## AMC 10 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 of

$$1 - (-2) - 3 - (-4) - 5 - (-6)?$$

$\textbf{(A)}\ -20 \qquad\textbf{(B)}\ -3 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 21$

$\textbf{D}$

$$1 - (-2) - 3 - (-4) - 5 - (-6)=1+2-3+4-5+6=5$$

Carl has $5$ cubes each having side length $1$, and Kate has $5$ cubes each having side length $2$. What is the total volume of these $10$ cubes?

$\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 25 \qquad\textbf{(C)}\ 28 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 45$

$\textbf{E}$

A cube with side length $1$ has volume $1^3=1$, so $5$ of these will have a total volume of $5\cdot1=5$.

A cube with side length $2$ has volume $2^3=8$, so $5$ of these will have a total volume of $5\cdot8=40$.

Adding together, we have $5+40=45$.

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.

How many distinguishable arrangements are there of $1$ brown tile, $1$ purple tile, $2$ green tiles, and $3$ yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)

$\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050$

$\textbf{B}$

There are $7!$ ways to order $7$ objects. However, since there's $3!=6$ ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and $2!=2$ ways to order the green tiles, we have to divide out these possibilities: $\dfrac{7!}{3! \cdot2!}=420$.

Driving along a highway, Megan noticed that her odometer showed $15951$ (miles). This number is a palindrome-it reads the same forward and backward. Then $2$ hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this $2$-hour period?

$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 55 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 65 \qquad\textbf{(E)}\ 70$

$\textbf{B}$

In order to get the smallest palindrome greater than $15951$, we need to raise the middle digit. Since the middle digit is 9, we have to raise a digit before the middle to keep it a palindrome. Hence, the 15951 turns to 160xx. To keep it a palindrome, our number is now $16061$.

So Megan drove $16061-15951=110$ miles. Since this happened over $2$ hours, she drove at $110/2=55$ mph.

How many positive even multiples of $3$ less than $2020$ are perfect squares?

$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 10 \qquad\textbf{(E)}\ 12$

$\textbf{A}$

Any even multiple of $3$ is a multiple of $6$, so we need to find multiples of $6$ that are perfect squares and less than $2020$. Any solution that we want will be in the form $(6n)^2$, where $n$ is a positive integer. Hence, we get $$(6n)^2<2020$$The smallest possible value is at $n=1$, and the largest is at $n=7$ (where the expression equals $1764$). Therefore, there are a total of 7 possible numbers.

Points $P$ and $Q$ lie in a plane with $PQ=8$. How many locations for point $R$ in this plane are there such that the triangle with vertices $P$, $Q$, and $R$ is a right triangle with area $12$ square units?

$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$

$\textbf{D}$

Let the brackets denote areas. We are given that\[[PQR]=\frac12\cdot PQ\cdot h_R=12.\]Since $PQ=8,$ it follows that $h_R=3.$

We construct a circle with diameter $\overline{PQ}.$ All such locations for $R$ are shown below:

We apply casework to the right angle of $\triangle PQR:$

If $\angle P=90^\circ,$ then $R\in\{R_1,R_5\}$ by the tangent.

If $\angle Q=90^\circ,$ then $R\in\{R_4,R_8\}$ by the tangent.

If $\angle R=90^\circ,$ then $R\in\{R_2,R_3,R_6,R_7\}$ by the Inscribed Angle Theorem.

Together, there are 8 such locations for $R.$

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 $\dfrac{1}{3}\cdot\pi\cdot3^2\cdot\sqrt7=3 \pi \sqrt7$.

Ms. Carr asks her students to read any $5$ of the $10$ books on a reading list. Harold randomly selects $5$ books from this list, and Betty does the same. What is the probability that there are exactly $2$ books that they both select?

$\textbf{(A)}\ \dfrac{1}{8} \qquad\textbf{(B)}\ \dfrac{5}{36} \qquad\textbf{(C)}\ \dfrac{14}{45} \qquad\textbf{(D)}\ \dfrac{25}{63} \qquad\textbf{(E)}\ \dfrac{1}{2}$

$\textbf{D}$

We don't care about which books Harold selects. We just care that Betty picks $2$ books from Harold's list and $3$ that aren't on Harold's list.

The total amount of combinations of books that Betty can select is $\dbinom{10}{5}=252$.

There are $\dbinom{5}{2}=10$ ways for Betty to choose $2$ of the books that are on Harold's list.

From the remaining $5$ books that aren't on Harold's list, there are $\dbinom{5}{3}=10$ ways to choose $3$ of them.

So the probability is $\dfrac{10\cdot10}{252}=\dfrac{25}{63}$.

The decimal representation of $$\frac{1}{20^{20}}$$ consists of a string of zeros after the decimal point, followed by a $9$ and then several more digits. How many zeros are in that initial string of zeros after the decimal point?

$\textbf{(A)}\ 23 \qquad\textbf{(B)}\ 24 \qquad\textbf{(C)}\ 25 \qquad\textbf{(D)}\ 26 \qquad\textbf{(E)}\ 27$

$\textbf{D}$

We have $$20^{20}=10^{20}\cdot2^{20}=10^{20}\cdot\left(2^{10}\right)^2=10^{20}\cdot1024^2$$ 1024 is slightly more than 1000, so $$20^{20}=10^{20}\cdot1024^2>10^{20}\cdot\left(10^3\right)^2=10^{26}$$ The original expression is slightly less than $\dfrac{1}{10^{26}}$, which means there are 26 zeros right after the decimal point.

Andy the Ant lives on a coordinate plane and is currently at $(-20, 20)$ facing east (that is, in the positive $x$-direction). Andy moves $1$ unit and then turns $90^{\circ}$ left. From there, Andy moves $2$ units (north) and then turns $90^{\circ}$ left. He then moves $3$ units (west) and again turns $90^{\circ}$ left. Andy continues his progress, increasing his distance each time by $1$ unit and always turning left. What is the location of the point at which Andy makes the $2020$th left turn?

$\textbf{(A)}\ (-1030, -994)\qquad\newline$

$\textbf{(B)}\ (-1030, -990)\qquad\newline$

$\textbf{(C)}\ (-1026, -994)\qquad\newline$

$\textbf{(D)}\ (-1026, -990)\qquad\newline$

$\textbf{(E)}\ (-1022, -994)$

$\textbf{B}$

Andy makes a total of $2020$ moves: $1010$ horizontal ($505$ left and $505$ right) and $1010$ vertical ($505$ up and $505$ down).

The $x$-coordinate of Andy's final position is\[-20+\overbrace{\underbrace{1-3}_{-2}+\underbrace{5-7}_{-2}+\underbrace{9-11}_{-2}+\cdots+\underbrace{2017-2019}_{-2}}^{\text{1010 terms, 505 pairs}}=-20-2\cdot505=-1030\]The $y$-coordinate of Andy's final position is\[20+\overbrace{\underbrace{2-4}_{-2}+\underbrace{6-8}_{-2}+\underbrace{10-12}_{-2}+\cdots+\underbrace{2018-2020}_{-2}}^{\text{1010 terms, 505 pairs}}=20-2\cdot505=-990\]Together, we have $(x,y)=(-1030, -990).$

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:\[6\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)=3\sqrt{3}-\pi\]

Steve wrote the digits $1$, $2$, $3$, $4$, and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$rd, $6$th, $9$th, $\ldots$ digits from the left), then erased every fourth digit from the resulting list (that is, the $4$th, $8$th, $12$th, $\ldots$ digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions $2019, 2020, 2021$?

$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 12$

$\textbf{D}$

Note that cycles exist initially and after each round of erasing.

Let the parentheses denote cycles. It follows that:

Initially, the list has cycles of length $5:$\[(12345)=12345123451234512345\cdots.\]To find one cycle after the first round of erasing, we need one cycle of length $\operatorname{lcm}(3,5)=15$ before erasing. So, we first group $\dfrac{15}{5}=3$ copies of the current cycle into one, then erase:\begin{align*} (12345)&\longrightarrow(123451234512345) \\ &\longrightarrow(12\cancel{3}45\cancel{1}23\cancel{4}51\cancel{2}34\cancel{5}) \\ &\longrightarrow(1245235134). \end{align*}As a quick confirmation, one cycle should have length $15\cdot\left(1-\dfrac{1}{3}\right)=10$ at this point.

To find one cycle after the second round of erasing, we need one cycle of length $\operatorname{lcm}(4,10)=20$ before erasing. So, we first group $\dfrac{20}{10}=2$ copies of the current cycle into one, then erase:\begin{align*} (1245235134)&\longrightarrow(12452351341245235134) \\ &\longrightarrow(124\cancel{5}235\cancel{1}341\cancel{2}452\cancel{3}513\cancel{4}) \\ &\longrightarrow(124235341452513). \end{align*}As a quick confirmation, one cycle should have length $20\cdot\left(1-\dfrac{1}{4}\right)=15$ at this point.

To find one cycle after the third round of erasing, we need one cycle of length $\operatorname{lcm}(5,15)=15$ before erasing. We already have it here, so we erase:\begin{align*} (124235341452513)&\longrightarrow(1242\cancel{3}5341\cancel{4}5251\cancel{3}) \\ &\longrightarrow(124253415251). \end{align*}As a quick confirmation, one cycle should have length $15\cdot\left(1-\dfrac{1}{5}\right)=12$ at this point.

Since $2019,2020,2021$ are congruent to $3,4,5$ modulo $12,$ respectively, the three digits in the final positions $2019,2020,2021$ are $4,2,5,$ respectively:\[(12\underline{425}3415251).\]Therefore, the answer is $4+2+5=11.$

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 $2\cdot\dfrac{1}{2}(\dfrac{2}{3}\cdot\dfrac{1}{10}+\dfrac{1}{3}\cdot\dfrac{2}{5})=\dfrac{1}{15}+\dfrac{2}{15}=\dfrac{1}{5}$.

In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$. What is the digit $A$?

$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7$

$\textbf{A}$

$$\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43$$Since this number is divisible by $4$ but not $8$, the last $2$ digits must be divisible by $4$ but the last $3$ digits cannot be divisible by $8$. This narrows the options down to $2$ and $6$.

Also, the number cannot be divisible by $3$. Adding up the digits, we get $18+4A$. If $A=6$, then the expression equals $42$, a multiple of $3$. This would mean that the entire number would be divisible by $3$, which is not what we want. Therefore, the only option is 2.

Let $B$ be a right rectangular prism (box) with edges lengths $1,$ $3,$ and $4$, together with its interior. For real $r\geq0$, let $S(r)$ be the set of points in $3$-dimensional space that lie within a distance $r$ of some point in $B$. The volume of $S(r)$ can be expressed as $ar^{3} + br^{2} + cr +d$, where $a,$ $b,$ $c,$ and $d$ are positive real numbers. What is $\dfrac{bc}{ad}?$

$\textbf{(A) } 6 \qquad\textbf{(B) } 19 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 26 \qquad\textbf{(E) } 38$

$\textbf{B}$

Split $S(r)$ into 4 regions:

1. The rectangular prism itself

2. The extensions of the faces of $B$

3. The quarter cylinders at each edge of $B$

4. The one-eighth spheres at each corner of $B$

Region 1: The volume of $B$ is $1 \cdot 3 \cdot 4 = 12$, so $d=12$.

Region 2: This volume is equal to the surface area of $B$ times $r$ (these "extensions" are just more boxes). The volume is then $\text{SA} \cdot r=2(1 \cdot 3 + 1 \cdot 4 + 3 \cdot 4)r=38r$ to get $c=38$.

Region 3: We see that there are 12 quarter-cylinders, 4 of each type. We have 4 quarter-cylinders of height 1, 4 quarter-cylinders of height 3, 4 quarter-cylinders of height 4. Since 4 quarter-cylinders make a full cylinder, the total volume is $4 \cdot \dfrac{1\pi r^2}{4} + 4 \cdot \dfrac{3\pi r^2}{4} + 4 \cdot \dfrac{4 \pi r^2}{4}=8 \pi r^2$. Therefore, $b=8\pi$.

Region 4: There is an eighth-sphere of radius $r$ at each corner of $B$. Since there are 8 corners, these eighth-spheres add up to 1 full sphere of radius $r$. The volume of this sphere is then $\dfrac{4}{3}\pi \cdot r^3$, so $a=\dfrac{4\pi}{3}$.

Using these values, $\dfrac{bc}{ad}=\dfrac{(8\pi)(38)}{(4\pi/3)(12)} = 19$.

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 $$\dfrac{1}{2}FI^2=1+3-2\sqrt2\rightarrow FI^2=8-4\sqrt2$$

What is the remainder when $2^{202} +202$ is divided by $2^{101}+2^{51}+1$?

$\textbf{(A) } 100 \qquad\textbf{(B) } 101 \qquad\textbf{(C) } 200 \qquad\textbf{(D) } 201 \qquad\textbf{(E) } 202$

$\textbf{D}$

Let $x=2^{50}$. We are now looking for the remainder of $\dfrac{4x^4+202}{2x^2+2x+1}$.\begin{align*}

\dfrac{4x^4+202}{2x^2+2x+1}&=\dfrac{(4x^4+4x^3+2x^2)-4x^3-2x^2+202}{2x^2+2x+1}\\

&=2x^2+\dfrac{(-4x^3-4x^2-2x)+2x^2+2x+202}{2x^2+2x+1}\\

&=2x^2-2x+\dfrac{(2x^2+2x+1)+201}{2x^2+2x+1}\\

&=2x^2-2x+1+\dfrac{201}{2x^2+2x+1}

\end{align*} Apparently the denominator $2x^2+2x+1$ is much greater than the numerator 201. So the reminder is 201.

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 $4^{19}=2^{38}.$

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.

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 $1+10+30+40+25+6=112$.