2} \left( x - 600 \right) & \mbox{ if } 400 < x \leq 800 \\ -\frac{1}{2} \left( x - 900 \right) & \mbox{ if } x > 800 \end{array} \right.. \] Therefore, the maximum value of $f(f(800)) - f(f(400))$ is $\boxed{50}$

Solution:
Divide both sides by $|x - y|$ to get $\frac{|f(x) - f(y)|}{|x - y|} \leq \frac{1}{2}$ . This means that when we take any two points on $f$ , the absolute value of the slope between the two points is at most $\frac{1}{2}$
Let $f(300) = f(900) = c$ , and since we want to find the maximum value of $|f(800) - f(400)|$ , we can take the most extreme case and draw a line with slope $-\frac{1}{2}$ down from $f(300)$ to $f(400)$ and a line with slope $\frac{1}{2}$ up from $f(800)$ to $f(900)$ . Then $f(400) = c - 50$ and $f(800) = c + 50$ , so $|f(800) - f(400)| = |c + 50 - (c - 50)| = 100$ , and this is attainable because the slope of the line connecting $f(400)$ and $f(800)$ still has absolute value less than $\frac{1}{2}$
Therefore, $|f(f(800)) - f(f(400))| \leq \frac{1}{2}|f(800) - f(400)| = \frac{1}{2}(100) = \boxed{50}$

Solution: Consider $g(x) = f(x)-f(300)$ . Then $g(x)$ satisfies all the conditions and $g(300) = g(900) = 0$ . We would want $g(400)$ and $g(800)$ as distant from each other as possible. So assign $g(400) = -50$ and $g(800) = 50$ , the possible lower and upper bounds respectively. It follows that one can obtain the upper bound for $|g(g(800)) - g(g(400))| = |g(50) - g(-50  )| \leq \frac{1}{2}(100) = \boxed{50}$ as the answer.


Answer: 220.0

Consider polynomials $P(x)$ of degree at most $3$ , each of whose coefficients is an element of $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ . How many such polynomials satisfy $P(-1) = -9$
$\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286$

Solution:
Suppose that $P(x)=ax^3+bx^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that \[P(-1)=-a+b-c+d=-9.\] Let $a'=9-a$ and $c'=9-c.$ Note that both of $a'$ and $c'$ are integers from $0$ through $9.$ Moreover, the ordered quadruples $(a,b,c,d)$ and the ordered quadruples $(a',b,c',d)$ have one-to-one correspondence.
We rewrite the given equation as $(9-a)+b+(9-c)+d=9,$ or \[a'+b+c'+d=9.\] By the stars and bars argument, there are $\binom{9+4-1}{4-1}=\boxed{220}$ ordered quadruples $(a',b,c',d).$

Solution: Suppose that $P(x)=ax^3+bx^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that $P(-1)=-a+b-c+d=-9,$ which rearranges to \[b+d+9=a+c.\] Note that $b+d+9$ is an integer from $9$ through $27,$ and $a+c$ is an integer from $0$ through $18.$ Therefore, both of $b+d+9$ and $a+c$ are integers from $9$ through $18.$ We construct the following table: \[\begin{array}{c|c|c|c||c} & & & & \\ [-2.5ex] \boldsymbol{b+d} & \boldsymbol{\#}\textbf{ of Ordered Pairs }\boldsymbol{(b,d)} & \boldsymbol{a+c} & \boldsymbol{\#}\textbf{ of Ordered Pairs }\boldsymbol{(a,c)} & \boldsymbol{\#}\textbf{ of Ordered Quadruples }\boldsymbol{(a,b,c,d)} \\ [0.5ex] \hline & & & & \\ [-2ex] 0 & 1 & 9 & 10 & 1\cdot10=10 \\ 1 & 2 & 10 & 9 & \phantom{0}2\cdot9=18 \\ 2 & 3 & 11 & 8 & \phantom{0}3\cdot8=24 \\ 3 & 4 & 12 & 7 & \phantom{0}4\cdot7=28 \\ 4 & 5 & 13 & 6 & \phantom{0}5\cdot6=30 \\ 5 & 6 & 14 & 5 & \phantom{0}6\cdot5=30 \\ 6 & 7 & 15 & 4 & \phantom{0}7\cdot4=28 \\ 7 & 8 & 16 & 3 & \phantom{0}8\cdot3=24 \\ 8 & 9 & 17 & 2 & \phantom{0}9\cdot2=18 \\ 9 & 10 & 18 & 1 & 10\cdot1=10 \end{array}\] We sum up the counts in the last column to get the answer $2\cdot(10+18+24+28+30)=\boxed{220}.$


Answer: 4.0

Consider sequences of positive real numbers of the form $x, 2000, y, \dots$ in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of $x$ does the term $2001$ appear somewhere in the sequence?
$\text{(A) }1 \qquad \text{(B) }2 \qquad \text{(C) }3 \qquad \text{(D) }4 \qquad \text{(E) more than }4$

Solution:
It never hurts to compute a few terms of the sequence in order to get a feel how it looks like. In our case, the definition is that $\forall$ (for all) $n>1:~ a_n = a_{n-1}a_{n+1} - 1$ . This can be rewritten as $a_{n+1} = \frac{a_n +1}{a_{n-1}}$ . We have $a_1=x$ and $a_2=2000$ , and we compute:
\begin{align*} a_3  & = \frac{a_2+1}{a_1} = \frac{2001}x \\ a_4  & = \frac{a_3+1}{a_2} = \frac{ \dfrac{2001}x + 1 }{ 2000 } = \frac{2001 + x}{2000x} \\ a_5 & = \frac{a_4+1}{a_3} = \frac{ \frac{2001 + x}{2000x} + 1 }{ \frac{2001}x } = \frac{ \frac{2001 + 2001x}{2000x} }{ \frac{2001}x } = \frac{1+x}{2000} \\ a_6 & = \frac{a_5+1}{a_4} = \frac{ \frac{1+x}{2000} + 1 }{ \frac{2001 + x}{2000x} } =  \frac{ \frac{2001+x}{2000} }{ \frac{2001 + x}{2000x} } = x \\ a_7 & = \frac{a_6+1}{a_5} = \frac{ x+1 }{ \frac{1+x}{2000} } = 2000 \end{align*}
At this point we see that the sequence will become periodic: we have $a_6=a_1$ $a_7=a_2$ , and each subsequent term is uniquely determined by the previous two.
Hence if $2001$ appears, it has to be one of $a_1$ to $a_5$ . As $a_2=2000$ , we only have four possibilities left. Clearly $a_1=2001$ for $x=2001$ , and $a_3=2001$ for $x=1$ . The equation $a_4=2001$ solves to $x = \frac{2001}{2000\cdot 2001 - 1}$ , and the equation $a_5=2001$ to $x=2000\cdot 2001 - 1$
No two values of $x$ we just computed are equal, and therefore there are $\boxed{4}$ different values of $x$ for which the sequence contains the value $2001$


Answer: 172.0

Consider sequences that consist entirely of $A$ 's and $B$ 's and that have the property that every run of consecutive $A$ 's has even length, and every run of consecutive $B$ 's has odd length. Examples of such sequences are $AA$ $B$ , and $AABAA$ , while $BBAB$ is not such a sequence. How many such sequences have length 14?

Solution: Let $a_n$ and $b_n$ denote, respectively, the number of sequences of length $n$ ending in $A$ and $B$ . If a sequence ends in an $A$ , then it must have been formed by appending two $A$ s to the end of a string of length $n-2$ . If a sequence ends in a $B,$ it must have either been formed by appending one $B$ to a string of length $n-1$ ending in an $A$ , or by appending two $B$ s to a string of length $n-2$ ending in a $B$ . Thus, we have the recursions \begin{align*} a_n &= a_{n-2} + b_{n-2}\\ b_n &= a_{n-1} + b_{n-2}  \end{align*} By counting, we find that $a_1 = 0, b_1 = 1, a_2 = 1, b_2 = 0$ \[\begin{array}{|r||r|r|||r||r|r|} \hline n & a_n & b_n & n & a_n & b_n\\ \hline 1&0&1& 8&6&10\\ 2&1&0& 9&11&11\\ 3&1&2& 10&16&21\\ 4&1&1& 11&22&27\\ 5&3&3& 12&37&43\\ 6&2&4& 13&49&64\\ 7&6&5& 14&80&92\\ \hline \end{array}\] Therefore, the number of such strings of length $14$ is $a_{14} + b_{14} = \boxed{172}$

Solution:
Let $a_n$ and $b_n$ denote, respectively, the number of sequences of length $n$ ending in $A$ and $B$
Additionally, let $t_n$ denote the total number of sequences of length $n$ . Then, $t_n=a_n+b_n$ , as the total amount of sequences of length $n$ consists of the sequences of length $n$ ending in $A$ and the sequences of length $n$ ending in $B$ \begin{align*} a_n &= a_{n-2} + b_{n-2}\\ b_n &= a_{n-1} + b_{n-2}  \end{align*} The recursion for $a_n$ tells us that $a_n=a_{n-2}+b_{n-2}$ . However, this is also the definition for $t_{n-2}$ . Therefore, $a_n=t_{n-2}$
We also know from our recursion for $b_n$ that $b_n=a_{n-1}+b_{n-2}$ . Substituting for $a_n$ and $b_n$ into our recursion for $t_n$ gives us $t_n=t_{n-2}+a_{n-1}+b_{n-2}$
Furthermore, note that since $a_n=t_{n-2}$ $a_{n-1}=t_{n-3}$ . Furthermore, using our definition for $t_{n-2}$ , we can rewrite $b_{n-2}$ as $t_{n-2}-a_{n-2}$ . Substituting for $a_{n-1}$ and $b_{n-2}$ into our recursion for $t_n$ gives us $t_n=t_{n-2}+t_{n-3}+t_{n-2}-a_{n-2}$
Finally, note that since $a_n=t_{n-2}$ $a_{n-2}=t_{n-4}$ . Substituting for $a_{n-2}$ into our recursion for $t_n$ gives us $t_n=2t_{n-2}+t_{n-3}-t_{n-4}$ . We now have a recursion only in terms of $t$
By counting, we find that $t_1=1$ $t_2=1$ $t_3=3$ , and $t_4=2$ \[\begin{array}{|r|r||r|r|}	 \hline	 n & t_n & n & t_n\\	 \hline	 1&1&8&16\\	 2&1&9&22\\ 3&3&10&37\\	 4&2&11&49\\	 5&6&12&80\\	 6&6&13&113\\	 7&11&14&172\\	 \hline \end{array}\]
Therefore, the number of such sequences of length 14 is $\boxed{172}$

Solution:
There must be an even amount of runs of consecutive $B$ s due to parity. Thus, we can split this sequence into the following cases: $A$ $BAAB$ $AABAAB$ $BAABAA$ $AABAABAA$ $BAABAABAAB$ $AABAABAABAAB$ $BAABAABAABAA$ , and $AABAABAABAABAA$ , in which the amount of letters in one run does not necessarily represent the amount of letters there can be.
For the first case and the last case, there is only one possible sequence of letters.
For all other cases, we can insert two of the same letter at a time into a run that has the exact same letter. For example, for the second case, we can insert two $A$ s and make the sequence $BAAAAB$ . There are three "slots" in which we can insert two additional letters in, and we must insert five groups of new letters. By stars and bars , the number of ways for the second case is $\binom{7}{2}=21$
Applying this logic to all of the other cases gives us $\binom{7}{3}$ $\binom{7}{3}$ $\binom{7}{4}$ $\binom{8}{6}$ $\binom{8}{1}$ , and $\binom{8}{1}$ . Adding 1+ $\binom{7}{2}$ $\binom{7}{3}$ $\binom{7}{3}$ $\binom{7}{4}$ $\binom{8}{6}$ $\binom{8}{1}$ $\binom{8}{1}$ gives us the answer $\boxed{172}$


Answer: 338.0

Consider systems of three linear equations with unknowns $x$ $y$ , and $z$ \begin{align*} a_1 x + b_1 y + c_1 z & = 0 \\ a_2 x + b_2 y + c_2 z & = 0 \\ a_3 x + b_3 y + c_3 z & = 0 \end{align*} where each of the coefficients is either $0$ or $1$ and the system has a solution other than $x=y=z=0$ .
For example, one such system is \[\{ 1x + 1y + 0z = 0, 0x + 1y + 1z = 0, 0x + 0y + 0z = 0 \}\] with a nonzero solution of $\{x,y,z\} = \{1, -1, 1\}$ . How many such systems of equations are there?
(The equations in a system need not be distinct, and two systems containing the same equations in a
different order are considered different.)
$\textbf{(A)}\ 302 \qquad\textbf{(B)}\ 338 \qquad\textbf{(C)}\ 340 \qquad\textbf{(D)}\ 343 \qquad\textbf{(E)}\ 344$

Solution:
Let $M_1=\begin{bmatrix}a_1 & b_1 & c_1\end{bmatrix}, M_2=\begin{bmatrix}a_2 & b_2 & c_2\end{bmatrix},$ and $M_3=\begin{bmatrix}a_3 & b_3 & c_3\end{bmatrix}.$
We wish to count the ordered triples $(M_1,M_2,M_3)$ of row matrices. We perform casework:
There are $9+3=12$ ordered triples $(M_1,M_2,M_3).$
Similarly, for each of $M_1+M_3=M_2$ and $M_2+M_3=M_1,$ there are $12$ ordered triples $(M_1,M_2,M_3).$
In this subcase, we have $\boldsymbol{12\cdot3=36}$ ordered triples $\boldsymbol{(M_1,M_2,M_3).}$
Together, the answer is $8+168+126+36=\boxed{338}.$

Solution:
We will use complementary counting and do casework on the equations.
There are $8$ possible equations:
Equation 1: $0 = 0$
Equation 2: $x = 0$
Equation 3: $y = 0$
Equation 4: $z = 0$
Equation 5: $x + y = 0$
Equation 6: $x + z = 0$
Equation 7: $y + z = 0$
Equation 8: $x + y + z = 0$
We will continue to refer to the equations by their number on this list.
$8^3 = 512$ total systems. Note that no two equations by themselves can force $x = y = z = 0$ . Therefore no system with Equation 1 or with repeated equations can force $x = y = z = 0$
Case 1: Equation 8 ( $x + y + z = 0$ ) is present.
Case 1a: Equation 8, and two equations from $\{5, 6, 7\}$
There are $\binom{3}{2} = 3$ ways to choose two equations from $\{5, 6, 7\}$ and $3! = 6$ ways to arrange each case. The number of options that force $x = y = z = 0$ is $3 \cdot 3! = 18$
Case 1b: Equation 8, one equation from $\{5, 6, 7\}$ , and one equation from $\{2, 3, 4\}$
There are $\binom{3}{1} = 3$ ways to choose one equation from $\{5, 6, 7\}$ . WLOG let us choose Equation 7. Given $x + y + z = 0$ and $y + z = 0$ , we conclude that $x = 0$ . The third equation can be either $y = 0$ or $z = 0$ . There are $3!$ ways to arrange each case. The number of options that force $x = y = z = 0$ is $3 \cdot 2 \cdot 3! = 36$
Case 1c: Equation 8, and two equations from $\{2, 3, 4\}$
There are $\binom{3}{2} = 3$ ways to choose two equations from $\{2, 3, 4\}$ and $3! = 6$ ways to arrange each case. Each of these cases forces $x = y = z = 0$ $3 \cdot 3! = 18$ total options.
Case 2: Equation 8 is $\textbf{not}$ present, at least one equation from $\{5, 6, 7\}$ is present.
Case 2a: Equations $\{5, 6, 7\}$ are all present.
There are $3!$ ways to arrange the three equations. $6$ options.
Case 2b: Two equations from $\{5, 6, 7\}$ are present. One equation from $\{2, 3, 4\}$ is present.
There are $\binom{3}{2}$ ways to choose two equations from $\{5, 6, 7\}$ . WLOG let Equations 5 and 6 be in our system: $x + y = 0$ and $x + z = 0$ . Any equation from $\{2, 3, 4\}$ will force $x = y = z = 0$ . There are $3!$ ways to arrange the equations. The number of options that force $x = y = z = 0$ is $\binom{3}{2} \cdot \binom{3}{1} \cdot 3! = 54$
Case 2c: One equation from $\{5, 6, 7\}$ is present. Two equations from $\{2, 3, 4\}$ are present.
There are $\binom{3}{1}$ ways to choose one equation from $\{5, 6, 7\}$ . WLOG let Equation 5 ( $x + y = 0$ ) be present. One of the two equations from $\{2, 3, 4\}$ must be Equation 4, $z = 0$ , since it is the only equation that restricts $z$ . The last equation can be either 2 or 3. There are $3!$ ways to arrange the equations. The number of options that force $x = y = z = 0$ is $\binom{3}{1} \cdot \binom{2}{1} \cdot 3! = 36$
Case 3: Only equations $\{2, 3, 4\}$ are present.
There are $3!$ ways to arrange the three equations. $6$ options.
We add up the cases: $18 + 36 + 18 + 6 + 54 + 36 + 6 = 174$ total systems force $x = y = z = 0$ . Thus $512 - 174 = \boxed{338}$ do not.

Solution: The total number of possible systems is $2^9 = 512$ , with $8$ possible sets of coefficients per equation. We will use complementary counting to find the number of systems which only have the solution $(0, 0, 0)$ and subtract that from the total. Similar to what is observed in Solution 2, if any equation is repeated or $0x + 0y + 0z = 0$ , there will only be two or fewer equations for three variables, making one unique solution impossible. Therefore, we must choose $3$ different equations from $7$ possible ones, giving $7 \cdot 6 \cdot 5 = 210$ systems. However, there are two exceptions to consider, which will have more than one solution. The first is of the form $x + y + z = 0$ $x + y = 0$ $z = 0$ ; the second is of the form $x = 0$ $y = 0$ $x + y = 0$ . In both cases, there are $3$ ways to choose the variables in the equations, and then $6$ ways to arrange them, giving $2 \cdot 3 \cdot 6 = 36$ exceptions. Subtracting this gives $210 - 36 = 174$ systems with only one solution, and the answer is then $512 - 174 = \boxed{338}$

Solution:
Denote vector $\overrightarrow{i} = \left( i_1, i_2, i_3 \right)^T$ for $i \in \left\{ a, b, c \right\}$ .
Thus, we need to count how many vector tuples $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ are linearly dependent.
We do complementary counting.
First, the total number of vector tuples $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ is $\left( 2^3 \right)^3 = 512$
Second, we count how many many vector tuples $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ are linearly independent.
To meet this condition, no vector can be a zero vector $\overrightarrow{0} = \left( 0, 0, 0 \right)^T$
Next, we do the casework analysis.
Case $1^c$ : Three vectors are all on axes.
In this case, the number of $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ is $3!$
Case $2^c$ : Two vectors are on axes and the third vector is not.
We construct such an instance in the following steps.
Step 1: We determine which two vectors lie on axes.
The number of ways is $3$
Step 2: For two vectors selected in Step 1, we determine which two axes they lie on.
The number of ways is $3 \cdot 2$
Step 3: For the third unselected vector, we determine its value.
To make three vectors linear independent, the third vector cannot be on the plane formed by the first two vectors.
So the number of ways is $3$
Following from the rule of product, the number of $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ in this case is $3 \cdot 3 \cdot 2 \cdot 3$
Case $3^c$ : One vector is on an axis and the other two are not.
We construct such an instance in the following steps.
Step 1: We determine which vector lies on an axis.
The number of ways is $3$
Step 2: For the selected vector, we determine which axis it lies on.
The number of ways is $3$
Step 3: We determine the values of the two unselected vectors.
First, to be linearly independent, these two vectors are distinct.
Second, to be linearly independent, we cannot have one vector $(1,1,1)$ and another one that is a diagonal vector on the plane that is perpendicular to the first selected vector.
Thus, the number or ways in this step is $4 \cdot 3-2 = 10$
Following from the rule of product, the number of $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ in this case is $3 \cdot 3 \cdot 10$
Case $(4.4)^c$ : No vector is on any axis.
In this case, any three distinct vectors are linearly independent.
So the number of $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ in this case is $4 \cdot 3 \cdot 2$
Putting all cases together, the number of vector tuples $\left( \overrightarrow{a} , \overrightarrow{b} , \overrightarrow{c} \right)$ that are linearly independent is \[ 8^3 - 3!  - 3 \cdot 3 \cdot 2 \cdot 3 - 3 \cdot 3 \cdot 10 - 4 \cdot 3 \cdot 2 = \boxed{338}. \] ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)


Answer: 800.0

Consider the dark square in an array of unit squares, part of which is shown. The first ring of squares around this center square contains $8$ unit squares. The second ring contains $16$ unit squares. If we continue this process, the number of unit squares in the $100^\text{th}$ ring is
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$\textbf{(A)}\ 396 \qquad \textbf{(B)}\ 404 \qquad \textbf{(C)}\ 800 \qquad \textbf{(D)}\ 10,\!000 \qquad \textbf{(E)}\ 10,\!404$

Solution:
We can partition the $n^\text{th}$ ring into $4$ rectangles: two containing $2n+1$ unit squares and two containing $2n-1$ unit squares.
There are $2(2n+1)+2(2n-1)=4n+2+4n-2=8n$ unit squares in the $n^\text{th}$ ring.
Thus, the $100^\text{th}$ ring has $8 \times 100 = \boxed{800}$ unit squares.

Solution:
We can make the $n^\text{th}$ ring by removing a square of side length $2n-1$ from a square of side length $2n+1$
This ring contains $(2n+1)^2-(2n-1)^2=(4n^2+4n+1)-(4n^2-4n+1)=8n$ unit squares.
Thus, the $100^\text{th}$ ring has $8 \times 100 = \boxed{800}$ unit squares.

Solution: Notice that the first ring around the center square contains $8$ unit squares, the second ring contains $16$ unit squares, the third contains $24$ unit squares, and so on. The number of squares in the $n^\text{th}$ ring is determined by the expression $8 \times n$ . Thus, the number of unit squares in the $100^\text{th}$ ring is equal to $8 \times 100$ , which equals $\boxed{800}$ unit squares.


Answer: 42.0

Consider the following $100$ sets of $10$ elements each: \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} How many of these sets contain exactly two multiples of $7$
$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 42\qquad\textbf{(C)}\ 43\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 50$

Solution:
We apply casework to this problem. The only sets that contain two multiples of seven are those for which:
Each case has $\left\lfloor\frac{100}{7}\right\rfloor=14$ sets. Therefore, the answer is $14\cdot3=\boxed{42}.$

Solution:
We find a pattern. \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} We can figure out that the first set has $1$ multiple of $7$ . The second set also has $1$ multiple of $7$ . The third set has $2$ multiples of $7$ . The fourth set has $1$ multiple of $7$ . The fifth set has $2$ multiples of $7$ . The sixth set has $1$ multiple of $7$ . The seventh set has $2$ multiples of $7$ . Calculating this pattern further, we can see (reasonably) that it repeats for each $7$ sets.
We see that the pattern for the number of multiples per $7$ sets goes: $1,1,2,1,2,1,2.$ So, for every $7$ sets, there are three sets with $2$ multiples of $7$ . We calculate $\left\lfloor\frac{100}{7}\right\rfloor$ and multiply that by $3$ . (We also disregard the remainder of $2$ since it doesn't add any extra sets with $2$ multiples of $7$ .). We get $14\cdot3= \boxed{42}$

Solution:
Each set contains exactly $1$ or $2$ multiples of $7$
There are $\dfrac{1000}{10}=100$ total sets and $\left\lfloor\dfrac{1000}{7}\right\rfloor = 142$ multiples of $7$
Thus, there are $142-100=\boxed{42}$ sets with $2$ multiples of $7$


Answer: 42.0

Solution:
We apply casework to this problem. The only sets that contain two multiples of seven are those for which:
Each case has $\left\lfloor\frac{100}{7}\right\rfloor=14$ sets. Therefore, the answer is $14\cdot3=\boxed{42}.$

Solution:
We find a pattern. \begin{align*} &\{1,2,3,\ldots,10\}, \\ &\{11,12,13,\ldots,20\},\\ &\{21,22,23,\ldots,30\},\\ &\vdots\\ &\{991,992,993,\ldots,1000\}. \end{align*} We can figure out that the first set has $1$ multiple of $7$ . The second set also has $1$ multiple of $7$ . The third set has $2$ multiples of $7$ . The fourth set has $1$ multiple of $7$ . The fifth set has $2$ multiples of $7$ . The sixth set has $1$ multiple of $7$ . The seventh set has $2$ multiples of $7$ . Calculating this pattern further, we can see (reasonably) that it repeats for each $7$ sets.
We see that the pattern for the number of multiples per $7$ sets goes: $1,1,2,1,2,1,2.$ So, for every $7$ sets, there are three sets with $2$ multiples of $7$ . We calculate $\left\lfloor\frac{100}{7}\right\rfloor$ and multiply that by $3$ . (We also disregard the remainder of $2$ since it doesn't add any extra sets with $2$ multiples of $7$ .). We get $14\cdot3= \boxed{42}$

Solution:
Each set contains exactly $1$ or $2$ multiples of $7$
There are $\dfrac{1000}{10}=100$ total sets and $\left\lfloor\dfrac{1000}{7}\right\rfloor = 142$ multiples of $7$
Thus, there are $142-100=\boxed{42}$ sets with $2$ multiples of $7$


Answer: 4.0

Consider the graphs of $y = Ax^2$ and $y^2+3 = x^2+4y$ , where $A$ is a positive constant and $x$ and $y$ are real variables. In how many points do the two graphs intersect?
$\mathrm{(A) \ }\text{exactly }4 \qquad \mathrm{(B) \ }\text{exactly }2 \qquad$
$\mathrm{(C) \  }\text{at least }1,\text{ but the number varies for different positive values of }A \qquad$
$\mathrm{(D) \  }0\text{ for at least one positive value of }A \qquad \mathrm{(E) \ }\text{none of these}$

Solution: Substituting $y = Ax^2$ into the equation $y^2+3 = x^2+4y$ gives \begin{align*}\left(Ax^2\right)+3 = x^2+4\left(Ax^2\right) &\iff A^2x^4+3 = x^2+4Ax^2 \\ &\iff A^2x^4-\left(4A+1\right)x^2+3 = 0 \\ &\iff x^2 = \frac{4A+1 \pm \sqrt{4A^2+8A+1}}{2A^2} \\ &\text{(using the quadratic formula)}.\end{align*} Now observe that since $A$ is positive, $4A^2+8A+1$ is also positive, so the square root will always give two distinct real values. Moreover, \[\left(4A+1\right)^2 = 16A^2+8A+1 > 4A^2+8A+1,\] so $4A+1-\sqrt{4A^2+8A+1} > 0$ , meaning that both solutions for $x^2$ are positive. Hence both solutions will give $2$ distinct values of $x$ (the positive and negative square roots), and each of these will correspond to a distinct point of intersection of the graphs, so there are $2 \cdot 2 = \boxed{4}$ points of intersection.

Solution:
Firstly, note that $y = Ax^2$ is an upward-facing parabola (since $A > 0$ ) whose vertex is at the origin. We now manipulate the equation of the second graph as follows: \begin{align*}y^2+3 = x^2+4y &\iff y^2-4y+3-x^2 = 0 \\ &\iff y^2-4y+4-x^2 = 1 \\ &\iff \frac{(y-2)^2}{1}-\frac{(x-0)^2}{1} = 1,\end{align*} showing that it is a vertical (upward- and downward-opening) hyperbola with center $(0,2)$ and asymptotes $y=x+1$ and $y=-x+1$ . It therefore remains to consider graphically where the parabola will intersect the hyperbola.
On the lower branch of the hyperbola, the maximum point is $(0,2-1) = (0,1)$ , which is above the vertex of the parabola. Therefore, by continuity and the symmetry of both the parabola and the hyperbola in the $y$ -axis, there are always exactly $2$ intersection points here.
For the top branch, as it approaches the asymptote $y = x+1$ , its slope also approaches that of this asymptote, which is $1$ . However, for any upward-opening parabola, the slope approaches infinity as $x$ does, so no matter how small $A$ is (i.e. how 'flat' the parabola is), the parabola will eventually overtake the hyperbola, giving a point of intersection with positive $x$ -coordinate. As above, symmetry gives another point of intersection with negative $x$ -coordinate, so that there are $2$ intersection points with this branch too.
Thus there are a total of $2+2 = \boxed{4}$ intersection points.


Answer: 342.0

Consider the integer \[N = 9 + 99 + 999 + 9999 + \cdots + \underbrace{99\ldots 99}_\text{321 digits}.\] Find the sum of the digits of $N$

Solution: Let's express the number in terms of $10^n$ . We can obtain $(10-1)+(10^2-1)+(10^3-1)+\cdots+(10^{321}-1)$ . By the commutative and associative property, we can group it into $(10+10^2+10^3+\cdots+10^{321})-321$ . We know the former will yield $1111....10$ , so we only have to figure out what the last few digits are. There are currently $321$ 1's. We know the last four digits are $1110$ , and that the others will not be affected if we subtract $321$ . If we do so, we get that $1110-321=789$ . This method will remove three $1$ 's, and add a $7$ $8$ and $9$ . Therefore, the sum of the digits is $(321-3)+7+8+9=\boxed{342}$

Solution: Observe how adding results in the last term but with a $1$ concatenated in front and also a $1$ subtracted ( $09$ $108$ $1107$ $11106$ ). Then for any index of terms, $n$ , the sum is $11...10-n$ , where the first term is of length $n+1$ . Here, that is $\boxed{342}$


Answer: 4.0

Consider the operation $*$ defined by the following table:
\[\begin{tabular}{c|cccc} * & 1 & 2 & 3 & 4 \\ \hline 1 & 1 & 2 & 3 & 4 \\ 2 & 2 & 4 & 1 & 3 \\ 3 & 3 & 1 & 4 & 2 \\ 4 & 4 & 3 & 2 & 1 \end{tabular}\]
For example, $3*2=1$ .  Then $(2*4)*(1*3)=$
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$

Solution: Using the chart, $(2*4)=3$ and $(1*3)=3$ . Therefore, $(2*4)*(1*3)=3*3=\boxed{4}$

Solution: By the chart, we can see that the " $*$ " operation is actually multiplication modulo $5$ . Thus, we can do $(2*4)*(1*3)\rightarrow(2\cdot4)\cdot(1\cdot3)=8\cdot3=24\rightarrow\boxed{4}$


Answer: 408.0

Consider the paper triangle whose vertices are $(0,0), (34,0),$ and $(16,24).$ The vertices of its midpoint triangle are the midpoints of its sides.  A triangular pyramid is formed by folding the triangle along the sides of its midpoint triangle.  What is the volume of this pyramid?

Solution:
As shown in the image above, let $D$ $E$ , and $F$ be the midpoints of $\overline{BC}$ $\overline{CA}$ , and $\overline{AB}$ , respectively.  Suppose $P$ is the apex of the tetrahedron, and let $O$ be the foot of the altitude from $P$ to $\triangle ABC$ .  The crux of this problem is the following lemma.
Lemma: The point $O$ is the orthocenter of $\triangle ABC$
Proof. Observe that \[OF^2 - OE^2 = PF^2 - PE^2 = AF^2 - AE^2;\] the first equality follows by the Pythagorean Theorem, while the second follows from $AF = FP$ and $AE = EP$ .  Thus, by the Perpendicularity Lemma, $AO$ is perpendicular to $FE$ and hence $BC$ .  Analogously, $O$ lies on the $B$ -altitude and $C$ -altitude of $\triangle ABC$ , and so $O$ is, indeed, the orthocenter of $\triangle ABC$
To find the coordinates of $O$ , we need to find the intersection point of altitudes $BE$ and $AD$ . The equation of $BE$ is simply $x=16$ $AD$ is perpendicular to line $BC$ , so the slope of $AD$ is equal to the negative reciprocal of the slope of $BC$ $BC$ has slope $\frac{24-0}{16-34}=-\frac{4}{3}$ , therefore $y=\frac{3}{4} x$ . These two lines intersect at $(16,12)$ , so that's the base of the height of the tetrahedron.
Let $S$ be the foot of altitude $BS$ in $\triangle BPQ$ . From the Pythagorean Theorem $h=\sqrt{BS^2-SO^2}$ . However, since $S$ and $O$ are, by coincidence, the same point, $SO=0$ and $h=12$
The area of the base is $102$ , so the volume is $\frac{102*12}{3}=\boxed{408}$

Solution:
Consider the diagram provided in the previous solution. We first note that the medial triangle has coordinates $(17, 0, 0)$ $(8, 12, 0)$ , and $(25, 12, 0)$ . We can compute the area of this triangle as $102$ . Suppose $(x, y, z)$ are the coordinates of the vertex of the resulting pyramid. Call this point $V$ . Clearly, the height of the pyramid is $z$ . The desired volume is thus $\frac{102z}{3} = 34z$
We note that when folding the triangle to form the pyramid, some side lengths must stay the same. In particular, $VR = RA$ $VP = PB$ , and $VQ = QC$ . We then use distance formula to find the distances from $V$ to each of the vertices of the medial triangle. We thus arrive at a fairly simple system of equations, yielding $z = 12$ . The desired volume is thus $34 \times 12 = \boxed{408}$

Solution:
The formed tetrahedron has pairwise parallel planar and oppositely equal length ( $4\sqrt{13},15,17$ ) edges and can be inscribed in a parallelepiped (rectangular box) with the six tetrahedral edges as non-intersecting diagonals of the box faces.  Let the edge lengths of the parallelepiped be $p,q,r$ and solve (by Pythagoras)
$p^2+q^2=4^2\cdot{13}$
$q^2+r^2=15^2$
$r^2+p^2=17^2$
to find that $(p^2,q^2,r^2)=(153,136,72)=(3^2\cdot{17},2^3\cdot{17},2^3\cdot{3^2}).$
Use the fact that the ratio of volumes between an inscribed tetrahedron and its circumscribing parallelepiped is $\tfrac{1}{3}$ and then the volume is
$\tfrac{1}{3}pqr=\tfrac{1}{3}\sqrt{2^6\cdot{3^4}\cdot{17^2}}=\boxed{408}$

Solution:
Let $A = (0,0), B = (16, 24), C = (34,0).$ Then define $D,E,F$ as the midpoints of $BC, AC, AB$ . By Pythagorean theorem, $EF = \frac{1}{2} BC = 15, DE = \frac{1}{2}AB = 4 \sqrt{13}, DF = \frac{1}{2} AC = 17.$ Then let $P$ be the point in space which is the vertex of the tetrahedron with base $DEF$
Note that $\triangle DEP \cong \triangle EDF$ . Create point $F'$ on the plane of $DEF$ such that $\triangle DEP \cong \triangle DEF'$ (i.e by reflecting $F$ over the perpendicular bisector of $DE$ ). Project $F, P$ onto $DE$ as $X, Y$ . Note by the definition of $F'$ then $\angle PYF'$ is the dihedral angle between planes $DEP, DEF$
Now see that by Heron's, \[[DEP] = [DEF] = \sqrt{(16 + 2 \sqrt{13})(16 - 2 \sqrt{13})(1 + 2 \sqrt{13})(-1 + 2 \sqrt{13})} = 102.\] So $PY$ , the hypotenuse $DEP$ has length $\frac{102 \cdot 2}{4 \sqrt{13}} = \frac{51}{\sqrt{13}}$ . Similarly $F'Y = \frac{51}{\sqrt{13}}.$ Further from Pythagoras $DY = \sqrt{DP^2 - PY^2} = \frac{18}{\sqrt{13}}.$ Symmetrically $EX = \frac{18}{\sqrt{13}}.$ Therefore $XY = DE - DY - EX = \frac{16}{\sqrt{13}}.$
By Law of Cosines on $\triangle PYF'$ \begin{align*} PF'^2 &= PY^2 + F'Y^2 - 2 \cdot PY \cdot F'Y \cos{\angle PYF'} \\ PF^2 - XY^2 &= 2 (\frac{51}{\sqrt{13}})^2 \cos{\angle PYF'} \\ (4\sqrt{13})^2 - (\frac{16}{\sqrt{13}})^2 &= 2 (\frac{51}{\sqrt{13}})^2 \cos{\angle PYF'}  \\ \cos{\angle PYF'} &= \frac{9}{17} \\ \sin{\angle PYF'} &= \frac{4 \sqrt{13}}{17}. \end{align*}.
Therefore the altitude of the tetrahedron from vertex $P$ to base $DEF$ is $PY \sin{\angle PYF'} = \frac{51}{\sqrt{13}} \frac{4 \sqrt{13}}{17} = 12.$ So the area is $\frac{1}{3}bh = \frac{1}{3} 12 \cdot 102 = \boxed{408}.$


Answer: 118.0

Consider the parallelogram with vertices $(10,45)$ $(10,114)$ $(28,153)$ , and $(28,84)$ .  A line through the origin cuts this figure into two congruent polygons .  The slope of the line is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers .  Find $m+n$

Solution: Let the first point on the line $x=10$ be $(10,45+a)$ where a is the height above $(10,45)$ .  Let the second point on the line $x=28$ be $(28, 153-a)$ . For two given points, the line will pass the origin if the coordinates are proportional (such that $\frac{y_1}{x_1} = \frac{y_2}{x_2}$ ). Then, we can write that $\frac{45 + a}{10} = \frac{153 - a}{28}$ . Solving for $a$ yields that $1530 - 10a = 1260 + 28a$ , so $a=\frac{270}{38}=\frac{135}{19}$ . The slope of the line (since it passes through the origin) is $\frac{45 + \frac{135}{19}}{10} = \frac{99}{19}$ , and the solution is $m + n = \boxed{118}$

Solution: You can clearly see that a line that cuts a parallelogram into two congruent pieces must go through the center of the parallelogram. Taking the midpoint of $(10,45)$ , and $(28,153)$ gives $(19,99)$ , which is the center of the parallelogram. Thus the slope of the line must be $\frac{99}{19}$ , and the solution is $\boxed{118}$

Solution:
Note that the area of the parallelogram is equivalent to $69 \cdot 18 = 1242,$ so the area of each of the two trapezoids with congruent area is $621.$ Therefore, since the height is $18,$ the sum of the bases of each trapezoid must be $69.$
The points where the line in question intersects the long side of the parallelogram can be denoted as $(10, \frac{10m}{n})$ and $(28, \frac{28m}{n}),$ respectively. We see that $\frac{10m}{n} - 45 + \frac{28m}{n} - 84 = 69,$ so $\frac{38m}{n} = 198 \implies \frac{m}{n} = \frac{99}{19} \implies \boxed{118}.$

Solution: $(\Sigma x_i /4, \Sigma y_i /4)$ is the centroid, which generates $(19,99)$ , so the answer is $\boxed{118}$ . This is the fastest way because you do not need to find the opposite vertices by drawing.


Answer: 294.0

Consider the paths of length $16$ that follow the lines from the lower left corner to the upper right corner on an $8\times 8$ grid. Find the number of such paths that change direction exactly four times, as in the examples shown below.
[asy] size(10cm); usepackage("tikz");label("\begin{tikzpicture}[scale=.5]\draw(0,0)grid(8,8);\draw[line width=2,red](0,0)--(2,0)--(2,3)--(5,3)--(5,8)--(8,8);\end{tikzpicture}",origin); label("\begin{tikzpicture}[scale=.5]\draw(0,0)grid(8,8);\draw[line width=2,red](0,0)--(0,3)--(3,3)--(3,5)--(8,5)--(8,8);\end{tikzpicture}",E); [/asy]

Solution:
We divide the path into eight “ $R$ ” movements and eight “ $U$ ” movements. Five sections of alternative $RURUR$ or $URURU$ are necessary in order to make four “turns.” We use the first case and multiply by $2$
For $U$ , we have seven ordered pairs of positive integers $(a,b)$ such that $a+b=8$
For $R$ , we subtract $1$ from each section (to make the minimum stars of each section $0$ ) and we use Stars and Bars to get ${7 \choose 5}=21$
Thus our answer is $7\cdot21\cdot2=\boxed{294}$

Solution:
Notice that the $RURUR$ case and the $URURU$ case is symmetrical. WLOG, let's consider the RURUR case.
Now notice that there is a one-to-one correspondence between this problem and the number of ways to distribute 8 balls into 3 boxes and also 8 other balls into 2 other boxes, such that each box has a nonzero amount of balls.
There are ${8+2-3 \choose 2}$ ways for the first part, and ${8+1-2 \choose 1}$ ways for the second part, by stars and bars.
The answer is $2\cdot {7 \choose 2} \cdot {7 \choose 1} = \boxed{294}$

Solution: Starting at the origin, you can either first go up or to the right. If you go up first, you will end on the side opposite to it (the right side) and if you go right first, you will end up on the top. It can then be observed that if you choose the turning points in the middle $7 \times 7$ grid, that will automatically determine your start and ending points. For example, in the diagram if you choose the point $(3,2)$ and $(5,3)$ , you must first move three up or two right, determining your first point, and move 5 up or 3 right, determining your final point. Knowing this is helpful because if we first move anywhere horizontally, we have $7$ points on each column to choose from and starting from left to right, we have $6,5,4,3,2,1$ points on that row to choose from. This gives us $7(6)+7(5)+7(4)+7(3)+7(2)+7(1)$ which simplifies to $7\cdot21$ . The vertical case is symmetrical so we have $7\cdot21\cdot2 = \boxed{294}$

Solution: As in Solution 1, there are two cases: $RURUR$ or $URURU$ . We will work with the first case and multiply by $2$ at the end. We use stars and bars; we can treat the $R$ s as the stars and the $U$ s as the bars. However, we must also use stars and bars on the $U$ s to see how many different patterns of bars we can create for the reds. We must have $1$ bar in $8$ blacks, so we use stars and bars on the equation \[x + y = 8\] . However, each divider must have at least one black in it, so we do the change of variable $x' = x-1$ and $y' = x-1$ . Our equation becomes \[x' + y' = 6\] . By stars and bars, this equation has $\binom{6 + 2 - 1}{1} = 7$ valid solutions. Now, we use stars and bars on the reds. We must distribute two bars amongst the reds, so we apply stars and bars to \[x + y + z = 8\] . Since each group must have one red, we again do a change of variables with $x' = x-1$ $y' = y-1$ , and $z' = z-1$ . We are now working on the equation \[x' + y' + z' = 5\] . By stars and bars, this has $\binom{5 + 3 - 1}{2} = 21$ solutions. The number of valid paths in this case is the number of ways to create the bars times the number of valid arrangements of the stars given fixed bars, which equals $21 \cdot 7 = 147$ . We must multiply by two to account for both cases, so our final answer is $147 \cdot 2 = \boxed{294}$


Answer: 936.0

Consider the points $A(0,12), B(10,9), C(8,0),$ and $D(-4,7).$ There is a unique square $S$ such that each of the four points is on a different side of $S.$ Let $K$ be the area of $S.$ Find the remainder when $10K$ is divided by $1000$

Solution: Let $(a,b)$ denote a normal vector of the side containing $A$ . Note that $\overline{AC}, \overline{BD}$ intersect and hence must be opposite vertices of the square. The lines containing the sides of the square have the form $ax+by=12b$ $ax+by=8a$ $bx-ay=10b-9a$ , and $bx-ay=-4b-7a$ . The lines form a square, so the distance between $C$ and the line through $A$ equals the distance between $D$ and the line through $B$ , hence $8a+0b-12b=-4b-7a-10b+9a$ , or $-3a=b$ . We can take $a=-1$ and $b=3$ . So the side of the square is $\frac{44}{\sqrt{10}}$ , the area is $K=\frac{1936}{10}$ , and the answer to the problem is $\boxed{936}$


Answer: 6.0

Consider the polynomial
\[P(x)=\prod_{k=0}^{10}(x^{2^k}+2^k)=(x+1)(x^2+2)(x^4+4)\cdots (x^{1024}+1024)\]
The coefficient of $x^{2012}$ is equal to $2^a$ . What is $a$
\[\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 24\]

Solution: The degree of $P(x)$ is $1024+512+256+\cdots+1=2047$ . We want to find the coefficient of $x^{2012}$ , so we need to omit the powers of $2$ that add up to $2047-2012=35$ . We find that $35=2^0+2^1+2^5$ . From here, we know that the answer is $2^0\cdot2^1\cdot2^5=2^6$ . Therefore, the answer is $\boxed{6.}$


Answer: 6.0

Solution:
When we use long division to divide $P(x)$ by $Q(x)$ , the remainder is $x^2-x+1$
So, since $z_1$ is a root, $P(z_1)=(z_1)^2-z_1+1$
Now this also follows for all roots of $Q(x)$ Now \[P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1\]
Now by Vieta's we know that $-z_4-z_3-z_2-z_1=-1$ ,
so by Newton's Sums we can find $z_1^2+z_2^2+z_3^2+z_4^2$
$a_ns_2+a_{n-1}s_1+2a_{n-2}=0$
$(1)(s_2)+(-1)(1)+2(-1)=0$
$s_2-1-2=0$
$s_2=3$
So finally $P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{006}.$

Solution:
Let $S_k=z_1^k+z_2^k+z_3^k+z_4^k$ then by Vieta's Formula we have \[S_{-1}=\frac{z_1z_2z_3+z_1z_3z_4+z_1z_2z_4+z_1z_2z_3}{z_1z_2z_3z_4}=0\] \[S_0=4\] \[S_1=1\] \[S_2=3\] By Newton's Sums we have \[a_4S_k+a_3S_{k-1}+a_2S_{k-2}+a_1S_{k-1}+a_0S_{k-4}=0\]
Applying the formula couples of times yields $P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\boxed{6}$


Answer: 572.0

Consider the region $A$ in the complex plane that consists of all points $z$ such that both $\frac{z}{40}$ and $\frac{40}{\overline{z}}$ have real and imaginary parts between $0$ and $1$ , inclusive. What is the integer that is nearest the area of $A$

Solution:
Let $z=a+bi \implies \frac{z}{40}=\frac{a}{40}+\frac{b}{40}i$ . Since $0\leq \frac{a}{40},\frac{b}{40}\leq 1$ we have the inequality \[0\leq a,b \leq 40\] which is a square of side length $40$
Also, $\frac{40}{\overline{z}}=\frac{40}{a-bi}=\frac{40a}{a^2+b^2}+\frac{40b}{a^2+b^2}i$ so we have $0\leq a,b \leq \frac{a^2+b^2}{40}$ , which leads to: \[(a-20)^2+b^2\geq 20^2\] \[a^2+(b-20)^2\geq 20^2\]
We graph them:
To find the area outside the two circles but inside the square, we want to find the unique area of the two circles. We can do this by adding the area of the two circles and then subtracting out their overlap. There are two methods of finding the area of overlap:
1. Consider that the area is just the quarter-circle with radius $20$ minus an isosceles right triangle with base length $20$ , and then doubled (to consider the entire overlapped area)
2. Consider that the circles can be converted into polar coordinates, and their equations are $r = 40sin\theta$ and $r = 40cos\theta$ . Using calculus with the appropriate bounds, we can compute the overlapped area.
Using either method, we compute the overlapped area to be $200\pi + 400$ , and so the area of the intersection of those three graphs is $40^2-(200\pi + 400) \Rightarrow 1200 - 200\pi \approx 571.68$
$\boxed{572}$


Answer: 0.5

Consider the sequence
$1,-2,3,-4,5,-6,\ldots,$
whose $n$ th term is $(-1)^{n+1}\cdot n$ . What is the average of the first $200$ terms of the sequence?
$\textbf{(A)}-\!1\qquad\textbf{(B)}-\!0.5\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 0.5\qquad\textbf{(E)}\ 1$

Solution:
The average of a list is the sum of all numbers divided by the size of the list.
The sum of the list can be found by adding the numbers in pairs: $(1 + -2) + (3 + -4) + ... + (199 + -200)$
The sum of each pair is $-1$ , and there are $100$ pairs, so the total sum is $-100$
There are $200$ numbers on the list, so the average is $\frac{-100}{200} = -0.5$ , and the answer is $\boxed{0.5}$