AMC10 2020 B

Because subtracting a negative number is equivalent to adding its opposite, the value of this expression is $$1 + 2 - 3 + 4 - 5 + 6 = 13 - 8 = 5.$$

The volume of each of the cubes of side length 1 is $1^3 = 1$. The volume of each of the cubes of side length 2 is $2^3 = 8$. Because there are 5 cubes of each type, the sum of the volumes is $5(1+8) = 45$.

Answer (E): Because \[ \frac{w}{y}=\frac{w}{x}\cdot\frac{x}{z}\cdot\frac{z}{y}=\frac{4}{3}\cdot\frac{6}{1}\cdot\frac{2}{3}=\frac{16}{3}, \] the requested ratio is $16:3$.

Consider in order the first five prime numbers that are possible values of $b$, namely 2, 3, 5, 7, and 11. The corresponding values of $a = 90 - b$ are 88, 87, 85, 83, and 79. The first prime in this latter list is 83, so $b = 7$ is the least value for $b$ for which both $a$ and $b$ are prime and $a > b$.

If all the tiles were distinguishable from each other, there would be $7! = 5040$ possible arrangements. However, this total must be divided by $2! = 2$ to account for the 2 different equivalent orderings of the 2 green tiles and also divided by $3! = 6$ to account for the 6 different equivalent orderings of the 3 yellow tiles. Therefore the answer is $$\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{2 \cdot 6} = 420.$$\n\nOR\n\nThere are $\binom{7}{3}$ ways to choose the positions for the yellow tiles, $\binom{4}{2}$ ways to choose the positions for the green tiles among the remaining positions, and then 2 ways to choose the positions for the brown and purple tiles. Therefore the answer is $$\binom{7}{3} \cdot \binom{4}{2} \cdot 2 = 420.$$

A five-digit palindrome is determined by its leftmost three digits, so the next palindrome after 15951 is 16061. This occurs in $16061 - 15951 = 110$ miles. Therefore her average speed over those 2 hours was $110 \div 2 = 55$ miles per hour.

In order for a perfect square $n^2$ to be even, $n$ must be a multiple of 2. Similarly, in order for $n^2$ to be a multiple of 3, $n$ must be a multiple of 3. Therefore $n$ must itself be a multiple of 6. Because $(7 \cdot 6)^2 = 1764 < 2020$ and $(8 \cdot 6)^2 = 2304 > 2020$, there are 7 such squares, namely $6^2, 12^2, 18^2, 24^2, 30^2, 36^2$, and $42^2$.

Answer (D): In order for $\triangle PQR$ to have area 12, point $R$ must lie on one of the two lines parallel to and 3 units away from line $\overline{PQ}$. If $\overline{PQ}$ is to be a leg of the right triangle, then there are 2 possible right triangles if $P$ is the right angle, and there are an additional 2 possible right triangles if $Q$ is the right angle. These 4 triangles are shown in the figure on the left. If instead $\overline{PQ}$ is the hypotenuse of the right triangle, then $R$ must lie on the circle with diameter $\overline{PQ}$. Because the radius of this circle is 4, each of the two lines parallel to and 3 units from $\overline{PQ}$ intersects the circle in exactly 2 points. These 4 triangles are shown in the figure on the right. The figures below show the 8 possible locations for $R$ that make $\triangle PQR$ a right triangle.

Answer (D): The given equation is equivalent to $x^{2020}+y^2-2y+1=1$, which is equivalent to $x^{2020}+(y-1)^2=1$. The only way to write 1 as the sum of the squares of two integers is as 0 + 1 or 1 + 0, so $x$ must be 0 or $\pm 1$. The complete set of solutions is $\{(0,0),(0,2),(1,1),(-1,1)\}$, which has 4 elements.

Answer (C): The length of the circular portion of the sector is $\frac{3}{4}\cdot 2\pi \cdot 4 = 6\pi$. That arc forms the circumference of the base of the cone, so the radius of the base of the cone is $r=\frac{6\pi}{2\pi}=3$. The slant height of the cone is the radius of the sector. By the Pythagorean Theorem the cone’s height is $h=\sqrt{4^2-3^2}=\sqrt{7}$. The volume of the cone is $$ \frac{1}{3}\pi r^2h=\frac{1}{3}\pi\cdot 3^2\cdot \sqrt{7}=3\pi\sqrt{7}. $$

Answer (D): Once Harold has selected his books, there are $\binom{5}{2}$ ways for Betty to choose 2 of Harold’s books and $\binom{5}{3}$ ways for her to choose 3 books that were not selected by Harold. There are $\binom{10}{5}$ ways in all for Betty to select her books. Thus the required probability is \[ \frac{\binom{5}{2}\cdot\binom{5}{3}}{\binom{10}{5}}=\frac{25}{63}. \]

Answer (D): Observe that $20^{20}=2^{20}\cdot 10^{20}=1024^{2}\cdot 10^{20}$. Because $10^{3}<1024<2000<10^{3.5}$, it follows that $10^{6}<1024^{2}<10^{7}$. Therefore $\frac{1}{20^{20}}$ is strictly between $10^{-27}$, whose decimal representation consists of 26 zeros after the decimal point followed by a 1, and $10^{-26}$, whose decimal representation consists of 25 zeros after the decimal point followed by a 1. Hence there are 26 zeros to the right of the decimal point preceding the first nonzero digit. The exact value is in fact \[ \frac{1}{20^{20}}=0.00000\ 00000\ 00000\ 00000\ 00000\ 09536\ 74316\ 40625. \]

Answer (B): Note that Andy’s moves east and west cause changes in his $x$-coordinate, and moves north and south cause changes in his $y$-coordinate. Summing all the changes gives the $x$- and $y$-coordinates of Andy’s final location: \[ x=-20+(1-3)+(5-7)+\cdots+(2017-2019) \] \[ =-20+505(-2) \] \[ =-1030 \] and \[ y=20+(2-4)+(6-8)+\cdots+(2018-2020) \] \[ =20+505(-2) \] \[ =-990. \] The final location is $(-1030,-990)$.

Answer (D): Consider the semicircles on two adjacent sides of the hexagon shown in the figure below, where $F$ and $G$ are the centers of the semicircles and the midpoints of their respective sides. Because $\angle FBG = 120^\circ$, it follows that $\angle FBD = 60^\circ$ and $\triangle FBD$ is equilateral. Thus the segment of the circle centered at $F$ determined by arc $BD$ (that is, the region, shaded in the figure, that lies inside the sector spanned by arc $BD$ but outside $\triangle FBD$) has area $\frac{\pi}{6}-\frac{\sqrt{3}}{4}$. The requested area is then the area of the hexagon, minus the areas of the six semicircles, plus 12 times the area of that segment, which is $$ 6\left(\frac{\sqrt{3}}{4}\cdot 2^2\right)-6\left(\frac{\pi}{2}\right)+12\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)=3\sqrt{3}-\pi. $$

Answer (D): The original list had period 5, with 12345 repeated. After every third number was deleted, the resulting list began 124523513412452..., which has period 10. After every fourth number was deleted from that list, the resulting list began 12423534145251312423..., which has period 15. After every fifth number was deleted from that list, the resulting list began 12425341525112425..., which has period 12. Because $2019 \bmod 12 = 3$, the numbers in positions 2019, 2020, and 2021 at that point were the same as the numbers in positions 3, 4, and 5, respectively. The requested sum is $4+2+5=11$.

Bela has a straightforward strategy that will guarantee winning. At his first turn he chooses the number $\frac{n}{2}$. This splits the playing field into two parts—the numbers less than $\frac{n}{2}$ and the numbers greater than $\frac{n}{2}$. Thereafter, whatever legal move Jenn makes, Bela makes the symmetric move in the other half of the playing field. Specifically, if Jenn chooses $x$, then Bela chooses $n - x$. By symmetry, if Jenn's move is possible, so is Bela's. Thus the first player to be unable to make a legal move will be Jenn, and she will lose. The game will end after a finite number of moves, because of the restriction that the chosen numbers are at least one unit apart.

Answer (C): This problem can be modeled by a regular decagon with the 5 diagonals connecting each point to the point opposite to it. A line segment between two vertices indicates that the people represented by those vertices know each other. See the figure below. The required pairings are then sets of 5 of the 15 line segments in this figure such that no 2 line segments in a set share an endpoint. (This is known as a perfect matching in graph theory.) The number of pairings can be counted by focusing on the number of diagonals used in the pairing. • There is 1 pairing using all 5 diagonals, and any pairing that includes 4 of the diagonals must use all of them. • Suppose a pairing uses exactly 3 diagonals. If 3 of the endpoints of these diagonals are consecutive points on the decagon, then a pairing can be formed by including sides of the decagon as the other 2 segments. There are 5 different pairings of this form. On the other hand, any pairing that includes 3 of the diagonals whose endpoints are not consecutive points on the decagon must use all the diagonals. • There are no pairings using exactly 2 of the diagonals, because any pairing that includes 2 of the diagonals will force at least 3 of the diagonals to be used. • If a pairing uses exactly 1 diagonal, then the rest of the pairing is uniquely determined. There are 5 different pairings of this form. • If a pairing uses no diagonals, then it must use nonadjacent sides of the decagon, and there are 2 different pairings of this form. As shown in the figure, the requested number of pairings is $1+5+5+2=13$.

By symmetry it may be assumed without loss of generality that George's first draw is red and therefore the urn contains two red balls and one blue ball before the second draw. With probability $\frac{2}{3}$, George will draw a red ball next. In that case the only way for the urn to end up with three balls of each color is for the next two draws to be blue, which will happen with probability $\frac{1}{4} \cdot \frac{2}{5} = \frac{1}{10}$. On the other hand, George will get a blue ball on his second draw with probability $\frac{1}{3}$, and the urn will then have two balls of each color. In that case no matter what color he gets on his third draw, for the urn to end up with three balls of each color the fourth draw must be different from the third draw, which will happen with probability $\frac{2}{5}$. Therefore, the probability that the urn will end up with three balls of each color is $$\frac{2}{3} \cdot \frac{1}{10} + \frac{1}{3} \cdot \frac{2}{5} = \frac{1}{5}.$$

The number of ways of dealing 10 cards from 52 is $\binom{52}{10}$. Expanding and simplifying this quantity gives $$\frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot 43}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2} = 2^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 23 \cdot 43 \cdot 47.$$ Dividing this quantity by 10 gives $$2 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 23 \cdot 43 \cdot 47.$$ The value of A can then be found by computing this quantity modulo 10: $$2 \cdot ((-3) \cdot 1 \cdot 3) \cdot ((-3) \cdot 3) \cdot (3 \cdot (-3)) \equiv 2 \cdot 1 \cdot 1 \cdot 1 = 2.$$

The solid $S(r)$ consists of the original prism, a prism with height $r$ on each face, a quarter cylinder of radius $r$ on each edge, and $\frac{1}{8}$ of a sphere of radius $r$ at each vertex, together with their interiors. Therefore the volume is equal to $$4 \cdot 3 \cdot 1 + 2(4 \cdot 3 \cdot r + 4 \cdot 1 \cdot r + 3 \cdot 1 \cdot r) + 4 \left( \frac{1}{4} \pi r^2 (4 + 3 + 1) \right) + 8 \left( \frac{1}{8} \cdot \frac{4}{3} \pi r^3 \right),$$ which simplifies to $\frac{4}{3}\pi r^3 + 8\pi r^2 + 38r + 12$. Thus the requested fraction is $$\frac{bc}{ad} = \frac{8\pi \cdot 38}{\frac{4}{3}\pi \cdot 12} = 19.$$

Square $ABCD$ has area $4$, so its side length is $AB=2$. Isosceles right triangle $HAE$ has area $1$, so leg length $AE=\sqrt{2}$, and thus $EB=2-\sqrt{2}$. Extend $\overline{EH}$ and $\overline{BC}$ to meet at $K$. Isosceles right triangle $KBE$ has area $\frac{1}{2}(2-\sqrt{2})^2=3-2\sqrt{2}$, so isosceles right triangle $FIK$ has area $4-2\sqrt{2}=\frac{1}{2}FI^2$. Hence $FI^2=8-4\sqrt{2}$.

Observe that $$\begin{align*}2^{202} + 202 &= (2^{101})^2 + 2 \cdot 2^{101} + 1 - 2 \cdot 2^{101} + 201 \\&= (2^{101} + 1)^2 - 2^{102} + 201 \\&= (2^{101} - 2^{51} + 1)(2^{101} + 2^{51} + 1) + 201.\end{align*}$$ Therefore the remainder is $201$.

Answer (C): Label the sides of the square $p$, $q$, $r$, and $s$, with $p$ being the side that starts as the upper horizontal side, $q$ being the side that starts as the left vertical side, $r$ being the side that starts as the bottom horizontal side, $s$ being the side that starts as the right vertical side. The position of the square after a sequence of transformations can be represented by a list of these four letters, starting with the side that is in the top horizontal position and moving counterclockwise. Thus $L$ takes $pqrs$ to $spqr$, $R$ takes $pqrs$ to $qrsp$, $H$ takes $pqrs$ to $psrq$, and $V$ takes $pqrs$ to $rqps$. Let $E=\{pqrs, qpsr, rspq, srqp\}$ and let $O=\{psrq, qrsp, rqps, spqr\}$. Each of the given transformations $L$, $R$, $H$, and $V$ will map each element of $E$ to a different element of $O$, and vice versa. Because $19$ is an odd number, any sequence of $19$ transformations will map the square to a position in set $O$, and exactly one of the given transformations will map this position back to position $pqrs$. Therefore there are $4^{19}=2^{38}$ possible sequences that will return the square to its original position.

Answer (C): Each positive integer $n$ can be uniquely expressed as $k^2+j$, where $j$ and $k$ are integers with $k\ge 1$ and $0\le j\le 2k$. When $n$ is so expressed, $\lfloor\sqrt{n}\rfloor=k$, and the given equation can be written as $$ k^2-(70k-j)+1000=0. $$ For a fixed value of $k$, there is a solution for $j$ if and only if $$ k^2-70k+1000\le 0 \quad\text{and}\quad k^2-68k+1000\ge 0. $$ Solving the first inequality gives $20\le k\le 50$, and solving the second gives $k\le 34-\sqrt{156}\approx 21\frac12$ or $k\ge 34+\sqrt{156}\approx 46\frac12$. Thus the suitable values of $k$ are $20, 21, 47, 48, 49,$ and $50$, and, for each of these, the unique value of $j$ is $(k-20)(50-k)$. Hence the given equation has 6 solutions for $n$, namely $400, 470, 2290, 2360, 2430,$ and $2500$.

Answer (A): Clearly $D(1)=0$. Suppose $n>1$. The first factor $f_1$ in a representation is a divisor of $n$ with $f_1>1$. If $f_1=n$, then there is a representation with just one factor. Suppose that $f_1$ is a divisor of $n$ and $1<f_1<n$. Then the further factors after $f_1$ are exactly the representations of $\frac{n}{f_1}$. This gives a recurrence relation: If $n>1$, then \[ D(n)=1+\sum_{\substack{f\mid n\\1<f<n}} D\!\left(\frac{n}{f}\right). \] Thus a table of values of $D(2^j3^k)$ can be constructed for $j=0,1,\ldots,5$, $k=0,1$. \[ \begin{array}{c|cc} j\backslash k & 0 & 1\\ \hline 0 & 0 & 1\\ 1 & 1 & 3\\ 2 & 2 & 8\\ 3 & 4 & 20\\ 4 & 8 & 48\\ 5 & 16 & 112 \end{array} \] From the table, it can be seen that $D(96)=112$.