The vertices V of a centrally symmetric hexagon in the complex plane are given by V=\\begingathered \sqrt2 i, -\sqrt2 i, \ (1)/(\sqrt8)(1+i), \ (1)/(\sqrt8)(-1+i), \ (1)/(\sqrt8)(1-i), \ (1)/(\sqrt8)(-1-i) \endgathered\. For each j, 1≤ j\le12, an element z_j is chosen from V at random, independently of the other choices. Let P=\prod_j=1^12z_j be the product of the 12 numbers selected. What is the probability that P=-1?

(D) \dfrac5\cdot7\cdot112\cdot3^10

2017 AMC 12A — Problems & Online Timed Exam

Timed

1:15:00

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 $\$8?$

$8$

$11$

$12$

$13$

$15$

Answer: D

Concepts:optimization rate

Difficulty rating: 890

Small Hint:

The best value is the $5$ -popsicle box, at $\$3$ for $5$

Big Hint:

Spend $\$6$ on two $5$ -boxes, then use the remaining $\$2$ on a $3$ -box

Solution:

The cheapest popsicles come from the $5$ -popsicle box, at $\dfrac{\$3}{5}=\$0.60$ each. Even at that rate, $14$ popsicles would cost $14\cdot\$0.60=\$8.40,$ more than $\$8.$

So Pablo can buy at most $13,$ and he achieves this with two $5$ -boxes for $\$6$ and one $3$ -box for $\$2,$ giving $2\cdot5+3=13$ popsicles.

Thus, the correct answer is D.

2.

The sum of two nonzero real numbers is $4$ times their product. What is the sum of the reciprocals of the two numbers?

$1$

$2$

$4$

$8$

$12$

Answer: C

Concepts:algebraic manipulation fraction

Difficulty rating: 1020

Small Hint:

The sum of reciprocals is $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}$

Big Hint:

Divide the equation $x+y=4xy$ by $xy$

Solution:

Let the numbers be $x$ and $y,$ so $x+y=4xy.$

Dividing both sides by $xy$ gives $\dfrac{x+y}{xy}=4,$ and the left side is exactly $\dfrac{1}{y}+\dfrac{1}{x}.$ So the sum of the reciprocals is $4.$

Thus, the correct answer is C.

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?

If Lewis did not receive an A, then he got all of the multiple choice questions wrong.

If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong.

If Lewis got at least one of the multiple choice questions wrong, then he did not receive an A.

If Lewis received an A, then he got all of the multiple choice questions right.

If Lewis received an A, then he got at least one of the multiple choice questions right.

Answer: B

Concepts:logical deduction

Difficulty rating: 1100

Small Hint:

An implication is logically equivalent to its contrapositive, not to its converse or inverse

Big Hint:

Negating “got all right” gives “got at least one wrong”

Solution:

The promise is “all right $\Rightarrow$ A.” An implication is equivalent only to its contrapositive: “not A $\Rightarrow$ not all right.”

“Not all right” means at least one question was wrong, which is exactly statement B. The converse and inverse do not follow, and getting “all wrong” is a much stronger claim than the negation.

Thus, the correct answer is B.

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?

$30\%$

$40\%$

$50\%$

$60\%$

$70\%$

Answer: A

Concepts:Pythagorean Theorem percentage

Difficulty rating: 1200

Small Hint:

Let the side be $x$ ; Jerry walks $2x$ and Silvia walks the diagonal

Big Hint:

The diagonal is $x\sqrt2,$ so the fraction saved is $\dfrac{2x-x\sqrt2}{2x}=1-\dfrac{\sqrt2}{2}$

Solution:

If the square has side $x,$ Jerry walks $x+x=2x,$ while Silvia walks the diagonal $\sqrt{x^2+x^2}=x\sqrt2.$

The fraction by which Silvia’s trip is shorter is $\begin{aligned} &\dfrac{2x-x\sqrt2}{2x}=1-\dfrac{\sqrt2}{2} \\ &\approx 1-0.707=0.293. \end{aligned}$

This is closest to $30\%.$

Thus, the correct answer is A.

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?

$240$

$245$

$290$

$480$

$490$

Answer: B

Concepts:graph theory basic counting

Difficulty rating: 1270

Small Hint:

The $20$ acquaintances only shake hands with the $10$ strangers

Big Hint:

Add up each person’s handshakes and divide by $2$

Solution:

Each of the $20$ people who know each other shakes hands with only the $10$ strangers. Each of the $10$ strangers shakes hands with all $29$ other people.

Summing handshake counts and dividing by $2$ (each handshake involves two people) gives $\begin{aligned} &\dfrac{1}{2}(20\cdot10+10\cdot29) \\ &=\dfrac{1}{2}(200+290)=245. \end{aligned}$

Thus, the correct answer is B.

6.

Joy has $30$ thin rods, one each of every integer length from $1$ cm through $30$ cm. She places the rods with lengths $3$ cm, $7$ cm, and $15$ 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?

$16$

$17$

$18$

$19$

$20$

Answer: B

Concepts:triangle inequality counting integers in a range

Difficulty rating: 1350

Small Hint:

Four lengths form a quadrilateral exactly when the longest is less than the sum of the other three

Big Hint:

This gives $5\lt n\lt 25$ ; then remove the rods already on the table

Solution:

Four lengths form a quadrilateral with positive area if and only if the longest is strictly less than the sum of the other three. With a fourth rod of length $n,$ this requires $15\lt 3+7+n$ and $n\lt 3+7+15,$ so $5\lt n\lt 25.$

The integers from $6$ to $24$ give $19$ values, but the rods of length $7$ and $15$ are already on the table, leaving $19-2=17$ choices.

Thus, the correct answer is B.

7.

Define a function on the positive integers recursively by $f(1)=2,$ $f(n)=f(n-1)+1$ if $n$ is even, and $f(n)=f(n-2)+2$ if $n$ is odd and greater than $1.$ What is $f(2017)?$

$2017$

$2018$

$4034$

$4035$

$4036$

Answer: B

Concepts:recursion pattern recognition

Difficulty rating: 1380

Small Hint:

Compute $f(1),f(2),f(3),\ldots$ and look for a pattern

Big Hint:

Check that $f(n)=n+1$ satisfies both recursive rules

Solution:

Listing values: $f(1)=2,$ $f(2)=f(1)+1=3,$ $f(3)=f(1)+2=4,$ $f(4)=f(3)+1=5,$ suggesting $f(n)=n+1.$

Both rules are consistent with $f(n)=n+1:$ for even $n,$ $(n-1)+1+1=n+1,$ and for odd $n,$ $(n-2)+1+2=n+1.$ Since the recursion determines $f$ uniquely, $f(2017)=2018.$

Thus, the correct answer is B.

8.

The region consisting of all points in three-dimensional space within $3$ units of line segment $AB$ has volume $216\pi.$ What is the length $AB?$

$6$

$12$

$18$

$20$

$24$

Answer: D

Concepts:cylinder sphere volume

Difficulty rating: 1440

Small Hint:

The region is a cylinder of radius $3$ with a hemisphere capping each end

Big Hint:

The two hemispheres combine into a full sphere: $9\pi h+\dfrac{4}{3}\pi 3^3=216\pi$

Solution:

Let $h=AB.$ The region is a cylinder of radius $3$ and height $h$ with a hemisphere of radius $3$ on each end.

The cylinder has volume $\pi\cdot3^2\cdot h=9\pi h,$ and the two hemispheres together form a sphere of volume $\dfrac{4}{3}\pi\cdot3^3=36\pi.$ So $9\pi h+36\pi=216\pi,$ giving $h=20.$

Thus, the correct answer is D.

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 this common value. Which of the following is a correct description of $S?$

a single point

two intersecting lines

three lines whose pairwise intersections are three distinct points

a triangle

three rays with a common endpoint

Answer: E

Concepts:coordinate geometry casework

Difficulty rating: 1500

Small Hint:

Set two of the three quantities equal and require the third to be at most that common value

Big Hint:

Each of the three pairings produces a ray, and all three start at the same point $(1,7)$

Solution:

Consider which two of $3,$ $x+2,$ $y-4$ are the (equal) larger pair.

If $3=x+2\ge y-4:$ then $x=1$ and $y\le7,$ a downward ray from $(1,7).$ If $3=y-4\ge x+2:$ then $y=7$ and $x\le1,$ a leftward ray from $(1,7).$ If $x+2=y-4\ge3:$ then $y=x+6$ and $x\ge1,$ a ray from $(1,7)$ going up and to the right.

All three rays share the endpoint $(1,7),$ so $S$ is three rays with a common endpoint.

Thus, the correct answer is E.

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 Chloé’s number?

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$\dfrac{5}{6}$

$\dfrac{7}{8}$

Answer: C

Concepts:geometric probability symmetry complementary probability

Difficulty rating: 1560

Small Hint:

Split on whether Laurent’s number is above or below $2017$

Big Hint:

If it is above $2017$ he always wins; if below, it is a symmetric coin flip

Solution:

With probability $\dfrac{1}{2},$ Laurent’s number lies in $[2017,4034],$ which exceeds any number Chloé could choose, so he wins for certain.

With the other probability $\dfrac{1}{2},$ Laurent’s number lies in $[0,2017],$ matching Chloé’s interval; by symmetry he is larger half the time. The total probability is $\dfrac{1}{2}\cdot1+\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}.$

Thus, the correct answer is C.

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?

$37$

$63$

$117$

$143$

$163$

Answer: D

Concepts:angle sum bounding to limit cases

Difficulty rating: 1570

Small Hint:

The true angle sum $(n-2)\cdot180$ is a multiple of $180$ just above $2017$

Big Hint:

The forgotten angle is that multiple minus $2017,$ and it must be less than $180$

Solution:

If the polygon has $n$ sides and the forgotten angle is $\alpha,$ then $(n-2)180=2017+\alpha.$ Since $0\lt\alpha\lt180,$ $2017\lt(n-2)180\lt2197.$

The only multiple of $180$ in this range is $2160=(14-2)180,$ so $n=14$ and $\alpha=2160-2017=143.$

Thus, the correct answer is D.

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 $S\gt0,$ in minutes, at which all $10$ horses will again simultaneously be at the starting point is $S=2520.$ Let $T\gt0$ 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 $T?$

$2$

$3$

$4$

$5$

$6$

Answer: B

Concepts:least common multiple divisibility

Difficulty rating: 1630

Small Hint:

Horse $k$ is back at the start at time $t$ exactly when $k$ divides $t$

Big Hint:

Find the smallest $t$ having at least $5$ divisors among $1$ through $10$

Solution:

Horse $k$ is at the starting point at time $t$ precisely when $t$ is a multiple of $k.$ So we want the smallest $t$ with at least $5$ divisors among $1,2,\ldots,10.$

The positive integers below $12$ have at most $4$ divisors, while $12$ is divisible by $1,2,3,4,$ and $6.$ Thus $T=12,$ and the sum of its digits is $1+2=3.$

Thus, the correct answer is B.

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 $\dfrac{1}{3}$ 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?

$132$

$135$

$138$

$141$

$144$

Answer: B

Concepts:distance rate and time system of equations

Difficulty rating: 1660

Small Hint:

Her usual drive is $3$ hours, so the distance equals $3$ times her usual speed

Big Hint:

The final two-thirds takes $276-\dfrac13\cdot180=216$ minutes at the reduced speed

Solution:

Let the distance be $d$ miles and the usual speed $r$ mph. Since the usual trip is $3$ hours, $d=3r.$

The first $\dfrac{1}{3}$ of the drive takes $\dfrac{1}{3}\cdot180=60$ minutes at speed $r,$ so the remaining $\dfrac{2}{3}$ takes $276-60=216$ minutes $=\dfrac{18}{5}$ hours at speed $r-20.$

That final portion covers $\dfrac{2}{3}d=2r$ miles, so $2r=(r-20)\cdot\dfrac{18}{5}.$ Solving gives $10r=18r-360,$ so $r=45$ and $d=3\cdot45=135.$

Thus, the correct answer is B.

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?

$12$

$16$

$28$

$32$

$40$

Answer: C

Concepts:arrangements with restrictions inclusion-exclusion

Difficulty rating: 1730

Small Hint:

Start from all $5!=120$ seatings and subtract the forbidden adjacencies

Big Hint:

Treat each forbidden pair as a block (with $2$ internal orders) and apply inclusion-exclusion

Solution:

Let $X,$ $Y,$ $Z$ be the seatings where Alice-Bob, Alice-Carla, and Derek-Eric are adjacent, respectively. The answer is $5!-|X\cup Y\cup Z|.$

By inclusion-exclusion, $|X\cup Y\cup Z|=(48\cdot3)$ $-(12+24+24)$ $+8=92,$ so the answer is $120-92=28.$

Thus, the correct answer is C.

15.

Let $f(x)=\sin x+2\cos x+3\tan x,$ using radian measure for the variable $x.$ In what interval does the smallest positive value of $x$ for which $f(x)=0$ lie?

$(0,1)$

$(1,2)$

$(2,3)$

$(3,4)$

$(4,5)$

Answer: D

Concepts:trigonometry

Difficulty rating: 1800

Small Hint:

For $0\lt x\lt\dfrac{\pi}{2}$ all three terms are positive, so there is no root there

Big Hint:

Just past $\dfrac{\pi}{2}$ the term $3\tan x$ makes $f$ negative; locate the first sign change after $\pi$

Solution:

For $0\lt x\lt\dfrac{\pi}{2}$ all three terms are positive, so $f(x)\gt0.$ For $\dfrac{\pi}{2}\lt x\lt\pi,$ set $s=\sin x\gt0$ and $t=-\cos x\gt0.$ Since $0\lt t\le1,$ $\begin{aligned} f(x)&=s-2t-\dfrac{3s}{t} \\ &=-2t-s\left(\dfrac{3}{t}-1\right)\lt0. \end{aligned}$ Thus no root occurs before $x=\pi.$

At $x=\pi,$ $f(\pi)=0+2(-1)+0=-2\lt0.$ At $x=\dfrac{5\pi}{4},$ $\tan x=1$ so $f=-\dfrac{\sqrt2}{2}+3\gt0.$ By the intermediate value theorem the smallest positive root lies in $\left(\pi,\dfrac{5\pi}{4}\right).$

Since $\pi\gt3$ and $\dfrac{5\pi}{4}\lt4,$ this interval sits inside $(3,4).$

Thus, the correct answer is D.

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 $\overline{JK}.$ 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 $P?$

$\dfrac{3}{4}$

$\dfrac{6}{7}$

$\dfrac{1}{2}\sqrt3$

$\dfrac{5}{8}\sqrt2$

$\dfrac{11}{12}$

Answer: B

Concepts:tangent circles Pythagorean Theorem coordinate geometry

Difficulty rating: 1840

Small Hint:

Let $C$ be the center of the large semicircle; then $PA=2+r,$ $PB=1+r,$ $PC=3-r$

Big Hint:

Drop a perpendicular from $P$ to $\overline{JK}$ and set the three expressions for its squared height equal

Solution:

The large semicircle has radius $3$ and center $C,$ the midpoint of $\overline{JK}.$ Placing $J$ at the origin, $A=2,$ $B=5,$ $C=3,$ $K=6$ along the base. Let $r$ be the radius of the circle at $P.$

By tangency, $PA=2+r,$ $PB=1+r,$ and $PC=3-r.$ Dropping a perpendicular from $P$ to the base at horizontal position $3+x$ with height $h,$ the Pythagorean theorem gives $\begin{aligned} h^2 &=(2+r)^2-(1+x)^2 \\ &=(3-r)^2-x^2 \\ &=(1+r)^2-(2-x)^2. \end{aligned}$

Equating the first expression with the middle one gives $5r-x=3,$ while equating the last expression with the middle one gives $2r+x=3.$ Adding yields $7r=6,$ so $r=\dfrac{6}{7}.$

Thus, the correct answer is B.

17.

There are $24$ different complex numbers $z$ such that $z^{24}=1.$ For how many of these is $z^6$ a real number?

$0$

$4$

$6$

$12$

$24$

Answer: D

Concepts:roots of unity complex number

Difficulty rating: 1910

Small Hint:

Write $z=e^{\frac{\pi i k}{12}}$ for $k=0,1,\ldots,23$

Big Hint:

Then $z^6=e^{\frac{\pi i k}{2}},$ which is real exactly when $k$ is even

Solution:

The $24$ solutions are the $24$ th roots of unity, $z=e^{\frac{\pi i k}{12}}$ for $k=0,1,\ldots,23.$

Then $z^6=e^{\frac{\pi i k}{2}}=\cos\dfrac{k\pi}{2}+i\sin\dfrac{k\pi}{2},$ which is real exactly when $\sin\dfrac{k\pi}{2}=0,$ i.e. when $k$ is even. There are $12$ even values of $k$ in the range.

Thus, the correct answer is D.

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 $S(n+1)?$

$1$

$3$

$12$

$1239$

$1265$

Answer: D

Concepts:digits modular arithmetic

Difficulty rating: 1990

Small Hint:

If $n$ ends in exactly $k$ nines, adding $1$ drops the digit sum by $9k$ and adds $1$

Big Hint:

So $S(n+1)=1275-9k;$ check which choice has this form

Solution:

Adding $1$ to $n$ increases the digit sum by $1,$ except that each trailing $9$ turns into a $0,$ losing $9.$ If $n$ ends in exactly $k$ nines, then $S(n+1)=S(n)+1-9k$ $=1275-9k.$

So the possible values are $1275,1266,1257,\ldots$ Among the choices, only $1239=1275-9\cdot4$ fits (for example, $n$ ending in four $9$ s preceded by enough $1$ s).

Thus, the correct answer is D.

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 $\dfrac{x}{y}?$

$\dfrac{12}{13}$

$\dfrac{35}{37}$

$1$

$\dfrac{37}{35}$

$\dfrac{13}{12}$

Answer: D

Concepts:similarity right triangle

Difficulty rating: 2040

Small Hint:

A square with its corner at the right angle has side $x=\dfrac{3\cdot4}{3+4}$

Big Hint:

A square with a side on a base of length $b$ and height $h$ to that base has side $\dfrac{bh}{b+h};$ use the hypotenuse as the base

Solution:

For the first square, the two smaller triangles it cuts off are similar to the whole triangle, giving $\dfrac{x}{3-x}=\dfrac{4-x}{x},$ so $x=\dfrac{12}{7}.$ (Equivalently, a square in the right angle has side $\dfrac{3\cdot4}{3+4}.$ )

For the second square, take the hypotenuse of length $5$ as base; the altitude to it is $h=\dfrac{3\cdot4}{5}=\dfrac{12}{5}.$ A square with a side on a base $b$ and height $h$ has side $\dfrac{bh}{b+h},$ so $y=\dfrac{5\cdot\tfrac{12}{5}}{5+\tfrac{12}{5}}=\dfrac{12}{\tfrac{37}{5}}=\dfrac{60}{37}.$

Therefore $\dfrac{x}{y}=\dfrac{12}{7}\cdot\dfrac{37}{60}=\dfrac{37}{35}.$

Thus, the correct answer is D.

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 $(\log_b a)^{2017}=\log_b(a^{2017})?$

$198$

$199$

$398$

$399$

$597$

Answer: E

Concepts:logarithm basic counting

Difficulty rating: 2110

Small Hint:

Let $u=\log_b a;$ the equation becomes $u^{2017}=2017u$

Big Hint:

Either $u=0$ or $u^{2016}=2017;$ count the $a$ values for each of the $199$ allowed bases

Solution:

Let $u=\log_b a.$ Since $\log_b(a^{2017})=2017\log_b a,$ the equation is $u^{2017}=2017u,$ so $u=0$ or $u^{2016}=2017.$

If $u=0,$ then $a=1,$ valid for every one of the $199$ bases. If $u^{2016}=2017,$ then $u=\pm2017^{\frac{1}{2016}},$ giving $2$ values of $a$ for each base, i.e. $2\cdot199=398$ pairs.

In total there are $199+398=597$ ordered pairs.

Thus, the correct answer is E.

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 $a_nx^n+a_{n-1}x^{n-1}$ $+\cdots+a_1x+a_0$ for some $n\ge1,$ 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?

$4$

$5$

$7$

$9$

$11$

Answer: D

Concepts:polynomial divisibility systematic listing

Difficulty rating: 2130

Small Hint:

By the Rational Root Theorem, any integer root of a polynomial with coefficients in $S$ divides its constant term

Big Hint:

Starting from $\{0,10\},$ reach $\pm1,\pm2,\pm5,\pm10;$ show nothing else can enter

Solution:

Using $10x+10,$ the root $-1$ enters $S.$ Then $1$ enters as a root of $-x^{10}-x^9-\cdots-x+10,$ and $-10$ enters from $x+10.$

Now $x^3+x-10$ has root $2,$ and $x+2$ gives $-2;$ then $2x-10$ and $2x+10$ give $\pm5.$ At this point $S=\{0,\pm1,\pm2,\pm5,\pm10\}.$

No further integer can appear. Inductively, every nonzero member of $S$ divides $10.$ If a polynomial used in the rule has constant term $0,$ factor out the largest possible power of $x;$ any nonzero root is then a root of a polynomial whose constant term is the first nonzero original coefficient. The Rational Root Theorem shows that the root divides this coefficient, which by the inductive hypothesis divides $10.$ So $S$ has $9$ elements.

Thus, the correct answer is D.

22.

A square is drawn in the Cartesian coordinate plane with vertices at $(2,2),$ $(-2,2),$ $(-2,-2),$ and $(2,-2).$ 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 $\dfrac{1}{8}$ 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 $\dfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m+n?$

$4$

$5$

$7$

$15$

$39$

Answer: E

Concepts:random walk recursive probability symmetry

Difficulty rating: 2270

Small Hint:

By symmetry, track only three interior states: the center, the four edge-midpoint neighbors, and the four diagonal neighbors

Big Hint:

Write hitting-probability equations for each state and solve the linear system for the center’s value

Solution:

By symmetry, group the relevant interior points into three types: $C=\{(0,0)\},$ the “axis” points $A=\{(\pm1,0),(0,\pm1)\},$ and the “diagonal” points $I=\{(\pm1,\pm1)\}.$ Let $a,c,i$ be the probabilities of eventually hitting a corner starting from a point of type $A,C,I.$

Reading off the transition probabilities (a point in $A$ goes to $A$ with prob $\tfrac28,$ to $C$ with $\tfrac18,$ to $I$ with $\tfrac28,$ and to a side interior with $\tfrac38,$ etc.) gives $\begin{aligned} &a=\tfrac28 a+\tfrac18 c+\tfrac28 i,\quad \\ &c=\tfrac48 a+\tfrac48 i,\quad \\ &i=\tfrac28 a+\tfrac18 c+\tfrac18. \end{aligned}$

Solving yields $a=\dfrac{1}{14},$ $c=\dfrac{4}{35},$ $i=\dfrac{11}{70}.$ The required probability is $c=\dfrac{4}{35},$ so $m+n=4+35=39.$

Thus, the correct answer is E.

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 $\begin{aligned} &f(x)=x^4+x^3+bx^2 \\ &\quad {}+100x+c. \end{aligned}$ What is $f(1)?$

$-9009$

$-8008$

$-7007$

$-6006$

$-5005$

Answer: C

Concepts:polynomial factoring

Difficulty rating: 2380

Small Hint:

Since the three roots of $g$ are roots of $f,$ we have $f(x)=(x-q)\,g(x)$ for one extra root $q$

Big Hint:

Match the $x^1$ coefficients to find $q,$ then $f(1)=(1-q)\,g(1)$

Solution:

Since $g$ has three distinct roots all shared by the quartic $f,$ we can write $f(x)=(x-q)g(x)$ for some remaining root $q.$ Expanding, $\begin{aligned} &f(x)=x^4+(a-q)x^3 \\ &\quad {}+(1-qa)x^2 \\ &\quad {}+(10-q)x-10q. \end{aligned}$

Matching the $x$ coefficient, $10-q=100,$ so $q=-90.$ Matching the $x^3$ coefficient, $a-q=1,$ so $a=-89.$

Then $g(1)=1+a+1+10$ $=12-89=-77$ and $1-q=91,$ so $\begin{aligned} &f(1)=(1-q)g(1) \\ &=91\cdot(-77)=-7007. \end{aligned}$

Thus, the correct answer is C.

24.

Quadrilateral $ABCD$ is inscribed in circle $O$ and has sides $AB=3,$ $BC=2,$ $CD=6,$ and $DA=8.$ Let $X$ and $Y$ be points on $BD$ such that $\dfrac{DX}{BD}=\dfrac{1}{4}$ and $\dfrac{BY}{BD}=\dfrac{11}{36}.$ Let $E$ be the intersection of line $AX$ and the line through $Y$ parallel to $AD.$ Let $F$ be the intersection of line $CX$ and the line through $E$ parallel to $AC.$ Let $G$ be the point on circle $O$ other than $C$ that lies on line $CX.$ What is $XF\cdot XG?$

$17$

$\dfrac{59-5\sqrt2}{3}$

$\dfrac{91-12\sqrt3}{4}$

$\dfrac{67-10\sqrt2}{3}$

$18$

Answer: A

Difficulty rating: 2520

Small Hint:

The parallels give $\triangle XEY\sim\triangle XAD$ and $\triangle XEF\sim\triangle XAC$

Big Hint:

Combine those ratios with Power of a Point to get $XF\cdot XG=XB\cdot XY;$ find $BD^2$ via Law of Cosines

Solution:

Because $YE\parallel AD$ and $EF\parallel AC,$ we get $\triangle XEY\sim\triangle XAD$ and $\triangle XEF\sim\triangle XAC,$ giving $\dfrac{XY}{XE}=\dfrac{XD}{XA}$ and $\dfrac{XF}{XE}=\dfrac{XC}{XA}.$ Hence $\dfrac{XC}{XD}=\dfrac{XF}{XY},$ so $XF\cdot XD=XC\cdot XY.$

Power of a Point at $X$ gives $XC\cdot XG=XD\cdot XB,$ and combining yields $XF\cdot XG=XB\cdot XY.$ With $d=BD,$ $DX=\dfrac14 d$ and $BY=\dfrac{11}{36}d,$ so $\begin{aligned} XF\cdot XG &=\left(d-\tfrac14 d\right) \\ &\quad {}\cdot\left(d-\tfrac14 d-\tfrac{11}{36}d\right) \\ &=\dfrac34 d\cdot\dfrac49 d=\dfrac{d^2}{3}. \end{aligned}$

Since $ABCD$ is cyclic, $\angle BAD$ and $\angle BCD$ are supplementary. The Law of Cosines on $\triangle ABD$ and $\triangle CBD$ gives $\dfrac{73-d^2}{48}=\dfrac{d^2-40}{24},$ so $d^2=51.$ Therefore $XF\cdot XG=\dfrac{51}{3}=17.$

Thus, the correct answer is A.

25.

The vertices $V$ of a centrally symmetric hexagon in the complex plane are given by $V=\left\{\begin{gathered} \sqrt2 i,\;-\sqrt2 i, \\ \tfrac{1}{\sqrt8}(1+i), \\ \tfrac{1}{\sqrt8}(-1+i), \\ \tfrac{1}{\sqrt8}(1-i), \\ \tfrac{1}{\sqrt8}(-1-i) \end{gathered}\right\}.$ For each $j,$ $1\le j\le12,$ 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 $P=-1?$

$\dfrac{5\cdot11}{3^{10}}$

$\dfrac{5^2\cdot11}{2\cdot3^{10}}$

$\dfrac{5\cdot11}{3^9}$

$\dfrac{5\cdot7\cdot11}{2\cdot3^{10}}$

$\dfrac{2^2\cdot5\cdot11}{3^{10}}$

Answer: E

Concepts:complex number binomial probability

Difficulty rating: 2650

Small Hint:

The two elements $\pm\sqrt2 i$ have magnitude $\sqrt2;$ the other four have magnitude $\dfrac12$

Big Hint:

For $|P|=1$ you need exactly $8$ of the large-magnitude factors and $4$ of the small ones, then get the phase right

Solution:

Let $A=\{\sqrt2 i,-\sqrt2 i\}$ (each of magnitude $\sqrt2$ ) and $B$ be the other four elements (each of magnitude $\dfrac12$ ). Since $|P|=(\sqrt2)^{\#A}\left(\tfrac12\right)^{\#B}=1$ forces $\#A=8$ and $\#B=4,$ exactly $8$ factors must come from $A$ and $4$ from $B.$

A product of $8$ elements of $A$ equals $\pm16$ (real), and a product of $4$ elements of $B$ equals one of $\pm\tfrac{1}{16},\pm\tfrac{i}{16}.$ Their product is one of $\pm1,\pm i,$ each equally likely, so exactly $\tfrac14$ of these configurations give $P=-1.$

The chance of landing in the $8$ -from- $A$ , $4$ -from- $B$ pattern is $\binom{12}{4}\left(\tfrac13\right)^8\left(\tfrac23\right)^4=\dfrac{880}{3^{10}}.$ Multiplying by $\tfrac14$ gives $\begin{aligned} P&=\dfrac{1}{4}\cdot\dfrac{880}{3^{10}} \\ &=\dfrac{220}{3^{10}}=\dfrac{2^2\cdot5\cdot11}{3^{10}}. \end{aligned}$

Thus, the correct answer is E.