## AMC 12 2020 Test A

**If you like our free learning materials, please click the advertisement below or anywhere. No paying, no donation, just a simple click. The advertising revenues will be used to provide more and better learning materials. Thank you!**

**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.

Carlos took $70\%$ of a whole pie. Maria took one third of the remainder. What portion of the whole pie was left?

$\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%$

$\textbf{C}$

We have\[\left(100\%-70\%\right)\cdot\left(1-\frac13\right)=30\%\cdot\frac23=20\%\]of the whole pie left.

The acronym AMC is shown in the rectangular grid below with grid lines spaced $1$ unit apart. In units, what is the sum of the lengths of the line segments that form the acronym AMC$?$

$\textbf{(A) } 17 \qquad \textbf{(B) } 15 + 2\sqrt{2} \qquad \textbf{(C) } 13 + 4\sqrt{2} \qquad \textbf{(D) } 11 + 6\sqrt{2} \qquad \textbf{(E) } 21$

$\textbf{C}$

Note that the length of a diagonal is $\sqrt2$. Hence, the sum of the lengths of all lines is $(3+2\sqrt2)+(4+2\sqrt2)+6=13+4\sqrt2$.

A driver travels for $2$ hours at $60$ miles per hour, during which her car gets $30$ miles per gallon of gasoline. She is paid $\$0.50$ per mile, and her only expense is gasoline at $\$2.00$ per gallon. What is her net rate of pay, in dollars per hour, after this expense?

$\textbf{(A) }20 \qquad\textbf{(B) }22 \qquad\textbf{(C) }24 \qquad\textbf{(D) } 25\qquad\textbf{(E) } 26$

$\textbf{E}$

Since the driver travels $60$ miles per hour and each hour she uses $2$ gallons of gasoline, she spends $\$4$ per hour on gas. If she gets $\$0.50$ per mile, then she gets $\$30$ per hour of driving. Subtracting the gas cost, her net rate of money earned per hour is $ 26$.

How many $4$-digit positive integers (that is, integers between $1000$ and $9999$, inclusive) having only even digits are divisible by $5?$

$\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500$

$\textbf{B}$

The units digit, for all numbers divisible by 5, must be either $0$ or $5$. However, since all digits are even, the units digit must be $0$. The middle two digits can be 0, 2, 4, 6, or 8, but the thousands digit can only be 2, 4, 6, or 8 since it cannot be zero. There is 1 choice for the units digit, 5 choices for each of the middle 2 digits, and 4 choices for the thousands digit, so there is a total of $4\cdot5\cdot5\cdot1 = 100$ numbers.

The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$-by-$5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?

$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$

$\textbf{C}$

Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by $5$ is the total value per row. The sum of the $25$ integers is $-10+-9+\cdots+14=11+12+13+14=50$. So the common sum is $\dfrac{50}{5}=10$.

In the plane figure shown below, $3$ of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetry$?$

$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

$\textbf{D}$

The two lines of symmetry must be horizontally and vertically through the middle. We can then fill the boxes in like so:

where the light gray boxes are the ones we have filled. Counting these, we get $7$ total boxes.

Seven cubes, whose volumes are $1$, $8$, $27$, $64$, $125$, $216$, and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?

$\textbf{(A) } 644 \qquad \textbf{(B) } 658 \qquad \textbf{(C) } 664 \qquad \textbf{(D) } 720 \qquad \textbf{(E) } 749$

$\textbf{B}$

The volume of each cube follows the pattern of $n^3$, for $n$ is between $1$ and $7$.

First, we will calculate the total surface area of the cubes, ignoring overlap. This value is $$6 ( 1^2 + 2^2 + \cdots + 7^2 ) = 6\sum_{n=1}^{7} n^2 = 6 \left( \frac{7(7 + 1)(2 \cdot 7 + 1)}{6} \right) = 7 \cdot 8 \cdot 15 = 840$$ Then, we need to subtract out the overlapped parts of the cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of the overlapped parts of the cubes is thus equal to $$2\sum_{n=1}^{6} n^2 = 182$$ Subtracting the overlapped surface area from the total surface area, we get $840 - 182 = 658$.

What is the median of the following list of $4040$ numbers$?$\[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\]

$\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$

$\textbf{C}$

If we sort the sequence in order, it is clearly that 2020 is not the 2020th number because some perfect squares less than 2020 will appear twice. The largest perfect square less than 2020 is $44^2=1936$. Hence, we see that the 2020th number is $2020-44=1976$, and the 2021th number is 1977. Then we get the median $$\dfrac{1976+1977}{2}=1976.5$$

How many solutions does the equation $\tan{(2x)} = \cos\left(\dfrac{x}{2}\right)$ have on the interval $[0, 2\pi]?$

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

$\textbf{E}$

We count the intersections of the graphs of $y=\tan(2x)$ and $y=\cos\left(\dfrac x2\right):$

The graph of $y=\tan(2x)$ has a period of $\dfrac{\pi}{2},$ asymptotes at $x=\dfrac{\pi}{4}+\dfrac{k\pi}{2},$ and zeros at $x=\dfrac{k\pi}{2}$ for some integer $k.$

On the interval $[0,2\pi],$ the graph has five branches:\[\biggl[0,\frac{\pi}{4}\biggr),\left(\frac{\pi}{4},\frac{3\pi}{4}\right),\left(\frac{3\pi}{4},\frac{5\pi}{4}\right),\left(\frac{5\pi}{4},\frac{7\pi}{4}\right),\left(\frac{7\pi}{4},2\pi\right].\]Note that $\tan(2x)\in[0,\infty)$ for the first branch, $\tan(2x)\in(-\infty,\infty)$ for the three middle branches, and $\tan(2x)\in(-\infty,0]$ for the last branch. Moreover, all branches are strictly increasing.

The graph of $y=\cos\left(\dfrac x2\right)$ has a period of $4\pi$ and zeros at $x=\pi+2k\pi$ for some integer $k.$

On the interval $[0,2\pi],$ note that $\cos\left(\dfrac x2\right)\in[-1,1].$ Moreover, the graph is strictly decreasing.

The graphs of $y=\tan(2x)$ and $y=\cos\left(\dfrac x2\right)$ intersect once on each of the five branches of $y=\tan(2x),$ as shown below:

Therefore, the answer is $5.$

There is a unique positive integer $n$ such that\[\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.\]What is the sum of the digits of $n?$

$\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13$

$\textbf{E}$

Since $\log_{a^b} c = \dfrac{1}{b} \log_a c$, the original equation $\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}$ becomes\[\log_2\left({\frac{1}{4}\log_{2}{n}}\right) = \frac{1}{2}\log_2\left({\frac{1}{2}\log_2{n}}\right).\]Using log property of addition, we expand both sides and then simplify:\begin{align*} \log_2{\frac{1}{4}}+\log_2{(\log_{2}{n}}) &= \frac{1}{2}\left[\log_2{\frac{1}{2}} +\log_{2}{(\log_2{n})}\right] \\ \log_2{\frac{1}{4}}+\log_2{(\log_{2}{n}}) &= \frac{1}{2}\left[-1 +\log_{2}{(\log_2{n})}\right] \\ -2+\log_2{(\log_{2}{n}}) &= -\frac{1}{2}+ \frac{1}{2}(\log_{2}{(\log_2{n})})\\

\frac{1}{2}(\log_{2}{(\log_2{n})}) &= \frac{3}{2}\\ \log_{2}{(\log_2{n})} &= 3 \\ \log_2{n}&=8 \\ n&=256. \end{align*}Adding the digits together, we have $2+5+6=13.$

A frog sitting at the point $(1, 2)$ begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length $1$, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices $(0,0), (0,4), (4,4),$ and $(4,0)$. What is the probability that the sequence of jumps ends on a vertical side of the square$?$

$\textbf{(A) } \dfrac{1}{2} \qquad \textbf{(B) } \dfrac{5}{8} \qquad \textbf{(C) } \dfrac{2}{3} \qquad \textbf{(D) } \dfrac{3}{4} \qquad \textbf{(E) } \dfrac{7}{8}$

$\textbf{B}$

Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical end of the square. Therefore, the probability of this happening is $\dfrac{1}{4}$. If the frog goes to the right, it will be in the center of the square at $(2,2)$, and by symmetry (since the frog is equidistant from all sides of the square), the chance it will hit a vertical side of a square is $\dfrac{1}{2}$. The probability of this happening is $\dfrac{1}{4} \cdot \dfrac{1}{2} = \dfrac{1}{8}$.

If the frog goes either up or down, it will hit a line of symmetry along the corner it is closest to and furthest to, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. The probability for it to hit a vertical wall is $\dfrac{1}{2}$. Because there's a $\dfrac{1}{2}$ chance of the frog going up or down, the total probability for this case is $\dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{1}{4}$.

Summing up all the cases, the answer is $\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{4} = \dfrac{5}{8}$.

We have another way to think about this question. Let $P_{(x,y)}$ denote the probability of the frog's sequence of jumps ends with it hitting a vertical edge when it is at $(x,y)$. Note that $P_{(1,2)}=P_{(3,2)}$ by reflective symmetry over the line $x=2$. Similarly, $P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}$, and $P_{(2,1)}=P_{(2,3)}$. Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point:\[P_{(1,2)}=\frac{1}{4}+\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}\]\[P_{(2,2)}=\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\]\[P_{(1,1)}=\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\]\[P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}\]We have a system of $4$ equations in $4$ variables, so we can solve for each of these probabilities. The answer is $P_{(1,2)}=\dfrac{5}{8}$.

Line $\ell$ in the coordinate plane has the equation $3x - 5y + 40 = 0$. This line is rotated $45^{\circ}$ counterclockwise about the point $(20, 20)$ to obtain line $k$. What is the $x$-coordinate of the $x$-intercept of line $k?$

$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$

$\textbf{B}$

Since the slope of the line is $\dfrac{3}{5}$, and the angle we are rotating around is x, then $$\tan x = \dfrac{3}{5}$$ $$\tan(x+45^{\circ}) = \frac{\tan x + \tan(45^{\circ})}{1-\tan x*\tan(45^{\circ})} = \frac{0.6+1}{1-0.6}= 4$$ Hence, the slope of the rotated line is $4$. Since we know the line intersects the point $(20,20)$, then we know the line is $y=4x-60$. Set $y=0$ to find the x-intercept, and so $x=15$.

There are integers $a$, $b$, and $c$, each greater than 1, such that\[\sqrt[a]{N \sqrt[b]{N \sqrt[c]{N}}} = \sqrt[36]{N^{25}}\]for all $N > 1$. What is $b$?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

$\textbf{B}$

\begin{align*}

\sqrt[a]{N \sqrt[b]{N \sqrt[c]{N}}} &=\sqrt[a]{N \sqrt[b]{N\cdot N^{\frac1c}}} \\

&=\sqrt[a]{N \sqrt[b]{N^{\frac1c+1}}}\\

&=\sqrt[a]{N \cdot N^{\frac1{bc}+\frac1b}}\\

&=\sqrt[a]{N^{\frac1{bc}+\frac1b+1}}\\

&=N^{\frac1{abc}+\frac1{ab}+\frac1a}

\end{align*} The equation is then $$N^{\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}}=N^{\frac{25}{36}}$$ which implies that $$\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}$$ Since $a$, $b$, and $c$ are integers greater than 1, we have $$\dfrac1a+\dfrac1{ab}+\dfrac1{abc}\le\dfrac1a+\dfrac1{a\cdot2}+\dfrac{1}{a\cdot2\cdot2}=\dfrac7{4a}$$ Hence, we have $$\dfrac{25}{36}\le\dfrac{7}{4a}\rightarrow a\le\dfrac{63}{25}$$ So we know $a=2$. Then the equation can be simplified as $$\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\dfrac12+\dfrac{1}{2b}+\dfrac1{2bc}=\frac{25}{36}\rightarrow \dfrac1b+\dfrac{1}{bc}=\dfrac{7}{18}$$ Similarly, we have $$\dfrac1b+\dfrac1{bc}\le\dfrac1b+\dfrac{1}{2b}=\dfrac3{2b}$$ So $$\dfrac7{18}\le\dfrac3{2b}\rightarrow b\le\dfrac{27}{7}$$ Therefore, we have $b=2\ \text{or}\ 3$. If $b=2$, we get $c=-\dfrac92$, which is not an integer. So the only possible solution is $(b,c)=(3,6)$.

The answer is 3.

Regular octagon $ABCDEFGH$ has area $n$. Let $m$ be the area of quadrilateral $ACEG$. What is $\dfrac{m}{n}?$

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

$\textbf{B}$

$ACEG$ is a square. WLOG $AB = 1,$ then using Law of Cosines, $AC^2 = [ACEG] = 1^2 + 1^2 - 2 \cos135^\circ = 2 + \sqrt{2}.$ The area of the octagon is just $[ACEG]$ plus the area of the four congruent (by symmetry) isosceles triangles, all an angle of $135$ in between two sides of length 1. Now,\[\dfrac{m}{n} = \dfrac{2 + \sqrt{2}}{2 + \sqrt{2} + 4 \cdot \tfrac{1}{2} \sin{135}} = \dfrac{2 + \sqrt{2}}{2 + 2 \sqrt{2}} = \dfrac{\sqrt{2}}{2}\]

In the complex plane, let $A$ be the set of solutions to $z^3 - 8 = 0$ and let $B$ be the set of solutions to $z^3 - 8z^2 - 8z + 64 = 0$. What is the greatest distance between a point of $A$ and a point of $B?$

$\textbf{(A) } 2\sqrt{3} \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 2\sqrt{21} \qquad \textbf{(E) } 9 + \sqrt{3}$

We solve each equation separately:

We solve $z^{3}-8=0$ by De Moivre's Theorem. Let $z=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,$ where $r$ is the magnitude of $z$ such that $r\geq0,$ and $\theta$ is the argument of $z$ such that $0\leq\theta<2\pi.$

We have\[z^3=r^3\operatorname{cis}(3\theta)=8(1),\]from which $$r^3=8$$ so $r=2.$

And $$\begin{cases} \begin{aligned} \cos(3\theta) &= 1 \\ \sin(3\theta) &= 0 \end{aligned} \end{cases}$$ so $3\theta=0,2\pi,4\pi,$ or $\theta=0,\dfrac{2\pi}{3},\dfrac{4\pi}{3}.$ The set of solutions to $z^{3}-8=0$ is $${A=\left\{2,-1+\sqrt{3}i,-1-\sqrt{3}i\right\}}$$ In the complex plane, the solutions form the vertices of an equilateral triangle whose circumcircle has center $0$ and radius $2.$

We solve $z^{3}-8z^{2}-8z+64=0$ by factoring by grouping.

We have\begin{align*} z^2(z-8)-8(z-8)&=0 \\ \bigl(z^2-8\bigr)(z-8)&=0. \end{align*}The set of solutions to $z^{3}-8z^{2}-8z+64=0$ is $${B=\left\{2\sqrt{2},-2\sqrt{2},8\right\}}$$ In the graph below, the points in set $A$ are shown in red, and the points in set $B$ are shown in blue. The greatest distance between a point of $A$ and a point of $B$ is the distance between $-1\pm\sqrt{3}i$ to $8,$ as shown in the dashed line segments.

By the Distance Formula, the answer is\[\sqrt{(-1-8)^2+\left(\pm\sqrt{3}-0\right)^2}=\sqrt{84}=2\sqrt{21}\]

A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0)$, $(2020, 0)$, $(2020, 2020),$ and $(0, 2020)$. The probability that the point is within $d$ units of a lattice point is $\dfrac{1}{2}$. (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth$?$

$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$

$\textbf{B}$

The diagram represents each unit square of the given $2020 \times 2020$ square. If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius $d$, the area covered by the circles should be $0.5$. Because of this, and the fact that there are four circles, we write \[4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}\] Solving for $d$, we obtain $d = \dfrac{1}{\sqrt{2\pi}}$, where with $\pi \approx 3$, we get $d \approx \dfrac{1}{\sqrt{6}} \approx \dfrac{1}{2.5} =0.4$.

The vertices of a quadrilateral lie on the graph of $y = \ln x$, and the $x$-coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln \dfrac{91}{90}$. What is the $x$-coordinate of the leftmost vertex?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 13$

$\textbf{D}$

Let the coordinates of the quadrilateral be $(n,\ln(n)),(n+1,\ln(n+1)),(n+2,\ln(n+2)),(n+3,\ln(n+3))$.

Therefore $[BCDE]=[ABCH]+[HCDG]+[GDEF]-[ABEF]$. All quadrilaterals on the right side of the equation are trapezoids, so

\begin{align*}

[BCDE]&=\frac{\ln(n+1)+\ln n}{2}+\frac{\ln(n+2)+\ln(n+1)}{2}+\frac{\ln(n+3)+\ln(n+2)}{2}-\frac{3(\ln(n+3)+\ln n)}{2}\\

&=\frac{2\ln(n+1)+2\ln(n+2)-2\ln(n+3)-2\ln n}{2}\\

&=\ln(n+1)+\ln(n+2)-\ln(n+3)-\ln n\\

&=\ln\tfrac{(n+1)(n+2)}{n(n+3)}\\

&=\ln\frac{91}{90}

\end{align*} Hence, we get $$\frac{(n+1)(n+2)}{n(n+3)}=\frac{91}{90}$$ Solving, we find $n=12$.

Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC = 20$, and $CD = 30$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E$, and $AE = 5$. What is the area of quadrilateral $ABCD$?

$\textbf{(A) } 330 \qquad\textbf{(B) } 340 \qquad\textbf{(C) } 350 \qquad\textbf{(D) } 360 \qquad\textbf{(E) } 370$

$\textbf{D}$

Since $AC=20$ and $CD=30$, we get $[ACD]=300$. Now we need to find $[ABC]$ to get the area of the whole quadrilateral. Drop an altitude from $B$ to $AC$ and call the point of intersection $F$. Let $FE=x$. Since $AE=5$, then $AF=5-x$.

By dropping this altitude, we see two similar triangles, $\triangle BFE \sim \triangle DCE$. Since $EC$ is $20-5=15$, and $DC=30$, we get that $BF=2x$.

We also see that $\triangle ABF\sim\triangle BCF$. So $\dfrac{BF}{CF}=\dfrac{AF}{BF}$, or $\dfrac{2x}{15+x}=\dfrac{5-x}{2x}$. Hence, we get $x=3$. Then $BF=2\cdot3=6$, $[ABC]=60$. Thus $[ABCD]=[ACD]+[ABC]=300+60=360$.

There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that\[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\]What is $k?$

$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$

$\textbf{C}$

First, substitute $2^{17}$ with $x$. Then, the given equation becomes $$\frac{x^{17}+1}{x+1}=x^{16}-x^{15}+x^{14}...-x^1+x^0$$ Now consider only $x^{16}-x^{15}$. This equals $x^{15}(x-1)=x^{15} \cdot (2^{17}-1)$. Note that $2^{17}-1$ equals $2^{16}+2^{15}+...+1$. Thus, we can see that $x^{16}-x^{15}$ forms the sum of 17 different powers of 2. Applying the same method to each of $x^{14}-x^{13}$, $x^{12}-x^{11}$, ... , $x^{2}-x^{1}$, we can see that each of the pairs forms the sum of 17 different powers of 2. This gives us $17 \cdot 8=136$. But we must count also the $x^0$ term. Thus, Our answer is $136+1=137$.

Let $T$ be the triangle in the coordinate plane with vertices $\left(0,0\right)$, $\left(4,0\right)$, and $\left(0,3\right)$. Consider the following five isometries (rigid transformations) of the plane: rotations of $90^{\circ}$, $180^{\circ}$, and $270^{\circ}$ counterclockwise around the origin, reflection across the $x$-axis, and reflection across the $y$-axis. How many of the $125$ sequences of three of these transformations (not necessarily distinct) will return $T$ to its original position? (For example, a $180^{\circ}$ rotation, followed by a reflection across the $x$-axis, followed by a reflection across the $y$-axis will return $T$ to its original position, but a $90^{\circ}$ rotation, followed by a reflection across the $x$-axis, followed by another reflection across the $x$-axis will not return $T$ to its original position.)

$\textbf{(A) } 12\qquad\textbf{(B) } 15\qquad\textbf{(C) }17 \qquad\textbf{(D) }20 \qquad\textbf{(E) }25$

$\textbf{A}$

First, any combination of motions we can make must reflect $T$ an even number of times. This is because every time we reflect $T$, it changes orientation. Once $T$ has been flipped once, no combination of rotations will put it back in place because it is the mirror image; however, flipping it again changes it back to the original orientation. Since we are only allowed $3$ transformations and an even number of them must be reflections, we either reflect $T$ $0$ times or $2$ times.

$\textbf{Case 1: 0 reflections on $T$.}$

In this case, we must use $3$ rotations to return $T$ to its original position. This could be $90+90+180$ or $180+270+270$. Each of the combinations above has 3 ways to arrange the order. So we have 6 ways for case 1.

$\textbf{Case 2: 2 reflections on $T$.}$

In this case, we first eliminate the possibility of having two of the same reflection. Since two reflections across the x-axis maps $T$ back to itself, inserting a rotation before, between, or after these two reflections would change $T$'s final location, meaning that any combination involving two reflections across the x-axis would not map $T$ back to itself. The same applies to two reflections across the y-axis.

Therefore, we must use one reflection about the x-axis, one reflection about the y-axis, and one rotation. Since a reflection about the x-axis changes the sign of the y component, a reflection about the y-axis changes the sign of the x component, and a $180^\circ$ rotation changes both signs, these three transformation composed (in any order) will suffice. It is therefore only a question of arranging the three, giving us $3! = 6$ combinations for case 2.

In conclusion, the answer is $6+6= 12$.

How many positive integers $n$ are there such that $n$ is a multiple of $5$, and the least common multiple of $5!$ and $n$ equals $5$ times the greatest common divisor of $10!$ and $n?$

$\textbf{(A) } 12 \qquad \textbf{(B) } 24 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 72$

$\textbf{D}$

We set up the following equation as the problem states:\[\text{lcm}{(5!, n)} = 5\text{gcd}{(10!, n)}\]Breaking each number into its prime factorization, we see that the equation becomes\[\text{lcm}{(2^3\cdot 3 \cdot 5, n)} = 5\text{gcd}{(2^8\cdot 3^4 \cdot 5^2 \cdot 7, n)}\]We can now determine the prime factorization of $n$. We know that its prime factors belong to the set $\{2, 3, 5, 7\}$. Since $n$ is a multiple of 5, we have $n=2^a\cdot3^b\cdot5^{c+1}\cot7^d$, where $a$, $b$, $c$ and $d$ are integers no less than 0. Then we have $$\text{lcm}(2^3\cdot3\cdot5,n)=2^{\max(a,3)}\cdot3^{\max(b,1)}\cdot5^{c+1}\cdot7^d$$ $$\text{gcd}{(2^8\cdot 3^4 \cdot 5^2 \cdot 7, n)}=2^{\min(a,8)}\cdot3^{\min(b,4)}\cdot5^{\min(c+1,2)}\cdot7^{\min(d,1)}$$ So the equation $\text{lcm}{(2^3\cdot 3 \cdot 5, n)} = 5\text{gcd}{(2^8\cdot 3^4 \cdot 5^2 \cdot 7, n)}$ becomes $$\max(a,3)=\min(a,8)\rightarrow a=3,4,5,6,7,8$$ $$\max(b,1)=\min(b,4)\rightarrow b=1,2,3,4$$ $$\min(c+1,2)+1=c+1\rightarrow c=2$$ $$\min(d,1)=d\rightarrow d=0,1$$ Therefore, we have $6\cdot4\cdot1\cdot2=48$ possible values of $n$.

Let $(a_n)$ and $(b_n)$ be the sequences of real numbers such that\[ (2 + i)^n = a_n + b_ni \]for all integers $n\geq 0$, where $i = \sqrt{-1}$. What is\[\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?\]

$\textbf{(A) }\dfrac 38\qquad\textbf{(B) }\dfrac7{16}\qquad\textbf{(C) }\dfrac12\qquad\textbf{(D) }\dfrac9{16}\qquad\textbf{(E) }\dfrac47$

$\textbf{B}$

Square the given equality to yield\[(3 + 4i)^n = (a_n + b_ni)^2 = (a_n^2 - b_n^2) + 2a_nb_ni,\]so $a_nb_n = \dfrac12\operatorname{Im}((3+4i)^n)$ and\[\sum_{n\geq 0}\frac{a_nb_n}{7^n} = \frac12\operatorname{Im}\left(\sum_{n\geq 0}\frac{(3+4i)^n}{7^n}\right) = \frac12\operatorname{Im}\left(\frac{1}{1 - \frac{3 + 4i}7}\right) =\frac 7{16}\]

Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (possibly empty, possibly all three dice) to reroll. After rerolling, he wins if and only if the sum of the numbers face up on the three dice is exactly $7$. Jason always plays to optimize his chances of winning. What is the probability that he chooses to reroll exactly two of the dice?

$\textbf{(A) } \dfrac{7}{36} \qquad\textbf{(B) } \dfrac{5}{24} \qquad\textbf{(C) } \dfrac{2}{9} \qquad\textbf{(D) } \dfrac{17}{72} \qquad\textbf{(E) } \dfrac{1}{4}$

$\textbf{A}$

We conclude all of the following after the initial roll:

$\textbf{Case 1:}$ Jason rerolls exactly zero dice if and only if the sum of the three dice is $7,$ in which the probability of winning is always $1.$

$\textbf{Case 2:}$ If Jason rerolls exactly one die, then the sum of the two other dice must be $2,3,4,5,$ or $6.$ The probability of winning is always $\dfrac16,$ as exactly $1$ of the $6$ possible outcomes of the die rerolled results in a win.

$\textbf{Case 3:}$ If Jason rerolls exactly two dice, then the outcome of the remaining die must be $1,2,3,4,$ or $5.$ Applying casework to the remaining die produces the following table:

The probability of winning is at most $\dfrac{5}{36}.$

$\textbf{Case 4:}$ If Jason (re)rolls all three dice, then the probability of winning is always $\dfrac{\binom62}{6^3}=\dfrac{15}{216}=\dfrac{5}{72}.$

For the denominator, rolling three dice gives a total of $6^3=216$ possible outcomes.

For the numerator, this is the same as counting the ordered triples of positive integers $(a,b,c)$ for which $a+b+c=7.$ Suppose that $7$ balls are lined up in a row. There are $6$ gaps between the balls, and placing dividers in $2$ of the gaps separates the balls into $3$ piles. From left to right, the numbers of balls in the piles correspond to $a,b,$ and $c,$ respectively. There are $\binom62=15$ ways to place the dividers. Note that the dividers' positions and the ordered triples have one-to-one correspondence, and $1\leq a,b,c\leq6$ holds for all such ordered triples.

The optimal strategy is that:

$\quad\bullet$ If Jason needs to reroll at least zero dice to win, then he rerolls exactly zero dice.

$\quad\bullet$ If Jason needs to reroll at least one die to win, then he rerolls exactly one die.

$\quad\bullet$ If Jason needs to reroll at least two dice to win, then he rerolls exactly two dice if and only if the probability of winning is greater than $\dfrac{5}{72},$ the probability of winning for rerolling all three dice.

The first three cases in the table above satisfy this requirement. We will analyze these cases by considering the initial outcomes of the two dice rerolled. Note that any of the three dice can be the remaining die, so we need a factor of $3$ for all counts in the third column:

Finally, the requested probability is\[\frac{3+12+27}{6^3}=\frac{42}{216}=\frac{7}{36}\]

Suppose that $\triangle ABC$ is an equilateral triangle of side length $s$, with the property that there is a unique point $P$ inside the triangle such that $AP = 1$, $BP = \sqrt{3}$, and $CP = 2$. What is $s?$

$\textbf{(A) } 1 + \sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \dfrac{8}{3} \qquad \textbf{(D) } \sqrt{5 + \sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$

$\textbf{B}$

We begin by rotating $\triangle{ APB}$ counterclockwise by $60^{\circ}$ about $A$, such that $P\mapsto Q$ and $B\mapsto C$. We see that $\triangle{ APQ}$ is equilateral with side length $1$, meaning that $\angle APQ = 60^{\circ}$. We also see that $\triangle{CPQ}$ is a $30$-$60$-$90$ right triangle, meaning that $\angle CPQ= 60^{\circ}$. Thus, by adding the two together, we see that $\angle APC = 120^{\circ}$.

We can now use the law of cosines as following:\begin{align*} s^2 &= (AP)^2 + (CP)^2 - 2\cdot AP\cdot CP\cdot \cos{\angle{APC}} \\ &= 1 + 4 - 2\cdot 1\cdot 2\cdot \cos{120^{\circ}} \\ &= 5 - 4\left(-\frac{1}{2}\right) \\ &= 7 \end{align*}giving us that $s=\sqrt{7}$.

The number $a = \dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers, has the property that the sum of all real numbers $x$ satisfying\[\lfloor x \rfloor \cdot \{x\} = a \cdot x^2\]is $420$, where $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$ and $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part of $x$. What is $p + q?$

$\textbf{(A) } 245 \qquad \textbf{(B) } 593 \qquad \textbf{(C) } 929 \qquad \textbf{(D) } 1331 \qquad \textbf{(E) } 1332$

$\textbf{C}$

Let $x_n$ be a root in the interval $(n,n+1)$. In this interval, $\lfloor x_n \rfloor = n$ and $\{x_n\}=x_n-n$, so we must have $ax_n^2 = nx_n-n^2$, i.e., $ax_n^2-nx_n+n^2=0$. We can homogenize this equation by setting $x_n=n\zeta$; then $x_1=\zeta$, and $\zeta$ is a root of $a\zeta^2-\zeta+1=0$.

Suppose $N$ is the largest integer for which there is such a root; we have, for $n=1,2,\ldots , N$,\[n < x_n = n\zeta < n+1\]Summing over $n\in \{1,2,\ldots , N\}$ we get\[\frac 12 N(N+1) < 420 = \frac 12 N(N+1)\zeta < \frac 12 N(N+3)\]From the right inequality we get $27< N$ and from the left one we get $N<29$. Thus $N=28$. Using this in the middle equality we get $\zeta = \dfrac{30}{29}$. Since $\zeta$ satisfies $a\zeta^2-\zeta+1=0$, we get\[a = \zeta^{-2}(\zeta-1)= \frac{29^2}{30^2}\cdot \frac 1{29}= \frac{29}{900}\]The answer is $29+900= 929.$