## AMC 12 2020 Test B

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**Instructions**

- This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
- You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
- No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator).
- Figures are not necessarily drawn to scale.
- You will have
**75 minutes**working time to complete the test.

What is the value in simplest form of the following expression?\[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}\]

$\textbf{(A) }5 \qquad \textbf{(B) }4 + \sqrt{7} + \sqrt{10} \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 4 + 3\sqrt{3} + 2\sqrt{5} + \sqrt{7}$

$\textbf{C}$

\[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7} = \sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16}\ = 1 + 2 + 3 + 4 =10\]

What is the value of the following expression?\[\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}\]

$\textbf{(A) } 1 \qquad \textbf{(B) } \dfrac{9951}{9950} \qquad \textbf{(C) } \dfrac{4780}{4779} \qquad \textbf{(D) } \dfrac{108}{107} \qquad \textbf{(E) } \dfrac{81}{80}$

$\textbf{A}$

\[\frac{100^2-7^2}{70^2-11^2} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)} = \frac{(100-7)(100+7)}{(70-11)(70+11)} \cdot \frac{(70-11)(70+11)}{(100-7)(100+7)}=1\]

The ratio of $w$ to $x$ is $4 : 3$, the ratio of $y$ to $z$ is $3 : 2$, and the ratio of $z$ to $x$ is $1 : 6$. What is the ratio of $w$ to $y$?

$\textbf{(A) }4:3 \qquad \textbf{(B) }3:2 \qquad \textbf{(C) } 8:3 \qquad \textbf{(D) } 4:1 \qquad \textbf{(E) } 16:3$

$\textbf{E}$

We have\[\frac wy = \frac wx \cdot \frac xz \cdot \frac zy = \frac43\cdot\frac61\cdot\frac23=\frac{16}{3}\]from which $w:y=16:3.$

The acute angles of a right triangle are $a^{\circ}$ and $b^{\circ}$, where $a>b$ and both $a$ and $b$ are prime numbers. What is the least possible value of $b$?

$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }7\qquad\textbf{(E) }11$

$\textbf{D}$

In a right triangle we have $a+b=90$.

The greatest prime number less than $90$ is $89$. If $a=89$, then $b=1$, which is not prime.

The next greatest prime number less than $90$ is $83$. If $a=83$, then $b=7$, which is prime, so the answer is 7.

Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\dfrac{2}{3}$ of its games and team $B$ has won $\dfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?

$\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$

$\textbf{C}$

If we consider the number of games team $B$ has played as $x$ and the number of games that team $A$ has played as $y$, then we can set up the following system of equations:\begin{align*} \frac{5}{8}x &= \frac{2}{3}y+7 \\ \frac{3}{8}x &= \frac{1}{3}y+7 \end{align*}The first system equated the number of wins of each team, while the second system equates the number of losses by each team. By multiplying the second equation by $2$ and solving the system, we get $y = 42$. So the answer is C.

For all integers $n \geq 9,$ the value of\[\frac{(n+2)!-(n+1)!}{n!}\]is always which of the following?

$\textbf{(A) } \textrm{a multiple of 4} \qquad\newline \textbf{(B) } \textrm{a multiple of 10} \qquad\newline \textbf{(C) } \textrm{a prime number} \qquad\newline \textbf{(D) } \textrm{a perfect square} \qquad\newline \textbf{(E) } \textrm{a perfect cube}$

$\textbf{D}$

\begin{align*}

\frac{(n+2)!-(n+1)!}{n!} &= \frac{(n+2)(n+1)n!-(n+1)n!}{n!}\\

&=(n+2)(n+1)-(n+1)\\

&=(n+1)^2

\end{align*} So the value must be a perfect square.

Two nonhorizontal, non vertical lines in the $xy$-coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?

$\textbf{(A)}\ \dfrac16 \qquad\textbf{(B)}\ \dfrac23 \qquad\textbf{(C)}\ \dfrac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6$

$\textbf{C}$

Let the slope of the lower line be $\tan\theta$. So the slope of the upper line is $$\tan\left(\theta+45^\circ\right)=\dfrac{\tan\theta+\tan45^\circ}{1-\tan\theta\tan45^\circ}=\dfrac{\tan\theta+1}{1-\tan\theta}$$ which is 6 times the slope of the lower line $$\dfrac{\tan\theta+1}{1-\tan\theta}=6\tan\theta$$ Hence, we get $\tan\theta=\dfrac12\ \text{or}\ \dfrac13$. To get a larger product, we take $\tan\theta=\dfrac12$. So the slope of the upper line is $6\cdot\dfrac12=3$. The answer is $3\cdot\dfrac12=\dfrac32$.

How many ordered pairs of integers $(x, y)$ satisfy the equation\[x^{2020}+y^2=2y?\]

$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \textrm{infinitely many}$

$\textbf{D}$

Rearranging the terms and and completing the square for $y$ yields the result $x^{2020}+(y-1)^2=1$. Hence, the value of $x$ could be 0, 1, or -1. Therefore, plugging in the above values for $x$ gives the ordered pairs $(0,0)$, $(1,1)$, $(-1,1)$, and $(0,2)$. The answer is 4.

A three-quarter sector of a circle of radius $4$ inches together with its interior can be rolled up to form the lateral surface of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches?

$\textbf{(A)}\ 3\pi \sqrt5 \qquad\textbf{(B)}\ 4\pi \sqrt3 \qquad\textbf{(C)}\ 3 \pi \sqrt7 \qquad\textbf{(D)}\ 6\pi \sqrt3 \qquad\textbf{(E)}\ 6\pi \sqrt7$

$\textbf{C}$

Notice that when the cone is created, the 2 shown radii when merged will become the slant height of the cone and the intact circumference of the circle will become the circumference of the base of the cone.

We can calculate that the intact circumference of the circle is $8\pi\cdot\dfrac{3}{4}=6\pi$. Since that is also equal to the circumference of the cone, the radius of the cone is $3$. We also have that the slant height of the cone is $4$. Therefore, we use the Pythagorean Theorem to calculate that the height of the cone is $\sqrt{4^2-3^2}=\sqrt7$. The volume of the cone is $\dfrac{1}{3}\cdot\pi\cdot3^2\cdot\sqrt7=3 \pi \sqrt7$.

In unit square $ABCD,$ the inscribed circle $\omega$ intersects $\overline{CD}$ at $M,$ and $\overline{AM}$ intersects $\omega$ at a point $P$ different from $M.$ What is $AP?$

$\textbf{(A) } \dfrac{\sqrt5}{12} \qquad \textbf{(B) } \dfrac{\sqrt5}{10} \qquad \textbf{(C) } \dfrac{\sqrt5}{9} \qquad \textbf{(D) } \dfrac{\sqrt5}{8} \qquad \textbf{(E) } \dfrac{2\sqrt5}{15}$

$\textbf{B}$

Call the midpoint of $\overline{AB}$ point $N.$ Draw in $\overline{NM}$ and $\overline{NP}.$ Note that $\angle{NPM}=90^{\circ}$ due to Thales's Theorem.

Using the Pythagorean theorem, $AM=\dfrac{\sqrt{5}}{2}.$ Now we just need to find $AP$ using similar triangles $\triangle APN\sim\triangle ANM:$\begin{align*} \frac{AP}{AN}&=\frac{AN}{AM} \\ \frac{AP}{1/2}&=\frac{1/2}{\sqrt5/2} \\ AP&=\frac{\sqrt5}{10} \end{align*}

As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length $2$ so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region—inside the hexagon but outside all of the semicircles?

$\textbf {(A) } 6\sqrt{3}-3\pi \qquad \textbf {(B) } \dfrac{9\sqrt{3}}{2} - 2\pi\ \qquad \textbf {(C) } \dfrac{3\sqrt{3}}{2} - \dfrac{\pi}{3} \qquad \textbf {(D) } 3\sqrt{3} - \pi \qquad \textbf {(E) } \dfrac{9\sqrt{3}}{2} - \pi$

$\textbf{D}$

First, subdivide the hexagon into 24 equilateral triangles with side length 1:

Now note that the entire shaded region is just 6 times this part:

The entire rhombus is just 2 equilatrial triangles with side lengths of 1, so it has an area of:\[2\cdot\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}\]The arc that is not included has an area of:\[\frac16 \cdot\pi \cdot1^2 = \frac{\pi}{6}\]Hence, the area of the shaded region in that section is\[\frac{\sqrt{3}}{2}-\frac{\pi}{6}\]For a final area of:\[6\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)=3\sqrt{3}-\pi\]

Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$

$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$

$\textbf{E}$

Let $O$ be the center of the circle, and $X$ be the midpoint of $\overline{CD}$. Let $CX=a$ and $EX=b$. This implies that $DE = a - b$. Since $CE = CX + EX = a + b$, we now want to find $(a+b)^2+(a-b)^2=2(a^2+b^2)$. Since $\angle CXO$ is a right angle, by Pythagorean theorem $a^2 + b^2 = CX^2 + OX^2 = (5\sqrt{2})^2=50$. Thus, our answer is $2\times50=100$.

Which of the following is the value of $\sqrt{\log_2{6}+\log_3{6}}?$

$\textbf{(A) } 1 \qquad\textbf{(B) } \sqrt{\log_5{6}} \qquad\textbf{(C) } 2 \qquad\textbf{(D) } \sqrt{\log_2{3}}+\sqrt{\log_3{2}} \qquad\textbf{(E) } \sqrt{\log_2{6}}+\sqrt{\log_3{6}}$

$\textbf{D}$

Recall that:$$\log_b{(uv)}=\log_b u + \log_b v$$ $$\log_b u\cdot\log_u b=1$$ We use these properties of logarithms to rewrite the original expression:\begin{align*} \sqrt{\log_2{6}+\log_3{6}}&=\sqrt{(\log_2{2}+\log_2{3})+(\log_3{2}+\log_3{3})} \\ &=\sqrt{2+\log_2{3}+\log_3{2}} \\ &=\sqrt{\left(\sqrt{\log_2{3}}+\sqrt{\log_3{2}}\right)^2} \\ &= \sqrt{\log_2{3}}+\sqrt{\log_3{2}} \end{align*}

Bela and Jenn play the following game on the closed interval $[0, n]$ of the real number line, where $n$ is a fixed integer greater than $4$. They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval $[0, n]$. Thereafter, the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?

$\textbf{(A)} \textrm{ Bela will always win.}\newline$

$\textbf{(B)} \textrm{ Jenn will always win.}\newline$

$\textbf{(C)} \textrm{ Bela will win if and only if }n \textrm{ is odd.}\newline$

$\textbf{(D)} \textrm{ Jenn will win if and only if }n \textrm{ is odd.} \qquad \newline$

$\textbf{(E)} \textrm { Jenn will win if and only if } n>8.$

$\textbf{A}$

If Bela selects the middle number in the range $[0, n]$ and then mirror whatever number Jenn selects, then if Jenn can select a number within the range, so can Bela. Jenn will always be the first person to run out of a number to choose, so the answer is A.

There are 10 people standing equally spaced around a circle. Each person knows exactly 3 of the other 9 people: the 2 people standing next to her or him, as well as the person directly across the circle. How many ways are there for the 10 people to split up into 5 pairs so that the members of each pair know each other?

$\textbf{(A) } 11 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$

$\textbf{C}$

Consider the $10$ people to be standing in a circle, where two people opposite each other form a diameter of the circle.

Let us use casework on the number of pairs that form a diameter of the circle.

$\textbf{Case 1:}$ $0$ diameters

There are $2$ ways: either $1$ pairs with $2$, $3$ pairs with $4$, and so on or $10$ pairs with $1$, $2$ pairs with $3$, etc.

$\textbf{Case 2:}$ $1$ diameter

There are $5$ possible diameters to draw (everyone else pairs with the person next to them).

Note that there cannot be $2$ diameters since there would be one person on either side that will not have a pair adjacent to them. The only scenario forced is when the two people on either side would be paired up across a diameter. Thus, a contradiction will arise.

$\textbf{Case 3:}$ $3$ diameters

There are $5$ possible sets of $3$ diameters to draw, as all 3 diameters are adjacent.

Note that there cannot be a case with $4$ diameters because then there would have to be $5$ diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises.

$\textbf{Case 4:}$ $5$ diameters

There is only $1$ way to do this.

Thus, in total there are $2+5+5+1=13$ possible ways.

An urn contains one red ball and one blue ball. A box of extra red and blue balls lie nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?

$\textbf{(A) } \dfrac16 \qquad \textbf{(B) }\dfrac15 \qquad \textbf{(C) } \dfrac14 \qquad \textbf{(D) } \dfrac13 \qquad \textbf{(E) } \dfrac12$

$\textbf{B}$

First, notice that when George chooses a ball he just adds another ball of the same color. On George's first move, he either chooses the red or the blue with a $\dfrac{1}{2}$ chance each. We can assume he chooses Red(chance $\dfrac{1}{2}$), and then multiply the final answer by two for symmetry. Now, there are two red balls and one blue ball in the urn. Then, he can either choose another Red(chance $\dfrac{2}{3}$), in which case he must choose two blues to get three of each with probability $\dfrac{1}{4}\cdot\dfrac{2}{5}=\dfrac{1}{10}$, or a blue for two blue and two red in the urn, with chance $\dfrac{1}{3}$. If he chooses blue next, he can either choose a red then a blue, or a blue then a red. Each of these has a $\dfrac{1}{2}\cdot\dfrac{2}{5}=\dfrac15$ for total of $2\cdot\dfrac{1}{5}=\dfrac{2}{5}$. The total probability that he ends up with three red and three blue is $2\cdot\dfrac{1}{2}(\dfrac{2}{3}\cdot\dfrac{1}{10}+\dfrac{1}{3}\cdot\dfrac{2}{5})=\dfrac{1}{15}+\dfrac{2}{15}=\dfrac{1}{5}$.

How many polynomials of the form $x^5 + ax^4 + bx^3 + cx^2 + dx + 2020$, where $a$, $b$, $c$, and $d$ are real numbers, have the property that whenever $r$ is a root, so is $\dfrac{-1+i\sqrt{3}}{2} \cdot r$? (Note that $i=\sqrt{-1}$)

$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$

$\textbf{C}$

Let $x_1=r$, then\[x_2=\frac{-1+i\sqrt{3}}{2} r=re^{i\frac{2\pi}{3}}\]\[x_3=\left( \frac{-1+i\sqrt{3}}{2} \right) ^2 r =re^{i\frac{4\pi}{3}}\]\[x_4=\left( \frac{-1+i\sqrt{3}}{2} \right) ^3 r=r\]which means $x_4$ is the same as $x_1$.

Now we have 3 different roots of the polynomial, $x_1$, $x_2$, and $x_3$. Next, we will prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there is one root $x_4=p$ which is different from the three roots we already know, then there must be two other roots,\[x_5=\frac{-1+i\sqrt{3}}{2} p=pe^{i\frac{2\pi}{3}}\]\[x_6=\left( \frac{-1+i\sqrt{3}}{2} \right) ^2 p =pe^{i\frac{4\pi}{3}}\]different from all known roots. We get 6 different roots for the polynomial, which contradicts the limit of 5 roots. Therefore the assumption of a different root is wrong, thus the roots must be chosen from $x_1$, $x_2$, and $x_3$.

The polynomial then can be written like $f(x)=(x-x_1)^m (x-x_2)^n (x-x_3)^q$, where $m$, $n$, and $q$ are positive integers and $m+n+q=5$. Since $a$, $b$, $c$ and $d$ are real numbers, then $n$ must be equal to $q$. Therefore $(m,n,q)$ can only be $(1,2,2)$ or $(3,1,1)$, so the answer is 2.

In square $ABCD$, points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA}$, respectively, so that $AE=AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD}$, respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$. See the figure below. Triangle $AEH$, quadrilateral $BFIE$, quadrilateral $DHJG$, and pentagon $FCGJI$ each has area $1.$ What is $FI^2$?

$\textbf{(A) } \dfrac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \dfrac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2$

$\textbf{B}$

Since the total area is $4$, the side length of square $ABCD$ is $2$. We see that since triangle $HAE$ is a right isosceles triangle with area 1, we can determine sides $HA$ and $AE$ both to be $\sqrt{2}$. Now, consider extending $FB$ and $IE$ until they intersect. Let the point of intersection be $K$. We note that $EBK$ is also a right isosceles triangle with side $2-\sqrt{2}$ and find its area to be $3-2\sqrt{2}$. Now, we notice that $FIK$ is also a right isosceles triangle (because $\angle EKB=45^\circ$) and find it's area to be $$\dfrac{1}{2}FI^2=1+3-2\sqrt2\rightarrow FI^2=8-4\sqrt2=$$

Square $ABCD$ in the coordinate plane has vertices at the points $A(1,1), B(-1,1), C(-1,-1),$ and $D(1,-1).$ Consider the following four transformations:

$\quad\bullet$ $L,$ a rotation of $90^{\circ}$ counterclockwise around the origin;

$\quad\bullet$ $R,$ a rotation of $90^{\circ}$ clockwise around the origin;

$\quad\bullet$ $H,$ a reflection across the $x$-axis; and

$\quad\bullet$ $V,$ a reflection across the $y$-axis.

Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying $R$ and then $V$ would send the vertex $A$ at $(1,1)$ to $(-1,-1)$ and would send the vertex $B$ at $(-1,1)$ to itself. How many sequences of $20$ transformations chosen from $\{L, R, H, V\}$ will send all of the labeled vertices back to their original positions? (For example, $R, R, V, H$ is one sequence of $4$ transformations that will send the vertices back to their original positions.)

$\textbf{(A)}\ 2^{37} \qquad\textbf{(B)}\ 3\cdot 2^{36} \qquad\textbf{(C)}\ 2^{38} \qquad\textbf{(D)}\ 3\cdot 2^{37} \qquad\textbf{(E)}\ 2^{39}$

$\textbf{C}$

For each transformation:

$\quad\bullet$ Each labeled vertex will move to an adjacent position.

$\quad\bullet$ The labeled vertices will maintain the consecutive order $ABCD$ in either direction (clockwise or counterclockwise).

$\quad\bullet$ $L$ and $R$ will retain the direction of the labeled vertices, but $H$ and $V$ will alter the direction of the labeled vertices.

After the $19$th transformation, vertex $A$ will be at either $(1,-1)$ or $(-1,1)$ because 19 is an odd. All possible configurations of the labeled vertices are shown below:

Each sequence of $19$ transformations generates one valid sequence of $20$ transformations. Therefore, the answer is $4^{19}=2^{38}.$

Two different cubes of the same size are to be painted, with the color of each face being chosen independently and at random to be either black or white. What is the probability that after they are painted, the cubes can be rotated to be identical in appearance?

$\textbf{(A)}\ \dfrac{9}{64} \qquad\textbf{(B)}\ \dfrac{289}{2048} \qquad\textbf{(C)}\ \dfrac{73}{512} \qquad\textbf{(D)}\ \dfrac{147}{1024} \qquad\textbf{(E)}\ \dfrac{589}{4096}$

$\textbf{D}$

Define two ways of painting to be in the same $class$ if one can be rotated to form the other. We can count the number of ways of painting for each specific $class$.

$\textbf{Case 1:}$ Black-white color distribution is 0-6 (out of 6 total faces)

Trivially $1^2 = 1$ way to paint the cubes.

$\textbf{Case 2:}$ Black-white color distribution is 1-5

Trivially all $\dbinom{6}{5} = 6$ ways belong to the same $class$, so $6^2$ ways to paint the cubes.

$\textbf{Case 3:}$ Black-white color distribution is 2-4

There are two $classes$ for this case: the $class$ where the two black faces are touching and the other $class$ where the two black faces are on opposite faces. There are $3$ members of the latter $class$ since there are $3$ unordered pairs of $2$ opposite faces of a cube. Thus, there are $\dbinom{6}{2} - 3 = 12$ members of the former $class$. Thus, $12^2 + 3^2$ ways to paint the cubes for this case.

$\textbf{Case 4:}$ Black-white color distribution is 3-3

By simple intuition, there are also two $classes$ for this case, the $class$ where the three black faces meet at a single vertex, and the other class where the three red faces are in a "straight line" along the edges of the cube. Note that since there are $8$ vertices in a cube, there are $8$ members of the former class and $\dbinom{6}{3} - 8 = 12$ members of the latter class. Thus, $12^2 + 8^2$ ways to paint the cubes for this case.

Note that by symmetry (since we are only switching the colors), the number of ways to paint the cubes for black-white color distributions 4-2, 5-1, and 6-0 is 2-4, 1-5, and 0-6 (respectively).

Thus, our total answer is\[\frac{2( 1^2 +6^2+ 12^2 + 3^2) + 12^2 + 8^2}{2^{12}} = \frac{588}{4096} = \frac{147}{1024}\]

How many positive integers $n$ satisfy\[\frac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\](Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$.)

$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$

$\textbf{C}$

We are given that\[\frac{n+1000}{70}=\lfloor\sqrt{n}\rfloor\] $\lfloor\sqrt{n}\rfloor$ must be an integer, which means that $n+1000$ is divisible by $70$. As $1000\equiv 20\pmod{70}$, this means that $n\equiv 50\pmod{70}$, so we can write $n=70k+50$ for non-negative integer $k$.

Therefore,\[\frac{n+1000}{70}=\frac{70k+1050}{70}=k+15=\lfloor\sqrt{70k+50}\rfloor\] Also, we can say that $\sqrt{70k+50}-1 < k+15$ and $k+15\leq\sqrt{70k+50}$.

Squaring the second inequality, we get $$k^{2}+30k+225\leq70k+50\implies k^{2}-40k+175\leq 0\implies 5\leq k\leq 35$$ Similarly solving the first inequality gives us $k < 19-\sqrt{155}$ or $k > 19+\sqrt{155}$. $\sqrt{155}$ is larger than $12$ and smaller than $13$, so instead, we can say $k\leq 6$ or $k\geq 32$.

Combining this with $5\leq k\leq 35$, we get $k=5,6,32,33,34,35$ are all solutions for $k$ that give a valid solution for $n$, meaning that our answer is 6.

What is the maximum value of $\dfrac{(2^t-3t)t}{4^t}$ for real values of $t?$

$\textbf{(A)}\ \dfrac{1}{16} \qquad\textbf{(B)}\ \dfrac{1}{15} \qquad\textbf{(C)}\ \dfrac{1}{12} \qquad\textbf{(D)}\ \dfrac{1}{10} \qquad\textbf{(E)}\ \dfrac{1}{9}$

$\textbf{C}$

Notice that $$\left(2^t-3t\right)\cdot3t\leq\left(\dfrac{2^t-3t+3t}{2}\right)^2=4^{t-1}$$ The original expression $$\dfrac{\left(2^t-3t\right)t}{4^t}=\dfrac{\left(2^t-3t\right)3t}{3\cdot4^t}\leq\dfrac{4^{t-1}}{3\cdot4^t}=\dfrac1{12}$$

How many integers $n \geq 2$ are there such that whenever $z_1, z_2, ..., z_n$ are complex numbers such that \[|z_1| = |z_2| = ... = |z_n| = 1 \textrm{ and } z_1 + z_2 + ... + z_n = 0,\]then the numbers $z_1, z_2, ..., z_n$ are equally spaced on the unit circle in the complex plane?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5$

$\textbf{B}$

For $n=2$, we see that if $z_{1}+z_{2}=0$, then $z_{1}=-z_{2}$, so they are evenly spaced along the unit circle.

For $n=3$, WLOG, we can set $z_{1}=1$. Notice that now $\Re(z_{2}+z_{3})=-1$ and $\Im\{z_{2}\}=-\Im\{z_{3}\}$. This forces $z_{2}$ and $z_{3}$ to be equal to $e^{i\frac{2\pi}{3}}$ and $e^{-i\frac{2\pi}{3}}$, meaning that all three are equally spaced along the unit circle.

We can now show that we can construct complex numbers when $n\geq 4$ that do not equally spaced on the unit circle.

Suppose that the condition in the problem holds for some $n=k$. We can now add two points $z_{k+1}$ and $z_{k+2}$ anywhere on the unit circle such that $z_{k+1}=-z_{k+2}$, which will break the condition of equally spaced unit circle. Now that we have shown that $n=2$ and $n=3$ works, by this construction, any $n\geq 4$ does not work. So the answer is 2.

Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product\[n = f_1\cdot f_2\cdots f_k,\]where $k\ge1$, the $f_i$ are integers strictly greater than $1$, and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$, $2\cdot 3$, and $3\cdot2$, so $D(6) = 3$. What is $D(96)$?

$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$

$\textbf{A}$

Note that $96 = 2^5 \cdot 3$. Since there are at most six not necessarily distinct factors $>1$ multiplying to $96$, we have six cases: $k=1, 2, ..., 6.$ Now we look at each of the six cases.

$k=1$: We see that there is $1$ way, merely $96$.

$k=2$: This way, we have a $3$ in one slot a $2$ in another, and symmetry. The four other $2$'s leave us with $5$ ways and symmetry doubles us so we have $10$.

$k=3$: We have $3, 2, 2$ as our baseline. We need to multiply by $2$ in $3$ places, and see that we can split the remaining three powers of $2$ in a manner that is $3-0-0$, $2-1-0$ or $1-1-1$. A $3-0-0$ split has $3 + 6 = 9$ ways of happening ($24-2-2$ and symmetry; $2-3-16$ and symmetry), a $2-1-0$ split has $6 \cdot 3 = 18$ ways of happening (due to all being distinct) and a $1-1-1$ split has $3$ ways of happening ($6-4-4$ and symmetry) so in this case we have $9+18+3=30$ ways.

$k=4$: We have $3, 2, 2, 2$ as our baseline, and for the two other $2$'s, we have a $2-0-0-0$ or $1-1-0-0$ split. The former grants us $4+12=16$ ways ($12-2-2-2$ and symmetry and $3-8-2-2$ and symmetry) and the latter grants us also $12+12=24$ ways ($6-4-2-2$ and symmetry and $3-4-4-2$ and symmetry) for a total of $16+24=40$ ways.

$k=5$: We have $3, 2, 2, 2, 2$ as our baseline and one place to put the last $2$: on another two or on the three. On the three gives us $5$ ways due to symmetry and on another two gives us $5 \cdot 4 = 20$ ways due to symmetry. Thus, we have $5+20=25$ ways.

$k=6$: We have $3, 2, 2, 2, 2, 2$ and symmetry and no more $2$s to multiply, so by symmetry, we have $6$ ways.

Thus, adding, we have $1+10+30+40+25+6=112$.

For each real number $a$ with $0 \leq a \leq 1$, let numbers $x$ and $y$ be chosen independently at random from the intervals $[0, a]$ and $[0, 1]$, respectively, and let $P(a)$ be the probability that\[\sin^2{(\pi x)} + \sin^2{(\pi y)} > 1\]What is the maximum value of $P(a)?$

$\textbf{(A)}\ \dfrac{7}{12} \qquad\textbf{(B)}\ 2 - \sqrt{2} \qquad\textbf{(C)}\ \dfrac{1+\sqrt{2}}{4} \qquad\textbf{(D)}\ \dfrac{\sqrt{5}-1}{2} \qquad\textbf{(E)}\ \dfrac{5}{8}$

$\textbf{B}$

Let's start first by manipulating the given inequality. \[\sin^{2}{(\pi x)}+\sin^{2}{(\pi y)}>1\]\[\sin^{2}{(\pi x)}>1-\sin^{2}{(\pi y)}=\cos^{2}{(\pi y)}\] Let's consider the boundary cases: $\sin{(\pi x)}=\pm \cos{(\pi y)}$.\[\sin{(\pi x)}=\pm \cos{(\pi y)}=\sin{(\pm\dfrac 12 {\pi}- \pi y)}\] If $\sin{(\pi x)}=\sin{(\dfrac 12 {\pi}- \pi y)}$, we have $\pi x=\dfrac12\pi-\pi y$ or $\pi x+\dfrac12\pi-\pi y=\pi$, which leads to \[y=\dfrac{1}{2}-x \quad \textrm{and} \quad y=x-\dfrac{1}{2}\]If $\sin{(\pi x)}=\sin{(-\dfrac 12 {\pi}- \pi y)}=\sin{(\dfrac32 {\pi}- \pi y)}$, we have $\pi x=\dfrac32\pi-\pi y$ or $\pi x+\dfrac32\pi-\pi y=\pi$, which leads to\[y=\dfrac{3}{2}-x\quad \textrm{and}\quad y=x+\dfrac{1}{2}\]If we graph these equations in $[0,1]\times[0,1]$, we get a rhombus shape.

Testing points in each section tells us that the inside of the rhombus satisfies the inequality in the problem statement.

From the region graph, notice that in order to maximize $P(a)$, $a\geq\dfrac{1}{2}$. We can solve the rest with geometric probability.

Instead of maximizing $P(a)$, we minimize $Q(a)=1-P(a)$. $Q(a)$ consists of two squares (each broken into two triangles), one of area $\dfrac{1}{4}$ and another of area $\left(a-\dfrac 12\right)^2$. To calculate $Q(a)$, we divide this area by $a$, so\[Q(a) = \frac{1}{a}\left[\dfrac{1}{4}+\left(a-\dfrac 12\right)^2\right] = \frac{1}{a}\left(\frac{1}{2}+a^2-a\right)= a+\frac 1{2a}-1\geq 2\sqrt{a\cdot\frac{1}{2a}}-1=\sqrt{2}-1\]The minimum value of $Q(a)$ can be achieved when $$a=\dfrac{1}{2a}\rightarrow a=\frac{\sqrt{2}}{2}$$ Therefore, the maximum value of $P(a)$ is $1-\min(Q(a))$, which is $1-(\sqrt{2}-1) =2 - \sqrt{2}$.