Answer: 761.0

Solution:
Split $F_n$ into $4$ congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to $4$ times the $(n-2)$ th triangular number (i.e. the diamonds within the triangles or between the diagonals) and $4(n-1)+1$ (the diamonds on sides of the triangles or on the diagonals). The $n$ th triangular number is $\frac{n(n+1)}{2}$ . Putting this together for $F_{20}$ this gives:
$\frac{4(18)(19)}{2}+4(19)+1=\boxed{761}$

Solution:
Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:

In the figure $F_n$ , the blue diamonds form a $n\times n$ square, and the red diamonds form a $(n-1)\times(n-1)$ square.
Hence the total number of diamonds in $F_{20}$ is $20^2 + 19^2 = \boxed{761}$

Solution:
When constructing $F_n$ from $F_{n-1}$ , we add $4(n-1)$ new diamonds. Let $d_n$ be the number of diamonds in $F_n$ . We now know that $d_1=1$ and $\forall n>1:~ d_n=d_{n-1} + 4(n-1)$
Hence we get: \begin{align*} d_{20} & = d_{19} + 4\cdot 19 \\ & = d_{18} + 4\cdot 18 + 4\cdot 19 \\ & = \cdots \\ & = 1 + 4(1+2+\cdots+18+19) \\ & = 1 + 4\cdot\frac{19\cdot 20}2 \\ & = \boxed{761}


Answer: 761.0

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

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


Answer: 761.0

Solution:
Split $F_n$ into $4$ congruent triangles by its diagonals (like in the pictures in the problem). This shows that the number of diamonds it contains is equal to $4$ times the $(n-2)$ th triangular number (i.e. the diamonds within the triangles or between the diagonals) and $4(n-1)+1$ (the diamonds on sides of the triangles or on the diagonals). The $n$ th triangular number is $\frac{n(n+1)}{2}$ . Putting this together for $F_{20}$ this gives:
$\frac{4(18)(19)}{2}+4(19)+1=\boxed{761}$

Solution:
Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:

In the figure $F_n$ , the blue diamonds form a $n\times n$ square, and the red diamonds form a $(n-1)\times(n-1)$ square.
Hence the total number of diamonds in $F_{20}$ is $20^2 + 19^2 = \boxed{761}$

Solution:
When constructing $F_n$ from $F_{n-1}$ , we add $4(n-1)$ new diamonds. Let $d_n$ be the number of diamonds in $F_n$ . We now know that $d_1=1$ and $\forall n>1:~ d_n=d_{n-1} + 4(n-1)$
Hence we get: \begin{align*} d_{20} & = d_{19} + 4\cdot 19 \\ & = d_{18} + 4\cdot 18 + 4\cdot 19 \\ & = \cdots \\ & = 1 + 4(1+2+\cdots+18+19) \\ & = 1 + 4\cdot\frac{19\cdot 20}2 \\ & = \boxed{761}


Answer: 1979.0

The first AMC $8$ was given in $1985$ and it has been given annually since that time. Samantha turned $12$ years old the year that she took the seventh AMC $8$ . In what year was Samantha born?
$\textbf{(A) }1979\qquad\textbf{(B) }1980\qquad\textbf{(C) }1981\qquad\textbf{(D) }1982\qquad \textbf{(E) }1983$

Solution:
The seventh AMC 8 would have been given in $1991$ . If Samantha was 12 then, that means she was born 12 years ago, so she was born in $1991-12=1979$
Our answer is $\boxed{1979}$ corrections made by DrDominic

Solution:
Since she was 12 when she took the seventh AMC 8, she should be $12-6=6$ years old when the first AMC 8 occurred. Therefore, she was born or was 'age 0' in $1985-6=\boxed{1979}$ ~SweetMango77
corrections made by DrDominic


Answer: 8041.0

The first four terms of an arithmetic sequence are $p$ $9$ $3p-q$ , and $3p+q$ . What is the $2010^\text{th}$ term of this sequence?
$\textbf{(A)}\ 8041 \qquad \textbf{(B)}\ 8043 \qquad \textbf{(C)}\ 8045 \qquad \textbf{(D)}\ 8047 \qquad \textbf{(E)}\ 8049$

Solution:
$3p-q$ and $3p+q$ are consecutive terms, so the common difference is $(3p+q)-(3p-q) = 2q$
\begin{align*}p+2q &= 9\\ 9+2q &= 3p-q\\ q&=2\\ p&=5\end{align*}
The common difference is $4$ . The first term is $5$ and the $2010^\text{th}$ term is
\[5+4(2009) = \boxed{8041}\]

Solution: Since all the answer choices are around $2010 \cdot 4 = 8040$ , the common difference must be $4$ . The first term is therefore $9 - 4 = 5$ , so the $2010^\text{th}$ term is $5 + 4 \cdot 2009 = \boxed{8041}$


Answer: 250.0

The first term of a sequence is $2005$ . Each succeeding term is the sum of the cubes of the digits of the previous term. What is the ${2005}^{\text{th}}$ term of the sequence?
$\textbf{(A) } 29 \qquad \textbf{(B) } 55 \qquad \textbf{(C) } 85 \qquad \textbf{(D) } 133 \qquad \textbf{(E) } 250$

Solution: Performing this operation several times yields the results of $133$ for the second term, $55$ for the third term, and $250$ for the fourth term. The sum of the cubes of the digits of $250$ equal $133$ , a complete cycle. The cycle is, excluding the first term, the $2^{\text{nd}}$ $3^{\text{rd}}$ , and $4^{\text{th}}$ terms will equal $133$ $55$ , and $250$ , following the fourth term. Any term number that is equivalent to $1\ (\text{mod}\ 3)$ will produce a result of $250$ . It just so happens that $2005\equiv 1\ (\text{mod}\ 3)$ , which leads us to the answer of $\boxed{250}$


Answer: 250.0

Solution: Performing this operation several times yields the results of $133$ for the second term, $55$ for the third term, and $250$ for the fourth term. The sum of the cubes of the digits of $250$ equal $133$ , a complete cycle. The cycle is, excluding the first term, the $2^{\text{nd}}$ $3^{\text{rd}}$ , and $4^{\text{th}}$ terms will equal $133$ $55$ , and $250$ , following the fourth term. Any term number that is equivalent to $1\ (\text{mod}\ 3)$ will produce a result of $250$ . It just so happens that $2005\equiv 1\ (\text{mod}\ 3)$ , which leads us to the answer of $\boxed{250}$


Answer: 0.0

The first three terms of an arithmetic progression are $x - 1, x + 1, 2x + 3$ , in the order shown. The value of $x$ is:
$\textbf{(A)}\ -2\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{undetermined}$

Solution:
Let $y$ represent the common difference between the terms. We have $(x+1)-y=(x-1)\implies y=2$
Substituting gives us $(2x+3)-2=(x+1)\implies 2x+1=x+1\implies x=0$
Therefore, our answer is $\boxed{0}$


Answer: 502.0

The first three terms of an arithmetic sequence are $2x - 3$ $5x - 11$ , and $3x + 1$ respectively. The $n$ th term of the sequence is $2009$ . What is $n$
$\textbf{(A)}\ 255 \qquad \textbf{(B)}\ 502 \qquad \textbf{(C)}\ 1004 \qquad \textbf{(D)}\ 1506 \qquad \textbf{(E)}\ 8037$

Solution: As this is an arithmetic sequence, the difference must be constant: $(5x-11) - (2x-3) = (3x+1) - (5x-11)$ . This solves to $x=4$ . The first three terms then are $5$ $9$ , and $13$ . In general, the $n$ th term is $1+4n$ . Solving $1+4n=2009$ , we get $n=\boxed{502}$


Answer: 0.0

The first two terms of a sequence are $a_1 = 1$ and $a_2 = \frac {1}{\sqrt3}$ . For $n\ge1$
What is $|a_{2009}|$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2 - \sqrt3\qquad \textbf{(C)}\ \frac {1}{\sqrt3}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2 + \sqrt3$

Solution:
Consider another sequence $\{\theta_1, \theta_2, \theta_3...\}$ such that $a_n = \tan{\theta_n}$ , and $0 \leq \theta_n < 180$
The given recurrence becomes
It follows that $\theta_{n + 2} \equiv \theta_{n + 1} + \theta_{n} \pmod{180}$ . Since $\theta_1 = 45, \theta_2 = 30$ , all terms in the sequence $\{\theta_1, \theta_2, \theta_3...\}$ will be a multiple of $15$
Now consider another sequence $\{b_1, b_2, b_3...\}$ such that $b_n = \theta_n/15$ , and $0 \leq b_n < 12$ . The sequence $b_n$ satisfies $b_1 = 3, b_2 = 2, b_{n + 2} \equiv b_{n + 1} + b_n \pmod{12}$
As the number of possible consecutive two terms is finite, we know that the sequence $b_n$ is periodic. Write out the first few terms of the sequence until it starts to repeat.
$\{b_n\} = \{3,2,5,7,0,7,7,2,9,11,8,7,3,10,1,11,0,11,11,10,9,7,4,11,3,2,5,7,...\}$
Note that $b_{25} = b_1 = 3$ and $b_{26} = b_2 = 2$ . Thus $\{b_n\}$ has a period of $24$ $b_{n + 24} = b_n$
It follows that $b_{2009} = b_{17} = 0$ and $\theta_{2009} = 15 b_{2009} = 0$ . Thus $a_{2009} = \tan{\theta_{2009}} = \tan{0} = 0.$
Our answer is $|a_{2009}| = \boxed{0}$

Solution:
First, some intuition. The given recurrence relation: $a_{n + 2} = \frac {a_n + a_{n + 1}}{1 - a_na_{n + 1}}$ looks much like the tangent addition formula. So, we let $a_n=\tan{\theta_n}$ . Then, we get: \[\tan{\theta_n}=\tan{(\theta_{n-1}+\theta_{n-2})}\] This gives us: \[\theta_n \equiv \theta_{n-1}+\theta_{n-2} \pmod{180}\] Now, observe that \[\theta_1=45\] \[\theta_2=30\] We know that this sequence of values for $\theta_n$ will repeat eventually because it is mod $180$ .
It is just a matter of when, so we start bashing:
\[\theta_3=75\] \[\theta_4=105\] \[\theta_5=0\] \[\theta_6=105\] \[\theta_7=105\] \[\theta_8=30\] \[\theta_9=135\] \[\theta_{10}=165\] \[\theta_{11}=120\] \[\theta_{12}=105\] \[\theta_{13}=45\] \[\theta_{14}=150\] \[\theta_{15}=15\] \[\theta_{16}=165\] \[\theta_{17}=0\] \[\theta_{18}=165\] \[\theta_{19}=165\] \[\theta_{20}=150\] \[\theta_{21}=135\] \[\theta_{22}=105\] \[\theta_{23}=60\] \[\theta_{24}=165\] \[\theta_{25}=45\] \[\theta_{26}=30\] And there is the repetition. So, this series has a period of 24. $2009 \equiv 17 \pmod{24}$ , so $|a_{2009}|=|\tan{\theta_{17}}|=|\tan{0}|=|0|= \boxed{0}$ ~Firebolt360


Answer: 11.0

The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is $\tfrac{a-\sqrt{2}}{b}$ , where $a$ and $b$ are positive integers. What is $a+b$
[asy] real x=.369; draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); filldraw((0,0)--(0,x)--(x,x)--(x,0)--cycle, gray); filldraw((0,1)--(0,1-x)--(x,1-x)--(x,1)--cycle, gray); filldraw((1,1)--(1,1-x)--(1-x,1-x)--(1-x,1)--cycle, gray); filldraw((1,0)--(1,x)--(1-x,x)--(1-x,0)--cycle, gray); filldraw((.5,.5-x*sqrt(2)/2)--(.5+x*sqrt(2)/2,.5)--(.5,.5+x*sqrt(2)/2)--(.5-x*sqrt(2)/2,.5)--cycle, gray); [/asy]
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$

Solution:
Let $s$ be the side length of the small squares.
The diagonal of the big square can be written in two ways: $\sqrt{2}$ and $s \sqrt{2} + s + s \sqrt{2}$
Solving for $s$ , we get $s = \frac{4 - \sqrt{2}}{7}$ , so our answer is $4 + 7 \Rightarrow \boxed{11}$

Solution: The diagonal of the small square can be written in two ways: $s \sqrt(2)$ and $2*(1-2s).$ Equating and simplifying gives $s = \frac{4 - \sqrt{2}}{7}$ . Hence our answer is $4 + 7 \Rightarrow \boxed{11}.$


Answer: 7.0

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

Solution:
The solutions to this equation are $z = 1$ $z = -1 \pm i\sqrt 3$ , and $z = -2\pm i\sqrt 2$ . Consider the five points $(1,0)$ $\left(-1,\pm\sqrt 3\right)$ , and $\left(-2,\pm\sqrt 2\right)$ ; these are the five points which lie on $\mathcal E$ . Note that since these five points are symmetric about the $x$ -axis, so must $\mathcal E$
Now let $r=b/a$ denote the ratio of the length of the minor axis of $\mathcal E$ to the length of its major axis. Remark that if we perform a transformation of the plane which scales every $x$ -coordinate by a factor of $r$ $\mathcal E$ is sent to a circle $\mathcal E'$ . Thus, the problem is equivalent to finding the value of $r$ such that $(r,0)$ $\left(-r,\pm\sqrt 3\right)$ , and $\left(-2r,\pm\sqrt 2\right)$ all lie on a common circle; equivalently, it suffices to determine the value of $r$ such that the circumcenter of the triangle formed by the points $P_1 = (r,0)$ $P_2 = \left(-r,\sqrt 3\right)$ , and $P_3 = \left(-2r,\sqrt 2\right)$ lies on the $x$ -axis.
Recall that the circumcenter of a triangle $ABC$ is the intersection point of the perpendicular bisectors of its three sides. The equations of the perpendicular bisectors of the segments $\overline{P_1P_2}$ and $\overline{P_1P_3}$ are \[y = \tfrac{\sqrt 3}2 + \tfrac{2r}{\sqrt 3}x\qquad\text{and}\qquad y = \tfrac{\sqrt 2}2 + \tfrac{3r}{\sqrt 2}\left(x + \tfrac r2\right)\] respectively. These two lines have different slopes for $r\neq 0$ , so indeed they will intersect at some point $(x_0,y_0)$ ; we want $y_0 = 0$ . Plugging $y = 0$ into the first equation yields $x = -\tfrac{3}{4r}$ , and so plugging $y=0$ into the second equation and simplifying yields \[-\tfrac{1}{3r} = x + \tfrac r2 = -\tfrac{3}{4r} + \tfrac{r}{2}.\] Solving yields $r=\sqrt{\tfrac 56}$
Finally, recall that the lengths $a$ $b$ , and $c$ (where $c$ is the distance between the foci of $\mathcal E$ ) satisfy $c = \sqrt{a^2 - b^2}$ . Thus the eccentricity of $\mathcal E$ is $\tfrac ca = \sqrt{1 - \left(\tfrac ba\right)^2} = \sqrt{\tfrac 16}$ and the requested answer is $\boxed{7}$

Solution:
Completing the square in the original equation, we have \[(z-1)\left((z+1)^2+3\right)\left((z+2)^2+2\right)=0,\] from which $z=1,-1\pm\sqrt{3}i,-2\pm\sqrt{2}i.$
Now, we will find the equation of an ellipse $\mathcal E$ that passes through $(1,0),\left(-1,\pm\sqrt3\right),$ and $\left(-2,\pm\sqrt2\right)$ in the $xy$ -plane. By symmetry, the center of $\mathcal E$ must be on the $x$ -axis.
The formula of $\mathcal E$ is \[\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1, \hspace{44.5mm} (\bigstar)\] with the center $(h,0)$ and the axes' lengths $2a$ and $2b.$
Plugging the points $(1,0),\left(-1,\sqrt3\right),$ and $\left(-2,\sqrt2\right)$ into $(\bigstar),$ respectively, we have the following system of equations: \begin{align*} \frac{(1-h)^2}{a^2}&=1, \\ \frac{(-1-h)^2}{a^2}+\frac{{\sqrt3}^2}{b^2}&=1, \\ \frac{(-2-h)^2}{a^2}+\frac{{\sqrt2}^2}{b^2}&=1. \end{align*} Since $t^2=(-t)^2$ holds for all real numbers $t,$ we clear fractions and simplify: \begin{align*} (1-h)^2&=a^2, \hspace{30.25mm} &(1)\\ b^2(1+h)^2 + 3a^2 &= a^2b^2, &(2)\\ b^2(2+h)^2 + 2a^2 &= a^2b^2. &(3) \end{align*} Applying the Transitive Property to $(2)$ and $(3),$ we isolate $a^2:$ \begin{align*} b^2(1+h)^2 + 3a^2 &= b^2(2+h)^2 + 2a^2 \\ a^2 &= b^2\left((2+h)^2-(1+h)^2\right) \\ a^2 &= b^2(2h+3). \hspace{26.75mm} (*) \end{align*} Substituting $(1)$ and $(*)$ into $(2),$ we solve for $h:$ \begin{align*} b^2(1+h)^2 + 3\underbrace{b^2(2h+3)}_{\text{by }(*)} &= \underbrace{(1-h)^2}_{\text{by }(1)}b^2 \\ (1+h)^2+3(2h+3)&=(1-h)^2 \\ 1+2h+h^2+6h+9&=1-2h+h^2 \\ 10h&=-9 \\ h&=-\frac{9}{10}. \end{align*} Substituting this into $(1),$ we get $a^2=\frac{361}{100}.$
Substituting the current results into $(*),$ we get $b^2=\frac{361}{120}.$
Finally, we obtain \[c^2 = a^2-b^2 = 361\left(\frac{1}{100}-\frac{1}{120}\right) = \frac{361}{600},\] from which \[\frac{c}{a}=\sqrt{\frac{c^2}{a^2}}=\sqrt{\frac{361/600}{361/100}}=\sqrt{\frac 16}.\] The answer is $1+6=\boxed{7}.$

Solution:
Starting from this system of equations from Solution 2: \begin{align*} \frac{(1-h)^2}{a^2}&=1, \\ \frac{(-1-h)^2}{a^2}+\frac{{\sqrt3}^2}{b^2}&=1, \\ \frac{(-2-h)^2}{a^2}+\frac{{\sqrt2}^2}{b^2}&=1. \end{align*} Let $A=a^{-2}$ and $B=b^{-2}$ . Therefore, the system can be rewritten as: \begin{align*} (h^2-2h+1)A&=1, &(1)\\ (h^2+2h+1)A+3B&=1, &(2)\\ (h^2+4h+4)A+2B&=1. &(3) \end{align*} Subtracting $(1)$ from $(2)$ and $(3)$ , we get \[4hA+3B=0\quad\text{and}\quad 3A-6hA+3B=0.\] Plugging the former into the latter and simplifying yields $6A=5B$ . Hence $a^2:b^2=6:5$ . Since $c^2=a^2-b^2$ , we get $a^2=6c^2$ , so the eccentricity is $\frac ca=\sqrt{\frac16}$
Therefore, the answer is $1+6=\boxed{7}$

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

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


Answer: 24000.0

The five tires of a car (four road tires and a full-sized spare) were rotated so that each tire was used the same number of miles during the first $30,000$ miles the car traveled.  For how many miles was each tire used?
$\text{(A)}\ 6000 \qquad \text{(B)}\ 7500 \qquad \text{(C)}\ 24,000 \qquad \text{(D)}\ 30,000 \qquad \text{(E)}\ 37,500$

Solution: In the $30,000$ miles, four tires were always used at one time, so the amount of miles the five tires were used in total is $30,000 \times 4=120,000$ . Five tires were used and each was used equally, so each tire was used for $\frac{120,000}{5}=\boxed{24,000}$


Answer: 608.0

The following analog clock has two hands that can move independently of each other. [asy]             unitsize(2cm);             draw(unitcircle,black+linewidth(2));              for (int i = 0; i < 12; ++i) {                 draw(0.9*dir(30*i)--dir(30*i));             }             for (int i = 0; i < 4; ++i) {                 draw(0.85*dir(90*i)--dir(90*i),black+linewidth(2));             }             for (int i = 1; i < 13; ++i) {                 label("\small" + (string) i, dir(90 - i * 30) * 0.75);             }             draw((0,0)--0.6*dir(90),black+linewidth(2),Arrow(TeXHead,2bp));             draw((0,0)--0.4*dir(90),black+linewidth(2),Arrow(TeXHead,2bp)); [/asy] Initially, both hands point to the number $12$ . The clock performs a sequence of hand movements so that on each movement, one of the two hands moves clockwise to the next number on the clock face while the other hand does not move.
Let $N$ be the number of sequences of $144$ hand movements such that during the sequence, every possible positioning of the hands appears exactly once, and at the end of the $144$ movements, the hands have returned to their initial position. Find the remainder when $N$ is divided by $1000$

Solution:
This problem is, in essence, the following:  A $12\times12$ coordinate grid is placed on a (flat) torus; how many loops are there that pass through each point while only moving up or right?  In other words, Felix the frog starts his journey at $(0,0)$ in the coordinate plane.  Every second, he jumps either to the right or up, until he reaches an $x$ - or $y$ -coordinate of $12$ .  At this point, if he tries to jump to a coordinate outside the square from $(0,0)$ to $(11,11)$ , he "wraps around" and ends up at an $x$ - or $y$ - coordinate of $0$ .  How many ways are there for Felix to jump on every grid point in this square, so that he ends at $(0,0)$ ?  This is consistent with the construction of the flat torus as $\mathbb Z^2/12\mathbb Z^2$ (2-dimensional modular arithmetic. $(\mathbb{Z}_{12})^2$
Moving on, define a $\textit{path}$ from point $A$ to point $B$ to be a sequence of "up"s and "right"s that takes Felix from $A$ to $B$ .  The $\textit{distance}$ from $A$ to $B$ is the length of the shortest path from $A$ to $B$ .  At the crux of this problem is the following consideration:  The points $A_i=(i,12-i), i\in{0,...,11}$ are pairwise equidistant, each pair having distance of $12$ in both directions.

A valid complete path then joins two $A_i$ 's, say $A_i$ and $A_j$ .  In fact, a link between some $A_i$ and $A_j$ fully determines the rest of the cycle, as the path from $A_{i+1}$ must "hug" the path from $A_i$ , to ensure that there are no gaps.  We therefore see that if $A_0$ leads to $A_k$ , then $A_i$ leads to $A_{i+k}$ .  Only the values of $k$ relatively prime to $12$ result in solutions, though, because otherwise $A_0$ would only lead to $\{A_i:\exists n\in \mathbb Z:i\equiv kn\quad\text{mod 12}\}$ .  The number of paths from $A_0$ to $A_k$ is ${12\choose k}$ , and so the answer is
\[{12\choose1}+{12\choose5}+{12\choose7}+{12\choose11}=1\boxed{608}.\]

Solution:
This is more of a solution sketch and lacks rigorous proof for interim steps, but illustrates some key observations that lead to a simple solution.
Note that one can visualize this problem as walking on a $N \times N$ grid where the edges warp. Your goal is to have a single path across all nodes on the grid leading back to $(0,\ 0)$ . For convenience, any grid position are presumed to be in $\mod N$
Note that there are exactly two ways to reach node $(i,\ j)$ , namely $(i - 1,\ j)$ and $(i,\ j - 1)$
As a result, if a path includes a step from $(i,\ j)$ to $(i + 1,\ j)$ , there cannot be a step from $(i,\ j)$ to $(i,\ j + 1)$ . However, a valid solution must reach $(i,\ j + 1)$ , and the only valid step is from $(i - 1,\ j + 1)$
So a solution that includes a step from $(i,\ j)$ to $(i + 1,\ j)$ dictates a step from $(i - 1,\ j + 1)$ to $(i,\ j + 1)$ and by extension steps from $(i - a,\ j + a)$ to $(i - a + 1,\ j + a)$ . We observe the equivalent result for steps in the orthogonal direction.
This means that in constructing a valid solution, taking one step in fact dictates N steps, thus it's sufficient to count valid solutions with $N = a + b$ moves of going right $a$ times and $b$ times up the grid. The number of distinct solutions can be computed by permuting 2 kinds of indistinguishable objects $\binom{N}{a}$
Here we observe, without proof, that if $\gcd(a, b) \neq 1$ , then we will return to the origin prematurely. For $N = 12$ , we only want to count the number of solutions associated with $12 = 1 + 11 = 5 + 7 = 7 + 5 = 11 + 1$
(For those attempting a rigorous proof, note that $\gcd(a, b) = \gcd(a + b, b) = \gcd(N, b) = \gcd(N, a)$ ).
The total number of solutions, noting symmetry, is thus
\[2\cdot\left(\binom{12}{1} + \binom{12}{5}\right) = 1608\]
This yields $\boxed{608}$ as our desired answer.

Solution:
Define a $12 \times 12$ matrix $X$ .
Each entry $x_{i, j}$ denotes the number of movements the longer hand moves, given that two hands jointly make $12 \left( i - 1 \right) + \left( j - 1 \right)$ movements.
Thus, the number of movements the shorter hand moves is $12 \left( i - 1 \right) + \left( j - 1 \right) - x_{i, j}$
Denote by $r_{i, j}$ the remainder of $x_{i, j}$ divided by 12.
Denote by $R$ this remainder matrix.
If two hands can return to their initial positions after 144 movements, then $r_{12, 12} = 0$ or 11.
Denote by $S_0$ (resp. $S_{11}$ ) the collection of feasible sequences of movements, such that $r_{12, 12} = 0$ (resp. $r_{12, 12} = 11$ ).
Define a function $f :  S_0 \rightarrow S_{11}$ , such that for any $\left\{ x_{i,j} , \ \forall \ i, j \in \left\{ 1, 2, \cdots , 12 \right\} \right\} \in S_0$ , the functional value of the entry indexed as $\left( i, j \right)$ is $12 \left( i - 1 \right) + \left( j - 1 \right) - x_{i, j}$ .
Thus, function $f$ is bijective. This implies $| S_0 | = | S_{11} |$
In the rest of analysis, we count $| S_0 |$
We make the following observations:
\begin{enumerate}
\item $x_{1, 1} = 0$ and $12 | x_{12, 12}$
These follow from the definition of $S_0$
\item Each column of $R$ is a permutation of $\left\{ 0, 1, \cdots , 11 \right\}$
The reasoning is as follows. Suppose there exist $i < i'$ $j$ , such that $r_{i, j} = r_{i', j}$ . Then this entails that the positions of two hands after the $\left( 12 \left( i' - 1 \right) + \left( j - 1 \right) \right)$ th movement coincide with their positions after the $\left( 12 \left( i - 1 \right) + \left( j - 1 \right) \right)$ th movement.
\item For any $j \in \left\{ 1, 2 ,\cdots , 11 \right\}$ $x_{i, j+1} - x_{i, j}$ is equal to either 0 for all $i$ or 1 for all $i$
The reasoning is as follows. If this does not hold and the $j$ th column in $R$ is a permutation of $\left\{ 0, 1, \cdots , 12 \right\}$ , then the $j+1$ th column is no longer a permutation of $\left\{ 0, 1, \cdots , 12 \right\}$ . This leads to the infeasibility of the movements.
\item $x_{i+1, 1} = x_{i, 12}$ for any $i \in \left\{ 1, 2, \cdots , 11 \right\}$
This follows from the conditions that the $12$ th column in $R$ excluding $r_{12, 12}$ and the first column in $R$ excluding $x_{1, 1}$ are both permutations of $\left\{ 1, 2, \cdots , 11 \right\}$
\end{enumerate}
All observations jointly imply that $x_{i, 12} = i \cdot x_{1, 12}$ .
Thus, $\left\{ r_{1, 12}, r_{2, 12} , \cdots , r_{11, 12} \right\}$ is a permutation of $\left\{ 1, 2, \cdots , 11 \right\}$ .
Thus, $x_{1, 12}$ is relatively prime to 12.
Because $x_{1, 1} = 0$ and $x_{1, 12} - x_{1, 1} \leq 11$ , we have $x_{1, 12} = 1$ , 5, 7, or 11.
Recall that when we move from $x_{1, 1}$ to $x_{1, 12}$ , there are 11 steps of movements. Each movement has $x_{1, j+1} - x_{i, j} = 0$ or 1.
Thus, for each given $x_{1, 12}$ , the number of feasible movements from $x_{1, 1}$ to $x_{1, 12}$ is $\binom{11}{x_{1, 12}}$
Therefore, the total number of feasible movement sequences in this problem is \begin{align*} | S_0 | + | S_{11} | & = 2 | S_0 | \\ & = 2 \cdot \sum_{x_{1, 12} = 1, 5, 7, 11} \binom{11}{x_{1, 12}} \\ & = 2 \left( 11 + 462 + 330 + 1 \right) \\ & = 1608 . \end{align*}
Therefore, the answer is $\boxed{608}$


Answer: 4.0

Solution: We first notice that the median name will be the $(19+1)/2=10^{\mbox{th}}$ name. The $10^{\mbox{th}}$ name is $\boxed{4}$

Solution: To find the median length of a name from a bar graph, we must add up the number of names. Doing so gives us $7 + 3 + 1 + 4 + 4 = 19$ . Thus the index of the median length would be the 10th name. Since there are $7$ names with length $3$ , and $3$ names with length $4$ , the $10$ th name would have $4$ letters. Thus our answer is $\boxed{4}$


Answer: 3.0

The following four statements, and only these are found on a card: [asy] pair A,B,C,D,E,F,G; A=(0,1); B=(0,5); C=(11,5); D=(11,1); E=(0,4); F=(0,3); G=(0,2); draw(A--B--C--D--cycle); label("On this card exactly one statement is false.", B, SE); label("On this card exactly two statements are false.", E, SE); label("On this card exactly three statements are false.", F, SE); label("On this card exactly four statements are false.", G, SE); [/asy]
(Assume each statement is either true or false.) Among them the number of false statements is exactly
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution: There can be at most one true statement on the card, eliminating $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}$ . If there are $0$ true on the card, statement $4$ ("On this card exactly four statements are false") will be correct, causing a contradiction. Therefore, the answer is $\boxed{3}$ , since $3$ are false and only the third statement ("On this card exactly three statements are false") is correct.

Solution: If all of them are false, that would mean that the $4$ th one is false too. Therefore, $E$ is not the correct answer. If exactly $3$ of them are false, that would mean that only $1$ statement is true. This is correct since if only $1$ statement is true, the card that is true is the one that has $3$ of these statements are false. If we have $1$ or $2$ false statements, that would mean that there is more than $1$ true statement. Therefore, our answer is $\boxed{3}$


Answer: 539.0

The formula for converting a Fahrenheit temperature $F$ to the corresponding Celsius temperature $C$ is $C = \frac{5}{9}(F-32).$ An integer Fahrenheit temperature is converted to Celsius, rounded to the nearest integer, converted back to Fahrenheit, and again rounded to the nearest integer
For how many integer Fahrenheit temperatures between 32 and 1000 inclusive does the original temperature equal the final temperature?

Solution:
Examine $F - 32$ modulo 9.
Generalizing this, we define that $9x + k = F - 32$ . Thus, $F = \left[\frac{9}{5}\left[\frac{5}{9}(9x + k)\right] + 32\right] \Longrightarrow F = \left[\frac{9}{5}(5x + \left[\frac{5}{9}k\right]) + 32\right] \Longrightarrow F = \left[\frac{9}{5} \left[\frac{5}{9}k \right] \right] + 9x + 32$ . We need to find all values $0 \le k \le 8$ that $\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k$ . Testing every value of $k$ shows that $k = 0, 2, 4, 5, 7$ , so $5$ of every $9$ values of $k$ work.
There are $\lfloor \frac{1000 - 32}{9} \rfloor = 107$ cycles of $9$ , giving $5 \cdot 107 = 535$ numbers that work. Of the remaining $6$ numbers from $995$ onwards, $995,\ 997,\ 999,\ 1000$ work, giving us $535 + 4 = \boxed{539}$ as the solution.

Solution:
Notice that $\left[ \frac{9}{5} \left[ \frac{5}{9} k \right] \right] = k$ holds if $k=\left[ \frac{9}{5}x\right]$ for some integer $x$ .
Thus, after translating from $F\to F-32$ we want count how many values of $x$ there are such that $k=\left[ \frac{9}{5}x\right]$ is an integer from $0$ to $968$ . This value is computed as $\left[968*\frac{5}{9}\right]+1 = \boxed{539}$ , adding in the extra solution corresponding to $0$

Solution: Notice that every $C$ value corresponds to exactly one $F$ value but multiple $F$ values can correspond to a $C$ value. Thus, the smallest $C$ value is $0$ and the largest $C$ value is $538$ yielding $\boxed{539}$ solutions.


Answer: 883.0

The fraction
\[\dfrac1{99^2}=0.\overline{b_{n-1}b_{n-2}\ldots b_2b_1b_0},\]
where $n$ is the length of the period of the repeating decimal expansion.  What is the sum $b_0+b_1+\cdots+b_{n-1}$
$\textbf{(A) }874\qquad \textbf{(B) }883\qquad \textbf{(C) }887\qquad \textbf{(D) }891\qquad \textbf{(E) }892\qquad$

Solution:
$\frac{1}{99^2}\\\\ =\frac{1}{99} \cdot \frac{1}{99}\\\\ =\frac{0.\overline{01}}{99}\\\\ =0.\overline{00010203...9799}$
So, the answer is $0+0+0+1+0+2+0+3+...+9+7+9+9=2\cdot10\cdot\frac{9\cdot10}{2}-(9+8)$ or $\boxed{883}$


Answer: 9.0

The fraction
\[\frac{\left(3^{2008}\right)^2-\left(3^{2006}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}\] simplifies to which of the following?
$\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ \frac{9}{4}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{9}{2}\qquad\mathrm{(E)}\ 9$

Solution:
Using Difference of Squares, $\frac{(3^{2008})^{2}-(3^{2006})^{2}}{(3^{2007})^{2}-(3^{2005}){^2}}$ becomes
$\frac{(3^{2008}+3^{2006})(3^{2008}-3^{2006})}{(3^{2007}+3^{2005})(3^{2007}-3^{2005})}$
$= \frac{3^{2006}(9+1) \cdot 3^{2006}(9-1)}{3^{2005}(9+1) \cdot 3^{2005}(9-1)}$
$= \boxed{9}$


Answer: 364.0

The function $f$ , defined on the set of ordered pairs of positive integers, satisfies the following properties: \[f(x, x) = x,\; f(x, y) = f(y, x), {\rm \ and\ } (x+y)f(x, y) = yf(x, x+y).\] Calculate $f(14,52)$

Solution: Let $z = x+y$ . By the substitution $z=x+y,$ we rewrite the third property in terms of $x$ and $z,$ then solve for $f(x,z):$ \begin{align*} zf(x,z-x) &= (z-x)f(x,z) \\ f(x,z) &= \frac{z}{z-x} \cdot f(x,z-x). \end{align*} Using the properties of $f,$ we have \begin{align*} f(14,52) &= \frac{52}{38} \cdot f(14,38) \\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot f(14,24) \\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot f(14,10)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot f(10,14)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot f(10,4)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot f(4,10)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot f(4,6)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot f(4,2)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot f(2,4)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot \frac{4}{2} \cdot f(2,2)\\ &= \frac{52}{38} \cdot \frac{38}{24} \cdot \frac{24}{10} \cdot \frac{14}{4} \cdot \frac{10}{6} \cdot \frac{6}{2} \cdot \frac{4}{2} \cdot 2\\ &=\boxed{364} ~MRENTHUSIASM (credit given to AoPS)

Solution:
Since all of the function's properties contain a recursive definition except for the first one, we know that $f(x,x) = x$ in order to obtain an integer answer. So, we have to transform $f(14,52)$ to this form by exploiting the other properties. The second one doesn't help us immediately, so we will use the third one.
Note that \[f(14,52) = f(14,14 + 38) = \frac{52}{38}\cdot f(14,38).\]
Repeating the process several times, \begin{align*} f(14,52) & = f(14,14 + 38) \\ & = \frac{52}{38}\cdot f(14,38) \\ & = \frac{52}{38}\cdot \frac{38}{24}\cdot f(14,14 + 24) \\ & = \frac{52}{24}\cdot f(14,24) \\ & = \frac{52}{10}\cdot f(10,14) \\ & = \frac{52}{10}\cdot \frac{14}{4}\cdot f(10,4) \\ & = \frac{91}{5}\cdot f(4,10) \\ & = \frac{91}{3}\cdot f(4,6) \\ & = 91\cdot f(2,4) \\ & = 91\cdot 2 \cdot f(2,2) \\ & = \boxed{364}

Solution:
Notice that $f(x,y) = \mathrm{lcm}(x,y)$ satisfies all three properties:
For the first two properties, it is clear that $\mathrm{lcm}(x,x) = x$ and $\mathrm{lcm}(x,y) = \mathrm{lcm}(y,x)$
For the third property, using the identities $\gcd(x,y) \cdot \mathrm{lcm}(x,y) = x\cdot y$ and $\gcd(x,x+y) = \gcd(x,y)$ gives \begin{align*} y \cdot \mathrm{lcm}(x,x+y) &= \dfrac{y \cdot x(x+y)}{\gcd(x,x+y)} \\ &= \dfrac{(x+y) \cdot xy}{\gcd(x,y)} \\ &= (x+y) \cdot \mathrm{lcm}(x,y). \end{align*} Hence, $f(x,y) = \mathrm{lcm}(x,y)$ is a solution to the functional equation.
Since this is an AIME problem, there is exactly one correct answer, and thus, exactly one possible value of $f(14,52)$
Therefore, we have \begin{align*} f(14,52) &= \mathrm{lcm}(14,52) \\ &= \mathrm{lcm}(2 \cdot 7,2^2 \cdot 13) \\ &= 2^2 \cdot 7 \cdot 13 \\ &= \boxed{364}


Answer: 58.0

The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$ , where $a$ $b$ $c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$ $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$ . Find the unique number that is not in the range of $f$

Solution: From $f(f(x))=x$ , it is obvious that $\frac{-d}{c}$ is the value not in the range. First notice that since $f(0)=\frac{b}{d}$ $f(\frac{b}{d})=0$ which means $a(\frac{b}{d})+b=0$ so $a=-d$ . Using $f(19)=19$ , we have that $b=361c+38d$ ; on $f(97)=97$ we obtain $b=9409c+194d$ . Solving for $d$ in terms of $c$ leads us to $d=-58c$ , so the answer is $\boxed{058}$

Solution: Begin by finding the inverse function of $f(x)$ , which turns out to be $f^{-1}(x)=\frac{19d-b}{a-19c}$ . Since $f(f(x))=x$ $f(x)=f^{-1}(x)$ , so substituting 19 and 97 yields the system, $\begin{array}{lcl} \frac{19a+b}{19c+d} & = & \frac{19d-b}{a-19c} \\ \frac{97a+b}{97c+d} & = & \frac{97d-b}{a-97c} \end{array}$ , and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get $116c=a-d$ . Coincidentally, then $116c+d=a$ , which is familiar because $f(116)=\frac{116a+b}{116c+d}$ , and since $116c+d=a$ $f(116)=\frac{116a+b}{a}$ . Also, $f(f(116))=\frac{a(\frac{116a+b}{a})+b}{c(\frac{116a+b}{a})+d}=116$ , due to $f(f(x))=x$ . This simplifies to $\frac{116a+2b}{c(\frac{116a+b}{a})+d}=116$ $116a+2b=116(c(\frac{116a+b}{a})+d)$ $116a+2b=116(c(116+\frac{b}{a})+d)$ $116a+2b=116c(116+\frac{b}{a})+116d$ , and substituting $116c+d=a$ and simplifying, you get $2b=116c(\frac{b}{a})$ , then $\frac{a}{c}=58$ . Looking at $116c=a-d$ one more time, we get $116=\frac{a}{c}+\frac{-d}{c}$ , and substituting, we get $\frac{-d}{c}=\boxed{58}$ , and we are done.

Solution: Because there are no other special numbers other than $19$ and $97$ , take the average to get $\boxed{58}$ . (Note I solved this problem the solution one way but noticed this and this probably generalizes to all $f(x)=x, f(y)=y$ questions like these)

Solution:
By the function definition, $f(f(x))$ $f$ is its own inverse, so the only value not in the range of $f$ is the value not in the domain of $f$ (which is $-d/c$ ).
Since $f(f(x))$ $f(f(0)=0$ (0 is a convenient value to use). $f(f(0))=f(f(\tfrac{b}{d})=\dfrac{a\cdot\tfrac{b}{d}+b}{c\cdot\tfrac{b}{d}+d}=\dfrac{ab+bd}{bc+d^2}=0 \Rightarrow ab+bd=0$
Then $ab+bd=b(a+d)=0$ and since $b$ is nonzero, $a=-d$
The answer we are searching for, $\dfrac{-d}{c}$ (the only value not in the range of $f$ ), can now be expressed as $\dfrac{a}{c}$
We are given $f(19)=19$ and $f(97)$ , and they satisfy the equation $f(x)=x$ , which simplifies to $\dfrac{ax+b}{cx+d}=x\Rightarrow x(cx+d)=ax+b\Rightarrow cx^2+(d-a)x+b=0$ . We have written this quadratic with roots $19$ and $97$
By Vieta, $\dfrac{-(d-a)}{c}=\dfrac{-(-a-a)}{c}=\dfrac{2a}{c}=19+97$
So our answer is $\dfrac{116}{2}=\boxed{058}$

Solution: Notice that the function is just an involution on the real number line. Since the involution has two fixed points, namely $19$ and $97$ , we know that the involution is an inversion with respect to a circle with a diameter from $19$ to $97$ . The only point that is undefined under an inversion is the center of the circle, which we know is $\frac{19+97}{2}=\boxed{58}$ in both $x$ and $y$ dimensions.


Answer: 11.0

The function $f$ has the property that for each real number $x$ in its domain, $1/x$ is also in its domain and
$f(x)+f\left(\frac{1}{x}\right)=x$
What is the largest set of real numbers that can be in the domain of $f$
$\mathrm{(A) \ } \{x|x\ne 0\}\qquad \mathrm{(B) \ } \{x|x<0\}$
$\mathrm{(C) \ } \{x|x>0\}$ $\mathrm{(D) \ } \{x|x\ne -1\;\rm{and}\; x\ne 0\;\rm{and}\; x\ne 1\}$
$\mathrm{(E) \ }  \{-1,1\}$

Solution:
We know that $f(x) + f \left(\frac{1}{x}\right) = x.$ Plugging in $x = \frac{1}{x}$ we get \[f \left(\frac{1}{x}\right) + f \left(\frac{1}{\frac{1}{x}}\right) = \frac{1}{x}\] \[f \left(\frac{1}{x}\right) + f(x) = \frac{1}{x}.\]
Also notice \[f \left(\frac{1}{x}\right) + f(x) = x\] by the commutative property(this is the same as the equation given in the problem. We are just rearranging). So we can set $\frac{1}{x} = x$ which gives us $x = \pm 1$ which is answer option $\boxed{1,1}.$


Answer: 2.0

The function $f$ is given by the table
\[\begin{tabular}{|c||c|c|c|c|c|}  \hline   x & 1 & 2 & 3 & 4 & 5 \\   \hline  f(x) & 4 & 1 & 3 & 5 & 2 \\  \hline \end{tabular}\]
If $u_0=4$ and $u_{n+1} = f(u_n)$ for $n \ge 0$ , find $u_{2002}$
$\text{(A) }1 \qquad \text{(B) }2 \qquad \text{(C) }3 \qquad \text{(D) }4 \qquad \text{(E) }5$

Solution:
We can guess that the series given by the problem is periodic in some way. Starting off, $u_0=4$ is given. $u_1=u_{0+1}=f(u_0)=f(4)=5,$ so $u_1=5.$ $u_2=u_{1+1}=f(u_1)=f(5)=2,$ so $u_2=2.$ $u_3=u_{2+1}=f(u_2)=f(2)=1,$ so $u_3=1.$ $u_4=u_{3+1}=f(u_3)=f(1)=4,$ so $u_4=4.$ Plugging in $4$ will give us $5$ as found before, and plugging in $5$ will give $2$ and so on. This means that our original guess of the series being periodic was correct. Summing up our findings in a nice table,
\[\begin{tabular}{|c||c|c|c|c|c|c|}  \hline   n & 0 & 1 & 2 & 3 & 4 & ...\\   \hline  un & 4 & 5 & 2 & 1 & 4 & ...\\  \hline \end{tabular}\]
in which the next $u_n$ is found by simply plugging in the number from the last box into $f(x).$ The function is periodic every $4$ terms. $2002 \equiv 2\pmod{4}$ , and counting $4$ starting from $u_1$ will give us our answer of $\boxed{2}$


Answer: 8.0

The function $f(x)$ satisfies $f(2+x)=f(2-x)$ for all real numbers $x$ . If the equation $f(x)=0$ has exactly four distinct real roots , then the sum of these roots is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$

Solution: Let one of the roots be $r_1$ . Also, define $x$ such that $2+x=r_1$ . Thus, we have $f(2+x)=f(r_1)=0$ and $f(2+x)=f(2-x)$ . Therefore, we have $f(2-x)=0$ , and $2-x$ is also a root. Let this root be $r_2$ . The sum $r_1+r_2=2+x+2-x=4$ . Similarly, we can let $r_3$ be a root and define $y$ such that $2+y=r_3$ , and we will find $2-y$ is also a root, say, $r_4$ , so $r_3+r_4=2+y+2-y=4$ . Therefore, $r_1+r_2+r_3+r_4=4+4=8, \boxed{8}$


Answer: 561.0

The function $f_{}^{}$ has the property that, for each real number $x,\,$
If $f(19)=94,\,$ what is the remainder when $f(94)\,$ is divided by $1000$

Solution:
\begin{align*}f(94)&=94^2-f(93)=94^2-93^2+f(92)=94^2-93^2+92^2-f(91)=\cdots \\ &= (94^2-93^2) + (92^2-91^2) +\cdots+ (22^2-21^2)+ 20^2-f(19) \\ &= 94+93+\cdots+21+400-94  \\ &= 4561 \end{align*}
So, the remainder is $\boxed{561}$

Solution:
Those familiar with triangular numbers and some of their properties will quickly recognize the equation given in the problem. It is well-known (and easy to show) that the sum of two consecutive triangular numbers is a perfect square; that is, \[T_{n-1} + T_n = n^2,\] where $T_n = 1+2+...+n = \frac{n(n+1)}{2}$ is the $n$ th triangular number.
Using this, as well as using the fact that the value of $f(x)$ directly determines the value of $f(x+1)$ and $f(x-1),$ we conclude that $f(n) = T_n + K$ for all odd $n$ and $f(n) = T_n - K$ for all even $n,$ where $K$ is a constant real number.
Since $f(19) = 94$ and $T_{19} = 190,$ we see that $K = -96.$ It follows that $f(94) = T_{94} - (-96) = \frac{94\cdot 95}{2} + 96 = 4561,$ so the answer is $\boxed{561}$


Answer: 2.0

The function $x^2+px+q$ with $p$ and $q$ greater than zero has its minimum value when:
$\textbf{(A) \ }x=-p  \qquad \textbf{(B) \ }x=\frac{p}{2} \qquad \textbf{(C) \ }x=-2p \qquad \textbf{(D) \ }x=\frac{p^2}{4q} \qquad$
$\textbf{(E) \ }x=\frac{-p}{2}$

Solution:
The minimum value of this parabola is found at its turning point, on the line $\boxed{2}$ .
Indeed, the turning point of any function of the form $ax^2+bx+c$ has an x-coordinate of $\frac{-b}{2a}$ . This can be seen at the average of the quadratic's two roots (whose sum is $\frac{-b}{a}$ ) or (using calculus) as the value of its derivative set equal to $0$


Answer: 997.0

The function f is defined on the set of integers and satisfies $f(n)=\begin{cases} n-3&\mbox{if}\ n\ge 1000\\ f(f(n+5))&\mbox{if}\ n<1000\end{cases}$
Find $f(84)$

Solution:
Define $f^{h} = f(f(\cdots f(f(x))\cdots))$ , where the function $f$ is performed $h$ times. We find that $f(84) = f(f(89)) = f^2(89) = f^3(94) = \ldots f^{y}(1004)$ $1004 = 84 + 5(y - 1) \Longrightarrow y = 185$ . So we now need to reduce $f^{185}(1004)$
Let’s write out a couple more iterations of this function: \begin{align*}f^{185}(1004)&=f^{184}(1001)=f^{183}(998)=f^{184}(1003)=f^{183}(1000)\\ &=f^{182}(997)=f^{183}(1002)=f^{182}(999)=f^{183}(1004)\end{align*} So this function reiterates with a period of 2 for $x$ . It might be tempting at first to assume that $f(1004) = 1001$ is the answer; however, that is not true since the solution occurs slightly before that. Start at $f^3(1004)$ \[f^{3}(1004)=f^{2}(1001)=f(998)=f^{2}(1003)=f(1000)=\boxed{997}\]

Solution:
Assume that $f(84)$ is to be performed $n+1$ times. Then we have \[f(84)=f^{n+1}(84)=f(f^n(84+5))\] In order to find $f(84)$ , we want to know the smallest value of \[f^n(84+5)\ge1000\] Because then \[f(84)=f(f^n(84+5))=(f^n(84+5))-3\] From which we'll get a numerical value for $f(84)$
Notice that the value of $n$ we expect to find is basically the smallest $n$ such that after $f(x)=f(f(x+5))$ is performed $\frac{n}{2}$ times and then $f(x)=x-3$ is performed back $\frac{n}{2}$ times, the result is greater than or equal to $1000$
In this case, the value of $n$ for $f(84)$ is $916$ , because \[84+\frac{916}{2}\cdot5-\frac{916}{2}\cdot3=1000\Longrightarrow f^{916}(84+5))=1000\] Thus \[f(84)=f(f^{916}(84+5))=f(1000)=1000-3=\boxed{997}\]


Answer: 125.0

Solution: Let the amount of coffee the maker will hold when full be $x$ .  Then, \[.36x=45 \Rightarrow x=125 \rightarrow \boxed{125}\]


Answer: 2.5

Solution:
Since we want to know how much money is spent in June, July and August, we need the difference between the amount of money spent by the beginning of June and the amount of money spent by the end of August.
We estimate these to be about $2.2$ million and $4.8$ million, respectively.  The difference is \[4.8-2.2=2.6\approx 2.5 \rightarrow \boxed{2.5}\]


Answer: 2.0

The graph of $x^2 + y = 10$ and the graph of $x + y = 10$ meet in two points. The distance between these two points is:
$\textbf{(A)}\ \text{less than 1} \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ \sqrt{2}\qquad \textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{more than 2}$

Solution:
We can merge the two equations to create $x^2+y=x+y$ . Using either the quadratic equation or factoring, we get two solutions with $x$ -coordinates $0$ and $1$
Plugging this into either of the original equations, we get $(0,10)$ and $(1,9)$ . The distance between those two points is $\boxed{2}$


Answer: 4.0

The graph of $y=x^6-10x^5+29x^4-4x^3+ax^2$ lies above the line $y=bx+c$ except at three values of $x$ , where the graph and the line intersect. What is the largest of these values?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 8$

Solution:
The $x$ values in which $y=x^6-10x^5+29x^4-4x^3+ax^2$ intersect at $y=bx+c$ are the same as the zeros of $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c$
Since there are $3$ zeros and the function is never negative, all $3$ zeros must be double roots because the function's degree is $6$
Suppose we let $p$ $q$ , and $r$ be the roots of this function, and let $x^3-ux^2+vx-w$ be the cubic polynomial with roots $p$ $q$ , and $r$
\begin{align*}(x-p)(x-q)(x-r) &= x^3-ux^2+vx-w\\ (x-p)^2(x-q)^2(x-r)^2 &= x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0\\ \sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} &= x^3-ux^2+vx-w = 0\end{align*}
In order to find $\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c}$ we must first expand out the terms of $(x^3-ux^2+vx-w)^2$
\[(x^3-ux^2+vx-w)^2\] \[= x^6-2ux^5+(u^2+2v)x^4-(2uv+2w)x^3+(2uw+v^2)x^2-2vwx+w^2\]
[Quick note: Since we don't know $a$ $b$ , and $c$ , we really don't even need the last 3 terms of the expansion.]
\begin{align*}&2u = 10\\ u^2+2v &= 29\\ 2uv+2w &= 4\\ u &= 5\\ v &= 2\\ w &= -8\\ &\sqrt{x^6-10x^5+29x^4-4x^3+ax^2-bx-c} = x^3-5x^2+2x+8\end{align*}
All that's left is to find the largest root of $x^3-5x^2+2x+8$
\begin{align*}&x^3-5x^2+2x+8 = (x-4)(x-2)(x+1)\\ &\boxed{4}

Solution:
The $x$ values in which $y=x^6-10x^5+29x^4-4x^3+ax^2$ intersect at $y=bx+c$ are the same as the zeros of $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c$ .
We also know that this graph has 3 places tangent to the x-axis, which means that each root has to have a multiplicity of 2.
Let the function be $(x-p)^2(x-q)^2(x-r)^2$
Applying Vieta's formulas, we get $2p+2q+2r = 10$ or $p+q+r = 5$ .
Applying it again, we get, after simplification, $p^2+q^2+r^2+4pq+4pr+4qr = 29$
Notice that squaring the first equation yields $p^2+q^2+r^2+2pq+2qr+2pr= 25$ , which is similar to the second equation.
Subtracting this from the second equation, we get $2pq+2pr+2qr = 4$ . Now that we have the $pq+pr+qr$ term, we can manpulate the equations to
yield the sum of squares. $2(p^2+q^2+r^2+2pq+2qr+2pr)-2pq-2pr-2qr= 25*2-4$ or $2p^2+2q^2+2r^2+2pq+2qr+2pr = 46$ . We finally reach $(p+q)^2+(q+r)^2+(p+r)^2 = 46$
Since the answer choices are integers, we can guess and check squares to get $\{(p+q)^2, (q+r)^2, (p+r)^2\} = \{1, 9, 36\}$ in some order. We can check that this works by adding then and seeing $2p+2q+2r = 10$ . We just need to take the lowest value in the set, square root it, and subtract the resulting value from 5 to get $\boxed{4}$

Solution:
First, $y=x^6-10x^5+29x^4-4x^3+ax^2-bx-c = 0$ has exactly $3$ roots. Therefore, $y = (kx^3+lx^2+mx+n)^2 = 0$
So, $k^2x^6+2klx^5+(2km+l^2)x^4+2(kn+lm)x^3+ax^2-bx-c = 0$
By matching the coefficients of the first $4$ terms, we have $k^2 = 1, 2kl = -10, 2km+l^2 = 29, 2kn+2lm = -4$
Solving the equations above, we have $2$ sets of solutions; first set of which is $k = 1, l = -5, m = 2, n = 8$ . Second set of which is $k = -1, l = 5, m = -2, n = -8$ . After squaring both sets, they are the same i.e. $x^3-5x^2+2x+8 = 0$
This is equal to $(x-4)(x-2)(x+1) = 0$ . The largest root is $\boxed{4}$


Answer: 800.0

The graph of $y^2 + 2xy + 40|x|= 400$ partitions the plane into several regions.  What is the area of the bounded region?

Solution:
The equation given can be rewritten as:
We can split the equation into a piecewise equation by breaking up the absolute value
Factoring the first one: (alternatively, it is also possible to complete the square
Hence, either $y = -20$ , or $2x = 20 - y \Longrightarrow y = -2x + 20$
Similarily, for the second one, we get $y = 20$ or $y = -2x - 20$ . If we graph these four equations, we see that we get a parallelogram with base 20 and height 40. Hence the answer is $\boxed{800}$


Answer: 6.0

Solution:
First of all, note that the equation $f(t)=6$ has two solutions: $t=-2$ and $t=1$
Given an $x$ , let $f(x)=t$ . Obviously, to have $f(f(x))=6$ , we need to have $f(t)=6$ , and we already know when that happens. In other words, the solutions to $f(f(x))=6$ are precisely the solutions to ( $f(x)=-2$ or $f(x)=1$ ).
Without actually computing the exact values, it is obvious from the graph that the equation $f(x)=-2$ has two and $f(x)=1$ has four different solutions, giving us a total of $2+4=\boxed{6}$ solutions.


Answer: 6.0

The graph shows the constant rate at which Suzanna rides her bike. If she rides a total of a half an hour at the same speed, how many miles would she have ridden?
[asy] import graph; /* this is a label */ Label f;  f.p=fontsize(0); xaxis(-0.9,20,Ticks(f, 5.0, 5.0));  yaxis(-0.9,20, Ticks(f, 22.0,5.0)); // real f(real x)  {  return x; }  draw(graph(f,-1,22),black+linewidth(1)); label("1", (-1,5), black);  label("2", (-1, 10), black); label("3", (-1, 15), black); label("4", (-1, 20), black); dot((5,5), black+linewidth(5)); dot((10,10), black+linewidth(5)); dot((15, 15), black+linewidth(5)); dot((20,20), black+linewidth(5)); label("MINUTES", (11,-5), S); label(rotate(90)*"MILES", (-5,11), W);[/asy]
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 5.5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 6.5\qquad\textbf{(E)}\ 7$

Solution: Suzanna's speed is $\frac{1}{5}$ . This means she runs $\frac{1}{5} \cdot 30 = \boxed{6}$

Solution: From the graph, we can see that every $5$ minutes Suzanna goes, her distance increases by $1$ . Since half an hour is $10$ minutes away, she would go $2$ more miles. $4+2$ is $6$ , so the answer is $\boxed{6}$


Answer: 6.0

Solution: Average the differences between each day. We get $10, -10,\text{ } 20,\text{ } 30,-20$ . We find the average of this list to get $\boxed{6}$

Solution: This solution may take longer to do than the first solution. In total, Asha studied for 400 minutes a week (80 minutes per day) and Sasha studied for 430 minutes a week (86 minutes per day).  86 - 80 = 6.  Therefore, the answer is $\boxed{6}$


Answer: 70.0

Solution:
The highest price was in Month 1, which was $17. The lowest price was in Month 3, which was $10. 17 is $\frac{17}{10}\cdot100=170\%$ of 10, and is $170-100=70\%$ more than 10.
Therefore, the answer is $\boxed{70}$


Answer: 36.0

The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$

Solution:
Begin by setting $x$ to 0, then set both equations to $h^2=\frac{2013-j}{3}$ and $h^2=\frac{2014-k}{2}$ , respectively. Notice that because the two parabolas have to have positive x-intercepts, $h\ge32$
We see that $h^2=\frac{2014-k}{2}$ , so we now need to find a positive integer $h$ which has positive integer x-intercepts for both equations.
Notice that if $k=2014-2h^2$ is -2 times a square number, then you have found a value of $h$ for which the second equation has positive x-intercepts. We guess and check $h=36$ to obtain $k=-578=-2(17^2)$
Following this, we check to make sure the first equation also has positive x-intercepts (which it does), so we can conclude the answer is $\boxed{036}$

Solution:
Let $x=0$ and $y=2013$ for the first equation, resulting in $j=2013-3h^2$ .  Substituting back in to the original equation, we get $y=3(x-h)^2+2013-3h^2$
Now we set $y$ equal to zero, since there are two distinct positive integer roots.  Rearranging, we get $2013=3h^2-3(x-h)^2$ , which simplifies to $671=h^2-(x-h)^2$ .  Applying difference of squares, we get $671=(2h-x)(x)$
Now, we know that $x$ and $h$ are both integers, so we can use the fact that $671=61\times11$ , and set $2h-x=11$ and $x=61$ (note that letting $x=11$ gets the same result).  Therefore, $h=\boxed{036}$

Solution:
Similar to the first two solutions, we deduce that $\text{(-)}j$ and $\text{(-)}k$ are of the form $3a^2$ and $2b^2$ , respectively, because the roots are integers and so is the $y$ -intercept of both equations. So the $x$ -intercepts should be integers also.
The first parabola gives \[3h^2+j=3\left(h^2-a^2\right)=2013\] \[h^2-a^2=671\] And the second parabola gives \[2h^2+k=2\left(h^2-b^2\right)=2014\] \[h^2-b^2=1007\]
We know that $671=11\cdot 61$ and that $1007=19\cdot 53$ . It is just a fitting coincidence that the average of $11$ and $61$ is the same as the average of $19$ and $53$ . That is $\boxed{036}$

Solution:
First, we expand both equations to get $y=3x^2-6hx+3h^2+j$ and $y=2x^2-4hx+2h^2+k$ . The $y$ -intercept for the first equation can be expressed as $3h^2+j$ . From this, the x-intercepts for the first equation can be written as
\[x=h \pm \sqrt{(-6h)^2-4*3(3h^2+j)}=h \pm \sqrt{36h^2-12(2013)}=h \pm \sqrt{36h^2-24156}\]
Since the $x$ -intercepts must be integers, $\sqrt{36h^2-24156}$ must also be an integer. From solution 1, we know $h$ must be greater than or equal to 32. We can substitute increasing integer values for $h$ starting from 32; we find that $h=36$
We can test this result using the second equation, whose $x$ -intercepts are \[x=h \pm \sqrt{(-4h)^2-4*2(2h^2+k)}=h \pm \sqrt{16h^2-8(2014)}=h \pm \sqrt{16h^2-16112}\] Substituting 36 in for $h$ , we get $h=36 \pm 68$ , which satisfies the requirement that all x-intercepts must be (positive) integers.
Thus, $h=\boxed{036}$

Solution:
We have the equation $y=3(x-h)^2 + j.$
We know: $(x,y):(0,2013)$ , so $h^2=2013/3 - j/3$ after plugging in the values and isolating $h^2$ . Therefore, $h^2=671-j/3$
Lets call the x-intercepts $x_1$ $x_2$ . Since both $x_1$ and $x_2$ are positive there is a relationship between $x_1$ $x_2$ and $h$ . Namely, $x_1+x_2=2h$ . The is because: $x_1-h=-(x_2-h)$
Similarly, we know: $(x,y):(x_1,0)$ , so $j=-3(x_1-h)^2$ . Combining the two equations gives us \[h^2=671+(x_1-h)^2\] \[h^2=671+x_1^2-2x_1h+h^2\] \[h=(671+x_1^2)/2x_1.\]
Now since we have this relationship, $2h=x_1+x_2$ , we can just multiply the last equation by 2(so that we get $2h$ on the left side) which gives us \[2h=671/x_1+x_1^2/x^1\] \[2h=671/x_1+x_1\] \[x_1+x_2=671/x_1+x_1\] \[x_2=671/x_1\] \[x_1x_2=671.\] Prime factorization of 671 gives 11 and 61. So now we know $x_1=11$ and $x_2=61$ . Lastly, we plug in the numbers,11 and 61, into $x_1+x_2=2h$ , so $\boxed{36}$

Solution: First, we start of exactly like solutions above and we find out that $j=2013-3h^2$ and $k=2014-2h^2$ We then plug j and k into $3(x-h)^2+j$ and $y=2(x-h)^2+k$ respectively. After that, we get two equations, $y=3x^2-6xh+2013$ and $y=2x^2-4xh+2014$ . We can apply Vieta's. Let the roots of the first equation be $a, b$ and the roots of the second equation be $c, d$ . Thus, we have that $a\cdot b=1007$ $a+b=2h$ and $c\cdot d=671$ $c+d=2h$ . Simple evaluations finds that $\boxed{36}$


Answer: 1.0

The graphs of $2x+3y-6=0, 4x-3y-6=0, x=2$ , and $y=\frac{2}{3}$ intersect in:
$\textbf{(A)}\ \text{6 points}\qquad\textbf{(B)}\ \text{1 point}\qquad\textbf{(C)}\ \text{2 points}\qquad\textbf{(D)}\ \text{no points}\\ \textbf{(E)}\ \text{an unlimited number of points}$

Solution:
We first convert each of the lines into slope-intercept form ( $y = mx + b$ ):
$2x+3y-6=0 ==> 3y = -2x + 6 ==> y = -\frac{2}{3}x + 2$
$4x - 3y - 6 = 0 ==> 4x - 6 = 3y ==> y = \frac{4}{3}x - 2$
$x = 2$ stays as is.
$y = \frac{2}{3}$ stays as is
We can graph the four lines here: [1]
When we do that, the answer turns out to be $\boxed{1}$


Answer: 140.0

The graphs of $x^2 + y^2 = 4 + 12x + 6y$ and $x^2 + y^2 = k + 4x + 12y$ intersect when $k$ satisfies $a \le k \le b$ , and for no other values of $k$ . Find $b-a$
$\mathrm{(A) \ }5 \qquad \mathrm{(B) \ }68 \qquad \mathrm{(C) \ }104 \qquad \mathrm{(D) \ }140 \qquad \mathrm{(E) \ }144$

Solution:
Both sets of points are quite obviously circles. To show this, we can rewrite each of them in the form $(x-x_0)^2 + (y-y_0)^2 = r^2$
The first curve becomes $(x-6)^2 + (y-3)^2 = 7^2$ , which is a circle centered at $(6,3)$ with radius $7$
The second curve becomes $(x-2)^2 + (y-6)^2 = 40+k$ , which is a circle centered at $(2,6)$ with radius $r=\sqrt{40+k}$
The distance between the two centers is $5$ , and therefore the two circles intersect if $2\leq r \leq 12$
From $\sqrt{40+k} \geq 2$ we get that $k\geq -36$ . From $\sqrt{40+k}\leq 12$ we get $k\leq 104$
Therefore $b-a = 104 - (-36) = \boxed{140}$


Answer: 10.0

The graphs of $y = -|x-a| + b$ and $y = |x-c| + d$ intersect at points $(2,5)$ and $(8,3)$ . Find $a+c$
$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 18$

Solution:
Each of the graphs consists of two orthogonal half-lines. In the first graph both point downwards at a $45^\circ$ angle, in the second graph they point upwards. One can easily find out that the only way how to get these graphs to intersect in two points is the one depicted below:
Obviously, the maximum of the first graph is achieved when $x=a$ , and its value is $-0+b=b$ . Similarly, the minimum of the other graph is $(c,d)$ . Therefore the two remaining vertices of the area between the graphs are $(a,b)$ and $(c,d)$
As the area has four right angles, it is a rectangle. Without actually computing $a$ and $c$ we can therefore conclude that $a+c=2+8=\boxed{10}$


Answer: 5.0

The graphs of $y=\log_3 x, y=\log_x 3, y=\log_\frac{1}{3} x,$ and $y=\log_x \dfrac{1}{3}$ are plotted on the same set of axes. How many points in the plane with positive $x$ -coordinates lie on two or more of the graphs?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

Solution:
Setting the first two equations equal to each other, $\log_3 x = \log_x 3$
Solving this, we get $\left(3, 1\right)$ and $\left(\frac{1}{3}, -1\right)$
Similarly with the last two equations, we get $\left(3, -1\right)$ and $\left(\frac{1}{3}, 1\right)$
Now, by setting the first and third equations equal to each other, we get $\left(1, 0\right)$
Pairing the first and fourth or second and third equations won't work because then $\log x \leq 0$
Pairing the second and fourth equations will yield $x = 1$ , but since you can't divide by $\log 1 = 0$ , it doesn't work.
After trying all pairs, we have a total of $5$ solutions $\rightarrow \boxed{5}$

Solution:
Note that $\log_b a =\log_c a / \log_c b$
Then $\log_b a = \log_a a / \log_a b = 1/ \log_a b$
$\log_\frac{1}{a} b = \log_a \frac{1}{a} / \log_a b = -1/ \log_a b$
$\log_\frac{1}{b} a = -\log_a b$
Therefore, the system of equations can be simplified to:
$y = t$
$y = -t$
$y = \frac{1}{t}$
$y = -\frac{1}{t}$
where $t = \log_3 x$ . Note that all values of $t$ correspond to exactly one positive $x$ value, so all $(t,y)$ intersections will correspond to exactly one $(x,y)$ intersection in the positive-x area.
Graphing this system of functions will generate a total of $5$ solutions $\rightarrow \boxed{5}$


Answer: 660.0

The graphs of the equations
are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side length $\tfrac{2}{\sqrt{3}}.\,$ How many such triangles are formed?

Solution:
We note that the lines partition the hexagon of the six extremal lines into disjoint unit regular triangles, and forms a series of unit regular triangles along the edge of the hexagon.

Solving the above equations for $k=\pm 10$ , we see that the hexagon in question is regular, with side length $\frac{20}{\sqrt{3}}$ . Then, the number of triangles within the hexagon is simply the ratio of the area of the hexagon to the area of a regular triangle. Since the ratio of the area of two similar figures is the square of the ratio of their side lengths, we see that the ratio of the area of one of the six equilateral triangles composing the regular hexagon to the area of a unit regular triangle is just $\left(\frac{20/\sqrt{3}}{2/\sqrt{3}}\right)^2 = 100$ . Thus, the total number of unit triangles is $6 \times 100 = 600$
There are $6 \cdot 10$ equilateral triangles formed by lines on the edges of the hexagon. Thus, our answer is $600+60 = \boxed{660}$

Solution:
There are three types of lines: horizontal, upward-slanting diagonal, and downward-slanting diagonal. There are $21$ of each type of line, and a unit equilateral triangle is determined by exactly one of each type of line. Given an upward-slanting diagonal and a downward-slanting diagonal, they determine exactly two unit equilateral triangles as shown below.  Therefore, if all horizontal lines are drawn, there will be a total of $2\cdot 21^2=882$ unit equilateral triangles. Of course, we only draw $21$ horizontal lines, so we are overcounting the triangles that are caused by the undrawn horizontal lines. In the below diagram, we draw the diagonal lines and the highest and lowest horizontal lines.
We see that the lines $y=-21,-20,\dots, -11$ and $y=11,12,\dots,21$ would complete several of the $882$ unit equilateral triangles. In fact, we can see that the lines $y=-21,-20,\dots,-11$ complete $1,2,(1+3),(2+4),(3+5),(4+6),\dots,(9+11)$ triangles, or $111$ triangles. The positive horizontal lines complete the same number of triangles, hence the answer is $882-2\cdot 111=\boxed{660}$

Solution: Picturing the diagram in your head should give you an illustration similar to the one above. The distance from parallel sides of the center hexagon is 20 units, and by extending horizontal lines to the sides of the hexagon. This shows that for every side of the hexagon there are 10 spaces. Therefore the side length of the biggest triangle (imagine one of the overlapping triangles in the Star of David) is 30. The total number of triangles in the hexagon can be found by finding the number of triangles in the extended triangle and subtracting the 3 corner triangles. This gives us $30^2 - 10^2 - 10^2- 10^2 = 600$ . That is the number of triangles in the hexagon. The remaining triangles form in groups of 10 on the exterior of each side. $600 + 6 * 10 = \boxed{660}$


Answer: 10.0

The greatest prime number that is a divisor of $16384$ is $2$ because $16384 = 2^{14}$ . What is the sum of the digits of the greatest prime number that is a divisor of $16383$
$\textbf{(A)} \: 3\qquad\textbf{(B)} \: 7\qquad\textbf{(C)} \: 10\qquad\textbf{(D)} \: 16\qquad\textbf{(E)} \: 22$

Solution:
We have \begin{align*} 16383 & = 2^{14} - 1 \\ & = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\ & = 129 \cdot 127 \\ \end{align*}
Since $129$ is composite, $127$ is the largest prime which can divide $16383$ . The sum of $127$ 's digits is $1+2+7=\boxed{10}$


Answer: 10.0

Solution:
We have \begin{align*} 16383 & = 2^{14} - 1 \\ & = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\ & = 129 \cdot 127 \\ \end{align*}
Since $129$ is composite, $127$ is the largest prime which can divide $16383$ . The sum of $127$ 's digits is $1+2+7=\boxed{10}$


Answer: 8.0

The grid below is to be filled with integers in such a way that the sum of the numbers in each row and the sum of the numbers in each column are the same. Four numbers are missing. The number $x$ in the lower left corner is larger than the other three missing numbers. What is the smallest possible value of $x$ [asy] unitsize(0.5cm); draw((3,3)--(-3,3)); draw((3,1)--(-3,1)); draw((3,-3)--(-3,-3)); draw((3,-1)--(-3,-1)); draw((3,3)--(3,-3)); draw((1,3)--(1,-3)); draw((-3,3)--(-3,-3)); draw((-1,3)--(-1,-3)); label((-2,2),"$-2$"); label((0,2),"$9$"); label((2,2),"$5$"); label((2,0),"$-1$"); label((2,-2),"$8$"); label((-2,-2),"$x$"); [/asy] $\textbf{(A) } -1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9 \qquad$

Solution: The sum of the numbers in each row is $12$ . Consider the second row. In order for the sum of the numbers in this row to equal $12$ , the two shaded numbers must add up to $13$  If two numbers add up to $13$ , one of them must be at least $7$ : If both shaded numbers are no more than $6$ , their sum can be at most $12$ . Therefore, for $x$ to be larger than the three missing numbers, $x$ must be at least $8$ . We can construct a working scenario where $x=8$  So, our answer is $\boxed{8}$

Solution:
The sum of the numbers in each row is $-2+9+5=12,$ and the sum of the numbers in each column is $5+(-1)+8=12.$
Let $y$ be the number in the lower middle. It follows that $x+y+8=12,$ or $x+y=4.$
We express the other two missing numbers in terms of $x$ and $y,$ as shown below:  We have $x>x-1, x>y+10,$ and $x>y.$ Note that the first inequality is true for all values of $x.$ We only need to solve the second inequality so that the third inequality is true for all values of $x.$ By substitution, we get $x>(4-x)+10,$ from which $x>7.$
Therefore, the smallest possible value of $x$ is $\boxed{8}.$

Solution:
This is based on the Solution 2 above and it is perhaps a little simpler than that.
Let $y$ be the number in the lower middle. Applying summation to first two columns yields the following.

Since $x$ is greater than the other three, we have $x>14-x,$ or $x>7.$
Therefore, the smallest possible value of $x$ is $\boxed{8}.$

Solution:
Note that the sum of the rows and columns must be $8+5-1=12$ . We proceed to test the answer choices.
Testing $\textbf{(A)}$ , when $x = -1$ , the number above $x$ must be $15$ , which contradicts the precondition that the numbers surrounding $x$ is less than $x$
Testing $\textbf{(B)}$ , the number above $x$ is $9$ , which does not work.
Testing $\textbf{(C)}$ , the number above $x$ is $8$ , which does not work.
Testing $\textbf{(D)}$ , the number above $x$ is $6$ , which does work. Hence, the answer is $\boxed{8}$

Solution:
The sum of the numbers in each column and row should be $5+(-1)+8=12$ . If we look at the $1^{\text{st}}$ column, the gray squares (shown below) sum to $12-(-2)=14$

If square $x$ has to be greater than or equal to the three blank squares, then the least $x$ can be is half the sum of the value of the gray squares, which is $14\div2=7$ . But square $x$ has to be greater than and not greater than or equal to the three blank squares, so the least $x$ can be is $7+1=8$ . Testing for the other rows and columns (it might be smaller than the other two squares!), we find that the smallest $x$ can be is indeed $8$ ; the other two squares are less than $8$ . Therefore, the answer is $\boxed{8}$


Answer: 2.0

The harmonic mean of two numbers can be calculated as twice their product divided by their sum. The harmonic mean of $1$ and $2016$ is closest to which integer?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 504 \qquad \textbf{(D)}\ 1008 \qquad \textbf{(E)}\ 2015$

Solution:
Since the harmonic mean is $2$ times their product divided by their sum, we get the equation
$\frac{2\times1\times2016}{1+2016}$
which is then
$\frac{4032}{2017}$
which is finally closest to $\boxed{2}$ .
-dragonfly


Answer: 799.0

The harmonic mean of two positive integers is the reciprocal of the arithmetic mean of their reciprocals. For how many ordered pairs of positive integers $(x,y)$ with $x<y$ is the harmonic mean of $x$ and $y$ equal to $6^{20}$

Solution: The harmonic mean of $x$ and $y$ is equal to $\frac{1}{\frac{\frac{1}{x}+\frac{1}{y}}2} = \frac{2xy}{x+y}$ , so we have $xy=(x+y)(3^{20}\cdot2^{19})$ , and by SFFT $(x-3^{20}\cdot2^{19})(y-3^{20}\cdot2^{19})=3^{40}\cdot2^{38}$ . Now, $3^{40}\cdot2^{38}$ has $41\cdot39=1599$ factors, one of which is the square root ( $3^{20}2^{19}$ ). Since $x<y$ , the answer is half of the remaining number of factors, which is $\frac{1599-1}{2}= \boxed{799}$


Answer: 8.0

The hundreds digit of a three-digit number is $2$ more than the units digit. The digits of the three-digit number are reversed, and the result is subtracted from the original three-digit number. What is the units digit of the result?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8$

Solution: Let the hundreds, tens, and units digits of the original three-digit number be $a$ $b$ , and $c$ , respectively. We are given that $a=c+2$ . The original three-digit number is equal to $100a+10b+c = 100(c+2)+10b+c = 101c+10b+200$ . The hundreds, tens, and units digits of the reversed three-digit number are $c$ $b$ , and $a$ , respectively. This number is equal to $100c+10b+a = 100c+10b+(c+2) = 101c+10b+2$ . Subtracting this expression from the expression for the original number, we get $(101c+10b+200) - (101c+10b+2) = 198$ . Thus, the units digit in the final result is $\boxed{8}$

Solution: The result must hold for any three-digit number with hundreds digit being $2$ more than the units digit. $301$ is such a number. Evaluating, we get $301-103=198$ . Thus, the units digit in the final result is $\boxed{8}$

Solution: Set the units digit of the original number as $x$ . Thus, its hundreds digit is $x+2$ . After the digits are reversed, the hundreds digit of the original number will be the units digit of the new number. Since $x-(x+2) = -2$ , we can do regrouping and "borrow" $1$ from the tens digit and bring it to the units digit as a $10$ . Therefore, the units digit will end up as $-2 + 10 = \boxed{8}$


Answer: 22.0

The hypotenuse of a right triangle is $10$ inches and the radius of the inscribed circle is $1$ inch. The perimeter of the triangle in inches is:
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 30$

Solution:
To begin, let's notice that the inscribed circle of the right triangle is its incircle , and that the radius of the incircle is the right triangle's inradius . In this case, the hypotenuse is 10, and the inradius is 1. The formula for the inradius of a right triangle is $r = (a+b-c)/2$ , where $r$ is the length of the inradius of the triangle, $c$ is the length hyptotenuse the the right triangle, and $a$ and $b$ are the lengths of the legs of the right triangle. From this, we can plug in values to notice that:
\begin{align*} r & = \frac{a+b-c}{2}\\\\ 1 & = \frac{a+b-10}{2}\\\\ 2 & = a+b-10\\\\ 12 &= a+b\\\\ \end{align*}
From this, we arrive at $a+b+c = 12+10 = 22$ . The answer is clearly $\boxed{22}$


Answer: 227.0

The incircle $\omega$ of triangle $ABC$ is tangent to $\overline{BC}$ at $X$ . Let $Y \neq X$ be the other intersection of $\overline{AX}$ with $\omega$ . Points $P$ and $Q$ lie on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $\overline{PQ}$ is tangent to $\omega$ at $Y$ . Assume that $AP = 3$ $PB = 4$ $AC = 8$ , and $AQ = \dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let the sides $\overline{AB}$ and $\overline{AC}$ be tangent to $\omega$ at $Z$ and $W$ , respectively. Let $\alpha = \angle BAX$ and $\beta = \angle AXC$ . Because $\overline{PQ}$ and $\overline{BC}$ are both tangent to $\omega$ and $\angle YXC$ and $\angle QYX$ subtend the same arc of $\omega$ , it follows that $\angle AYP = \angle QYX = \angle YXC = \beta$ . By equal tangents, $PZ = PY$ . Applying the Law of Sines to $\triangle APY$ yields \[\frac{AZ}{AP} = 1 + \frac{ZP}{AP} = 1 + \frac{PY}{AP} = 1 + \frac{\sin\alpha}{\sin\beta}.\] Similarly, applying the Law of Sines to $\triangle ABX$ gives \[\frac{AZ}{AB} = 1 - \frac{BZ}{AB} = 1 - \frac{BX}{AB} = 1 - \frac{\sin\alpha}{\sin\beta}.\] It follows that \[2 = \frac{AZ}{AP} + \frac{AZ}{AB} = \frac{AZ}3 + \frac{AZ}7,\] implying $AZ = \tfrac{21}5$ . Applying the same argument to $\triangle AQY$ yields \[2 = \frac{AW}{AQ} + \frac{AW}{AC} = \frac{AZ}{AQ} + \frac{AZ}{AC} = \frac{21}5\left(\frac{1}{AQ} + \frac 18\right),\] from which $AQ = \tfrac{168}{59}$ . The requested sum is $168 + 59 = \boxed{227}$

Solution:
Let the incircle of $ABC$ be tangent to $AB$ and $AC$ at $Z$ and $W$ . By Brianchon's theorem on tangential hexagons $QWCBZP$ and $PYQCXB$ , we know that $ZW,CP,BQ$ and $XY$ are concurrent at a point $O$ . Let $PQ \cap BC = M$ . Then by La Hire's $A$ lies on the polar of $M$ so $M$ lies on the polar of $A$ . Therefore, $ZW$ also passes through $M$ . Then projecting through $M$ , we have \[-1 = (A,O;Y,X) \stackrel{M}{=} (A,Z;P,B) \stackrel{M}{=} (A,W;Q,C).\] Therefore, $\frac{AP \cdot ZB}{MP \cdot AB} = 1 \implies \frac{3 \cdot ZB}{ZP \cdot 7} = 1$ . Since $ZB+ZP=4$ we know that $ZP = \frac{6}{5}$ and $ZB = \frac{14}{5}$ . Therefore, $AW = AZ = \frac{21}{5}$ and $WC = 8 - \frac{21}{5} = \frac{19}{5}$ . Since $(A,W;Q,C) = -1$ , we also have $\frac{AQ \cdot WC}{NQ \cdot AC} = 1 \implies \frac{AQ \cdot \tfrac{19}{5}}{(\tfrac{21}{5} - AQ) \cdot 8} = 1$ . Solving for $AQ$ , we obtain $AQ = \frac{168}{59} \implies m+n = \boxed{227}$ .
😃
-Vfire

Solution:
Let the center of the incircle of $\triangle ABC$ be $O$ . Link $OY$ and $OX$ . Then we have $\angle OYP=\angle OXB=90^{\circ}$
$\because$ $OY=OX$
$\therefore$ $\angle OYX=\angle OXY$
$\therefore$ $\angle PYX=\angle YXB$
$\therefore$ $\sin \angle PYX=\sin \angle YXB=\sin \angle YXC=\sin \angle PYA$
Let the incircle of $ABC$ be tangent to $AB$ and $AC$ at $M$ and $N$ , let $MP=YP=x$ and $NQ=YQ=y$
Use Law of Sine in $\triangle APY$ and $\triangle AXB$ , we have
$\frac{\sin \angle PAY}{PY}=\frac{\sin \angle PYA}{PA}$
$\frac{\sin \angle BAX}{BX}=\frac{\sin \angle AXB}{AB}$
therefore we have
$\frac{3}{x}=\frac{7}{4-x}$
Solve this equation, we have $x=\frac{6}{5}$
As a result, $MB=4-x=\frac{14}{5}=BX$ $AM=x+3=\frac{21}{5}=AN$ $NC=8-AN=\frac{19}{5}=XC$ $AQ=\frac{21}{5}-y$ $PQ=\frac{6}{5}+y$
So, $BC=\frac{14}{5}+\frac{19}{5}=\frac{33}{5}$
Use Law of Cosine in $\triangle BAC$ and $\triangle PAQ$ , we have
$\cos \angle BAC=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}$
$\cos \angle PAQ=\frac{AP^2+AQ^2-PQ^2}{2\cdot AP\cdot AQ}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}+y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}$
And we have
$\cos \angle BAC=\cos \angle PAQ$
So
$\frac{7^2+8^2-{(\frac{33}{5})}^2}{2\cdot 7\cdot 8}=\frac{3^2+{(\frac{21}{5}-y)}^2-{(\frac{6}{5}+y)}^2}{2\cdot {(\frac{21}{5}-y)}\cdot 3}$
Solve this equation, we have $y=\frac{399}{295}=QN$
As a result, $AQ=AN-QN=\frac{21}{5}-\frac{399}{295}=\frac{168}{59}$
So, the final answer of this question is $168+59=\boxed{227}$


Answer: 91.0

The increasing geometric sequence $x_{0},x_{1},x_{2},\ldots$ consists entirely of integral powers of $3.$ Given that
$\sum_{n=0}^{7}\log_{3}(x_{n}) = 308$ and $56 \leq \log_{3}\left ( \sum_{n=0}^{7}x_{n}\right ) \leq 57,$
find $\log_{3}(x_{14}).$

Solution:
Suppose that $x_0 = a$ , and that the common ratio between the terms is $r$
The first conditions tells us that $\log_3 a + \log_3 ar + \ldots + \log_3 ar^7 = 308$ . Using the rules of logarithms , we can simplify that to $\log_3 a^8r^{1 + 2 + \ldots + 7} = 308$ . Thus, $a^8r^{28} = 3^{308}$ . Since all of the terms of the geometric sequence are integral powers of $3$ , we know that both $a$ and $r$ must be powers of 3. Denote $3^x = a$ and $3^y = r$ . We find that $8x + 28y = 308$ . The possible positive integral pairs of $(x,y)$ are $(35,1),\ (28,3),\ (21,5),\ (14,7),\ (7,9),\ (0,11)$
The second condition tells us that $56 \le \log_3 (a + ar + \ldots ar^7) \le 57$ . Using the sum formula for a geometric series and substituting $x$ and $y$ , this simplifies to $3^{56} \le 3^x \frac{3^{8y} - 1}{3^y-1} \le 3^{57}$ . The fractional part $\approx \frac{3^{8y}}{3^y} = 3^{7y}$ . Thus, we need $\approx 56 \le x + 7y \le 57$ . Checking the pairs above, only $(21,5)$ is close.
Our solution is therefore $\log_3 (ar^{14}) = \log_3 3^x + \log_3 3^{14y} = x + 14y = \boxed{091}$

Solution: All these integral powers of $3$ are all different, thus in base $3$ the sum of these powers would consist of $1$ s and $0$ s. Thus the largest value $x_7$ must be $3^{56}$ in order to preserve the givens. Then we find by the given that $x_7x_6x_5\dots x_0 = 3^{308}$ , and we know that the exponents of $x_i$ are in an arithmetic sequence. Thus $56+(56-r)+(56-2r)+\dots +(56-7r) = 308$ , and $r = 5$ . Thus $\log_3 (x_{14}) = \boxed{091}$

Solution:
Like above, we use logarithmic identities to convert the problem into workable equations. We begin by labelling the powers of 3. Call $x_0$ $x_1$ $x_2$ ..., as $3^n$ $3^{n+m}$ , and $3^{n+2m}$ ... respectively. With this format we can rewrite the first given equation as $n + n + m + n+2m + n+3m+...+n+7m = 308$ . Simplify to get $2n+7m=77$ . (1)
Now, rewrite the second given equation as $3^{56} \leq \left( \sum_{n=0}^{7}x_{n} \right) \leq 3^{57}$ . Obviously, $x_7$ , aka $3^{n+7m}$ $<3^{57}$ because there are some small fractional change that is left over. This means $n+7m$ is $\leq56$ . Thinking about the geometric sequence, it's clear each consecutive value of $x_i$ will be either a power of three times smaller or larger. In other words, the earliest values of $x_i$ will be negligible compared to the last values of $x_i$ . Even in the best case scenario, where the common ratio is 3, the values left of $x_7$ are not enough to sum to a value greater than 2 times $x_7$ (amount needed to raise the power of 3 by 1). This confirms that $3^{n+7m} = 3^{56}$ . (2)
Use equations 1 and 2 to get $m=5$ and $n=21$ $\log_{3}{x_{14}} = \log_{3}{3^{21+14*5}} = 21+14*5 = \boxed{091}$

Solution:
Proceed as in Solution 3 for the first few steps. We have the sequence $3^{a},3^{a+n},3^{a+2n}...$ . As stated above, we then get that $a+a+n+...+a+7n=308$ , from which we simplify to $2a+7n=77$ . From here, we just go brute force using the second statement (that $3^{56}\leq 3^{a}+...+3^{a+7n}\leq 3^{57}$ ). Rearranging the equation from earlier, we get \[n=11-\frac{2a}{7}\] from which it is clear that $a$ is a multiple of $7$ . Testing the first few values of $a$ , we get:
Case 1 ( $a=7,n=9$ )
The sequence is then $3^{7}+...+3^{70}$ , which breaks the upper bound.
Case 2 ( $a=14,n=7$ )
The sequence is then $3^{14}+...+3^{63}$ , which also breaks the upper bound.
Case 3 ( $a=21,n=5$ )
This is the first reasonable one, giving $3^{21}+...+3^{56}$ . It seems like this would break the upper bound, but from some testing we get: \[3^{21}+...+3^{56}<3^{57}\] \[1+...+3^{35}<3^{36}\] \[1+...+3^{30}<2*3^{35}\] \[3^{5}+...+3^{30}<2*3^{35}-1\] \[1+...+3^{25}<2*3^{30}-3^{-5}\] Repeating this process over and over, we eventually get (while ignoring the extremely small fractions left over) \[1<2*3^{5}\] which confirms that this satisfies our upper bound. Thus $a=21,n=5$ , so $x_14=3^{a+14n}\rightarrow3^{91}$ . We then get the requested answer, $\log_3(3^{91})=\boxed{091}$ ~ amcrunner


Answer: 981.0

The increasing sequence $1,3,4,9,10,12,13\cdots$ consists of all those positive integers which are powers of 3 or sums of distinct powers of 3. Find the $100^{\mbox{th}}$ term of this sequence.

Solution: Rewrite all of the terms in base 3. Since the numbers are sums of distinct powers of 3, in base 3 each number is a sequence of 1s and 0s (if there is a 2, then it is no longer the sum of distinct powers of 3). Therefore, we can recast this into base 2 (binary) in order to determine the 100th number. $100$ is equal to $64 + 32 + 4$ , so in binary form we get $1100100$ . However, we must change it back to base 10 for the answer, which is $3^6 + 3^5 + 3^2 = 729 + 243 + 9 = \boxed{981}$

Solution: Notice that the first term of the sequence is $1$ , the second is $3$ , the fourth is $9$ , and so on. Thus the $64th$ term of the sequence is $729$ . Now out of $64$ terms which are of the form $729$ $'''S'''$ $32$ of them include $243$ and $32$ do not. The smallest term that includes $243$ , i.e. $972$ , is greater than the largest term which does not, or $854$ . So the $96$ th term will be $972$ , then $973$ , then $975$ , then $976$ , and finally $\boxed{981}$

Solution: After the $n$ th power of 3 in the sequence, the number of terms after that power but before the $(n+1)$ th power of 3 is equal to the number of terms before the $n$ th power, because those terms after the $n$ th power are just the $n$ th power plus all the distinct combinations of powers of 3 before it, which is just all the terms before it. Adding the powers of $3$ and the terms that come after them, we see that the $100$ th term is after $729$ , which is the $64$ th term. Also, note that the $k$ th term after the $n$ th power of 3 is equal to the power plus the $k$ th term in the entire sequence. Thus, the $100$ th term is $729$ plus the $36$ th term. Using the same logic, the $36$ th term is $243$ plus the $4$ th term, $9$ . We now have $729+243+9=\boxed{981}$

Solution: Writing out a few terms of the sequence until we reach the next power of 3 (27), we see that the $2^{nth}$ term is equal to $3^n$ . From here, we can ballpark the range of the 100th term. The 64th term is $3^6$ $729$ and the 128th term is $3^7$ $2187$ . Writing out more terms of the sequence until the next power of 3 again (81) we can see that the ( $2^n$ $2^{n+1}$ )/2 term is equal to $3^n$ $3^{n-1}$ . From here, we know that the 96th term is $3^6$ $3^5$ $972$ . From here, we can construct the 100th term by following the sequence in increasing order. The 97th term is $972 + 1 = 973$ , the 98th term is $972 + 3 = 975$ , the 99th term is $972 + 3 + 1 = 976$ , and finally the 100th term is $972 + 9 = \boxed{981}$

Solution:
The number of terms $3^n$ produces includes each power of 3 ( $1, 3^1, ..., 3^n$ ), the sums of two power of 3s(ex. $3^1 + 1$ ), three power of 3s (ex. $3^1 + 1 + 3^n$ ), all the way to the sum of them all. Since there are $n+1$ powers of 3, the one number sum gives us ${n+1\choose 1}$ terms, the two number ${n+1\choose 2}$ terms, all the way to the sum of all the powers which gives us ${n+1\choose n+1}$ terms. Summing all these up gives us $2^{n+1} - 1 ^ {*}$ according to the theorem
Since $2^6$ is the greatest power $<100$ , then $n=5$ and the sequence would look like { $3^0, ..., 3^5$ }, where $3^5$ or $243$ would be the $2^5 - 1 = 63$ rd number. The next largest power $729$ would be the 64th number. However, its terms contributed extends beyond 100, so we break it to smaller pieces.
Noting that $729$ plus any combination of lower powers ${1, 3^1 . . .3^4}$ is < $729 + 243$ , so we can add all those terms( $2^5 - 1 = 31$ ) into our sequence:
Our sequence now has $63 + 1 + 31 = 95$ terms. The remaining $5$ would just be the smallest sums starting with $729 + 243$ or $972$
Hence the 100th term would be $972 + 9 = \boxed{981}$ . ~SoilMilk


Answer: 528.0

The increasing sequence $2,3,5,6,7,10,11,\ldots$ consists of all positive integers that are neither the square nor the cube of a positive integer. Find the 500th term of this sequence.

Solution: Because there aren't that many perfect squares or cubes, let's look for the smallest perfect square greater than $500$ . This happens to be $23^2=529$ . Notice that there are $23$ squares and $8$ cubes less than or equal to $529$ , but $1$ and $2^6$ are both squares and cubes. Thus, there are $529-23-8+2=500$ numbers in our sequence less than $529$ .  Magically, we want the $500th$ term, so our answer is the biggest non-square and non-cube less than $529$ , which is $\boxed{528}$

Solution:
This solution is similar as Solution 1, but to get the intuition why we chose to consider $23^2 = 529$ , consider this:
We need $n - T = 500$ , where $n$ is an integer greater than 500 and $T$ is the set of numbers which contains all $k^2,k^3\le 500$
Firstly, we clearly need $n > 500$ , so we substitute n for the smallest square or cube greater than $500$ . However, if we use $n=8^3=512$ , the number of terms in $T$ will exceed $n-500$ . Therefore, $n=23^2=529$ , and the number of terms in $T$ is $23+8-2=29$ by the Principle of Inclusion-Exclusion , fulfilling our original requirement of $n-T=500$ .
As a result, our answer is $529-1 = \boxed{528}$


Answer: 63.0

The increasing sequence $3, 15, 24, 48, \ldots\,$ consists of those positive multiples of 3 that are one less than a perfect square .  What is the remainder when the 1994th term of the sequence is divided by 1000?

Solution: One less than a perfect square can be represented by $n^2 - 1 = (n+1)(n-1)$ . Either $n+1$ or $n-1$ must be divisible by 3. This is true when $n \equiv -1,\ 1 \equiv 2,\ 1 \pmod{3}$ . Since 1994 is even, $n$ must be congruent to $1 \pmod{3}$ . It will be the $\frac{1994}{2} = 997$ th such term, so $n = 4 + (997-1) \cdot 3 = 2992$ . The value of $n^2 - 1 = 2992^2 - 1 \pmod{1000}$ is $\boxed{063}$


Answer: 10.0

The infinite product \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots\] evaluates to a real number. What is that number?
$\textbf{(A) }\sqrt{10}\qquad\textbf{(B) }\sqrt[3]{100}\qquad\textbf{(C) }\sqrt[4]{1000}\qquad\textbf{(D) }10\qquad\textbf{(E) }10\sqrt[3]{10}$

Solution:
We can write $\sqrt[3]{10}$ as $10 ^ \frac{1}{3}$ . Similarly, $\sqrt[3]{\sqrt[3]{10}} = (10 ^ \frac{1}{3}) ^ \frac{1}{3} = 10 ^ \frac{1}{3^2}$
By continuing this, we get the form \[10 ^ \frac{1}{3} \cdot 10 ^ \frac{1}{3^2} \cdot 10 ^ \frac{1}{3^3} \cdots,\] which is \[10 ^ {\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots}.\] Using the formula for an infinite geometric series $S = \frac{a}{1-r}$ , we get \[\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2}.\] Thus, our answer is $10 ^ \frac{1}{2} = \boxed{10}$

Solution: We can write this infinite product as $L$ (we know from the answer choices that the product must converge): \[L = \sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots.\] If we raise everything to the third power, we get: \[L^3 =  10 \, \cdot \, \sqrt[3]{10} \, \cdot \, \sqrt[3]{\sqrt[3]{10}} \cdots = 10L \implies L^3 - 10L = 0 \implies L \in \left\{0, \pm \sqrt{10}\right\}.\] Since $L$ is positive (as it is an infinite product of positive numbers), it must be that $L = \boxed{10}.$


Answer: 10.0

Solution: Move the first term inside the second radical. We get \[\sqrt[3]{10} \cdot \sqrt[3]{\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots = \sqrt[3]{10\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots.\] Do this for the third radical as well: \[\sqrt[3]{10\sqrt[3]{10}} \cdot \sqrt[3]{\sqrt[3]{\sqrt[3]{10}}} \cdots = \sqrt[3]{10\sqrt[3]{10}\sqrt[3]{\sqrt[3]{10}}} \cdots = \sqrt[3]{10\sqrt[3]{10\sqrt[3]{10\cdots}}}.\] It is clear what the pattern is. Setting the answer as $P,$ we have \[P = \sqrt[3]{10P},\] from which $P = \boxed{10}.$


Answer: 345.0

The inscribed circle of triangle $ABC$ is tangent to $\overline{AB}$ at $P_{},$ and its radius is $21$ .  Given that $AP=23$ and $PB=27,$ find the perimeter of the triangle.

Solution:
Let $Q$ be the tangency point on $\overline{AC}$ , and $R$ on $\overline{BC}$ . By the Two Tangent Theorem $AP = AQ = 23$ $BP = BR = 27$ , and $CQ = CR = x$ . Using $rs = A$ , where $s = \frac{27 \cdot 2 + 23 \cdot 2 + x \cdot 2}{2} = 50 + x$ , we get $(21)(50 + x) = A$ . By Heron's formula $A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(50+x)(x)(23)(27)}$ . Equating and squaring both sides,
\begin{eqnarray*}  [21(50+x)]^2 &=& (50+x)(x)(621)\\ 441(50+x) &=& 621x\\ 180x = 441 \cdot 50 &\Longrightarrow & x = \frac{245}{2}   \end{eqnarray*}
We want the perimeter, which is $2s = 2\left(50 + \frac{245}{2}\right) = \boxed{345}$

Solution:
Let the incenter be denoted $I$ .  It is commonly known that the incenter is the intersection of the angle bisectors of a triangle.  So let $\angle ABI = \angle CBI = \alpha, \angle BAI = \angle CAI = \beta,$ and $\angle BCI = \angle ACI = \gamma.$
We have that \begin{eqnarray*} \tan \alpha & = & \frac {21}{27} \\ \tan \beta & = & \frac {21}{23} \\ \tan \gamma & = & \frac {21}x. \end{eqnarray*} So naturally we look at $\tan \gamma.$ But since $\gamma = \frac \pi2 - (\beta + \alpha)$ we have \begin{eqnarray*} \tan \gamma & = & \tan\left(\frac \pi2 - (\beta + \alpha)\right) \\ & = & \frac 1{\tan(\alpha + \beta)} \\ \Rightarrow \frac {21}x & = & \frac {1 - \frac {21\cdot 21}{23\cdot 27}}{\frac {21}{27} + \frac {21}{23}} \end{eqnarray*} Doing the algebra, we get $x = \frac {245}2.$
The perimeter is therefore $2\cdot\frac {245}2 + 2\cdot 23 + 2\cdot 27 = \boxed{345}.$


Answer: 592.0

The integer $n$ is the smallest positive multiple of $15$ such that every digit of $n$ is either $8$ or $0$ . Compute $\frac{n}{15}$

Solution:
Any multiple of 15 is a multiple of 5 and a multiple of 3.
Any multiple of 5 ends in 0 or 5; since $n$ only contains the digits 0 and 8, the units digit of $n$ must be 0.
The sum of the digits of any multiple of 3 must be divisible by 3.  If $n$ has $a$ digits equal to 8, the sum of the digits of $n$ is $8a$ .  For this number to be divisible by 3, $a$ must be divisible by 3. We also know that $a>0$ since $n$ is positive. Thus $n$ must have at least three copies of the digit 8.
The smallest number which meets these two requirements is 8880.  Thus the answer is $\frac{8880}{15} = \boxed{592}$


Answer: 240.0

The internal angles of quadrilateral $ABCD$ form an arithmetic progression. Triangles $ABD$ and $DCB$ are similar with $\angle DBA = \angle DCB$ and $\angle ADB = \angle CBD$ . Moreover, the angles in each of these two triangles also form an arithmetic progression. In degrees, what is the largest possible sum of the two largest angles of $ABCD$
$\textbf{(A)}\ 210 \qquad \textbf{(B)}\ 220 \qquad \textbf{(C)}\ 230 \qquad \textbf{(D)}\ 240 \qquad \textbf{(E)}\ 250$

Solution:
Since the angles of Quadrilateral $ABCD$ form an arithmetic sequence, we can assign each angle with the value $a$ $a+d$ $a+2d$ , and $a+3d$ . Also, since these angles form an arithmetic progression, we can reason out that $(a)+(a+3d)=(a+d)+(a+2d)=180$
For the sake of simplicity, lets rename the angles of each similar triangle. Let $\angle ADB = \angle CBD = \alpha$ $\angle DBA = \angle DCB = \beta$ $\angle CDB = \angle BAD = \gamma$
Now the four angles of $ABCD$ are $\beta$ $\alpha + \beta$ $\gamma$ , and $\alpha + \gamma$
As for the similar triangles, we can name their angles $y$ $y+b$ , and $y+2b$ . Therefore $y+y+b+y+2b=180$ and $y+b=60$ . Because these 3 angles are each equal to one of $\alpha, \beta, \gamma$ , we know that one of these three angles is equal to 60 degrees.
Now we we use trial and error. Let $\alpha = 60^{\circ}$ . Then the angles of ABCD are $\beta$ $60^{\circ} + \beta$ $\gamma$ , and $60^{\circ} + \gamma$ . Since these four angles add up to 360, then $\beta + \gamma= 120$ . If we list them in increasing value, we get $\beta$ $\gamma$ $60^{\circ} + \beta$ $60^{\circ} + \gamma$ . Note that this is the only sequence that works because the common difference is less than 45. So, this would give us the four angles 45, 75, 105, and 135. In this case, $\alpha, \beta, \gamma$ also form an arithmetic sequence with 45, 60, and 75, and the largest two angles of the quadrilateral add up to 240 which is an answer choice.
If we apply the same reasoning to $\beta$ and $\gamma$ , we would get the sum of the highest two angles as 220, which works but is lower than 240. Therefore, $\boxed{240}$ is the correct answer.


Answer: 100.0

The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $9$ trapezoids, let $x$ be the angle measure in degrees of the larger interior angle of the trapezoid. What is $x$
[asy] unitsize(4mm); defaultpen(linewidth(.8pt)); int i; real r=5, R=6;  path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0); for(i=0; i<9; ++i) { draw(rotate(20*i)*t); } draw((-r,0)--(R+1,0)); draw((-R,0)--(-R-1,0)); [/asy]
$\text{(A) } 100 \qquad \text{(B) } 102 \qquad \text{(C) } 104 \qquad \text{(D) } 106 \qquad \text{(E) } 108$

Solution:
Extend all the legs of the trapezoids. They will all intersect in the middle of the bottom side of the picture, forming the situation shown below.

Each of the angles at $X$ is $\frac{180^\circ}9 = 20^\circ$ . From $\triangle XYZ$ , the degree measure of the smaller interior angle of the trapezoid is $\frac{180^\circ - 20^\circ}2 = 80^\circ$ , hence the degree measure of the larger interior angle is $180^\circ - 80^\circ = \boxed{100}$

Solution: A decagon can be formed from the trapezoids and the base. The sum of the decagon's angles is $180(10-2)=1440^\circ$ . Letting the larger angle in each trapezoid be $x$ , the two angles formed by the line each measures $(180-x)^\circ$ . There are $8$ congruent angles left. Each of those angles measures $(360-2x)^\circ$ .  Putting it all together: $8(360-2x)+2(180-x)=1440\implies x=\boxed{100}$

Solution: If we reflect the arch across the line, we form an 18-gon. $\frac{180*(18-2)}{18} = 160^\circ$ so each interior angle of the 18-gon is $160^\circ$ . Let $x$ be the degree measure of the larger interior angle of a trapezoid. From the diagram, we see that $2x + 160 = 360$ , so $2x = 200$ and $x = 100$ , or $\boxed{100}$


Answer: 1.0

The large circle has diameter $\text{AC}$ .  The two small circles have their centers on $\text{AC}$ and just touch at $\text{O}$ , the center of the large circle.  If each small circle has radius $1$ , what is the value of the ratio of the area of the shaded region to the area of one of the small circles?
[asy] pair A=(-2,0), O=origin, C=(2,0); path X=Arc(O,2,0,180), Y=Arc((-1,0),1,180,0), Z=Arc((1,0),1,180,0), M=X..Y..Z..cycle; filldraw(M, black, black); draw(reflect(A,C)*M); draw(A--C, dashed);  label("A",A,W); label("C",C,E); label("O",O,SE); dot((-1,0)); dot(O); dot((1,0)); label("$1$",(-.5,0),N); label("$1$",(1.5,0),N); [/asy]
$\text{(A)}\ \text{between }\frac{1}{2}\text{ and 1} \qquad \text{(B)}\ 1 \qquad \text{(C)}\ \text{between 1 and }\frac{3}{2}$
$\text{(D)}\ \text{between }\frac{3}{2}\text{ and 2} \qquad \text{(E)}\ \text{cannot be determined from the information given}$

Solution:
The small circle has radius $1$ , thus its area is $\pi$
The large circle has radius $2$ , thus its area is $4\pi$
The area of the semicircle above $AC$ is then $2\pi$
The part that is not shaded are two small semicircles. Together, these form one small circle, hence their total area is $\pi$ . This means that the area of the shaded part is $2\pi-\pi=\pi$ .
This is equal to the area of a small circle, hence the correct answer is $\boxed{1}$


Answer: 20.0

Solution:
Clearly, no unit cube has more than one face painted, so the number of unit cubes with at least one face painted is equal to the number of painted unit squares.
There are $10$ painted unit squares on the half of the cube shown, so there are $10\cdot 2=20$ unit cubes with at least one face painted, thus our answer is $\boxed{20}$

Solution: Since it says at least one, we can count the number of unpainted cubes, and subtract from 27. There is 1 inner cube, 2 center cubes (see the face with 4 blacks) and 4 edge cubes (see the top two in the center top face), so 7 unpainted. Thus $27 - 7 = 20$ our answer is $\boxed{20}$


Answer: 11.0

The largest integer $n$ for which $n^{200}<5^{300}$ is
$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }9 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ }11 \qquad \mathrm{(E) \ } 12$

Solution: Since both sides are positive, we can take the $100th$ root of both sides to find the largest integer $n$ such that $n^2<5^3$ . Fortunately, this is simple to evaluate: $5^3=125$ , and the largest square less than $125$ is $11^2=121$ , so the largest $n$ is $11, \boxed{11}$


Answer: 6.0

The largest number by which the expression $n^3 - n$ is divisible for all possible integral values of $n$ , is:
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6$

Solution:
Factoring the polynomial gives $(n+1)(n)(n-1)$ According to the factorization, one of those factors must be a multiple of two because there are more than 2 consecutive integers. In addition, because there are three consecutive integers, one of the integers must be a multiple of 3.  Therefore $6$ must divide the given expression.
Plugging in $n=2$ yields $6$ .  So the largest possibility is $6$
Clearly the answer is $\boxed{6}$

Solution: In general, $r!$ $n(n+1)(n+2)...(n+r-1)$ were $r$ and $n$ are integers. So here $3!$ $n^3$ $n$ always for any integer $n$ .Hence,the correct answer is $6$ $\boxed{6}$


Answer: 14.0

The largest whole number such that seven times the number is less than 100 is
$\text{(A)} \ 12 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E)} \ 16$

Solution: We want to find the smallest integer $x$ so that $7x < 100$ . Dividing by 7 gets $x < 14\dfrac{2}{7}$ , so the answer is 14. $\boxed{14}$