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2017 AMC 12A 真题及答案解析

Category: AMC美国数学竞赛, 国际竞赛, 福利干货, 竞赛资料 Date: 2017年7月31日下午7:01

2017 AMC 12A真题

答案解析请参考文末

Problem 1

Pablo buys popsicles for his friends. The store sells single popsicles for $1 each, 3-popsicle boxes for $2, and 5-popsicle boxes for $3. What is the greatest number of popsicles that Pablo can buy with $8? $textbf{(A)} 8qquadtextbf{(B)} 11qquadtextbf{(C)} 12qquadtextbf{(D)} 13qquadtextbf{(E)} 15$

Problem 2

The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers? $textbf{(A)} 1qquadtextbf{(B)} 2qquadtextbf{(C)} 4qquadtextbf{(D)} 8qquadtextbf{(E)} 12$

Problem 3

Ms. Carroll promised that anyone who got all the multiple choice questions right on the upcoming exam would receive an A on the exam. Which one of these statements necessarily follows logically? $textbf{(A)} text{ If Lewis did not receive an A, then he got all of the multiple choice questions wrong.} qquadtextbf{(B)} text{ If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong.} qquadtextbf{(C)} text{ If Lewis got at least one of the multiple choice questions wrong, then he did not receive an A.} qquadtextbf{(D)} text{ If Lewis received an A, then he got all of the multiple choice questions right.} qquadtextbf{(E)} text{ If Lewis received an A, then he got at least one of the multiple choice questions right.}$

Problem 4

Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip? $textbf{(A)} 30%qquadtextbf{(B)} 40%qquadtextbf{(C)} 50%qquadtextbf{(D)} 60%qquadtextbf{(E)} 70%$

Problem 5

At a gathering of $30$ people, there are $20$ people who all know each other and $10$ people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur? $textbf{(A)} 240qquadtextbf{(B)} 245qquadtextbf{(C)} 290qquadtextbf{(D)} 480qquadtextbf{(E)} 490$

Problem 6

Joy has $30$ thin rods, one each of every integer length from $1 text{ cm}$ through $30 text{ cm}$ . She places the rods with lengths $3 text{ cm}$ , $7 text{ cm}$ , and $15 text{cm}$ on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod? $textbf{(A)} 16 qquadtextbf{(B)} 17 qquadtextbf{(C)} 18 qquadtextbf{(D)} 19 qquadtextbf{(E)} 20$

Problem 7

Define a function on the positive integers recursively by $f(1) = 2$ , $f(n) = f(n-1) + 2$ if $n$ is even, and $f(n) = f(n-2) + 2$ if $n$ is odd and greater than $1$ . What is $f(2017)$ ? $textbf{(A)} 2017 qquadtextbf{(B)} 2018 qquadtextbf{(C)} 4034 qquadtextbf{(D)} 4035 qquadtextbf{(E)} 4036$

Problem 8

The region consisting of all points in three-dimensional space within $3$ units of line segment $overline{AB}$ has volume $216 pi$ . What is the length $AB$ ? $textbf{(A)} 6 qquadtextbf{(B)} 12 qquadtextbf{(C)} 18 qquadtextbf{(D)} 20 qquadtextbf{(E)} 24$

Problem 9

Let $S$ be the set of points $(x,y)$ in the coordinate plane such that two of the three quantities $3$ , $x+2$ , and $y-4$ are equal and the third of the three quantities is no greater than the common value. Which of the following is a correct description of $S$ ? $textbf{(A)} text{a single point} qquadtextbf{(B)} text{two intersecting lines} qquadtextbf{(C)} text{three lines whose pairwise intersections are three distinct points} qquadtextbf{(D)} text{a triangle}qquadtextbf{(E)} text{three rays with a common point}$

Problem 10

Chloé chooses a real number uniformly at random from the interval $[ 0,2017 ]$ . Independently, Laurent chooses a real number uniformly at random from the interval $[ 0 , 4034 ]$ . What is the probability that Laurent's number is greater than Chloe's number? $textbf{(A)} dfrac{1}{2} qquadtextbf{(B)} dfrac{2}{3} qquadtextbf{(C)} dfrac{3}{4} qquadtextbf{(D)} dfrac{5}{6} qquadtextbf{(E)} dfrac{7}{8}$

Problem 11

Claire adds the degree measures of the interior angles of a convex polygon and arrives at a sum of $2017$ . She then discovers that she forgot to include one angle. What is the degree measure of the forgotten angle? $textbf{(A)} 37qquadtextbf{(B)} 63qquadtextbf{(C)} 117qquadtextbf{(D)} 143qquadtextbf{(E)} 163$

Problem 12

There are $10$ horses, named Horse 1, Horse 2, $ldots$ , Horse 10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse $k$ runs one lap in exactly $k$ minutes. At time 0 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds. The least time $S > 0$ , in minutes, at which all $10$ horses will gain simultaneously be at the starting point is $S = 2520$ . Let $T>0$ be the least time, in minutes, such that at least $5$ of the horses are again at the starting point. What is the sum of the digits of $T$ ? $textbf{(A)} 2qquadtextbf{(B)} 3qquadtextbf{(C)} 4qquadtextbf{(D)} 5qquadtextbf{(E)} 6$

Problem 13

Driving at a constant speed, Sharon usually takes $180$ minutes to drive from her house to her mother's house. One day Sharon begins the drive at her usual speed, but after driving $frac{1}{3}$ of the way, she hits a bad snowstorm and reduces her speed by $20$ miles per hour. This time the trip takes her a total of $276$ minutes. How many miles is the drive from Sharon's house to her mother's house? $textbf{(A)} 132 qquadtextbf{(B)} 135 qquadtextbf{(C)} 138 qquadtextbf{(D)} 141 qquadtextbf{(E)} 144$

Problem 14

Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of $5$ chairs under these conditions? $textbf{(A)} 12 qquad textbf{(B)} 16 qquadtextbf{(C)} 28 qquadtextbf{(D)} 32 qquadtextbf{(E)} 40$

Problem 15

Let $f(x) = sin{x} + 2cos{x} + 3tan{x}$ , using radian measure for the variable $x$ . In what interval does the smallest positive value of $x$ for which $f(x) = 0$ lie? $textbf{(A)} (0,1) qquad textbf{(B)} (1, 2) qquadtextbf{(C)} (2, 3) qquadtextbf{(D)} (3, 4) qquadtextbf{(E)} (4,5)$

Problem 16

In the figure below, semicircles with centers at $A$ and $B$ and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $JK$ . The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at $P$ is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at $P$ ? $[asy] size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (-1,0)+(2+6/7)*dir(36.86989); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot(P); [/asy]$ $textbf{(A)} frac{3}{4} qquad textbf{(B)} frac{6}{7} qquadtextbf{(C)} frac{1}{2}sqrt{3} qquadtextbf{(D)} frac{5}{8}sqrt{2} qquadtextbf{(E)} frac{11}{12}$

Problem 17

There are $24$ different complex numbers $z$ such that $z^{24}=1$ . For how many of these is $z^6$ a real number? $textbf{(A)} 0 qquadtextbf{(B)} 4 qquadtextbf{(C)} 6 qquadtextbf{(D)} 12 qquadtextbf{(E)} 24$

Problem 18

Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ , $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$ ? $textbf{(A)} 1 qquadtextbf{(B)} 3qquadtextbf{(C)} 12qquadtextbf{(D)} 1239qquadtextbf{(E)} 1265$

Problem 19

A square with side length $x$ is inscribed in a right triangle with sides of length $3$ , $4$ , and $5$ so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length $y$ is inscribed in another right triangle with sides of length $3$ , $4$ , and $5$ so that one side of the square lies on the hypotenuse of the triangle. What is $tfrac{x}{y}$ ? $textbf{(A)} frac{12}{13} qquad textbf{(B)} frac{35}{37} qquadtextbf{(C)} 1 qquadtextbf{(D)} frac{37}{35} qquadtextbf{(E)} frac{13}{12}$

Problem 20

How many ordered pairs $(a,b)$ such that $a$ is a positive real number and $b$ is an integer between $2$ and $200$ , inclusive, satisfy the equation $(log_b a)^{2017}=log_b(a^{2017})?$ $textbf{(A)} 198qquadtextbf{(B)} 199qquadtextbf{(C)} 398qquadtextbf{(D)} 399qquadtextbf{(E)} 597$

Problem 21

A set $S$ is constructed as follows. To begin, $S = {0,10}$ . Repeatedly, as long as possible, if $x$ is an integer root of some polynomial $a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{1}x + a_0$ for some $ngeq{1}$ , all of whose coefficients $a_i$ are elements of $S$ , then $x$ is put into $S$ . When no more elements can be added to $S$ , how many elements does $S$ have? $textbf{(A)} 4 qquad textbf{(B)} 5 qquadtextbf{(C)} 7 qquadtextbf{(D)} 9 qquadtextbf{(E)} 11$

Problem 22

A square is drawn in the Cartesian coordinate plane with vertices at $(2, 2)$ , $(-2, 2)$ , $(-2, -2)$ , $(2, -2)$ . A particle starts at $(0,0)$ . Every second it moves with equal probability to one of the eight lattice points (points with integer coordinates) closest to its current position, independently of its previous moves. In other words, the probability is $1/8$ that the particle will move from $(x, y)$ to each of $(x, y + 1)$ , $(x + 1, y + 1)$ , $(x + 1, y)$ , $(x + 1, y - 1)$ , $(x, y - 1)$ , $(x - 1, y - 1)$ , $(x - 1, y)$ , or $(x - 1, y + 1)$ . The particle will eventually hit the square for the first time, either at one of the 4 corners of the square or at one of the 12 lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is $m/n$ , where $m$ and $n$ are relatively prime positive integers. What is $m + n$ ? $textbf{(A)} 4 qquad textbf{(B)} 5 qquadtextbf{(C)} 7 qquadtextbf{(D)} 15 qquadtextbf{(E)} 39$

Problem 23

For certain real numbers $a$ , $b$ , and $c$ , the polynomial $[g(x) = x^3 + ax^2 + x + 10]$ has three distinct roots, and each root of $g(x)$ is also a root of the polynomial $[f(x) = x^4 + x^3 + bx^2 + 100x + c.]$ What is $f(1)$ ? $textbf{(A)} -9009 qquadtextbf{(B)} -8008 qquadtextbf{(C)} -7007 qquadtextbf{(D)} -6006 qquadtextbf{(E)} -5005$

Problem 24

Quadrilateral $ABCD$ is inscribed in circle $O$ and has side lengths $AB=3, BC=2, CD=6$ , and $DA=8$ . Let $X$ and $Y$ be points on $overline{BD}$ such that $frac{DX}{BD} = frac{1}{4}$ and $frac{BY}{BD} = frac{11}{36}$ . Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $overline{AD}$ . Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $overline{AC}$ . Let $G$ be the point on circle $O$ other than $C$ that lies on line $CX$ . What is $XFcdot XG$ ? $textbf{(A) }17qquadtextbf{(B) }frac{59 - 5sqrt{2}}{3}qquadtextbf{(C) }frac{91 - 12sqrt{3}}{4}qquadtextbf{(D) }frac{67 - 10sqrt{2}}{3}qquadtextbf{(E) }18$

Problem 25

The vertices $V$ of a centrally symmetric hexagon in the complex plane are given by $[V=left{ sqrt{2}i,-sqrt{2}i, frac{1}{sqrt{8}}(1+i),frac{1}{sqrt{8}}(-1+i),frac{1}{sqrt{8}}(1-i),frac{1}{sqrt{8}}(-1-i) right}.]$ For each $j$ , $1leq jleq 12$ , an element $z_j$ is chosen from $V$ at random, independently of the other choices. Let $P={prod}_{j=1}^{12}z_j$ be the product of the $12$ numbers selected. What is the probability that $P=-1$ ? $textbf{(A) } dfrac{5cdot11}{3^{10}} qquad textbf{(B) } dfrac{5^2cdot11}{2cdot3^{10}} qquad textbf{(C) } dfrac{5cdot11}{3^{9}} qquad textbf{(D) } dfrac{5cdot7cdot11}{2cdot3^{10}} qquad textbf{(E) } dfrac{2^2cdot5cdot11}{3^{10}}$

2017AMC12A详细解析

1.
By the greedy algorithm, we can take two 5-popsicle boxes and one 3-popsicle box with $$8$ . To prove that this is optimal, consider an upper bound as follows: at the rate of $$3$ per 5 popsicles, we can get $frac{40}{3}$ popsicles, which is less than 14. $boxed{textbf{D}}$ .

2.
Let $x, y$ be our two numbers. Then $x+y = 4xy$ . Thus, $frac{1}{x} + frac{1}{y} = frac{x+y}{xy} = 4$ . $boxed{ textbf{C}}$ .

3.
Taking the contrapositive of the statement "if he got all of them right, he got an A" yields "if he didn't get an A, he didn't get all of them right", yielding the answer $boxed{textbf{(B)}}$ .

4.
Let $j$ represent how far Jerry walked, and $s$ represent how far Sylvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides, $j = 2$ Since Silvia walked the diagonal, she walked the hypotenuse of a 45, 45, 90 triangle with leg length 1. Thus, $s = sqrt{2} = 1.414...$ We can then take $frac{j-s}{j} approx frac{2 - 1.4}{2} = 0.3 = 30%$ $boxed{ textbf{A}}$ .

5.
Let the group of people who all know each other be $A$ , and let the group of people who know no one be $B$ . Handshakes occur between each pair $(a,b)$ such that $ain A$ and $bin B$ , and between each pair of members in $B$ . Thus, the answer is $|A||B|+{|B|choose 2} = 20cdot 10+{10choose 2} = 200+45 = boxed{(B)= 245}$
6.
The quadrilateral cannot be a straight line. Thus, the fourth side must be longer than $15 - (3 + 7) = 5$ and shorter than $15 + 3 + 7$ = 25. This means Joy can use the 19 possible integer rod lengths that fall into $[6, 24]$ . However, she has already used the rods of length $7$ cm and $15$ cm so the answer is $19 - 2 = 17$ $boxed{textbf{(B)}}$
7.
This is a recursive function, which means the function is used to evaluate itself. To solve this, we must identify the base case, $f(1)=2$ . We also know that when $n$ is odd, $f(n)=f(n-2)+2$ . Thus we know that $f(2017)=f(2015)+2$ . Thus we know that n will always be odd in the recursion of $f(2017)$ , and we add $2$ each recursive cycle, which there are $1008$ of. Thus the answer is $1008*2+2=2018$ , which is answer $boxed{textbf{(B)}}$ .

8.
Let the length $AB$ be $L$ . Then, we see that the region is just the union of the cylinder with central axis $overline{AB}$ and radius $3$ and the two hemispheres connected to each face of the cylinder (also with radius $3$ ). Thus the volume is $9pi L + frac{4}{3}pi(3)^3 = 9pi L + 36pi ( = 216pi)$ $9pi L = 180pi$ $L = boxed{(D)= 20}$
9.
If the two equal values are $3$ and $x+2$ , then $x=1$ . Also, $y-4leqslant 3$ because 3 is the common value. Solving for $y$ , we get $yleqslant 7$ . Therefore the portion of the line $x=1$ where $yleqslant 7$ is part of $S$ . This is a ray with an endpoint of $(1, 7)$ .
Similar to the process above, we assume that the two equal values are $3$ and $y-4$ . Solving the equation $3=y-4$ then $y=7$ . Also, $x+2leqslant 3$ because 3 is the common value. Solving for $x$ , we get $xleqslant 1$ . Therefore the portion of the line $y=7$ where $xleqslant 1$ is also part of $S$ . This is another ray with the same endpoint as the above ray: $(1, 7)$ .
If $x+2$ and $y-4$ are the two equal values, then $x+2=y-4$ . Solving the equation for $y$ , we get $y=x+6$ . Also $3leqslant y-4$ because $y-4$ is one way to express the common value (using $x-2$ as the common value works as well). Solving for $y$ , we get $ygeqslant 7$ . Therefore the portion of the line $y=x+6$ where $ygeqslant 7$ is part of $S$ like the other two rays. The lowest possible value that can be achieved is also $(1, 7)$ .
Since $S$ is made up of three rays with common endpoint $(1, 7)$ , the answer is $boxed{E}$ .

10.
Suppose Laurent's number is in the interval $[ 0, 2017 ]$ . Then, by symmetry, the probability of Laurent's number being greater is $dfrac{1}{2}$ . Next, suppose Laurent's number is in the interval $[ 2017, 4034 ]$ . Then Laurent's number will be greater with probability $1$ . Since each case is equally likely, the probability of Laurent's number being greater is $dfrac{1 + frac{1}{2}}{2} = dfrac{3}{4}$ , so the answer is C.

11.
We know that the sum of the interior angles of the polygon is a multiple of $180$ . Note that $leftlceilfrac{2017}{180}rightrceil = 12$ and $180cdot 12 = 2160$ , so the angle Claire forgot is $equiv 2160-2017=143mod 180$ . Since the polygon is convex, the angle is $leq 180$ , so the answer is $boxed{(D) = 143}$ .

12.
We know that Horse $k$ will be at the starting point after $n$ minutes if $k|n$ . Thus, we are looking for the smallest $n$ such that at least $5$ of the numbers ${1,2,cdots,10}$ divide $n$ . Thus, $n$ has at least $5$ positive integer divisors.
We quickly see that $12$ is the smallest number with at least $5$ positive integer divisors, and that $1,2,3,4,6$ are each numbers of horses. Thus, our answer is $1+2=boxed{textbf{(B) } 3}$ .

13.
Let total distance be $x$ . Her speed in miles per minute is $tfrac{x}{180}$ . Then, the distance that she drove before hitting the snowstorm is $tfrac{x}{3}$ . Her speed in snowstorm is reduced $20$ miles per hour, or $tfrac{1}{3}$ miles per minute. Knowing it took her $276$ minutes in total, we create equation: $[text{Time before Storm}, + , text{Time after Storm} = text{Total Time} Longrightarrow]$ $[frac{text{Distance before Storm}}{text{Speed before Storm}} + frac{text{Distance in Storm}}{text{Speed in Storm}} = text{Total Time} Longrightarrow frac{tfrac{x}{3}}{tfrac{x}{180}} + frac{tfrac{2x}{3}}{tfrac{x}{180} - tfrac{1}{3}} = 276]$ Solving equation, we get $x=135$ $Longrightarrow boxed{B}$ .

14.
Alice may sit in the center chair, in an end chair, or in a next-to-end chair. Suppose she sits in the center chair. The 2nd and 4th chairs (next to her) must be occupied by Derek and Eric, in either order, leaving the end chairs for Bob and Carla in either order; this yields $2! * 2! = 4$ ways to seat the group.
Next, suppose Alice sits in one of the end chairs. Then the chair beside her will be occupied by either Derek or Eric. The center chair must be occupied by Bob or Carla, leaving the last two people to fill the last two chairs in either order. $2$ ways to seat Alice times $2$ ways to fill the next chair times $2$ ways to fill the center chair times $2$ ways to fill the last two chairs yields $2 * 2 * 2 * 2 = 16$ ways to fill the chairs.
Finally, suppose Alice sits in the second or fourth chair. Then the chairs next to her must be occupied by Derek and Eric in either order, and the other two chairs must be occupied by Bob and Carla in either order. This yields $2 * 2 * 2 = 8$ ways to fill the chairs.
In total, there are $4 + 8 + 16$ ways to fill the chairs, so the answer is $boxed{bold{(C)} 28}$ .

15.
We must first get an idea of what $f(x)$ looks like:
Between 0 and 1, $f(x)$ starts at $2$ and increases; clearly there is no zero here.
Between 1 and $frac{pi}{2}$ , $f(x)$ starts at a positive number and increases to $infty$ ; there is no zero here either.
Between $frac{pi}{2}$ and 3, $f(x)$ starts at $-infty$ and increases to some negative number; there is no zero here either.
Between 3 and $pi$ , $f(x)$ starts at some negative number and increases to -2; there is no zero here either.
Between $pi$ and $pi+frac{pi}{4} < 4$ , $f(x)$ starts at -2 and increases to $-frac{sqrt2}{2} + 2left(-frac{sqrt2}{2}right) + 3left(1right)=3left(1-frac{sqrt2}{2}right)>0$ . There is a zero here by the Intermediate Value Theorem. Therefore, the answer is $boxed{(D)}$ .

16.
Connect the centers of the tangent circles! (call the center of the large circle $C$ ) $[asy] size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label("$A$",A,SW); label("$C$",C,S); label("$B$",B,SE); label("$P$",P,N); [/asy]$ Notice that we don't even need the circles anymore; thus, draw triangle $Delta ABP$ with cevian $PC$ : $[asy] size(5cm); draw((-1,0)--(2,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--P); label("$A$",A,SW); label("$C$",C,S); label("$B$",B,SE); label("$P$",P,N); [/asy]$ and use Stewart's Theorem: $[AB cdot AC cdot BC + AB cdot {CP}^2 = AC cdot {BP}^2 + BC cdot {AP}^2]$ From what we learned from the tangent circles, we have $AB = 3$ , $AC = 1$ , $BC = 2$ , $AP = 2 + r$ , $BP = 1 + r$ , and $CP = 3 - r$ , where $r$ is the radius of the circle centered at $P$ that we seek.
Thus: $[3 cdot 1 cdot 2 + 3 {left(3-rright)}^2 = 1 {left(1+rright)}^2 + 2 {left(2+rright)}^2]$ $[6 + 3left(9 - 6r + r^2right) = left(1 + 2r + r^2right) + 2left(4 + 4r + r^2right)]$ $[33 - 18r + 3r^2 = 9 + 10r + 3r^2]$ $[28r = 24]$ $[r = boxed{frac{6}{7}} to boxed{textbf{(B)}}]$
17.
Note that these $z$ such that $z^{24}=1$ are $e^{frac{nipi}{12}}$ for integer $0leq n<24$ . So $z^6=e^{frac{nipi}{2}}$ This is real if $frac{n}{2}in mathbb{Z} Leftrightarrow (n$ is even $)$ . Thus, the answer is the number of even $0leq n<24$ which is $boxed{(D)= 12}$ .

18.
Note that $nequiv S(n)bmod 9$ , so $S(n+1)-S(n)equiv n+1-n = 1bmod 9$ . So, since $S(n)=1274equiv 5bmod 9$ , we have that $S(n+1)equiv 6bmod 9$ . The only one of the answer choices $equiv 6bmod 9$ is $boxed{(D)= 1239}$ .

19.
Analyze the first right triangle. $[asy] pair A,B,C; pair D, e, F; A = (0,0); B = (4,0); C = (0,3); D = (0, 12/7); e = (12/7 , 12/7); F = (12/7, 0); draw(A--B--C--cycle); draw(D--e--F); label("$x$", D/2, W); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$D$", D, W); label("$E$", e, NE); label("$F$", F, S); [/asy]$ Note that $triangle ABC$ and $triangle FBE$ are similar, so $frac{BF}{FE} = frac{AB}{AC}$ . This can be written as $frac{4-x}{x}=frac{4}{3}$ . Solving, $x = frac{12}{7}$ .
Now we analyze the second triangle. $[asy] pair A,B,C; pair q, R, S, T; A = (0,0); B = (4,0); C = (0,3); q = (1.297, 0); R = (2.27 , 1.297); S = (0.973, 2.27); T = (0, 0.973); draw(A--B--C--cycle); draw(q--R--S--T--cycle); label("$y$", (q+R)/2, NW); label("$A'$", A, SW); label("$B'$", B, SE); label("$C'$", C, N); label("$Q$", (q-(0,0.3))); label("$R$", R, NE); label("$S$", S, NE); label("$T$", T, W); [/asy]$ Similarly, $triangle A'B'C'$ and $triangle RB'Q$ are similar, so $RB' = frac{4}{3}y$ , and $C'S = frac{3}{4}y$ . Thus, $C'B' = C'S + SR + RB' = frac{4}{3}y + y + frac{3}{4}y = 5$ . Solving for $y$ , we get $y = frac{60}{37}$ . Thus, $frac{x}{y} = frac{37}{35}$ .

20.
By the properties of logarithms, we can rearrange the equation to read $x^{2017}=2017x$ with $x=log_b a$ . If $xneq 0$ , we may divide by it and get $x^{2016}=2017$ , which implies $x=pm root{2016}of{2017}$ . Hence, we have $3$ possible values $x$ , namely $[x=0,qquad x=2017^{frac1{2016}},, text{and}quad x=-2017^{frac1{2016}}.]$ Since $log_b a=x$ is equivalent to $a=b^x$ , each possible value $x$ yields exactly $199$ solutions $(b,a)$ , as we can assign $a=b^x$ to each $b=2,3,dots,200$ . In total, we have $3cdot 199=boxed{textbf{(E) } 597}$ solutions.

21.
At first, $S={0,10}$ . $[begin{tabular}{r c l c l} (10x+10) & has root & (x=-1) & so now & (S={-1,0,10}) (-x^{10}-x^9-x^8-x^7-x^6-x^5-x^4-x^3-x^2-x+10) & has root & (x=1) & so now & (S={-1,0,1,10}) (x+10) & has root & (x=-10) & so now & (S={-10,-1,0,1,10}) (x^4-x^2-x-10) & has root & (x=2) & so now & (S={-10,-1,0,1,2,10}) (x^4-x^2+x-10) & has root & (x=-2) & so now & (S={-10,-2,-1,0,1,2,10}) (2x-10) & has root & (x=5) & so now & (S={-10,-2,-1,0,1,2,5,10}) (2x+10) & has root & (x=-5) & so now & (S={-10,-5,-2,-1,0,1,2,5,10}) end{tabular}]$ At this point, no more elements can be added to $S$ . To see this, let $begin{align*} a_{n}x^n + a_{n-1}x^{n-1} + ... + a_{2}x^2 + a_{1}x + a_0 &= 0 x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) + a_0 &= 0 x(a_{n}x^{n-1} + a_{n-1}x^{n-2} + ... + a_{2}x + a_{1}) &= -a_0 end{align*}$ with each $a_i$ in $S$ . $x$ is a factor of $a_0$ , and $a_0$ is in $S$ , so $x$ has to be a factor of some element in $S$ . There are no such integers left, so there can be no more additional elements. ${-10,-5,-2,-1,0,1,2,5,10}$ has $9$ elements $to boxed{textbf{(D)}}$
22.
We let $c, e,$ and $m$ be the probability of reaching a corner before an edge when starting at an "inside corner" (e.g. $(1, 1)$ ), an "inside edge" (e.g. $(1, 0)$ ), and the middle respectively.
Starting in the middle, there is a $frac{4}{8}$ chance of moving to an inside edge and a $frac{4}{8}$ chance of moving to an inside corner, so $[m = frac{1}{2}e + frac{1}{2}c.]$ Starting at an inside edge, there is a $frac{2}{8}$ chance of moving to another inside edge, a $frac{2}{8}$ chance of moving to an inside corner, a $frac{1}{8}$ chance of moving into the middle, and a $frac{3}{8}$ chance of reaching an outside edge and stopping. Therefore, $[e = frac{1}{4}e + frac{1}{4}c + frac{1}{8}m + frac{3}{8}*0 = frac{1}{4}e + frac{1}{4}c + frac{1}{8}m.]$ Starting at an inside corner, there is a $frac{2}{8}$ chance of moving to an inside edge, a $frac{1}{8}$ chance of moving into the middle, a $frac{4}{8}$ chance of moving to an outside edge and stopping, and finally a $frac{1}{8}$ chance of reaching that elusive outside corner. This gives $[c = frac{1}{4}e + frac{1}{8}m + frac{1}{2}0 + frac{1}{8}*1 = frac{1}{4}e + frac{1}{8}m + frac{1}{8}.]$ Solving this system of equations gives $[m = frac{4}{35},]$ $[e = frac{1}{14},]$ $[c = frac{11}{70}.]$ Since the particle starts at $(0, 0),$ it is $m$ we are looking for, so the final answer is $[4 + 35 = boxed{textbf{(E) }39}.]$
23.
Let $r_1,r_2,$ and $r_3$ be the roots of $g(x)$ . Let $r_4$ be the additional root of $f(x)$ . Then from Vieta's formulas on the quadratic term of $g(x)$ and the cubic term of $f(x)$ , we obtain the following: $begin{align*} r_1+r_2+r_3&=-a r_1+r_2+r_3+r_4&=-1 end{align*}$ Thus $r_4=a-1$ .
Now applying Vieta's formulas on the constant term of $g(x)$ , the linear term of $g(x)$ , and the linear term of $f(x)$ , we obtain: $begin{align*} r_1r_2r_3 & = -10 r_1r_2+r_2r_3+r_3r_1 &= 1 r_1r_2r_3+r_2r_3r_4+r_3r_4r_1+r_4r_1r_2 & = -100 end{align*}$ Substituting for $r_1r_2r_3$ in the bottom equation and factoring the remainder of the expression, we obtain: $[-10+(r_1r_2+r_2r_3+r_3r_1)r_4=-10+r_4=-100]$ It follows that $r_4=-90$ . But $r_4=a-1$ so $a=-89$ Now we can factor $f(x)$ in terms of $g(x)$ as $[f(x)=(x-r_4)g(x)=(x+90)g(x)]$ Then $f(1)=91g(1)$ and $[g(1)=1^3-89cdot 1^2+1+10=-77]$ Hence $f(1)=91cdot(-77)=boxed{textbf{(C)},-7007}$ .

24.
Using the given ratios, note that $frac{XY}{BD} = 1 - frac{1}{4} - frac{11}{36} = frac{4}{9}.$ By AA Similarity, $triangle AXD sim triangle EXY$ with a ratio of $frac{DX}{XY} = frac{9}{16}$ and $triangle ACX sim triangle EFX$ with a ratio of $frac{AX}{XE} = frac{DX}{XY} = frac{9}{16}$ , so $frac{XF}{CX} = frac{16}{9}$ .
Now we find the length of $BD$ . Because the quadrilateral is cyclic, we can simply use the Law of Cosines. $[BD^2=3^2+8^2-48cosangle BAD=2^2+6^2-24cos (180-angle BAD)=2^2+6^2+24cosangle BAD]$ $[rightarrow cosangle BAD = frac{11}{24}]$ $[rightarrow BD=sqrt{51}]$ By Power of a Point, $CXcdot XG = DXcdot XB = frac{sqrt{51}}{4} frac{3sqrt{51}}{4}$ . Thus $XFcdot XG = frac{XF}{CX} CXcdot XG = frac{51}{3} = boxed{textbf{(A)} 17}.$
25.
It is possible to solve this problem using elementary counting methods. This solution proceeds by a cleaner generating function.
We note that $pm sqrt{2}i$ both lie on the imaginary axis and each of the $frac{1}{sqrt{8}}(pm 1pm i)$ have length $frac{1}{2}$ and angle of odd multiples of $pi/4$ , i.e. $pi/4,3pi/4,5pi,4,7pi/4$ . When we draw these 6 complex numbers out on the complex plane, we get a crystal-looking thing. Note that the total number of ways to choose 12 complex numbers is $6^{12}$ . Now we count the number of good combinations.
We first consider the lengths. When we multiply 12 complex numbers together, their magnitudes multiply. Suppose we have $n$ of the numbers $pm sqrt{2}i$ ; then we must have $left(sqrt{2}right)^ncdotleft(frac{1}{2}right)^{12-n}=1 Longrightarrow n=8$ . Having $n=8$ will take care of the length of the product; now we need to deal with the angle.
We require $sumthetaequivpi mod 2pi$ . Letting $z$ be $e^{ipi/4}$ , we see that the angles we have available are ${z^1,z^2,z^3,z^5,z^6,z^7}$ , where we must choose exactly 8 angles from the set ${z^2,z^6}$ and exactly 4 from the set ${z^1,z^3,z^5,z^7}$ . If we found a good combination where we had $a_i$ of each angle $z^i$ , then the amount this would contribute to our count would be $binom{12}{4,8}cdotbinom{8}{a_2,a_6}cdotbinom{4}{a_1,a_3,a_5,a_7}$ . We want to add these all up. We proceed by generating functions.
Consider $[(t_2x^2+t_6x^6)^8(t_1x^1+t_3x^3+t_5x^5+t_7x^7)^4.]$ The expansion will be of the form $sum_ileft(sum_{sum a=i} binom{8}{a_2,a_6}binom{4}{a_1,a_3,a_5,a_7}{t_1}^{a_1}{t_2}^{a_2}{t_3}^{a_3}{t_5}^{a_5}{t_6}^{a_6}{t_7}^{a_7}x^i right)$ . Note that if we reduced the powers of $x$ mod $8$ and fished out the coefficient of $x^4$ and plugged in $t_i=1 forall,i$ (and then multiplied by $binom{12}{4,8}$ ) then we would be done. Since plugging in $t_i=1$ doesn't affect the $x$ 's, we do that right away. The expression then becomes $[x^{20}(1+x^4)^8(1+x^2+x^4+x^6)^4=x^{20}(1+x^4)^{12}(1+x^2)^4=x^4(1+x^4)^{12}(1+x^2)^4,]$ where the last equality is true because we are taking the powers of $x$ mod $8$ . Let $[x^n]f(x)$ denote the coefficient of $x^n$ in $f(x)$ . Note $[x^4] x^4(1+x^4)^{12}(1+x^2)^4=[x^0](1+x^4)^{12}(1+x^2)^4$ . We use the roots of unity filter, which states $[text{terms of }f(x)text{ that have exponent congruent to }ktext{ mod }n=frac{1}{n}sum_{m=1}^n frac{f(z^mx)}{z^{mk}},]$ where $z=e^{ipi/n}$ . In our case $k=0$ , so we only need to find the average of the $f(z^mx)$ 's. $begin{align*} z^0 &Longrightarrow (1+x^4)^{12}(1+x^2)^4, z^1 &Longrightarrow (1-x^4)^{12}(1+ix^2)^4, z^2 &Longrightarrow (1+x^4)^{12}(1-x^2)^4, z^3 &Longrightarrow (1-x^4)^{12}(1-ix^2)^4, z^4 &Longrightarrow (1+x^4)^{12}(1+x^2)^4, z^5 &Longrightarrow (1-x^4)^{12}(1+ix^2)^4, z^6 &Longrightarrow (1+x^4)^{12}(1-x^2)^4, z^7 &Longrightarrow (1-x^4)^{12}(1-ix^2)^4. end{align*}$ We plug in $x=1$ and take the average to find the sum of all coefficients of $x^{text{multiple of 8}}$ . Plugging in $x=1$ makes all of the above zero except for $z^0$ and $z^4$ . Averaging, we get $2^{14}$ . Now the answer is simply $[frac{binom{12}{4,8}}{6^{12}}cdot 2^{14}=boxed{frac{2^2cdot 5cdot 11}{3^{10}}}.]$