Answer: 59.0

Consider the sequence $(a_k)_{k\ge 1}$ of positive rational numbers defined by $a_1 = \frac{2020}{2021}$ and for $k\ge 1$ , if $a_k = \frac{m}{n}$ for relatively prime positive integers $m$ and $n$ , then
\[a_{k+1} = \frac{m + 18}{n+19}.\] Determine the sum of all positive integers $j$ such that the rational number $a_j$ can be written in the form $\frac{t}{t+1}$ for some positive integer $t$

Solution: We know that $a_{1}=\tfrac{t}{t+1}$ when $t=2020$ so $1$ is a possible value of $j$ . Note also that $a_{2}=\tfrac{2038}{2040}=\tfrac{1019}{1020}=\tfrac{t}{t+1}$ for $t=1019$ . Then $a_{2+q}=\tfrac{1019+18q}{1020+19q}$ unless $1019+18q$ and $1020+19q$ are not relatively prime which happens when $q+1$ divides $18q+1019$ (by the Euclidean Algorithm), or $q+1$ divides $1001$ . Thus, the least value of $q$ is $6$ and $j=2+6=8$ . We know $a_{8}=\tfrac{1019+108}{1020+114}=\tfrac{1127}{1134}=\tfrac{161}{162}$ . Now $a_{8+q}=\tfrac{161+18q}{162+19q}$ unless $18q+161$ and $19q+162$ are not relatively prime which happens the first time $q+1$ divides $18q+161$ or $q+1$ divides $143$ or $q=10$ , and $j=8+10=18$ . We have $a_{18}=\tfrac{161+180}{162+190}=\tfrac{341}{352}=\tfrac{31}{32}$ . Now $a_{18+q}=\tfrac{31+18q}{32+19q}$ unless $18q+31$ and $19q+32$ are not relatively prime. This happens the first time $q+1$ divides $18q+31$ implying $q+1$ divides $13$ , which is prime so $q=12$ and $j=18+12=30$ . We have $a_{30}=\tfrac{31+216}{32+228}=\tfrac{247}{260}=\tfrac{19}{20}$ . We have $a_{30+q}=\tfrac{18q+19}{19q+20}$ , which is always reduced by EA, so the sum of all $j$ is $1+2+8+18+30=\boxed{059}$

Solution:
Let $a_{j_1}, a_{j_2}, a_{j_3}, \ldots, a_{j_u}$ be all terms in the form $\frac{t}{t+1},$ where $j_1<j_2<j_3<\cdots<j_u,$ and $t$ is some positive integer.
We wish to find $\sum_{i=1}^{u}{j_i}.$ Suppose $a_{j_i}=\frac{m}{m+1}$ for some positive integer $m.$
To find $\boldsymbol{a_{j_{i+1}},}$ we look for the smallest positive integer $\boldsymbol{k'}$ for which \[\boldsymbol{a_{j_{i+1}}=a_{j_i+k'}=\frac{m+18k'}{m+1+19k'}}\] is reducible:
If $\frac{m+18k'}{m+1+19k'}$ is reducible, then there exists a common factor $d>1$ for $m+18k'$ and $m+1+19k'.$ By the Euclidean Algorithm , we have \begin{align*} d\mid m+18k' \text{ and } d\mid m+1+19k' &\implies d\mid m+18k' \text{ and } d\mid k'+1 \\ &\implies d\mid m-18 \text{ and } d\mid k'+1. \end{align*} Since $m-18$ and $k'+1$ are not relatively prime, and $m$ is fixed, the smallest value of $k'$ such that $\frac{m+18k'}{m+1+19k'}$ is reducible occurs when $k'+1$ is the smallest prime factor of $m-18.$
We will prove that for such value of $\boldsymbol{k',}$ the number $\boldsymbol{a_{j_{i+1}}}$ can be written in the form $\boldsymbol{\frac{t}{t+1}:}$ \[a_{j_{i+1}}=a_{j_i+k'}=\frac{m+18k'}{m+1+19k'}=\frac{(m-18)+18(k'+1)}{(m-18)+19(k'+1)}=\frac{\frac{m-18}{k'+1}+18}{\frac{m-18}{k'+1}+19}, \hspace{10mm} (*)\] where $t=\frac{m-18}{k'+1}+18$ must be a positive integer.
We start with $m=2020$ and $a_{j_1}=a_1=\frac{2020}{2021},$ then find $a_{j_2}, a_{j_3}, \ldots, a_{j_u}$ by filling out the table below recursively: \[\begin{array}{c|c|c|c|c|c}  & & & & & \\ [-2ex] \boldsymbol{i} & \boldsymbol{m} & \boldsymbol{m-18} & \boldsymbol{k'+1} & \boldsymbol{k'} & \boldsymbol{a_{j_{i+1}} \left(\textbf{by } (*)\right)} \\ [0.5ex]  \hline   & & & & & \\ [-1.5ex]  1 & 2020 & 2002 & 2 & 1 & \hspace{4.25mm} a_2 = \frac{1019}{1020} \\ [1ex]      2 & 1019 & 1001 & 7 & 6 & \hspace{2.75mm} a_8 = \frac{161}{162} \\ [1ex]     3 & 161 & 143 & 11 & 10 & a_{18} = \frac{31}{32} \\ [1ex] 4 & 31 & 13 & 13 & 12 & a_{30} = \frac{19}{20} \\ [1ex] 5 & 19 & 1 & \text{N/A} & \text{N/A} & \text{N/A} \\ [1ex] \end{array}\] As $\left(j_1,j_2,j_3,j_4,j_5\right)=(1,2,8,18,30),$ the answer is $\sum_{i=1}^{5}{j_i}=\boxed{059}.$


Answer: 16.0

Consider the sequence defined recursively by $u_1=a$ (any positive number), and $u_{n+1}=-1/(u_n+1)$ $n=1,2,3,...$ For which of the following values of $n$ must $u_n=a$
$\mathrm{(A) \ 14 } \qquad \mathrm{(B) \ 15 } \qquad \mathrm{(C) \ 16 } \qquad \mathrm{(D) \ 17 } \qquad \mathrm{(E) \ 18 }$

Solution: Repeatedly applying the function, and simplifying, we get \[a,\quad-\frac1{a+1},\quad-\frac{a+1}a,\] and then $a$ again. So $a$ must appear at every third term after $u_1$ . The only option given of the form $1+3k$ is $\boxed{16}$


Answer: 15.0

Consider the sequence of numbers defined recursively by $t_1=1$ and for $n>1$ by $t_n=1+t_{(n/2)}$ when $n$ is even
and by $t_n=\frac{1}{t_{(n-1)}}$ when $n$ is odd. Given that $t_n=\frac{19}{87}$ , the sum of the digits of $n$ is
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 17 \qquad \textbf{(C)}\ 19 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 23$

Solution:
If $n$ is even, then $t_{(n/2)}$ would be negative, which is not possible. Therefore, $n$ is odd. With this function, backwards thinking is the key. If $t_x < 1$ , then $x$ is odd, and $t_{(x-1)} = \frac{1}{t_{x}}$ . Otherwise, you keep on subtracting 1 and halving x until $t_\frac{x}{2^{n}} < 1$ .
We can use this logic to go backwards until we reach $t_1 = 1$ , like so:
$t_n=\frac{19}{87}\\\\t_{n-1} = \frac{87}{19}\\\\t_{\frac{n-1}{2}} = \frac{68}{19}\\\\t_{\frac{n-1}{4}} = \frac{49}{19}\\\\t_{\frac{n-1}{8}} = \frac{30}{19}\\\\t_{\frac{n-1}{16}} = \frac{11}{19}\\\\t_{\frac{n-1}{16} - 1} = \frac{19}{11}\\\\t_{\frac{\frac{n-1}{16} - 1}{2}} = \frac{8}{11}\\\\t_{\frac{\frac{n-1}{16} - 1}{2} - 1} = \frac{11}{8}\\\\t_{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2}} = \frac{3}{8}\\\\t_{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1} = \frac{8}{3}\\\\t_{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{2}} = \frac{5}{3}\\\\t_{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4}} = \frac{2}{3}\\\\t_{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1} = \frac{3}{2}\\\\t_{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2}} = \frac{1}{2}\\\\t_{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1} = 2\\\\t_{\frac{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1}{2}} = t_1 = 1 \Rightarrow \frac{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1}{2} = 1 \Rightarrow n = 1905$ , so the answer is $\boxed{15}$


Answer: 1999.0

Consider the sequence of numbers: $4,7,1,8,9,7,6,\dots$ For $n>2$ , the $n$ -th term of the sequence is the units digit of the sum of the two previous terms. Let $S_n$ denote the sum of the first $n$ terms of this sequence. The smallest value of $n$ for which $S_n>10,000$ is:
$\text{(A) }1992 \qquad \text{(B) }1999 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$

Solution:
The sequence is infinite. As there are only $100$ pairs of digits, sooner or later a pair of consecutive digits will occur for the second time. As each next digit only depends on the previous two, from this point on the sequence will be periodic.
(Additionally, as every two consecutive digits uniquely determine the previous one as well, the first pair of digits that will occur twice must be the first pair $4,7$ .)
Hence it is a good idea to find the period. Writing down more terms of the sequence, we get:
\[4,7,1,8,9,7,6,3,9,2,1,3,4,7,\dots\]
and we found the period. The length of the period is $12$ , and its sum is $4+7+\cdots+1+3 = 60$ . Hence for each $k$ we have $S_{12k} = 60k$
We have $\lfloor 10000/60 \rfloor = 166$ and $166\cdot 12 = 1992$ , therefore $S_{1992} = 60\cdot 166 = 9960$ .
The rest can now be computed by hand, we get $S_{1998} = 9960+4+7+1+8+9+7= 9996$ , and $S_{1999}=9996 + 6 = 10002$ , thus the answer is $\boxed{1999}$


Answer: 810.0

Consider the set of 30 parabolas defined as follows: all parabolas have as focus the point (0,0) and the directrix lines have the form $y=ax+b$ with $a$ and $b$ integers such that $a\in \{-2,-1,0,1,2\}$ and $b\in \{-3,-2,-1,1,2,3\}$ . No three of these parabolas have a common point. How many points in the plane are on two of these parabolas?
$\textbf{(A)}\ 720\qquad\textbf{(B)}\ 760\qquad\textbf{(C)}\ 810\qquad\textbf{(D)}\ 840\qquad\textbf{(E)}\ 870$

Solution:
Being on two parabolas means having the same distance from the common focus and both directrices. In particular, you have to be on an angle bisector of the directrices, and clearly on the same "side" of the directrices as the focus. So it's easy to see there are at most two solutions per pair of parabolae. Convexity and continuity imply exactly two solutions unless the directrices are parallel and on the same side of the focus.
So out of $2\dbinom{30}{2}$ possible intersection points, only $2*5*2*\dbinom{3}{2}$ fail to exist. This leaves $870-60=810=\boxed{810}$ solutions.


Answer: 810.0

Solution: Through similar reasoning as above in Solution I, determine that two parabolas that have a common focus intersect zero times if there directrixes are parallel and the focus lies on the same side of both directrixes, and intersect twice otherwise. Thereby, as each parabola will intersect $30-3 = 27$ other parabolas twice, we see that the answer is \[2 \times \frac{30 \times 27}{2} = \boxed{810}.\]


Answer: 1.0

Consider the set of all fractions $\frac{x}{y}$ , where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by $1$ , the value of the fraction is increased by $10\%$
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{infinitely many}$

Solution:
You can create the equation $\frac{x+1}{y+1}=\frac{11x}{10y}$
Cross multiplying and combining like terms gives $xy + 11x - 10y = 0$
This can be factored into $(x - 10)(y + 11) = -110$
$x$ and $y$ must be positive, so $x > 0$ and $y > 0$ , so $x - 10> -10$ and $y + 11 > 11$
Using the factors of 110, we can get the factor pairs: $(-1, 110),$ $(-2, 55),$ and $(-5, 22).$
But we can't stop here because $x$ and $y$ must be relatively prime.
$(-1, 110)$ gives $x = 9$ and $y = 99$ $9$ and $99$ are not relatively prime, so this doesn't work.
$(-2, 55)$ gives $x = 8$ and $y = 44$ . This doesn't work.
$(-5, 22)$ gives $x = 5$ and $y = 11$ . This does work.
We found one valid solution so the answer is $\boxed{1}$

Solution:
The condition required is $\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}$
Observe that $x+1 > \frac{11}{10}\cdot x$ so $x$ is at most $9.$
By multiplying by $\frac{y+1}{x}$ and simplifying we can rewrite the condition as $y=\frac{11x}{10-x}$ .  Since $x$ and $y$ are integer, this only has solutions for $x\in\{5,8,9\}$ .  However, only the first yields a $y$ that is relative prime to $x$
There is only one valid solution so the answer is $\boxed{1}$


Answer: 600.0

Consider the set of all triangles $OPQ$ where $O$ is the origin and $P$ and $Q$ are distinct points in the plane with nonnegative integer coordinates $(x,y)$ such that $41x + y = 2009$ . Find the number of such distinct triangles whose area is a positive integer.

Solution:
Let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$
We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem).
$\det \left(\begin{array}{c} P \\ Q\end{array}\right)=\det \left(\begin{array}{cc}x_1 &y_1\\x_2&y_2\end{array}\right).$
Since the triangle has half the area of the parallelogram, we just need the determinant to be even.
The determinant is
\[(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))\]
Since $2009$ is not even, $((x_1)-(x_2))$ must be even, thus the two $x$ 's must be of the same parity. Also note that the maximum value for $x$ is $49$ and the minimum is $0$ . There are $25$ even and $25$ odd numbers available for use as coordinates and thus there are $(_{25}C_2)+(_{25}C_2)=\boxed{600}$ such triangles.

Solution:
As in the solution above, let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$
If the coordinates of $P$ and $Q$ have nonnegative integer coordinates, $P$ and $Q$ must be lattice points either
We can calculate the y-intercept of the line $41x+y=2009$ to be $(0,2009)$ and the x-intercept to be $(49,0)$
Using the point-to-line distance formula, we can calculate the height of $\triangle OPQ$ from vertex $O$ (the origin) to be:
$\dfrac{|41(0) + 1(0) - 2009|}{\sqrt{41^2 + 1^2}} = \dfrac{2009}{\sqrt{1682}} = \dfrac{2009}{29\sqrt2}$
Let $b$ be the base of the triangle that is part of the line $41x+y=2009$
The area is calculated as: $\dfrac{1}{2}\times b \times \dfrac{2009}{29\sqrt2} = \dfrac{2009}{58\sqrt2}\times b$
Let the numerical area of the triangle be $k$
So, $k = \dfrac{2009}{58\sqrt2}\times b$
We know that $k$ is an integer. So, $b = 58\sqrt2 \times z$ , where $z$ is also an integer.
We defined the points $P$ and $Q$ as $P=(x_1,y_1)$ and $Q=(x_2,y_2)$
Changing the y-coordinates to be in terms of x, we get:
$P=(x_1,2009-41x_1)$ and $Q=(x_2,2009-41x_2)$
The distance between them equals $b$
Using the distance formula, we get
$PQ = b = \sqrt{(-41x_2+ 41x_1)^2 + (x_2 - x_1)^2} = 29\sqrt2 \times |x_2 - x_1| = 58\sqrt2\times z$ $(*)$
WLOG, we can assume that $x_2 > x_1$
Taking the last two equalities from the $(*)$ string of equalities and putting in our assumption that $x_2>x_1$ , we get
$29\sqrt2\times (x_2-x_1) = 58\sqrt2\times z$
Dividing both sides by $29\sqrt2$ , we get
$x_2-x_1 = 2z$
As we mentioned, $z$ is an integer, so $x_2-x_1$ is an even integer. Also, $x_2$ and $x_1$ are both positive integers. So, $x_2$ and $x_1$ are between 0 and 49, inclusive. Remember, $x_2>x_1$ as well.
...
Summing them up, we get that there are $2+4+\dots + 48 = \boxed{600}$ triangles.


Answer: 21.0

Consider the set of complex numbers $z$ satisfying $|1+z+z^{2}|=4$ . The maximum value of the imaginary part of $z$ can be written in the form $\tfrac{\sqrt{m}}{n}$ , where $m$ and $n$ are relatively prime positive integers. What is $m+n$
$\textbf{(A)}~20\qquad\textbf{(B)}~21\qquad\textbf{(C)}~22\qquad\textbf{(D)}~23\qquad\textbf{(E)}~24$

Solution:
First, substitute in $z=a+bi$
\[|1+(a+bi)+(a+bi)^2|=4\] \[|(1+a+a^2-b^2)+ (b+2ab)i|=4\] \[(1+a+a^2-b^2)^2+ (b+2ab)^2=16\] \[(1+a+a^2-b^2)^2+ b^2(1+4a+4a^2)=16\]
Let $p=b^2$ and $q=1+a+a^2$
\[(q-p)^2+ p(4q-3)=16\] \[p^2-2pq+q^2   + 4pq -3p=16\]
We are trying to maximize $b=\sqrt p$ , so we'll turn the equation into a quadratic to solve for $p$ in terms of $q$
\[p^2+(2q-3)p+(q^2-16)=0\] \[p=\frac{(-2q+3)\pm \sqrt{-12q+73}}{2}\]
We want to maximize $p$ . Since $q$ is always negatively contributing to $p$ 's value, we want to minimize $q$
Due to the trivial inequality: $q=1+a+a^2=(a+\frac 12)^2+\frac{3}4 \geq \frac{3}4$
If we plug $q$ 's minimum value in, we get that $p$ 's maximum value is \[p=\frac{(-2(\frac 34)+3)+ \sqrt{-12(\frac 34)+73}}{2}=\frac{\frac 32+ 8}{2}=\frac{19}{4}\]
Then \[b=\frac{\sqrt{19}}{2}\] and \[m+n=\boxed{21}\]

Solution:
We are given that $1+z+z^2=c$ where $c$ is some complex number with magnitude $4$ . Rearranging the quadratic to standard form and applying the quadratic formula, we have \[z=\frac{-1\pm \sqrt{1^2-4(1)(1-c)}}{2}=\frac{-1\pm\sqrt{4c-3}}{2}.\]
The imaginary part of $z$ is maximized when $c=-4$ . (Why? See note below.)
Thus $z=i\sqrt{19}/2$ , and so the answer is $\boxed{21}$

Solution:
 We can write the given condition as \[\left|\left(z+\frac{1}{2}\right)^2 + \frac{3}{4}\right| = 4.\] Letting $u = \left(z+\frac{1}{2}\right)^2$ , the equation $\left|u + \frac{3}{4}\right| = 4$ equates to the circle centered at $-\frac{3}{4}$ with radius $4$ in the complex plane, call it $\omega$ . Thus the locus of $\left(z+\frac{1}{2}\right)^2$ is $\omega$ . Let $v = z+\frac{1}{2}$ , and since the $+\frac{1}{2}$ does not change $z$ 's imaginary part, we now need to find $v$ with the largest imaginary part such that $v^2$ lies on $\omega$
Note that the point on $\omega$ with largest magnitude is $19/4$ and has argument $\pi$ , call it $\phi$ (The leftmost point on $\omega$ ). The value $v'$ with positive imaginary part such that $(v')^2 = \phi$ has an argument of $\frac{\pi}{2}$ and a magnitude of $\frac{\sqrt{19}}{2}$
Since across all values of $v$ the imaginary part is given by $r\sin{\theta}$ and $v'$ has the largest possible $r$ and the largest possible value of $\sin{\theta},$ it must have the largest imaginary part.
This can non-rigorously be seen by sketching the oval which is the locus of $v$
This gives $19 + 2 \implies \boxed{21}$

Solution:
To start, we factor $1+z+z^2$ to get:
\[|(z-\frac{-1+\sqrt{3}i}{2})(z-\frac{-1-\sqrt{3}i}{2})|=4\]
Note that since the magnitude of a product of complex numbers is equal to the product of the magnitudes:
\[|(z+\frac{1-\sqrt{3}i}{2})||(z+\frac{1+\sqrt{3}i}{2})|=4\]
Now, we substitute $z=a+bi$ (Note that we are trying to maximize b):
\[|a+\frac{1}{2}+(b+\frac{\sqrt{3}}{2})i||a+\frac{1}{2}+(b-\frac{\sqrt{3}}{2})i|=4\]
Since we are trying to maximize $b$ , we want the real parts of the components to be as small as possible, which we can do by setting $a=-\frac{1}{2}$ . This leaves us with:
\[|(b+\frac{\sqrt{3}}{2})i||(b-\frac{\sqrt{3}}{2})i|=4\] \[(b+\frac{\sqrt{3}}{2})(b-\frac{\sqrt{3}}{2})=4\] \[b^2-\frac{3}{4}=4\] \[b^2=\frac{19}{4}\] \[b=\frac{\sqrt{19}}{2}\]
This gives us $19 + 2 \implies \boxed{21}$


Answer: 9.0

Consider the set of numbers $\{1, 10, 10^2, 10^3, \ldots, 10^{10}\}$ . The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)} 101$

Solution: The requested ratio is \[\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.\] Using the formula for a geometric series, we have \[10^9 + 10^8 + \ldots + 10 + 1 = \dfrac{10^{10} - 1}{10 - 1} = \dfrac{10^{10} - 1}{9},\] which is very close to $\dfrac{10^{10}}{9},$ so the ratio is very close to $\boxed{9}.$

Solution: The problem asks for the value of \[\dfrac{10^{10}}{10^9 + 10^8 + \ldots + 10 + 1}.\] Written in base 10, we can find the value of $10^9 + 10^8 + \ldots + 10 + 1$ to be $1111111111.$ Long division gives us the answer to be $\boxed{9}.$

Solution: Let $f(n)=\dfrac{10^n}{1+10+10^2+10^3+\cdots+10^{n-1}}$ . We are approximating $f(10)$ . Trying several small values of $n$ gives answers very close to $9$ , so our answer is $\boxed{9}\approx9.09.$ ~Technodoggo


Answer: 505.0

Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures $3$ by $4$ by $5$ units. Given that the volume of this set is $\frac{m + n\pi}{p},$ where $m, n,$ and $p$ are positive integers , and $n$ and $p$ are relatively prime , find $m + n + p.$

Solution:
The set can be broken into several parts: the big $3\times 4 \times 5$ parallelepiped, $6$ external parallelepipeds that each share a face with the large parallelepiped and have a height of $1$ , the $1/8$ spheres (one centered at each vertex of the large parallelepiped), and the $1/4$ cylinders connecting each adjacent pair of spheres.
The combined volume of these parts is $60+94+\frac{4}{3}\pi+12\pi = \frac{462+40\pi}{3}$ . Thus, the answer is $m+n+p = 462+40+3 = \boxed{505}$


Answer: 27.0

Consider the statement, "If $n$ is not prime, then $n-2$ is prime." Which of the following values of $n$ is a counterexample to this statement?
$\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27$

Solution: Since a counterexample must be a value of $n$ which is not prime, $n$ must be composite, so we eliminate $\text{A}$ and $\text{C}$ . Now we subtract $2$ from the remaining answer choices, and we see that the only time $n-2$ is not prime is when $n = \boxed{27}$


Answer: 27.0

Solution: Since a counterexample must be a value of $n$ which is not prime, $n$ must be composite, so we eliminate $\text{A}$ and $\text{C}$ . Now we subtract $2$ from the remaining answer choices, and we see that the only time $n-2$ is not prime is when $n = \boxed{27}$


Answer: 4.0

Consider the statements:
$\textbf{(1)}\ p\text{ }\wedge\sim q\wedge r\qquad\textbf{(2)}\ \sim p\text{ }\wedge\sim q\wedge r\qquad\textbf{(3)}\ p\text{ }\wedge\sim q\text{ }\wedge\sim r\qquad\textbf{(4)}\ \sim p\text{ }\wedge q\text{ }\wedge r$
where $p,q$ , and $r$ are propositions. How many of these imply the truth of $(p\rightarrow q)\rightarrow r$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution:
Statement $1$ states that $p$ is true and $q$ is false.  Therefore, $p \rightarrow q$ is false, because a premise being true and a conclusion being false is, itself, false.  This means that $(p \rightarrow q) \rightarrow X$ , where $X$ is any logical statement (or series of logical statements) must be true - if your premise is false, then the implication is automatically true.  So statement $1$ implies the truth of the given statement.
Statement $3$ similarly has $p$ as true and $q$ is false, so it also implies the truth of the given statement.
Statement $2$ states that $p$ and $q$ are both false.  This in turn means that $p \rightarrow q$ is true.  Since $r$ is also true from statement $2$ , this means that $(p \rightarrow q) \rightarrow r$ is true, since $T \rightarrow T$ is $T$ .  Thus statement $2$ implies the truth of the given statement.
Statement $4$ states that $p$ is false and $q$ is true.  In this case, $p \rightarrow q$ is true - your conclusion can be true even if your premise is false.  And, since $r$ is also true from statement $4$ , this means $(p \rightarrow q) \rightarrow r$ is true.  Thus, statement $4$ implies the truth of the given statement.
All four statements imply the truth of the given statement, so the answer is $\boxed{4}$


Answer: 100.0

Consider these two operations: \begin{align*} a \, \blacklozenge \, b &= a^2 - b^2\\ a \, \bigstar \, b &= (a - b)^2 \end{align*} What is the output of $(5 \, \blacklozenge \, 3) \, \bigstar \, 6?$
$\textbf{(A) } {-}20 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 100 \qquad \textbf{(E) } 220$

Solution: We can substitute $5$ $3$ , and $6$ into the functions' definitions: \begin{align*} (5 \, \blacklozenge \, 3) \, \bigstar \, 6 &= \left(5^2-3^2\right) \, \bigstar \, 6 \\ (5 \, \blacklozenge \, 3) \, \bigstar \, 6 &= \left(25-9\right) \, \bigstar \, 6 \\ &= 16 \, \bigstar \, 6 \\ &= (16-6)^2 \\ &= \boxed{100} ~pog ~MathFun1000 (Minor Edits)


Answer: 70.0

Solution: Since $81$ students took the test, the median is the score of the $41^{st}$ student. The five rightmost intervals include $2+3+6+12+16=39$ students, so the $41^{st}$ one must lie in the next interval, which is $\boxed{70}$


Answer: 24.0

Consider those functions $f$ that satisfy $f(x+4)+f(x-4) = f(x)$ for all real $x$ . Any such function is periodic, and there is a least common positive period $p$ for all of them. Find $p$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 32$

Solution:
Recall that $p$ is the fundamental period of function $f$ iff $p$ is the smallest positive $p$ such that $f(x) = f(x + p)$ for all $x$
In this case, we know that $f(x+ 4) + f(x - 4) = f(x)$ .  Plugging in $x+4$ in for $x$ to get the next equation in the recursion, we also get $f(x + 8) + f(x) = f(x + 4)$ .  Adding those two equations gives $f(x + 8) +  f(x - 4) = 0$ after cancelling out common terms.
Again plugging in $x + 4$ in for $x$ in that last equation (in order to get $f(x)$ ), we find that $f(x) = -f(x + 12)$ .  Now, plugging in $x+12$ for $x$ , we get $f(x + 12) = -f(x + 24)$ .  This proves that $f(x)  = f(x + 24)$ , so there is a period of $24$ , which gives answer $\boxed{24}$ .  We now eliminate answers $A$ through $C$


Answer: 13.0

Consider two solid spherical balls, one centered at $\left(0, 0,\frac{21}{2}\right)$ with radius $6$ , and the other centered at $(0, 0, 1)$ with radius $\frac{9}{2}$ . How many points with only integer coordinates (lattice points) are there in the intersection of the balls?
$\text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15$

Solution:
The two equations of the balls are
\[x^2 + y^2 + \left(z - \frac{21}{2}\right)^2 \le 36\]
\[x^2 + y^2 + (z - 1)^2 \le \frac{81}{4}\]
Note that along the $z$ axis, the first ball goes from $10.5 \pm 6$ , and the second ball goes from $1 \pm 4.5$ .  The only integer value that $z$ can be is $z=5$
Plugging that in to both equations, we get:
\[x^2 + y^2 \le \frac{23}{4}\]
\[x^2 + y^2 \le \frac{17}{4}\]
The second inequality implies the first inequality, so the only condition that matters is the second inequality.
From here, we do casework, noting that $|x|, |y| \le 3$
For $x=\pm 2$ , we must have $y=0$ .  This gives $2$ points.
For $x = \pm 1$ , we can have $y\in \{-1, 0, 1\}$ .  This gives $2\cdot 3 = 6$ points.
For $x = 0$ , we can have $y \in \{-2, -1, 0, 1, 2\}$ .  This gives $5$ points.
Thus, there are $\boxed{13}$

Solution:
Because both spheres have their centers on the x-axis, we can simplify the graph a bit by looking at a 2-dimensional plane (the previous z-axis is the new x-axis while the y-axis remains the same).
The spheres now become circles with centers at $(1,0)$ and $(\frac{21}{2},0)$ .  They have radii $\frac{9}{2}$ and $6$ , respectively.
Let circle $A$ be the circle centered on $(1,0)$ and circle $B$ be the one centered on $(\frac{21}{2},0)$
The point on circle $A$ closest to the center of circle $B$ is $(\frac{11}{2},0)$ .  The point on circle B closest to the center of circle $A$ is $(\frac{11}{2},0)$
Taking a look back at the 3-dimensional coordinate grid with the spheres, we can see that their intersection appears to be a circle with congruent "dome" shapes on either end.  Because the tops of the "domes" are at $(0,0,\frac{9}{2})$ and $(0,0,\frac{11}{2})$ , respectively, the lattice points inside the area of intersection must have z-value $5$ (because $5$ is the only integer between $\frac{9}{2}$ and $\frac{11}{2}$ ).  Thus, the lattice points in the area of intersection must all be on the 2-dimensional circle.  The radius of the circle will be the distance from the z-axis.
Now, looking at the 2-dimensional coordinate plane, we see that the radius of the circle (now the distance from the x-axis, because there is no more z-axis) is the altitude of a triangle with two points on centers of circles $A$ and $B$ and third point at the first quadrant intersection of the circles.  Let's call that altitude $h$
We know all three side lengths of this triangle: $\frac{9}{2}$ (the radius of circle $A$ ), $6$ (the radius of circle $B$ ), and $\frac{19}{2}$ (the distance between the centers of circles $A$ and $B$ ).  We can now find the area of the triangle using Heron's formula:
\[s=\frac{\frac{9}{2}+6+\frac{19}{2}}{2}=20\]
\[A=\sqrt{(20)(20-\frac{9}{2})(20-6)(20-\frac{19}{2})}=\sqrt{110}\]
Using the area of the triangle, we can find that altitude $h$ from the x-axis:
\[\sqrt{110}=\frac{h*\frac{19}{2}}{2}\]
\[h=\frac{4*\sqrt{110}}{19}\]
Remember, the altitude $h$ is also the radius of the circle containing all the solutions to the problem.
Going back to the 3-dimensional grid and looking at the circle, we can again make the figure 2-dimensional.
The radius $h$ of the circle in a 2-dimensional plane is $\frac{4\sqrt{110}}{19}$ , a little greater than $2$ .  We know that $h>2$ because $4\sqrt{110}>4*10$ , and $\frac{40}{19}>2$
Finally, looking at a circle with radius slightly larger than $2$ , we see that there are $\boxed{13}$

Solution:
Note that the spheres are on the same $x$ and $y$ axis. Therefore, we can draw the spheres so that only the $x$ and $z$ axis are featured.
$A = (1, 0, 5)$ $\sqrt{1^2+4^2} = \sqrt{17}<\frac92$ $A$ is inside the smaller sphere. $\sqrt{1^2+(\frac{11}{2})^2} = \frac{\sqrt{126}}{2}<6$ $A$ is inside the larger sphere. Point $A$ is inside the intersection.
$B = (2, 0, 5)$ $\sqrt{2^2+4^2} = \sqrt{20}<\frac92$ $B$ is inside the smaller sphere. $\sqrt{2^2+(\frac{11}{2})^2} = \frac{\sqrt{137}}{2}<6$ $B$ is inside the larger sphere. Point $B$ is inside the intersection.
$C = (3, 0, 5)$ $\sqrt{3^2+4^2} = \sqrt{25}>\frac92$ $C$ is outside the smaller sphere. $\sqrt{3^2+(\frac{11}{2})^2} = \frac{\sqrt{157}}{2}>6$ $C$ is outside the larger sphere. Point $C$ is outside the intersection.
The intersection of $2$ spheres is a circle, by drawing the circle flat we can see that there are $4$ more points within the intersection.
We can see that points with the same $x$ -coordinates with $A$ are still inside the circle. $B$ is very close to the circle, the points with the same $x$ -coordinates with $B$ are not necessarily inside the circle.
For example, $D = (2,1,5)$ $\sqrt{2^2+1^2+4^2}=\sqrt{21}>\frac92$ $D$ is outside the smaller sphere. $\sqrt{2^2 + 1^2 + (\frac{11}{2})^2} = \frac{\sqrt{141}}{2}<6$ $D$ is inside the larger sphere. Point $D$ is outside the intersection.
Therefore, the answer is $9+4 = \boxed{13}$


Answer: 23.0

Construct a square on one side of an equilateral triangle. On one non-adjacent side of the square, construct a regular pentagon, as shown. On a non-adjacent side of the pentagon, construct a hexagon. Continue to construct regular polygons in the same way, until you construct an octagon. How many sides does the resulting polygon have?
[asy] defaultpen(linewidth(0.6)); pair O=origin, A=(0,1), B=A+1*dir(60), C=(1,1), D=(1,0), E=D+1*dir(-72), F=E+1*dir(-144), G=O+1*dir(-108); draw(O--A--B--C--D--E--F--G--cycle); draw(O--D, dashed); draw(A--C, dashed);[/asy]
$\textbf{(A)}\ 21 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 29$

Solution: Of the six shapes used to create the polygon, the triangle and octagon are adjacent to the others on one side, and the others are adjacent on two sides. In the triangle and octagon $3+8-2(1)=9$ sides are on the outside of the final polygon. In the other shapes $4+5+6+7-4(2) = 14$ sides are on the outside. The resulting polygon has $9+14 = \boxed{23}$ sides.

Solution: We can quickly see a pattern if we draw out the other shapes. Every shape will have two of its sides taken out except the triangle and octagon. We can then make the expression $2+2+3+4+5+7$ which is $\boxed{23}$


Answer: 60.0

Convex pentagon $ABCDE$ has side lengths $AB=5$ $BC=CD=DE=6$ , and $EA=7$ . Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$

Solution: Assume the incircle touches $AB$ $BC$ $CD$ $DE$ $EA$ at $P,Q,R,S,T$ respectively. Then let $PB=x=BQ=RD=SD$ $ET=y=ES=CR=CQ$ $AP=AT=z$ . So we have $x+y=6$ $x+z=5$ and $y+z$ =7, solve it we have $x=2$ $z=3$ $y=4$ . Let the center of the incircle be $I$ , by SAS we can proof triangle $BIQ$ is congruent to triangle $DIS$ , and triangle $CIR$ is congruent to triangle $SIE$ . Then we have $\angle AED=\angle BCD$ $\angle ABC=\angle CDE$ . Extend $CD$ , cross ray $AB$ at $M$ , ray $AE$ at $N$ , then by AAS we have triangle $END$ is congruent to triangle $BMC$ . Thus $\angle M=\angle N$ . Let $EN=MC=a$ , then $BM=DN=a+2$ . So by law of cosine in triangle $END$ and triangle $ANM$ we can obtain \[\frac{2a+8}{2(a+7)}=\cos N=\frac{a^2+(a+2)^2-36}{2a(a+2)}\] , solved it gives us $a=8$ , which yield triangle $ANM$ to be a triangle with side length 15, 15, 24, draw a height from $A$ to $NM$ divides it into two triangles with side lengths 9, 12, 15, so the area of triangle $ANM$ is 108. Triangle $END$ is a triangle with side lengths 6, 8, 10, so the area of two of them is 48, so the area of pentagon is $108-48=\boxed{60}$

Solution:
Suppose that the circle intersects $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ $\overline{DE}$ , and $\overline{EA}$ at $P$ $Q$ $R$ $S$ , and $T$ respectively. Then $AT = AP = a$ $BP = BQ = b$ $CQ = CR = c$ $DR = DS = d$ , and $ES = ET = e$ . So $a + b = 5$ $b + c = 6$ $c + d = 6$ $d + e = 6$ , and $e + a = 7$ . Then $2a + 2b + 2c + 2d + 2e = 30$ , so $a + b + c + d + e= 15$ . Then we can solve for each individually. $a = 3$ $b = 2$ $c = 4$ $d = 2$ , and $e = 4$ . To find the radius, we notice that $4 \arctan(\frac{2}{r}) + 4 \arctan(\frac{4}{r}) + 2 \arctan (\frac{3}{r}) = 360 ^ \circ$ , or $2 \arctan(\frac{2}{r}) + 2 \arctan(\frac{4}{r}) + \arctan (\frac{3}{r}) = 180 ^ \circ$ . Each of these angles in this could be represented by complex numbers. When two complex numbers are multiplied, their angles add up to create the angle of the resulting complex number. Thus, $(r + 2i)^2 \cdot (r + 4i)^2 \cdot (r + 3i)$ is real. Expanding, we get:
\[(r^2 + 4ir - 4)(r^2 + 8ir -16)(r + 3i)\]
\[(r^4 + 12ir^3 - 52r^2 - 96ir + 64)(r + 3i)\]
On the last expanding, we only multiply the reals with the imaginaries and vice versa, because we only care that the imaginary component equals 0.
\[15ir^4 - 252ir^2 + 192i = 0\]
\[5r^4 - 84r^2 + 64 = 0\]
\[(5r^2 - 4)(r^2 - 16) = 0\]
$r$ must equal 4, as r cannot be negative or be approximately equal to 1.
Thus, the area of $ABCDE$ is $4 \cdot (a + b + c + d + e) = 4 \cdot 15 = \boxed{60}$

Solution: This pentagon is very close to a regular pentagon with side lengths $6$ . The area of a regular pentagon with side lengths $s$ is $\frac{5s^2}{4\sqrt{5-2\sqrt{5}}}$ $5-2\sqrt{5}$ is slightly greater than $\frac{1}{2}$ given that $2\sqrt{5}$ is slightly less than $\frac{9}{2}$ $4\sqrt{5-2\sqrt{5}}$ is then slightly greater than $2\sqrt{2}$ . We will approximate that to be $2.9$ . The area is now roughly $\frac{180}{2.9}$ , but because the actual pentagon is not regular, but has the same perimeter of the regular one that we are comparing to we can say that this is an overestimate on the area and turn the $2.9$ into $3$ thus turning the area into $\frac{180}{3}$ which is $60$ and since $60$ is a multiple of the semiperimeter $15$ , we can safely say that the answer is most likely $\boxed{60}$

Solution:
Let $\omega$ be the inscribed circle, $I$ be its center, and $r$ be its radius. The area of $ABCDE$ is equal to its semiperimeter, $15,$ times $r$ , so the problem is reduced to finding $r$ . Let $a$ be the length of the tangent segment from $A$ to $\omega$ , and analogously define $b$ $c$ $d$ , and $e$ . Then $a+b=5$ $b+c= c+d=d+e=6$ , and $e+a=7$ , with a total of $a+b+c+d+e=15$ . Hence $a=3$ $b=d=2$ , and $c=e=4$ . It follows that $\angle B= \angle D$ and $\angle C= \angle E$ . Let $Q$ be the point where $\omega$ is tangent to $\overline{CD}$ . Then $\angle IAE = \angle IAB =\frac{1}{2}\angle A$ . Now we claim that points $A, I, Q$ are collinear, which can be proved if $\angle{AIQ}=\angle{QIA}=180^{\circ}$ . The sum of the internal angles in polygons $ABCQI$ and $AIQDE$ are equal, so $\angle IAE + \angle AIQ + \angle IQD + \angle D + \angle E = \angle IAB + \angle B + \angle C + \angle CQI + \angle QIA$ , which implies that $\angle AIQ$ must be $180^\circ$ . Therefore points $A$ $I$ , and $Q$ are collinear.  Because $\overline{AQ} \perp \overline{CD}$ , it follows that \[AC^2-AD^2=CQ^2-DQ^2=c^2-d^2=12.\] Another expression for $AC^2-AD^2$ can be found as follows. Note that $\tan \left(\frac{\angle B}{2}\right) = \frac{r}{2}$ and $\tan \left(\frac{\angle E}{2}\right) = \frac{r}{4}$ , so \[\cos (\angle B) =\frac{1-\tan^2 \left(\frac{\angle B}{2}\right)}{1+\tan^2 \left(\frac{\angle B}{2}\right)} =  \frac{4-r^2}{4+r^2}\] and \[\cos (\angle E) = \frac{1-\tan^2 \left(\frac{\angle E}{2}\right)}{1+\tan^2 \left(\frac{\angle E}{2}\right)}= \frac{16-r^2}{16+r^2}.\] Applying the Law of Cosines to $\triangle ABC$ and $\triangle AED$ gives \[AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos (\angle B) = 5^2+6^2-2 \cdot 5 \cdot 6 \cdot \frac{4-r^2}{4+r^2}\] and \[AD^2=AE^2+DE^2-2 \cdot AE \cdot DE \cdot \cos(\angle E) = 7^2+6^2-2 \cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}.\] Hence
\[12=AC^2- AD^2= 5^2-2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2} -7^2+2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2},\] yielding \[2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}- 2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2}= 36;\] equivalently \[7(16-r^2)(4+r^2)-5(4-r^2)(16+r^2) = 3(16+r^2)(4+r^2).\] Substituting $x=r^2$ gives the quadratic equation $5x^2-84x+64=0$ , with solutions $\frac{42 - 38}{5}=\frac45$ , and $\frac{42 + 38}{5}= 16$ . The solution $r^2=\frac45$ corresponds to a five-pointed star, which is not convex. Indeed, if $r<3$ , then $\tan \left(\frac{\angle A}{2}\right)$ $\tan \left(\frac{\angle C}{2}\right)$ , and $\tan \left(\frac{\angle E}{2}\right)$ are less than $1,$ implying that $\angle A$ $\angle C$ , and $\angle E$ are acute, which cannot happen in a convex pentagon. Thus $r^2=16$ and $r=4$ . The requested area is $15\cdot4 = \boxed{60}$


Answer: 84.0

Convex quadrilateral $ABCD$ has $AB = 18, \angle{A} = 60^\circ,$ and $\overline{AB} \parallel \overline{CD}.$ In some order, the lengths of the four sides form an arithmetic progression, and side $\overline{AB}$ is a side of maximum length. The length of another side is $a.$ What is the sum of all possible values of $a$
$\textbf{(A) } 24 \qquad \textbf{(B) } 42 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 66 \qquad \textbf{(E) } 84$

Solution:
Let $E$ be a point on $\overline{AB}$ such that $BCDE$ is a parallelogram. Suppose that $BC=ED=b, CD=BE=c,$ and $DA=d,$ so $AE=18-c,$ as shown below.  We apply the Law of Cosines to $\triangle ADE:$ \begin{align*} AD^2 + AE^2 - 2\cdot AD\cdot AE\cdot\cos 60^\circ &= DE^2 \\ d^2 + (18-c)^2 - d(18-c) &= b^2 \\ (18-c)^2 - d(18-c) &= b^2 - d^2 \\ (18-c)(18-c-d) &= (b+d)(b-d). \hspace{15mm}(\bigstar) \end{align*} Let $k$ be the common difference of the arithmetic progression of the side-lengths. It follows that $b,c,$ and $d$ are $18-k, 18-2k,$ and $18-3k,$ in some order. It is clear that $0\leq k<6.$
If $k=0,$ then $ABCD$ is a rhombus with side-length $18,$ which is valid.
If $k\neq0,$ then we have six cases:
Together, the sum of all possible values of $a$ is $18+(13+3+8)+(14+12+16)=\boxed{84}.$

Solution:
Let $b, c$ , and $d$ denote the sides $BC, CD$ , and $AD$ respectively.  Since $AB\parallel CD$ , we get \[\tfrac{\sqrt 3}{2}\ d = b\sin\theta \quad \textrm{and}\quad \tfrac 12 d + c + b\cos\theta = 18.\] Using $b^2\sin^2\theta + b^2\cos^2\theta = b^2$ , we eliminate $\theta$ from above to get $(36-2c-d)^2+3d^2=4b^2$ , which rearranges to $(36-2c-d)^2-d^2=4(b^2-d^2)$ , and, upon factoring, yields \begin{align}     (18-c)(18-c-d)=(b+d)(b-d). \end{align} We divide into two cases, depending on whether $c$ is the smallest side.
If $c$ is not the smallest side then $18-c=\pm (b-d)$ . If $c=18$ , we get a rhombus of side $18$ , so  one possible value is $a=18$ . Otherwise, we can cancel the common factor from $(1)$ . After rearranging we get \[18-c=-b \quad \textrm{or}\quad  18-c=b+2d.\] The first condition is false because $-b< 0 <18-c$ ; the second condition is false because $b+2d > |b-d| = 18-c$
If $c$ is the smallest side, then $18-c = \pm 3(b-d)$ . Assuming $c<18$ we can cancel common factors in $(1)$ to get \[8b=13d \quad \textrm{or}\quad 8b=7d.\] The first condition yields the solution $(c,d,b)=(3,8,13)$ and the second condition yields the solution $(c,b,d)=(12,14,16)$
Together, the sum of all possible values of $a$ is $18+(3+8+13)+(12+14+16)=\boxed{84}.$

Solution:
Denote $x = AD$ $\theta = \angle B$ .
Hence, $BC = \frac{\sqrt{3}}{2} \cdot \frac{x}{\sin \theta}$ $DC = 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta$
$\textbf{Case 1}$ $DC = AD = BC = AB$
This is a rhombus. So each side has length $18$
For the following cases, we consider four sides that have distinct lengths.
To make their lengths an arithmetic sequence, we must have $\theta \neq 120^\circ$
Therefore, in the subsequent analysis, we exclude the solution $\theta = 120^\circ$
$\textbf{Case 2}$ $DC < AD < BC < AB$
Because the lengths of these sides form an arithmetic sequence, we have the following system of equations: \[ AB - BC = BC - AD = AD - DC . \]
Hence, \begin{eqnarray*} & 18 - \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta} = \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta} - x = x - \left( 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta \right) . & \end{eqnarray*}
By solving this system of equations, we get $\left( \cos \theta , \sin \theta , x\right) = \left( \frac{11}{13} , \frac{4 \sqrt{3}}{13} , 8 \right)$
Thus, in this case, $DC = 3$ $AD = 8$ $BC = 13$
$\textbf{Case 3}$ $DC < BC < AD < AB$
Because the lengths of these sides form an arithmetic sequence, we have the following system of equations: \[ AB - AD = AD - BC = BC - DC . \]
Hence, \begin{eqnarray*} & 18 - x = x - \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta} = \frac{\sqrt{3}}{2}\cdot\frac{x}{\sin \theta}  - \left( 18 - \frac{x}{2} - \frac{\sqrt{3}}{2} x \cot \theta \right) . & \end{eqnarray*}
By solving this system of equations, we get $\left( \cos \theta , \sin \theta , x\right) = \left( - \frac{1}{7} , \frac{4 \sqrt{3}}{7} , 16 \right)$
Thus, in this case, $DC = 12$ $AD = 16$ $BC = 14$
$\textbf{Case 4}$ $BC < CD < AD < AB$
By doing the similar analysis, we can show there is no solution in this case.
$\textbf{Case 5}$ $BC < AD < CD < AB$
By doing the similar analysis, we can show there is no solution in this case.
$\textbf{Case 6}$ $AD < CD < BC < AB$
By doing the similar analysis, we can show there is no solution in this case.
$\textbf{Case 7}$ $AD < BC < CD < AB$
By doing the similar analysis, we can show there is no solution in this case.
Therefore, the sum of all possible values of $a$ is \begin{align*} 18 + \left( 3 + 8 + 13 \right) + \left( 12 + 14 + 16 \right) & = 84 . \end{align*}
Therefore, the answer is $\boxed{84}$


Answer: 6.0

Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12$ .  Diagonals $AC$ and $BD$ intersect at $E$ $AC = 14$ , and $\triangle AED$ and $\triangle BEC$ have equal areas.  What is $AE$
$\textbf{(A)}\ \frac {9}{2}\qquad \textbf{(B)}\ \frac {50}{11}\qquad \textbf{(C)}\ \frac {21}{4}\qquad \textbf{(D)}\ \frac {17}{3}\qquad \textbf{(E)}\ 6$

Solution: The easiest way for the areas of the triangles to be equal would be if they were congruent [1] . A way for that to work would be if $ABCD$ were simply an isosceles trapezoid! Since $AC = 14$ and $AE:EC = 3:4$ (look at the side lengths and you'll know why!), $\boxed{6}$


Answer: 6.0

Solution:
Using the fact that $[AED] = [BEC]$ and the fact that $\triangle AEB \sim \triangle EDC$ (which should be trivial given the two equal triangles) we have that
\[\frac{AE}{DC} = \frac{BE}{EC} = \frac{9}{12}\]
We know that $DC=EC,$ so we have
\[\frac{AE}{EC} = \frac{BE}{EC} = \frac{3}{4}\]
Thus
\[\frac{AE}{EC} = \frac{3}{4}\]
But $EC = 14 - AE$ so we have
\[\frac{AE}{14 - AE} = \frac{3}{4}\]
Simplifying gives $AE = \boxed{6}.$


Answer: 6.0

Solution: The easiest way for the areas of the triangles to be equal would be if they were congruent [1] . A way for that to work would be if $ABCD$ were simply an isosceles trapezoid! Since $AC = 14$ and $AE:EC = 3:4$ (look at the side lengths and you'll know why!), $\boxed{6}$


Answer: 6.0

Solution:
Using the fact that $[AED] = [BEC]$ and the fact that $\triangle AEB \sim \triangle EDC$ (which should be trivial given the two equal triangles) we have that
\[\frac{AE}{DC} = \frac{BE}{EC} = \frac{9}{12}\]
We know that $DC=EC,$ so we have
\[\frac{AE}{EC} = \frac{BE}{EC} = \frac{3}{4}\]
Thus
\[\frac{AE}{EC} = \frac{3}{4}\]
But $EC = 14 - AE$ so we have
\[\frac{AE}{14 - AE} = \frac{3}{4}\]
Simplifying gives $AE = \boxed{6}.$


Answer: 36.0

Convex quadrilateral $ABCD$ has $AB=3$ $BC=4$ $CD=13$ $AD=12$ , and $\angle ABC=90^{\circ}$ , as shown. What is the area of the quadrilateral?
[asy] pair A=(0,0), B=(-3,0), C=(-3,-4), D=(48/5,-36/5); draw(A--B--C--D--A);  label("$A$",A,N); label("$B$",B,NW); label("$C$",C,SW); label("$D$",D,E); draw(rightanglemark(A,B,C,25)); [/asy]
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 58.5$

Solution: Note that by the pythagorean theorem, $AC=5$ .  Also note that $\angle CAD$ is a right angle because $\triangle CAD$ is a right triangle.  The area of the quadrilateral is the sum of the areas of $\triangle ABC$ and $\triangle CAD$ which is equal to \[\frac{3\times4}{2} + \frac{5\times12}{2} = 6 + 30 = \boxed{36}\]


Answer: 13.0

Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps.  However, instead of walking up the steps one at a time, both Cozy and Dash jump.  Cozy goes two steps up with each jump (though if necessary, he will just jump the last step).  Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than $5$ steps left).  Suppose Dash takes $19$ fewer jumps than Cozy to reach the top of the staircase.  Let $s$ denote the sum of all possible numbers of steps this staircase can have.  What is the sum of the digits of $s$
$\textbf{(A) }9\qquad\textbf{(B) }11\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }15$

Solution:
Let $n$ be the number of steps. We have
\[\left\lceil \frac{n}{2} \right\rceil - 19 = \left\lceil \frac{n}{5} \right\rceil\]
We will proceed to solve this equation via casework.
Case $1$ $\left\lceil \frac{n}{2} \right\rceil = \frac{n}{2}$
Our equation becomes $\frac{n}{2} - 19 = \frac{n}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $n$ is an integer, we quickly find that $j=1$ and $j=4$ yield $n=64$ and $n=66$ , respectively.
Case $2$ $\left\lceil \frac{n}{2} \right\rceil = \frac{n}{2}+\frac{1}{2}$
Our equation becomes $\frac{n}{2} +\frac{1}{2} - 19 = \frac{n}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $n$ is an integer, we quickly find that $j=2$ yields $n=63$ . Summing up we get $63+64+66=193$ . The sum of the digits is $\boxed{13}$

Solution:
We know from the problem that Dash goes $3$ steps further than Cozy per jump (assuming they aren't within $4$ steps from the top). That means that if Dash takes $19$ fewer jumps than Cozy to get to the top of the staircase, the staircase must be at least $3 \cdot 19=57$ steps high. We then start using guess-and-check:
$57$ steps: $\left \lceil {57/2} \right \rceil = 29$ jumps for Cozy, and $\left \lceil {57/5} \right \rceil = 12$ jumps for Dash, giving a difference of $17$ jumps.
$58$ steps: $\left \lceil {58/2} \right \rceil = 29$ jumps for Cozy, and $\left \lceil {58/5} \right \rceil = 12$ jumps for Dash, giving a difference of $17$ jumps.
$59$ steps: $\left \lceil {59/2} \right \rceil = 30$ jumps for Cozy, and $\left \lceil {59/5} \right \rceil = 12$ jumps for Dash, giving a difference of $18$ jumps.
$60$ steps: $\left \lceil {60/2} \right \rceil = 30$ jumps for Cozy, and $\left \lceil {60/5} \right \rceil = 12$ jumps for Dash, giving a difference of $18$ jumps.
$\vdots$
By the time we test $61$ steps, we notice that when the number of steps exceeds a multiple of $2$ , the difference in jumps increases. So, we have to find the next number that will increase the difference. $62$ doesn't because both both Cozy's and Dash's number of jumps increases, but $63$ does, and $64$ $65$ actually gives a difference of $20$ jumps, but $66$ goes back down to $19$ (because Dash had to take another jump when Cozy didn't). We don't need to go any further because the difference will stay above $19$ onward.
Therefore, the possible numbers of steps in the staircase are $63$ $64$ , and $66$ , giving a sum of $193$ . The sum of those digits is $13$ , so the answer is $\boxed{13}$


Answer: 13.0

Solution:
We're looking for natural numbers $x$ such that $\left \lceil{\frac{x}{5}}\right \rceil + 19 = \left \lceil{\frac{x}{2}}\right \rceil$
Let's call $x = 10a + b$ . We now have $2a  + \left \lceil{\frac{b}{5}}\right \rceil + 19 = 5a + \left  \lceil{\frac{b}{2}}\right \rceil$ , or
$19 - 3a = \left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil$
Obviously, since $b \le 10$ , this will not work for any value of $a$ under $6$ . In addition, since obviously $\frac{b}{2} \ge \frac{b}{5}$ , this will not work for any value over six, so we have $a = 6$ and $\left \lceil{\frac{b}{2}}\right \rceil - \left \lceil{\frac{b}{5}}\right \rceil = 1.$
This can be achieved when $\left \lceil{\frac{b}{5}}\right \rceil = 1$ and $\left \lceil{\frac{b}{2}}\right \rceil = 2$ , or when $\left \lceil{\frac{b}{5}}\right \rceil = 2$ and $\left \lceil{\frac{b}{2}}\right \rceil = 3$
Case One:
We have $b \le 5$ and $3 \le b \le 4$ , so $b = 3, 4$
Case Two:
We have $6 \le b \le 9$ and $5 \le b \le 6$ , so $b = 6$
We then have $63 + 64 + 66 = 193$ , which has a digit sum of $\boxed{13}$

Solution:
Translate the problem into following equation:
$n = 5D - \{0 \sim 4\} = 2C - \{0 \sim 1\}$
Since $C = D + 19$ , we have
$5D - \{0 \sim 4\} = 2D + 38 - \{0 \sim 1\}$
i.e.,
$3D = 38 + \{0 \sim 4\} - \{0 \sim 1\}$
We then have $D = 13$ when $\{1\} - \{0\}$ or $\{2\} - \{1\}$ (the dog's last jump has $2$ steps and the cat's last jump has $1$ step), which yields $n = 64$ and $n = 63$ respectively.
Another solution is $D = 14$ when $\{4\} - \{0\}$ , which yields $n = 66$
Therefore, with $63 + 64 + 66 = 193$ , the digit sum is $\boxed{13}$


Answer: 13.0

Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$

Solution:
We can translate this wordy problem into this simple equation:
\[\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil\]
We will proceed to solve this equation via casework.
Case 1: $\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}$
Our equation becomes $\frac{s}{2} - 19 = \frac{s}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $s$ is an integer, we quickly find that $j=1$ and $j=4$ yield $s=64$ and $s=66$ , respectively.
Case 2: $\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}+\frac{1}{2}$
Our equation becomes $\frac{s}{2} +\frac{1}{2} - 19 = \frac{s}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $s$ is an integer, we quickly find that $j=2$ yields $s=63$
Summing up we get $63+64+66=193$ . The sum of the digits is $\boxed{13}$

Solution:
It can easily be seen that the problem can be expressed by the equation: \[\left\lceil \frac{s}{2} \right\rceil - \left\lceil \frac{s}{5} \right\rceil = 19\]
However, because the ceiling function is difficult to work with, we can rewrite the previous equation as:
\[\frac{s+a}{2} - \frac{s+b}{5} = 19\] Where $a \in \{0,1\}$ and $b \in \{0,1,2,3,4\}$ Multiplying both sides by ten and simplifying, we get: \[5s+5a-2s-2b=190\] \[3s = 190+2b-5a\] \[s = 63 + \frac{1+2b-5a}{3}\]
Because s must be an integer, we need to find the values of $a$ and $b$ such that $2b-5a \equiv 2 \mod 3$ . We solve using casework.
Case 1: $a = 0$
If $a = 0$ , we have $2b \equiv 2 \mod 3$ . We can easily see that $b = 1$ or $b = 4$ , which when plugged into our original equation lead to $s = 64$ and $s=66$ respectively.
Case 2: $a = 1$
If $a = 1$ , we have $2b-5 \equiv 2 \mod 3$ , which can be rewritten as $2b \equiv 1 \mod 3$ . We can again easily see that $b = 2$ is the only solution, which when plugged into our original equation lead to $s = 63$
Adding these together we get $64+66+63=193$ . The sum of the digits is $\boxed{13}$

Solution:
As before, we write the equation:
\[\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil.\]
To get a ballpark estimate of where $s$ might lie, we remove the ceiling functions to find:
\[\frac{s}{2} - 19 = \frac{s}{5}.\]
This gives $\frac{3s}{10} = 19$ , and thus values for $s$ will be around $\frac{190}{3} = 63.\overline3$
Now, to establish some bounds around this estimated working value, we note that if $s=60$ , Cozy takes 30 steps while Dash takes 12, a difference of 18.  If $s=70$ , Cozy takes 35 steps while Dash takes 14, a difference of 21.  When $s$ increases from a multiple of ten, the difference will never decrease beyond what it is at the multiple of ten, and likewise, when it decreases, it never becomes greater than at the multiple of ten, so any working values of $s$ will be between $60$ and $70$
Then, by inspection, $s=63, 64,$ or $66$ , so $\sum s = 193 \implies \boxed{13}.$

Solution: Notice that the possible number of steps in the staircase is around 60 to 70. By testing all of the values between 60 and 70, we see that 63, 64 and 66 work. Adding those up gives 193, so the answer is $1+9+3=\boxed{13}.$

Solution:
We represent C's steps with $2c + a = n$ and D's steps with $5d + b = n$ , where $a \in \{1,2\}$ and $b \in \{1,2,3,4,5\}$ , where $n$ is the number of steps, $c$ is the number of jumps C takes bar the last one, and $d$ is the number of jumps D takes bar the last.
The reason for starting at 1 and ending at 5 instead of 0 through 4 is that the last step can be quite problematic to deal with, especially if it is possible to make it in one go, so we treat it as a different jump that can take all possible jump values. We know that D makes it in 19 fewer than C, so $c+1-(d+1) = 19 \implies c = 19+d \implies 5d + b = 2(19+d) + a \implies 3d = 38 + a - b$
Now that we have this nice equivalence, we can do the thing and it works.
If we take both sides mod 3 and rearrange, we get $b \equiv 2+a \pmod{3}$ This gives us the following satisfactory $(a,b)$ relational pairs: $\{(1,3),(2,1),(2,4)\}$ . We can now just find the corresponding $d$ value for each pair, sum it all up using $5d + b = n$ and sum the digits to reveal $\boxed{13}$ as the answer.


Answer: 3.0

Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ \sqrt{2} \qquad \mathrm{(C)}\ \sqrt{3} \qquad \mathrm{(D)}\ 2 \qquad \mathrm{(E)}\ 2\sqrt{2}$

Solution: Crystal runs north one mile, then her next two moves can be broken up into four individual moves: for her northeast section, it forms a $45-45-90$ triangle whose legs are each $\frac{\sqrt{2}}{2}$ . For her southeast section, it is also a $45-45-90$ triangle whose legs are each $\frac{\sqrt{2}}{2}$ . Notice that the two of the legs cancel each other out; she moves north $\frac{\sqrt{2}}{2}$ units and also south $\frac{\sqrt{2}}{2}$ units. So her net movement these two moves is $\sqrt{2}$ to the right. Finally, after the third move, she is at the corner of a right triangle with legs $1$ and $\sqrt{2}$ . Using the Pythagorean theorem, $d^{2}=1^{2}+\left(\sqrt{2}\right)^{2}=1+2=3$ and $d=\sqrt{3} \Longrightarrow \boxed{3}$


Answer: 89.0

Cube $ABCDEFGH,$ labeled as shown below, has edge length $1$ and is cut by a plane passing through vertex $D$ and the midpoints $M$ and $N$ of $\overline{AB}$ and $\overline{CG}$ respectively. The plane divides the cube into two solids. The volume of the larger of the two solids can be written in the form $\tfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution: Define a coordinate system with $D$ at the origin and $C,$ $A,$ and $H$ on the $x$ $y$ , and $z$ axes respectively. Then $D=(0,0,0),$ $M=(.5,1,0),$ and $N=(1,0,.5).$ It follows that the plane going through $D,$ $M,$ and $N$ has equation $2x-y-4z=0.$ Let $Q = (1,1,.25)$ be the intersection of this plane and edge $BF$ and let $P = (1,2,0).$ Now since $2(1) - 1(2) - 4(0) = 0,$ $P$ is on the plane. Also, $P$ lies on the extensions of segments $DM,$ $NQ,$ and $CB$ so the solid $DPCN = DMBCQN + MPBQ$ is a right triangular pyramid. Note also that pyramid $MPBQ$ is similar to $DPCN$ with scale factor $.5$ and thus the volume of solid $DMBCQN,$ which is one of the solids bounded by the cube and the plane, is $[DPCN] - [MPBQ] = [DPCN] - \left(\frac{1}{2}\right)^3[DPCN] = \frac{7}{8}[DPCN].$ But the volume of $DPCN$ is simply the volume of a pyramid with base $1$ and height $.5$ which is $\frac{1}{3} \cdot 1 \cdot .5 = \frac{1}{6}.$ So $[DMBCQN] = \frac{7}{8} \cdot \frac{1}{6} = \frac{7}{48}.$ Note, however, that this volume is less than $.5$ and thus this solid is the smaller of the two solids. The desired volume is then $[ABCDEFGH] - [DMBCQN] = 1 - \frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089.}$

Solution: Define a coordinate system with $D = (0,0,0)$ $M = (1, \frac{1}{2}, 0)$ $N = (0,1,\frac{1}{2})$ . The plane formed by $D$ $M$ , and $N$ is $z = \frac{y}{2} - \frac{x}{4}$ . It intersects the base of the unit cube at $y = \frac{x}{2}$ . The z-coordinate of the plane never exceeds the height of the unit cube for $0 \leq x \leq 1, 0 \leq y \leq 1$ . Therefore, the volume of one of the two regions formed by the plane is \[\int_0^1 \int_{\frac{x}{2}}^1 \int_0^{\frac{y}{2}-\frac{x}{4}}dz\,dy\,dx = \frac{7}{48}\] Since $\frac{7}{48} < \frac{1}{2}$ , our answer is $1-\frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089}$

Solution: Same coordinate system, but we will use domains instead of using triple integrals. By the way, the method to obtain the equation of the plane is the cross product ( $\vec{[1,.5,0]}\times\vec{[0,1,.5]}=\vec{[.25,-.5,1]}$ ). We can multiply this vector by $-4$ to make things look cleaner and get $\vec{[-1,2,-4]}$ . We then get the desired plane, $-x+2y-4z=0$ , or $z=\frac{2y-x}{4}$ . We use a double integral with a Type I domain. Observing the diagram, the domain is where $0 \leq x \leq 1, .5x \leq y \leq 1$ . The integral is then \[.25\int_0^1\int_{.5x}^1 2y - x\] which becomes \[.25\int_0^1 (y^2-xy)|_{.5x}^1\] which becomes \[.25\int_0^1 1-x+.25x^2\] which then becomes \[.25(1-\frac{1}{2}+\frac{1}{12})\] and finally \[\frac{7}{48}\] So our answer is $1^3-\frac{7}{48} = \frac{41}{48}, 41+48 = \boxed{089}$

Solution: Let $Q$ be the intersection of the plane with edge $FB,$ then $\triangle MQB$ is similar to $\triangle DNC$ and the volume $[DNCMQB]$ is a sum of areas of cross sections of similar triangles running parallel to face $ABFE.$ Let $x$ be the distance from face $ABFE,$ let $h$ be the height parallel to $AB$ of the cross-sectional triangle at that distance, and $a$ be the area of the cross-sectional triangle at that distance. $A(x)=\frac{h^2}{4},$ and $h=\frac{x+1}{2},$ then $A=\frac{(x+1)^2}{16}$ , and the volume $[DNCMQB]$ is $\int^1_0{A(x)}\,\mathrm{d}x=\int^1_0{\frac{(x+1)^2}{16}}\,\mathrm{d}x=\frac{7}{48}.$ Thus the volume of the larger solid is $1-\frac{7}{48}=\frac{41}{48} \rightarrow p+q = \boxed{089}$

Solution:
The volume of a frustum is $\frac{h_2b_2 -h_1b_1}3$ where $b_i$ is the area of the base and $h_i$ is the height from the chopped off apex to the base.
We can easily see that from symmetry, the area of the smaller front base is $\frac{1}{16}$ and the area of the larger back base is $\frac{1}4$
Now to find the height of the apex.
Extend the $DM$ and (call the intersection of the plane with $FB$ G) $NG$ to meet at $x$ . Now from similar triangles $XMG$ and $XDN$ we can easily find the total height of the triangle $XDN$ to be $2$
Now straight from our formula, the volume is $\frac{7}{48}$ Thus the answer is:
$1-\text{Volume} \Longrightarrow \boxed{089}$

Solution:
We will solve for the area of the smaller region, and then subtract it from 1.
Let $X$ be the point where plane $DMN$ intersects $FB$ . Then $DMBCNX$ can be split into triangular pyramid $DMBX$ and quadrilateral pyramid $BCNXD$
Pyramid $DMBX$ has base $DMB$ with area $\frac{1}2 \cdot 1 \div 2 = \frac{1}4$ . The height is $BX = \frac{1}4$ , so the volume of $DMBX$ is $\frac{1}4 \cdot \frac{1}4 \div 3 = \frac{1}{48}$
Similarly, pyramid $BCNXD$ has base $BCNX$ with area $(\frac{1}4 + \frac{1}2) \cdot 1 = \frac{3}8$ . The height is $CD = 1$ , so the volume of $BCNXD$ is $\frac{3}8 \cdot 1 \div 3 = \frac{1}8$
Adding up the volumes of $DMBX$ and $BCNXD$ , we find that the volume of $DMBCNX$ is $\frac{1}{48} + \frac{6}{48} = \frac{7}{48}$ . Therefore the volume of the larger solid is $1 - \frac{7}{48} = \frac{41}{48} \rightarrow p+q = \boxed{089}$

Solution: We use a coordinate system with $C = (0,0,0)$ $D = (1,0,0)$ $M = (\frac{1}{2},1,0)$ $N = (0,0,\frac{1}{2})$ . Then the plane going through $D$ $M$ , and $N$ has equation $z = \frac{1}{2} - \frac{1}{2}x - \frac{1}{4}y$ . We set up a double integral in this coordinate system. Consider the region to be integrated over in the $xy$ -plane. From $x=0$ to $x=\frac{1}{2}$ , the upper bound of the region is $y = 1$ . From $x= \frac{1}{2}$ to $x=1$ , the upper bound of the region is $y = 2 - 2x$ . In both cases, the lower bound of the region is $y = 0$ . Thus, we have the double integral $\int_{0}^{\frac{1}{2}} \int_{0}^{1} \frac{1}{2} - \frac{1}{2}x - \frac{1}{4}y dydx + \int_{\frac{1}{2}}^{1} \int_{0}^{2-2x} \frac{1}{2} - \frac{1}{2}x - \frac{1}{4}y dydx$ . We find that this sum evaluates to $\frac{7}{48}$ . However, this is the volume of the smaller region, so the larger region's volume is that of the cube minus that of the smaller region. Since the cube has side length $1$ , its volume is $1$ , so the volume of the larger region is $1 - \frac{7}{48} = \frac{41}{48}$ . Thus, our answer is $41 + 48 = \boxed{089}$


Answer: 37.0

Danica drove her new car on a trip for a whole number of hours, averaging $55$ miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a $3$ -digit number with $a\ge1$ and $a+b+c\le7$ . At the end of the trip, the odometer showed $cba$ miles. What is $a^2+b^2+c^2$
$\textbf {(A) } 26 \qquad \textbf {(B) } 27 \qquad \textbf {(C) } 36 \qquad \textbf {(D) } 37 \qquad \textbf {(E) } 41$

Solution:
We can set up an algebraic equation for this problem.
From what's given, we have that $100c+10b+a=55x+100a+10b+c$
This simplifies to be $0=55x+99a-99c\implies -55x=99a-99c$
Factoring, we get that $-55x=99(a-c)\implies x=-\frac{9(a-c)}{5}$
Hence, notice that we want $a-c=-5$ so that $x=9$
The only pair that works for this problem that satisfies the original requirements is $(1,6)$
Hence, $a=1, b=0, c=6$
Checking, we have that $106+55(9)=601\implies 601=601$
Hence, the answer is $1^2+0^2+6^2=37\implies\boxed{37}$


Answer: 37.0

Danica drove her new car on a trip for a whole number of hours, averaging 55 miles per hour. At the beginning of the trip, $abc$ miles was displayed on the odometer, where $abc$ is a 3-digit number with $a \geq{1}$ and $a+b+c \leq{7}$ . At the end of the trip, the odometer showed $cba$ miles. What is $a^2+b^2+c^2?$
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37\qquad\textbf{(E)}\ 41$

Solution: We know that the number of miles she drove is divisible by $5$ , so $a$ and $c$ must either be the equal or differ by $5$ . We can quickly conclude that the former is impossible, so $a$ and $c$ must be $5$ apart.  Because we know that $c > a$ and $a + c \le 7$ and $a \ge 1$ , we find that the only possible values for $a$ and $c$ are $1$ and $6$ , respectively.  Because $a + b + c \le 7$ $b = 0$ .  Therefore, we have \[a^2 + b^2 + c^2 = 36 + 0 + 1 = \boxed{37}\]

Solution: Let the number of hours Danica drove be $k$ . Then we know that $100a + 10b + c + 55k$ $100c + 10b + a$ . Simplifying, we have $99c - 99a = 55k$ , or $9c - 9a = 5k$ . Thus, k is divisible by $9$ . Because $55 * 18 = 990$ $k$ must be $9$ , and therefore $c - a = 5$ . Because $a + b + c \leq{7}$ and $a \geq{1}$ $a = 1$ $c = 6$ and $b = 0$ , and our answer is $a^2 + b^2 + c^2 = 6^2 + 0^2 + 1^2 = 37$ , or $\boxed{37}$


Answer: 1.0

Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the greatest number of additional cars she must buy in order to be able to arrange all her cars this way?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 533$

Solution:
The least multiple of 6 greater than 23 is 24. So she will need to add $24-23=\boxed{1}$ more model car.
~avamarora

Solution: 6 x 4 = 24, which is 1 more than 23. So, the answer is $\boxed{1}$


Answer: 18.0

Daniel finds a rectangular index card and measures its diagonal to be $8$ centimeters.
Daniel then cuts out equal squares of side $1$ cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be $4\sqrt{2}$ centimeters, as shown below. What is the area of the original index card? [asy] // Diagram by MRENTHUSIASM, edited by Djmathman size(200); defaultpen(linewidth(0.6)); draw((489.5,-213) -- (225.5,-213) -- (225.5,-185) -- (199.5,-185) -- (198.5,-62) -- (457.5,-62) -- (457.5,-93) -- (489.5,-93) -- cycle); draw((206.29,-70.89) -- (480.21,-207.11), linetype ("6 6"),Arrows(size=4,arrowhead=HookHead)); draw((237.85,-182.24) -- (448.65,-95.76),linetype ("6 6"),Arrows(size=4,arrowhead=HookHead)); label("$1$",(450,-80)); label("$1$",(475,-106)); label("$8$",(300,-103)); label("$4\sqrt 2$",(300,-173)); [/asy] $\textbf{(A) } 14 \qquad \textbf{(B) } 10\sqrt{2} \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 12\sqrt{2} \qquad \textbf{(E) } 18$

Solution:
 Label the bottom left corner of the larger rectangle (without the square cut out) as $A$ and the top right as $D$ $w$ is the width of the rectangle and $\ell$ is the length. Now we have vertices $E, F, G, H$ as vertices of the irregular octagon created by cutting out the squares. Let $I, J$ be the two closest vertices formed by the squares.
The distance between the two closest vertices of the squares is thus $IJ=\left(4\sqrt{2}\right).$ Substituting, we get
\[(IJ)^2 = (w-2)^2 + (\ell-2)^2 = \left(4\sqrt{2}\right)^2 = 32 \implies w^2+\ell^2-4w-4\ell = 24.\] Using the fact that the diagonal of the rectangle is $8,$ we get \[w^2+\ell^2 = 64.\] Subtracting the first equation from the second equation, we get \[4w+4\ell=40 \implies w+\ell = 10.\] Squaring yields \[w^2 + 2w\ell + \ell^2 = 100.\] Subtracting the second equation from this, we get $2w\ell = 36,$ and thus area of the original rectangle is $w\ell = \boxed{18}.$


Answer: 54.0

Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire.  Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day.  All three friends visited Daphne yesterday.  How many days of the next $365$ -day period will exactly two friends visit her?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72$

Solution:
The $365$ -day time period can be split up into $6$ $60$ -day time periods, because after $60$ days, all three of them visit again (Least common multiple of $3$ $4$ , and $5$ ).
You can find how many times each pair of visitors can meet by finding the LCM of their visiting days and dividing that number by 60.
Remember to subtract $1$ , because you do not wish to count the $60$ th day, when all three visit.
A and B visit $\frac{60}{3 \cdot 4}-1 = 4$ times.
A and C visit $\frac{60}{3 \cdot 5}-1 = 3$ times.
B and C visit $\frac{60}{4 \cdot 5}-1 = 2$ times.
This is a total of $9$ visits per $60$ day period.
Therefore, the total number of $2$ -person visits is $9 \cdot 6 = \boxed{54}$

Solution:
From the information above, we get that $A=3x$ $B=4x$ $C=5x$
Now, we want the days in which exactly two of these people meet up
The three pairs are $(A,B)$ $(B,C)$ $(A,C)$
Notice that we are trying to find the LCM of each pair.
Hence, $LCM(A,B)=12x$ $LCM(B,C)=20x$ $LCM(A,C)=15x$
Notice that we want to eliminate when all these friends meet up. By doing this, we will find the LCM of the three letters.
Hence, $LCM(A,B,C)=60x$
Now, we add all of the days up(including overcount).
We get $30+18+24=72$ .
Now, because $60(6)=360$ , we have to subtract $6$ days from every pair.
Hence, our answer is $72-18=\boxed{54}$


Answer: 52.0

Dave arrives at an airport which has twelve gates arranged in a straight line with exactly $100$ feet between adjacent gates. His departure gate is assigned at random. After waiting at that gate, Dave is told the departure gate has been changed to a different gate, again at random. Let the probability that Dave walks $400$ feet or less to the new gate be a fraction $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
There are $12 \cdot 11 = 132$ possible situations ( $12$ choices for the initially assigned gate, and $11$ choices for which gate Dave's flight was changed to). We are to count the situations in which the two gates are at most $400$ feet apart.
If we number the gates $1$ through $12$ , then gates $1$ and $12$ have four other gates within $400$ feet, gates $2$ and $11$ have five, gates $3$ and $10$ have six, gates $4$ and $9$ have have seven, and gates $5$ $6$ $7$ $8$ have eight. Therefore, the number of valid gate assignments is \[2\cdot(4+5+6+7)+4\cdot8 = 2 \cdot 22 + 4 \cdot 8 = 76\] so the probability is $\frac{76}{132} = \frac{19}{33}$ . The answer is $19 + 33 = \boxed{052}$

Solution:
As before, derive that there are $132$ possibilities for Dave's original and replacement gates.
Now suppose that Dave has to walk $100k$ feet to get to his new gate.  This means that Dave's old and new gates must be $k$ gates apart.  (For example, a $100$ foot walk would consist of the two gates being adjacent to each other.)  There are $12-k$ ways to pick two gates which are $k$ gates apart, and $2$ possibilities for gate assignments, for a total of $2(12-k)$ possible assignments for each $k$
As a result, the total number of valid gate arrangements is \[2\cdot 11 + 2\cdot 10 + 2\cdot 9 + 2\cdot 8 = 76\] and so the requested probability is $\tfrac{19}{33}$ for a final answer of $\boxed{052}$


Answer: 41.0

Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $m$ and $n$ be relatively prime positive integers such that $\dfrac mn$ is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find $m+n$

Solution:
There are many almost equivalent approaches that lead to summing a geometric series. For example, we can compute the probability of the opposite event. Let $p$ be the probability that Dave will make at least two more throws than Linda. Obviously, $p$ is then also the probability that Linda will make at least two more throws than Dave, and our answer will therefore be $1-2p$
How to compute $p$
Suppose that Linda made exactly $t$ throws. The probability that this happens is $(5/6)^{t-1}\cdot (1/6)$ , as she must make $t-1$ unsuccessful throws followed by a successful one. In this case, we need Dave to make at least $t+2$ throws. This happens if his first $t+1$ throws are unsuccessful, hence the probability is $(5/6)^{t+1}$
Thus for a fixed $t$ the probability that Linda makes $t$ throws and Dave at least $t+2$ throws is $(5/6)^{2t} \cdot (1/6)$
Then, as the events for different $t$ are disjoint, $p$ is simply the sum of these probabilities over all $t$ . Hence:
\begin{align*} p & = \sum_{t=1}^\infty \left(\frac 56\right)^{2t} \cdot \frac 16 \\ & = \frac 16 \cdot \left(\frac 56\right)^2 \cdot \sum_{x=0}^\infty \left(\frac{25}{36}\right)^x \\ & = \frac 16 \cdot \frac{25}{36} \cdot \frac 1{1 - \dfrac{25}{36}} \\ & = \frac 16 \cdot \frac{25}{36} \cdot \frac{36}{11} \\ & = \frac {25}{66} \end{align*}
Hence the probability we were supposed to compute is $1 - 2p = 1 - 2\cdot \frac{25}{66} = 1 - \frac{25}{33} = \frac 8{33}$ , and the answer is $8+33 = \boxed{041}$

Solution:
Let $p$ be the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. (We will call this event "a win", and the opposite event will be "a loss".)
Let both players roll their first die.
With probability $\frac 1{36}$ , both throw a six and we win.
With probability $\frac{10}{36}$ exactly one of them throws a six. In this case, we win if the remaining player throws a six in their next throw, which happens with probability $\frac 16$
Finally, with probability $\frac{25}{36}$ none of them throws a six. Now comes the crucial observation: At this moment, we are in exactly the same situation as in the beginning. Hence in this case we will win with probability $p$
We just derived the following linear equation: \[p = \frac 1{36} + \frac{10}{36} \cdot \frac 16 + \frac{25}{36} \cdot p\]
Solving for $p$ , we get $p=\frac 8{33}$ , hence the answer is $8+33 = \boxed{041}$

Solution:
Let's write out the probabilities with a set number of throws that Linda rolls before getting a 6. The probability of Linda rolling once and gets 6 right away is $\frac{1}{6}$ . The probability that Dave will get a six in the same, one less, or one more throw is $\frac{1}{6} + \frac{5}{6} * \frac{5}{6}$ . Thus the combined probability is $\frac{11}{216}$
Let's do the same with the probability that Linda rolls twice and getting a six. This time it is $\frac{5}{6} * \frac{1}{6}$ . The probability that Dave meets the requirements set is $\frac{1}{6} + \frac{5}{6} * \frac{1}{6} + \frac{5}{6} * \frac{5}{6} * \frac{1}{6}$ . Combine the probabilities again to get $\frac{455}{7776}$ . (or not, because you can simplify without calculating later)
It's clear that as the number of rolls before getting a six increases, the probability that Dave meets the requirements is multiplied by $\frac{5}{6} * \frac{5}{6}$ . We can use this pattern to solve for the sum of an infinite geometric series.
First, set the case where Linda rolls only once aside. It doesn't fit the same pattern as the rest, so we'll add it separately at the end. Next, let $a = (\frac{5}{6} * \frac{1}{6}) * (\frac{1}{6} + \frac{5}{6} * \frac{1}{6} + \frac{5}{6} * \frac{5}{6} * \frac{1}{6}) = \frac{455}{7776}$ as written above. Each probability where the number of tosses Linda makes increases by one will be $a * (\frac{25}{36})^{n+1}$ . Let $S$ be the sum of all these probabilities.
$S = a + a * \frac{25}{36} + a * (\frac{25}{36})^2...$
$S * \frac{25}{36} = a * \frac{25}{36} + a * (\frac{25}{36})^2 + a * (\frac{25}{36})^3...$
Subtract the second equation from the first to get
$S * \frac{11}{36} = a$
$S = a * \frac{36}{11}$
$S = \frac{455}{2376}$
Don't forget to add the first case where Linda rolls once.
$\frac{455}{2376} + \frac{11}{216} = \frac{8}{33}$
$8 + 33 = \boxed{41}$


Answer: 59.0

David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, $A,\text{ }B,\text{ }C$ , which can each be inscribed in a circle with radius $1$ . Let $\varphi_A$ denote the measure of the acute angle made by the diagonals of quadrilateral $A$ , and define $\varphi_B$ and $\varphi_C$ similarly. Suppose that $\sin\varphi_A=\tfrac{2}{3}$ $\sin\varphi_B=\tfrac{3}{5}$ , and $\sin\varphi_C=\tfrac{6}{7}$ . All three quadrilaterals have the same area $K$ , which can be written in the form $\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
Suppose our four sides lengths cut out arc lengths of $2a$ $2b$ $2c$ , and $2d$ , where $a+b+c+d=180^\circ$ . Then, we only have to consider which arc is opposite $2a$ . These are our three cases, so \[\varphi_A=a+c\] \[\varphi_B=a+b\] \[\varphi_C=a+d\] Our first case involves quadrilateral $ABCD$ with $\overarc{AB}=2a$ $\overarc{BC}=2b$ $\overarc{CD}=2c$ , and $\overarc{DA}=2d$
Then, by Law of Sines, $AC=2\sin\left(\frac{\overarc{ABC}}{2}\right)=2\sin(a+b)$ and $BD=2\sin\left(\frac{\overarc{BCD}}{2}\right)=2\sin(a+d)$ . Therefore,
\[K=\frac{1}{2}\cdot AC\cdot BD\cdot \sin(\varphi_A)=2\sin\varphi_A\sin\varphi_B\sin\varphi_C=\frac{24}{35},\] so our answer is $24+35=\boxed{059}$

Solution:
Suppose the four side lengths of the quadrilateral cut out arc lengths of $2a$ $2b$ $2c$ , and $2d$ $a+b+c+d=180^\circ$ .
Therefore, without losing generality,
\[\varphi_A=a+b\] \[\varphi_B=b+c\] \[\varphi_C=a+c\]
$(1)+(3)-(2)$ $(1)+(2)-(3)$ , and $(2)+(3)-(1)$ yields
\[2a=\varphi_A+\varphi_C-\varphi_B\] \[2b=\varphi_A+\varphi_B-\varphi_C\] \[2c=\varphi_B+\varphi_C-\varphi_A\]
Because $2d=360^\circ-2a-2b-2c,$ Therefore,
\[2d=360^\circ-\varphi_A-\varphi_B-\varphi_C\]
Using the sum-to-product identities , our area of the quadrilateral $K$ then would be
\begin{align*}  K&=\frac{1}{2}(\sin(2a)+\sin(2b)+\sin(2c)+\sin(2d))\\ &=\frac{1}{2}(\sin(\varphi_A+\varphi_B-\varphi_C)+\sin(\varphi_B+\varphi_C-\varphi_A)+\sin(\varphi_C+\varphi_A-\varphi_B)-\sin(\varphi_A+\varphi_B+\varphi_C))\\ &=\frac{1}{2}(2\sin\varphi_B\cos(\varphi_A-\varphi_C)-2\sin\varphi_B\cos(\varphi_A+\varphi_C))\\ &=\frac{1}{2}\cdot2\cdot2\sin\varphi_A\sin\varphi_B\sin\varphi_C\\ &=2\sin\varphi_A\sin\varphi_B\sin\varphi_C\\ &=\frac{24}{35}\\ \end{align*}
Therefore, our answer is $24+35=\boxed{059}$

Solution:
Let the four stick lengths be $a$ $b$ $c$ , and $d$ . WLOG, let’s say that quadrilateral $A$ has sides $a$ and $d$ opposite each other, quadrilateral $B$ has sides $b$ and $d$ opposite each other, and quadrilateral $C$ has sides $c$ and $d$ opposite each other. The area of a convex quadrilateral can be written as $\frac{1}{2} d_1 d_2 \sin{\theta}$ , where $d_1$ and $d_2$ are the lengths of the diagonals of the quadrilateral and $\theta$ is the angle formed by the intersection of $d_1$ and $d_2$ . By Ptolemy's theorem $d_1 d_2 = ad+bc$ for quadrilateral $A$ , so, defining $K_A$ as the area of $A$ \[K_A = \frac{1}{2} (ad+bc)\sin{\varphi_A}\] Similarly, for quadrilaterals $B$ and $C$ \[K_B = \frac{1}{2} (bd+ac)\sin{\varphi_B}\] and \[K_C = \frac{1}{2} (cd+ab)\sin{\varphi_C}\] Multiplying the three equations and rearranging, we see that \[K_A K_B K_C = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\sin{\varphi_A}\sin{\varphi_B}\sin{\varphi_B}\] \[K^3 = \frac{1}{8} (ab+cd)(ac+bd)(ad+bc)\left(\frac{2}{3}\right) \left(\frac{3}{5}\right) \left(\frac{6}{7}\right)\] \[\frac{70}{3}K^3 = (ab+cd)(ac+bd)(ad+bc)\] The circumradius $R$ of a cyclic quadrilateral with side lengths $a$ $b$ $c$ , and $d$ and area $K$ can be computed as $R = \frac{\sqrt{(ab+cd)(ac+bd)(ad+bc)}}{4K}$ .
Inserting what we know, \[1 = \frac{\sqrt{\frac{70}{3}K^3}}{4K}\quad \Rightarrow \quad 16K^2 = \frac{70}{3}K^3\quad \Rightarrow \quad \frac{24}{35} = K\] So our answer is $24 + 35 = \boxed{059}$

Solution: Let the sides of the quadrilaterals be $a,b,c,$ and $d$ in some order such that $A$ has $a$ opposite of $c$ $B$ has $a$ opposite of $b$ , and $C$ has $a$ opposite of $d$ . Then, let the diagonals of $A$ be $e$ and $f$ . Similarly to solution $2$ , we get that $\tfrac{2}{3}(ac+bd)=\tfrac{3}{5}(ab+cd)=\tfrac{6}{7}(ad+bc)=2K$ , but this is also equal to $2\cdot\tfrac{eab+ecd}{4(1)}=2\cdot\tfrac{fad+fbc}{4(1)}$ using the area formula for a triangle using the circumradius and the sides, so $\tfrac{e(ab+cd)}{2}=\tfrac{3}{5}(ab+cd)$ and $\tfrac{f(ad+bc)}{2}=\tfrac{6}{7}(ad+bc)$ . Solving for $e$ and $f$ , we get that $e=\tfrac{6}{5}$ and $f=\tfrac{12}{7}$ , but $K=\tfrac{1}{2}\cdot\tfrac{2}{3}\cdot{}ef$ , similarly to solution $2$ , so $K=\tfrac{24}{35}$ and the answer is $24+35=\boxed{059}$


Answer: 5.0

Define $L_n$ as the least common multiple of all the integers from $1$ to $n$ inclusive. There is a unique integer $h$ such that \[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{17}=\frac{h}{L_{17}}\] What is the remainder when $h$ is divided by $17$
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$

Solution:
Notice that $L_{17}$ contains the highest power of every prime below $17$ since higher primes cannot divide $L_{17}$ . Thus, $L_{17}=16\cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17$
When writing the sum under a common fraction, we multiply the denominators by $L_{17}$ divided by each denominator. However, since $L_{17}$ is a multiple of $17$ , all terms will be a multiple of $17$ until we divide out $17$ , and the only term that will do this is $\frac{1}{17}$ . Thus, the remainder of all other terms when divided by $17$ will be $0$ , so the problem is essentially asking us what the remainder of $\frac{L_{17}}{17} = L_{16}$ divided by $17$ is. This is equivalent to finding the remainder of $16 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ divided by $17$
We use modular arithmetic to simplify our answer:
This is congruent to $-1 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \pmod{17}$
Evaluating, we get: \begin{align*} (-1) \cdot 9 \cdot 35 \cdot 11 \cdot 13 &\equiv (-1) \cdot 9 \cdot 1 \cdot 11 \cdot 13 \pmod{17} \\ &\equiv 9 \cdot 11 \cdot (-13) \pmod{17} \\ &\equiv 9 \cdot 11 \cdot 4\pmod{17} \\ &\equiv 2 \cdot 11 \pmod{17} \\ &\equiv 5\pmod{17} \end{align*} Therefore the remainder is $\boxed{5}$

Solution:
As in solution 1, we express the LHS as a sum under one common denominator. We note that \[\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{17} = \frac{\frac{17!}{1}}{17!} + \frac{\frac{17!}{2}}{17!} + \frac{\frac{17!}{3}}{17!} + \dots + \frac{\frac{17!}{17}}{17!}\]
Now, we have $h = L_{17}\left(\frac{\frac{17!}{1} + \frac{17!}{2} + \frac{17!}{3} + \dots + \frac{17!}{17}}{17!}\right)$ . We'd like to find $h \pmod{17},$ so we can evaluate our expression $\pmod{17}.$ Since $\frac{\frac{17!}{1}}{17!}, \frac{\frac{17!}{2}}{17!}, \dots, \frac{\frac{17!}{16}}{17!}$ don't have a factor of $17$ in their denominators, and since $L_{17}$ is a multiple of $17,$ multiplying each of those terms and adding them will get a multiple of $17.$ $\pmod{17}$ , that result is $0.$ Thus, we only need to consider $L_{17}\cdot \frac{\frac{17!}{17}}{17!} = \frac{L_{17}}{17} \pmod{17}.$ Proceed with solution $1$ to get $\boxed{5}$

Solution:
Using Wolstenholmes' Theorem, we can rewrite $1 + \frac{1}{2} \dots + \frac{1}{16}$ as $\frac{17^2 n}{(17 - 1)!} = \frac{17^2 n}{16!}$ (for some $n \in \mathbb{Z}$ ). Adding the $\frac{1}{17}$ to $\frac{17^2 n}{16!}$ , we get $\frac{17^3 n + 16!}{17!}$
Now we have $\frac{17^3 n + 16!}{17!} = \frac{h}{L_{17}}$ and we want $h \pmod{17}$ . We find that $\frac{L_{17}(17^3 n + 16!)}{17!} = \frac{L_{16}(17^3 n + 16!)}{16!} = h$ . Taking $\pmod{17}$ and multiplying, we get $L_{16}(17^3 n + 16!) \equiv 16! \cdot h \pmod{17}$
Applying Wilson's Theorem on $16!$ and reducing, we simplify the congruence to $L_{16}(0 - 1) \equiv -L_{16} \equiv -h \pmod{17}$ . Now we proceed with Solution 1 and find that $L_{16} \equiv 5 \pmod{17}$ , so our answer is $\boxed{5}$


Answer: 385.0

Define $f(x)=|| x|-\tfrac{1}{2}|$ and $g(x)=|| x|-\tfrac{1}{4}|$ . Find the number of intersections of the graphs of \[y=4 g(f(\sin (2 \pi x))) \quad\text{ and }\quad x=4 g(f(\cos (3 \pi y))).\]

Solution: If we graph $4g(f(x))$ , we see it forms a sawtooth graph that oscillates between $0$ and $1$ (for values of $x$ between $-1$ and $1$ , which is true because the arguments are between $-1$ and $1$ ). Thus by precariously drawing the graph of the two functions in the square bounded by $(0,0)$ $(0,1)$ $(1,1)$ , and $(1,0)$ , and hand-counting each of the intersections, we get $\boxed{385}$

Solution:
We will denote $h(x)=4g(f(x))$ for simplicity. Denote $p(x)$ as the first equation and $q(y)$ as the graph of the second. We notice that both $f(x)$ and $g(x)$ oscillate between 0 and 1. The intersections are thus all in the square $(0,0)$ $(0,1)$ $(1,1)$ , and $(1,0)$ . Every $p(x)$ wave going up and down crosses every $q(y)$ wave. Now, we need to find the number of times each wave touches 0 and 1.
We notice that $h(x)=0$ occurs at $x=-\frac{3}{4}, -\frac{1}{4}, \frac{1}{4}, \frac{3}{4}$ , and $h(x)=1$ occurs at $x=-1, -\frac{1}{2}, 0,\frac{1}{2},1$ . A sinusoid passes through each point twice during each period, but it only passes through the extrema once. $p(x)$ has 1 period between 0 and 1, giving 8 solutions for $p(x)=0$ and 9 solutions for $p(x)=1$ , or 16 up and down waves. $q(y)$ has 1.5 periods, giving 12 solutions for $q(y)=0$ and 13 solutions for $q(y)=1$ , or 24 up and down waves. This amounts to $16\cdot24=384$ intersections.
However, we have to be very careful when counting around $(1, 1)$ . At this point, $q(y)$ has an infinite downwards slope and $p(x)$ is slanted, giving us an extra intersection; thus, we need to add 1 to our answer to get $\boxed{385}$

Solution:
We can easily see that only $x, y \in \left[0,1 \right]$ may satisfy both functions.
We call function $y = 4g \left( f \left( \sin \left( 2 \pi x \right) \right) \right)$ as Function 1 and function $x = 4g \left( f \left( \cos \left( 3 \pi y \right) \right) \right)$ as Function 2.
For Function 1, in each interval $\left[ \frac{i}{4} , \frac{i+1}{4} \right]$ with $i \in \left\{ 0, 1, \cdots, 3 \right\}$ , Function 1's value oscillates between 0 and 1. It attains 1 at $x = \frac{i}{4}$ $\frac{i+1}{4}$ and another point between these two.
Between two consecutive points whose functional values are 1, the function first decreases from 1 to 0 and then increases from 0 to 1.
So the graph of this function in this interval consists of 4 monotonic pieces.
For Function 2, in each interval $\left[ \frac{i}{6} , \frac{i+1}{6} \right]$ with $i \in \left\{ 0, 1, \cdots, 5 \right\}$ , Function 2's value oscillates between 0 and 1. It attains 1 at $y = \frac{i}{6}$ $\frac{i+1}{6}$ and another point between these two.
Between two consecutive points whose functional values are 1, the function first decreases from 1 to 0 and then increases from 0 to 1.
So the graph of this function in this interval consists of 4 monotonic curves.
Consider any region $\left[ \frac{i}{4} , \frac{i+1}{4} \right] \times \left[ \frac{j}{6} , \frac{j+1}{6} \right]$ with $i \in \left\{ 0, 1, \cdots, 3 \right\}$ and $j \in \left\{0, 1, \cdots , 5 \right\}$ but $\left( i, j \right) \neq \left( 3, 5 \right)$ .
Both functions have four monotonic pieces.
Because Function 1's each monotonic piece can take any value in $\left[ \frac{j}{6} , \frac{j+1}{6} \right]$ and Function 2' each monotonic piece can take any value in $\left[ \frac{i}{4} , \frac{i+1}{4} \right]$ , Function 1's each monotonic piece intersects with Function 2's each monotonic piece.
Therefore, in the interval $\left[ \frac{i}{4} , \frac{i+1}{4} \right] \times \left[ \frac{j}{6} , \frac{j+1}{6} \right]$ , the number of intersecting points is $4 \cdot 4 = 16$
Next, we prove that if an intersecting point is on a line $x = \frac{i}{4}$ for $i \in \left\{ 0, 1, \cdots, 4 \right\}$ , then this point must be $\left( 1, 1 \right)$
For $x = \frac{i}{4}$ , Function 1 attains value 1.
For Function 2, if $y = 1$ , then $x = 1$ .
Therefore, the intersecting point is $\left( 1, 1 \right)$
Similarly, we can prove that if an intersecting point is on a line $y = \frac{i}{6}$ for $i \in \left\{ 0, 1, \cdots, 6 \right\}$ , then this point must be $\left( 1, 1 \right)$
Therefore, in each region $\left[ \frac{i}{4} , \frac{i+1}{4} \right] \times \left[ \frac{j}{6} , \frac{j+1}{6} \right]$ with $i \in \left\{ 0, 1, \cdots, 3 \right\}$ and $j \in \left\{0, 1, \cdots , 5 \right\}$ but $\left( i, j \right) \neq \left( 3, 5 \right)$ , all 16 intersecting points are interior.
That is, no two regions share any common intersecting point.
Next, we study region $\left[ \frac{3}{4} , 1 \right] \times \left[ \frac{5}{6} , 1 \right]$ .
Consider any pair of monotonic pieces, where one is from Function 1 and one is from Function 2, except the pair of two monotonic pieces from two functions that attain $\left( 1 , 1 \right)$ .
Two pieces in each pair intersects at an interior point on the region.
So the number of intersecting points is $4 \cdot 4 - 1 = 15$
Finally, we compute the number intersection points of two functions' monotonic pieces that both attain $\left( 1, 1 \right)$
One trivial intersection point is $\left( 1, 1 \right)$ .
Now, we study whether they intersect at another point.
Define $x = 1 - x'$ and $y = 1 - y'$ .
Thus, for positive and sufficiently small $x'$ and $y'$ , Function 1 is reduced to \[ y' = 4 \sin 2 \pi x' \hspace{1cm} (1) \] and Function 2 is reduced to \[ x' = 4 \left( 1 - \cos 3 \pi y' \right) . \hspace{1cm} (2) \]
Now, we study whether there is a non-zero solution.
Because we consider sufficiently small $x'$ and $y'$ , to get an intuition and quick estimate, we do approximations of the above equations.
Equation (1) is approximated as \[ y' = 4 \cdot 2 \pi x' \] and Equation (2) is approximated as \[ x' = 2 \left( 3 \pi y' \right)^2  \]
To solve these equations, we get $x' = \frac{1}{8^2 \cdot 18 \pi^4}$ and $y' = \frac{1}{8 \cdot 18 \pi^3}$ .
Therefore, two functions' two monotonic pieces that attain $\left( 1, 1 \right)$ have two intersecting points.
Putting all analysis above, the total number of intersecting points is $16 \cdot 4 \cdot 6 + 1 = \boxed{385}$


Answer: 401.0

Define $n!!$ to be $n(n-2)(n-4)\cdots 3\cdot 1$ for $n$ odd and $n(n-2)(n-4)\cdots 4\cdot 2$ for $n$ even. When $\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}$ is expressed as a fraction in lowest terms, its denominator is $2^ab$ with $b$ odd. Find $\dfrac{ab}{10}$

Solution:
First, note that $(2n)!! = 2^n \cdot n!$ , and that $(2n)!! \cdot (2n-1)!! = (2n)!$
We can now take the fraction $\dfrac{(2i-1)!!}{(2i)!!}$ and multiply both the numerator and the denominator by $(2i)!!$ . We get that this fraction is equal to $\dfrac{(2i)!}{(2i)!!^2} = \dfrac{(2i)!}{2^{2i}(i!)^2}$
Now we can recognize that $\dfrac{(2i)!}{(i!)^2}$ is simply ${2i \choose i}$ , hence this fraction is $\dfrac{{2i\choose i}}{2^{2i}}$ , and our sum turns into $S=\sum_{i=1}^{2009} \dfrac{{2i\choose i}}{2^{2i}}$
Let $c = \sum_{i=1}^{2009} {2i\choose i} \cdot 2^{2\cdot 2009 - 2i}$ .
Obviously $c$ is an integer, and $S$ can be written as $\dfrac{c}{2^{2\cdot 2009}}$ .
Hence if $S$ is expressed as a fraction in lowest terms, its denominator will be of the form $2^a$ for some $a\leq 2\cdot 2009$
In other words, we just showed that $b=1$ .
To determine $a$ , we need to determine the largest power of $2$ that divides $c$
Let $p(i)$ be the largest $x$ such that $2^x$ that divides $i$
We can now return to the observation that $(2i)! = (2i)!! \cdot (2i-1)!! = 2^i \cdot i! \cdot (2i-1)!!$ . Together with the obvious fact that $(2i-1)!!$ is odd, we get that $p((2i)!)=p(i!)+i$
It immediately follows that $p\left( {2i\choose i} \right) = p((2i)!) - 2p(i!) = i - p(i!)$ ,
and hence $p\left( {2i\choose i} \cdot 2^{2\cdot 2009 - 2i} \right) = 2\cdot 2009 - i - p(i!)$
Obviously, for $i\in\{1,2,\dots,2009\}$ the function $f(i)=2\cdot 2009 - i - p(i!)$ is is a strictly decreasing function.
Therefore $p(c) = p\left( {2\cdot 2009\choose 2009} \right) = 2009 - p(2009!)$
We can now compute $p(2009!) = \sum_{k=1}^{\infty} \left\lfloor \dfrac{2009}{2^k} \right\rfloor = 1004 + 502 + \cdots + 3 + 1 = 2001$ .
Hence $p(c)=2009-2001=8$
And thus we have $a=2\cdot 2009 - p(c) = 4010$ , and the answer is $\dfrac{ab}{10} = \dfrac{4010\cdot 1}{10} = \boxed{401}$

Solution: Using the steps of the previous solution we get $c = \sum_{i=1}^{2009} {2i\choose i} \cdot 2^{2\cdot 2009 - 2i}$ and if you do the small cases(like $1, 2, 3, 4, 5, 6$ ) you realize that you can "thin-slice" the problem and simply look at the cases where $i=2009, 2008$ (they're nearly identical in nature but one has $4$ with it) since $\dbinom{2i}{I}$ hardly contains any powers of $2$ or in other words it's very inefficient and the inefficient cases hold all the power so you can just look at the highest powers of $2$ in $\dbinom{4018}{2009}$ and $\dbinom{4016}{2008}$ and you get the minimum power of $2$ in either expression is $8$ so the answer is $\frac{4010}{10} \implies \boxed{401}$ since it would violate the rules of the AIME and the small cases if $b>1$

Solution: We can logically deduce that the $b$ value will be 1. Listing out the first few values of odd and even integers, we have: $1, 3, 5, 7, 9, 11, 13...$ and $2, 4, 6, 8, 10, 12, 14, 16...$ . Obviously, none of the factors of $2$ in the denominator will cancel out, since the numerator is odd. Starting on the second term of the numerator, a factor of $3$ occurs every $3$ terms, and starting out on the third term of the denominator, a factor of $3$ appears also every $3$ terms. Thus, the factors of $3$ on the denominator will always cancel out. We can apply the same logic for every other odd factor, so once terms all cancel out, the denominator of the final expression will be in the form $1 \cdot 2^a$ . Since there will be no odd factors in the denominators, all the denominators will be in the form $2^a$ where $a$ is the number of factors of $2$ in $(2009 \cdot 2)! = 4018!$ . This is simply $\sum_{n=1}^{11} \left \lfloor{\frac{4018}{2^n}}\right \rfloor = 4010$ . Therefore, our answer is $\boxed{401}$

Solution:
Using the initial steps from Solution 1, $S=\sum_{i=1}^{2009} \dfrac{{2i\choose i}}{2^{2i}}$ . Clearly $b = 1$ as in all the summands there are no non-power of 2 factors in the denominator. So we seek to find $a$ . Note that $2^a$ would be the largest denominator in all the summands, so when they are summed it is the common denominator.
Taking the p-adic valuations of each term, the powers of 2 in the denominator for $i$ is $(2i) - v_2(\binom{2i}{i}).$ We can use Kummers theorem to see that $v_2(\binom{2i}{i})$ is the number of digits carried over when $i$ is added to $i$ in base $2$ . This is simply the number of $1$ 's in the binary representation of $i$
Looking at the binary representations of some of the larger $i$ we see $2009 = 11111011001_2$ having eight $1$ 's. So the power of two is $2 \cdot 2009 - 8 = 4010$ . Experimenting with $2008, 2007, 2006$ we see that the power of two are all $< 4010$ , and under $2005$ the power of two $2i -  v_2(\binom{2i}{i}) < 2i < 4010$ . Therefore $a = 4010$ and $\frac{ab}{10} = \boxed{401}.$


Answer: 0.0

Define $x \heartsuit y$ to be $|x-y|$ for all real numbers $x$ and $y$ . Which of the following statements is not true?
$\mathrm{(A) \ } x \heartsuit y = y \heartsuit x$ for all $x$ and $y$
$\mathrm{(B) \ } 2(x \heartsuit y) = (2x) \heartsuit (2y)$ for all $x$ and $y$
$\mathrm{(C) \ } x \heartsuit 0 = x$ for all $x$
$\mathrm{(D) \ } x \heartsuit x = 0$ for all $x$
$\mathrm{(E) \ } x \heartsuit y > 0$ if $x \neq y$

Solution:
We start by looking at the answers. Examining statement C, we notice:
$x \heartsuit 0 = |x-0| = |x|$
$|x| \neq x$ when $x<0$ , but statement C says that it does for all $x$
Therefore the statement that is not true is $\boxed{0}$


Answer: 0.0

Solution:
We start by looking at the answers. Examining statement C, we notice:
$x \heartsuit 0 = |x-0| = |x|$
$|x| \neq x$ when $x<0$ , but statement C says that it does for all $x$
Therefore the statement that is not true is $\boxed{0}$


Answer: 2.0

Define $x\diamond y$ to be $|x-y|$ for all real numbers $x$ and $y.$ What is the value of \[(1\diamond(2\diamond3))-((1\diamond2)\diamond3)?\]
$\textbf{(A)}\ {-}2 \qquad \textbf{(B)}\ {-}1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 2$

Solution: We have \begin{align*} (1\diamond(2\diamond3))-((1\diamond2)\diamond3) &= |1-|2-3|| - ||1-2|-3| \\ &= |1-1| - |1-3| \\ &= 0-2 \\ &= \boxed{2} ~MRENTHUSIASM

Solution: Observe that the $\diamond$ function is simply the positive difference between two numbers. Thus, we evaluate: the difference between $2$ and $3$ is $1;$ the difference between $1$ and $1$ is $0;$ the difference between $1$ and $2$ is $1;$ the difference between $1$ and $3$ is $2;$ and finally, $0-2=\boxed{2}.$


Answer: 2.0

Solution: We have \begin{align*} (1\diamond(2\diamond3))-((1\diamond2)\diamond3) &= |1-|2-3|| - ||1-2|-3| \\ &= |1-1| - |1-3| \\ &= 0-2 \\ &= \boxed{2} ~MRENTHUSIASM

Solution: Observe that the $\diamond$ function is simply the positive difference between two numbers. Thus, we evaluate: the difference between $2$ and $3$ is $1;$ the difference between $1$ and $1$ is $0;$ the difference between $1$ and $2$ is $1;$ the difference between $1$ and $3$ is $2;$ and finally, $0-2=\boxed{2}.$


Answer: 5100.0

Define \[P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).\] How many integers $n$ are there such that $P(n)\leq 0$
$\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100$

Solution:
We perform casework on $P(n)\leq0:$
Together, the answer is $100+5000=\boxed{5100}.$

Solution: Notice that $P(x)$ is nonpositive when $x$ is between $100^2$ and $99^2, 98^2$ and $97^2, \cdots$ $2^2$ and $1^2$ (inclusive), because there are an odd number of negatives, which means that the number of values equals \[((100+99)(100-99) + 1) + ((98+97)(98-97)+1) + \cdots + ((2+1)(2-1)+1).\] This reduces to \[200 + 196 + 192 + \cdots + 4 = 4(1+2+\cdots + 50) = 4 \cdot\frac{50 \cdot 51}{2} = \boxed{5100}.\] ~Zeric

Solution:
We know that $P(x)$ is a $100$ -degree function with a positive leading coefficient. That is, $P(x)=x^{100}+ax^{99}+bx^{98}+...+\text{(constant)}$
Since the degree of $P(x)$ is even, its end behaviors match. And since the leading coefficient is positive, we know that both ends approach $\infty$ as $x$ goes in either direction, from which \[\lim_{x\to-\infty} P(x)=\lim_{x\to\infty} P(x)=\infty.\] So the first time $P(x)$ is going to be negative is when it intersects the $x$ -axis at an $x$ -intercept and it's going to dip below. This happens at $1^2$ , which is the smallest intercept.
However, when it hits the next intercept, it's going to go back up again into positive territory, we know this happens at $2^2$ . And when it hits $3^2$ , it's going to dip back into negative territory. Clearly, this is going to continue to snake around the intercepts until $100^2$
To get the amount of integers below and/or on the $x$ -axis, we simply need to count the integers. For example, the amount of integers in between the $[1^2,2^2]$ interval we got earlier, we subtract and add one. $(2^2-1^2+1)=4$ integers, so there are four integers in this interval that produce a negative result.
Doing this with all of the other intervals, we have \[(2^2-1^2+1)+(4^2-3^2+1)+\cdots+(100^2-99^2+1)=\boxed{5100}\] from Solution 2's result.

Solution: We know $P(x) \leq 0$ when an odd number of its factors are positive and negative. For example, to make the first factor positive, $x \in [1^2, 2^2]$ . then there will be a even number of positive factors. We would do $2^2 - 1^2 + 1 (inclusive)$ to find all integers that work. In short we can generalize too: \begin{align*} x^2 - (x-1)^2 + 1 &= 2x \\ x^2 - (x^2 - 2x + 1) + 1 &= 2x \\ x^2 - x^2 + 2x - 1 + 1 &= 2x. \\ \end{align*} But remember this only works when $x \in \{2, 4, 6, 8 \cdots 98, 100\}$ because only then will there be a odd amount of positive and negative factors. So we can set $x = 2k$ , for $k \in \{1, 2, 3, 4, \cdots 49, 50\}$ Now we only have to solve: \[\sum_{k=1}^{k=50}2(2k) = 2\sum_{k = 1}^{k = 50}2k = 4\sum_{k = 1}^{k = 50}k = 4 \cdot \dfrac{(50)(51)}{2} = 2 \cdot (50)(51) = \boxed{5100}.\] ~Wiselion