1989 AMC 12 Solutions | LIVE by Po-Shen Loh

1.

$(-1)^{5^2}+1^{2^5}=$

$-7$

$-2$

$0$

$1$

$57$

Concepts:order of operations exponent

Difficulty rating: 1010

Small Hint:

Evaluate the exponents from the top down

Big Hint:

The parity of the exponent on $-1$ determines its sign

Solution:

Because $5^2=25$ is odd, $(-1)^{5^2}=-1.$ Also $1^{2^5}=1,$ so the sum is $0.$

Thus the correct answer is C.

2.

$\sqrt{\frac19+\frac1{16}}=$

$\frac15$

$\frac14$

$\frac27$

$\frac5{12}$

$\frac7{12}$

Concepts:fraction radical

Difficulty rating: 970

Small Hint:

Add the fractions before taking the square root

Big Hint:

Use $144$ as a common denominator

Solution:

We have $\frac19+\frac1{16}=\frac{16+9}{144}=\frac{25}{144}.$ Its positive square root is $\frac5{12}.$

Thus the correct answer is D.

3.

A square is cut into three rectangles along two lines parallel to a side, as shown. If the perimeter of each of the three rectangles is $24,$ then the area of the original square is

$24$

$36$

$64$

$81$

$96$

Concepts:perimeter square (geometry)rectangle

Difficulty rating: 1030

Small Hint:

Equal perimeters force the three rectangle widths to be equal

Big Hint:

Express one rectangle’s perimeter using the square’s side length

Solution:

Let the square have side $s.$ Since every rectangle has length $s$ and the same perimeter, their three widths are equal and are each $\frac{s}{3}.$ Thus $2\left(s+\frac{s}{3}\right)=24,$ so $s=9$ and the square’s area is $81.$

Thus the correct answer is D.

4.

In the figure, $ABCD$ is an isosceles trapezoid with side lengths $AD=BC=5,$ $AB=4,$ and $DC=10.$ The point $C$ is on $\overline{DF}$ and $B$ is the midpoint of hypotenuse $\overline{DE}$ in the right triangle $DEF.$ Then $CF=$

$3.25$

$3.5$

$3.75$

$4.0$

$4.25$

Concepts:trapezoid midpoint similarity

Difficulty rating: 1600

Small Hint:

Drop perpendiculars from $A$ and $B$ to $\overline{DF}$

Big Hint:

The segment through midpoint $B$ parallel to $EF$ reaches the midpoint of $DF$

Solution:

Drop perpendiculars $AG$ and $BH$ to $DF.$ In the isosceles trapezoid, $GH=AB=4$ and $DG=HC=\frac{DC-AB}{2}=3.$ Since $B$ is the midpoint of $DE$ and $BH\parallel EF,$ the midpoint theorem makes $H$ the midpoint of $DF.$ Hence $DH=DG+GH=7,$ so $DF=14$ and $CF=14-10=4.$

Thus the correct answer is D.

5.

Toothpicks of equal length are used to build a rectangular grid as shown. If the grid is $20$ toothpicks high and $10$ toothpicks wide, then the number of toothpicks used is

$30$

$200$

$410$

$420$

$430$

Concepts:counting shapes in figures rectangle systematic listing

Difficulty rating: 1060

Small Hint:

Count horizontal and vertical toothpicks separately

Big Hint:

A grid $10$ toothpicks wide has $11$ vertical grid lines

Solution:

There are $11$ vertical grid lines with $20$ toothpicks each, and $21$ horizontal grid lines with $10$ toothpicks each. The total is $11\cdot20+21\cdot10=430.$

Thus the correct answer is E.

6.

If $a,$ $b\gt0$ and the triangle in the first quadrant bounded by the coordinate axes and the graph of $ax+by=6$ has area $6,$ then $ab=$

$3$

$6$

$12$

$108$

$432$

Concepts:coordinate geometry triangle area linear equation

Difficulty rating: 1210

Small Hint:

Find the intercepts of the line on the two axes

Big Hint:

Use those intercepts as the base and height of the triangle

Solution:

The intercepts are $\frac{6}{a}$ and $\frac{6}{b}.$ Therefore the triangle’s area is $\frac12\cdot\frac6a\cdot\frac6b=\frac{18}{ab}=6,$ which gives $ab=3.$

Thus the correct answer is A.

7.

In $\triangle ABC,$ $\angle A=100^\circ,$ $\angle B=50^\circ,$ $\angle C=30^\circ,$ $AH$ is an altitude, and $BM$ is a median. Then $\angle MHC=$

$15^\circ$

$22.5^\circ$

$30^\circ$

$40^\circ$

$45^\circ$

Concepts:median (geometry)altitude angle chasing

Difficulty rating: 1560

Small Hint:

Compare the horizontal and vertical changes from $H$ to midpoint $M$

Big Hint:

Both changes are half the corresponding legs of right triangle $AHC$

Solution:

Because $M$ is the midpoint of $AC,$ the horizontal change from $H$ to $M$ is $\frac{HC}{2},$ while the vertical change is $\frac{AH}{2}.$ Thus $\begin{aligned} \tan\angle MHC &=\frac{\frac{AH}{2}}{\frac{HC}{2}}\\ &=\frac{AH}{HC}\\ &=\tan\angle ACH. \end{aligned}$ Hence $\angle MHC=\angle C=30^\circ.$

Thus the correct answer is C.

8.

For how many integers $n$ between $1$ and $100$ does $x^2+x-n$ factor into the product of two linear factors with integer coefficients?

$0$

$1$

$2$

$9$

$10$

Concepts:factoring counting integers in a range quadratic

Difficulty rating: 1530

Small Hint:

Write the factors as $(x-r)(x+s)$ with positive integers $r,s$

Big Hint:

The coefficient of $x$ forces the two integer parameters to differ by $1$

Solution:

An integer factorization must have the form $(x-k)(x+k+1)$ for some positive integer $k.$ Expanding gives $x^2+x-k(k+1),$ so $n=k(k+1).$ The values $k=1,2,\ldots,9$ give $n\le100,$ while $10\cdot11\gt100.$ There are $9$ values.

Thus the correct answer is D.

9.

Mr. and Mrs. Zeta want to name their baby Zeta so that its monogram (first, middle, and last initials) will be in alphabetical order with no letters repeated. How many such monograms are possible?

$276$

$300$

$552$

$600$

$15600$

Concepts:combinations arrangements with restrictions

Difficulty rating: 1470

Small Hint:

The last initial is already fixed as $Z$

Big Hint:

Choosing the other two distinct letters determines their alphabetical order

Solution:

The first two initials must be two distinct letters chosen from $A$ through $Y.$ Once chosen, their order is forced. Therefore the number of monograms is $\binom{25}{2}=300.$

Thus the correct answer is B.

10.

Consider the sequence defined recursively by $u_1=a$ (any positive number), and $u_{n+1}=-\frac{1}{u_n+1},$ $n=1,$ $2,$ $3,$ $\ldots.$ For which of the following values of $n$ must $u_n=a?$

$14$

$15$

$16$

$17$

$18$

Concepts:recursion pattern recognition rational equation

Difficulty rating: 1770

Small Hint:

Compute $u_2,u_3,$ and $u_4$ symbolically

Big Hint:

Look for the period of the transformation $x\mapsto-\frac{1}{x+1}$

Solution:

Direct substitution gives $\begin{aligned} u_2&=-\frac1{a+1},\\ u_3&=-\frac{a+1}{a},\\ u_4&=a. \end{aligned}$ The sequence therefore repeats every $3$ terms, so $u_n=a$ whenever $n\equiv1\pmod3.$ Among the choices, only $16$ has that form.

Thus the correct answer is C.

11.

Let $a,$ $b,$ $c$ and $d$ be integers with $a\lt2b,$ $b\lt3c,$ and $c\lt4d.$ If $d\lt100,$ the largest possible value for $a$ is

$2367$

$2375$

$2391$

$2399$

$2400$

Concepts:inequality extremal argument counting integers in a range

Difficulty rating: 1420

Small Hint:

Maximize the integers from $d$ backward to $a$

Big Hint:

Each strict inequality lowers the greatest allowable integer by $1$

Solution:

The largest possible values are $d=99,\ c=395,\ b=1184,$ and finally $a=2367.$ Each value satisfies its strict inequality, so this maximum is attainable.

Thus the correct answer is A.

12.

The traffic on a certain east-west highway moves at a constant speed of $60$ miles per hour in both directions. An eastbound driver passes $20$ westbound vehicles in a five-minute interval. Assume vehicles in the westbound lane are equally spaced. Which of the following is closest to the number of westbound vehicles present in a $100$ -mile section of highway?

$100$

$120$

$200$

$240$

$400$

Concepts:relative speed distance rate and time ratio and proportion

Difficulty rating: 1360

Small Hint:

The cars approach one another at the sum of their speeds

Big Hint:

Use the distance covered at relative speed to find the spacing between westbound vehicles

Solution:

The relative speed is $120$ miles per hour, so in five minutes the driver covers $10$ relative miles. Passing $20$ equally spaced vehicles in that distance means the spacing is about $\frac{10}{20}=\frac{1}{2}$ mile. Thus a $100$ -mile section contains about $\frac{100}{\frac{1}{2}}=200$ vehicles.

Thus the correct answer is C.

13.

Two strips of width $1$ overlap at an angle of $\alpha$ as shown. The area of the overlap (shown shaded) is

$\sin\alpha$

$\frac1{\sin\alpha}$

$\frac1{1-\cos\alpha}$

$\frac1{\sin^2\alpha}$

$\frac1{(1-\cos\alpha)^2}$

Concepts:parallelogram area trigonometry

Difficulty rating: 1810

Small Hint:

The overlap is a parallelogram with altitude $1$

Big Hint:

If its slanted side has length $L,$ then $L\sin\alpha=1$

Solution:

The overlap is a parallelogram. Taking a slanted side as its base, the perpendicular height is the width $1$ of the slanted strip. If that base has length $L,$ its vertical component is the width of the horizontal strip, so $L\sin\alpha=1.$ Hence $L=\frac{1}{\sin\alpha},$ and the area is $L\cdot1=\frac{1}{\sin\alpha}.$

Thus the correct answer is B.

14.

$\cot10+\tan5=$

$\csc5$

$\csc10$

$\sec5$

$\sec10$

$\sin15$

Concepts:trigonometry algebraic manipulation

Difficulty rating: 1720

Small Hint:

Rewrite both terms using sine and cosine

Big Hint:

Use $\sin10=2\sin5\cos5$ and $\cos10=\cos^25-\sin^25$

Solution:

Using $\sin10=2\sin5\cos5,$ $\begin{aligned} \cot10+\tan5 &=\frac{\cos10}{\sin10}\\ &\quad+\frac{\sin5}{\cos5}. \end{aligned}$ Combining the fractions gives numerator $\cos10+2\sin^25.$ Since $\cos10=\cos^25-\sin^25,$ this numerator is $1,$ and the expression is $\csc10.$

Thus the correct answer is B.

15.

In $\triangle ABC,$ $AB=5,$ $BC=7,$ $AC=9$ and $D$ is on $\overline{AC}$ with $BD=5.$ Find the ratio $AD:DC.$

$4:3$

$7:5$

$11:6$

$13:5$

$19:8$

Concepts:Stewart’s Theorem ratio and proportion

Difficulty rating: 2170

Small Hint:

Let $AD=m$ and $DC=9-m$

Big Hint:

Apply Stewart’s Theorem to cevian $BD$

Solution:

Let $AD=m$ and $DC=9-m.$ Stewart’s Theorem gives $\begin{gathered} 7^2m+5^2(9-m) \\ =9\bigl(5^2+m(9-m)\bigr). \end{gathered}$ Simplifying yields $9m^2-57m=0,$ so $m=\frac{19}{3}.$ Then $DC=\frac{8}{3},$ and $AD:DC=19:8.$

Thus the correct answer is E.

16.

A lattice point is a point in the plane with integer coordinates. How many lattice points are on the line segment whose endpoints are $(3,17)$ and $(48,281)?$ (Include both endpoints of the segment in your count.)

$2$

$4$

$6$

$16$

$46$

Concepts:lattice point greatest common divisor

Difficulty rating: 1560

Small Hint:

Compute the horizontal and vertical coordinate differences

Big Hint:

A segment with differences $\Delta x,\Delta y$ contains $\gcd(|\Delta x|,|\Delta y|)+1$ lattice points

Solution:

The coordinate differences are $45$ and $264,$ whose greatest common divisor is $3.$ The number of lattice points, including both endpoints, is therefore $\gcd(45,264)+1=4.$

Thus the correct answer is B.

17.

The perimeter of an equilateral triangle exceeds the perimeter of a square by $1989$ cm. The length of each side of the triangle exceeds the length of each side of the square by $d$ cm. The square has perimeter greater than $0.$ How many positive integers are not possible values for $d?$

$0$

$9$

$221$

$663$

infinitely many

Concepts:perimeter linear equation counting integers in a range

Difficulty rating: 1550

Small Hint:

Let the square’s side length be $s$

Big Hint:

Translate the positive-perimeter condition into a strict inequality for $d$

Solution:

If the square has side $s,$ the triangle has side $s+d.$ The perimeter condition gives $3(s+d)-4s=1989,$ so $s=3d-1989.$ The condition $s\gt0$ is equivalent to $d\gt663.$ Thus the positive integers $1$ through $663$ are precisely the impossible values.

Thus the correct answer is D.

18.

The set of all real numbers $x$ for which $x+\sqrt{x^2+1}-\frac1{x+\sqrt{x^2+1}}$ is a rational number is the set of all

integers $x$

rational $x$

real $x$

$x$ for which $\sqrt{x^2+1}$ is rational

$x$ for which $x+\sqrt{x^2+1}$ is rational

Concepts:rationalizing denominator radical algebraic manipulation

Difficulty rating: 1740

Small Hint:

Rationalize the reciprocal term

Big Hint:

The conjugate $\sqrt{x^2+1}-x$ is the reciprocal of the denominator

Solution:

Rationalizing gives $\frac1{x+\sqrt{x^2+1}}=\sqrt{x^2+1}-x.$ The entire expression is therefore $2x.$ It is rational exactly when $x$ is rational.

Thus the correct answer is B.

19.

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths $3,$ $4,$ and $5.$ What is the area of the triangle?

$6$

$\frac{18}{\pi^2}$

$\frac9{\pi^2}(\sqrt3-1)$

$\frac9{\pi^2}(\sqrt3+1)$

$\frac9{\pi^2}(\sqrt3+3)$

Concepts:arc circumcircle, circumcenter, and circumradius triangle area

Difficulty rating: 2340

Small Hint:

The circumference is $12,$ so first find the radius and the three central angles

Big Hint:

Split the triangle into three triangles having the circle’s center as a common vertex

Solution:

The radius is $R=\frac{12}{2\pi}=\frac{6}{\pi}.$ The arc lengths $3,4,5$ give central angles $\frac{\pi}{2},\frac{2\pi}{3},\frac{5\pi}{6}.$ Their sines sum to $1+\frac{\sqrt3}{2}+\frac{1}{2}=\frac{3+\sqrt3}{2}.$ Splitting the triangle at the center, its area is $\begin{aligned} K&=\frac12\cdot\frac{36}{\pi^2} \cdot\frac{3+\sqrt3}{2}\\ &=\frac9{\pi^2}(\sqrt3+3). \end{aligned}$

Thus the correct answer is E.

20.

Let $x$ be a real number selected uniformly at random between $100$ and $200.$ If $\lfloor\sqrt{x}\rfloor=12,$ find the probability that $\lfloor\sqrt{100x}\rfloor=120.$ ( $\lfloor v\rfloor$ means the greatest integer less than or equal to $v.$ )

$\frac2{25}$

$\frac{241}{2500}$

$\frac1{10}$

$\frac{96}{625}$

$1$

Concepts:conditional probability floor and ceiling functions geometric probability

Difficulty rating: 2150

Small Hint:

Convert each floor condition into an interval for $x$

Big Hint:

The conditional probability is the ratio of the two relevant interval lengths

Solution:

The given condition is $144\le x\lt169,$ an interval of length $25.$ The desired condition is $120^2\le100x\lt121^2,$ or $144\le x\lt146.41.$ Its length is $2.41=\frac{241}{100},$ so the conditional probability is $\frac{\frac{241}{100}}{25}=\frac{241}{2500}.$

Thus the correct answer is B.

21.

A square flag has a red cross of uniform width with a blue square in the center on a white background as shown. (The cross is symmetric with respect to each of the diagonals of the square.) If the entire cross (both the red arms and the blue center) takes up $36\%$ of the area of the flag, what percent of the area of the flag is blue?

$0.5$

$1$

$2$

$3$

$6$

Concepts:area decomposition percentage symmetry

Difficulty rating: 1980

Small Hint:

Let $h$ be half a diagonal of the central blue square after scaling the flag side to $1$

Big Hint:

The four white corner triangles together have area $(1-2h)^2$

Solution:

Scale the flag to a unit square, and let $h$ be the horizontal distance from its center to a vertex of the blue square. The four congruent white corner triangles have total area $(1-2h)^2.$ Since the cross occupies $0.36,$ $(1-2h)^2=0.64.$ With $0\lt h\lt\frac{1}{2},$ this gives $h=0.1.$ The blue square has perpendicular diagonals $2h$ and $2h,$ so its area is $\frac12(2h)^2=2h^2=0.02,$ or $2\%$ of the flag.

Thus the correct answer is C.

22.

A child has a set of $96$ distinct blocks. Each block is one of $2$ materials (plastic, wood), $3$ sizes (small, medium, large), $4$ colors (blue, green, red, yellow), and $4$ shapes (circle, hexagon, square, triangle). How many blocks in the set are different from the “plastic medium red circle” in exactly two ways? (The “wood medium red square” is such a block.)

$29$

$39$

$48$

$56$

$62$

Concepts:casework combinations multiplication principle

Difficulty rating: 1600

Small Hint:

For each attribute, count the alternatives different from the specified block

Big Hint:

Choose a pair of attributes to change, then multiply their alternative counts

Solution:

The numbers of alternatives for material, size, color, and shape are $1,2,3,3.$ Changing exactly two attributes gives $\begin{aligned} &1\cdot2+1\cdot3+1\cdot3\\ &\qquad+2\cdot3+2\cdot3+3\cdot3=29. \end{aligned}$

Thus the correct answer is A.

23.

A particle moves through the first quadrant as follows. During the first minute it moves from the origin to $(1,0).$ Thereafter, it continues to follow the directions indicated in the figure, going back and forth between the positive $x$ and $y$ axes, moving one unit of distance parallel to an axis in each minute. At which point will the particle be after exactly $1989$ minutes?

$(35,44)$

$(36,45)$

$(37,45)$

$(44,35)$

$(45,36)$

Concepts:process simulation pattern recognition square (geometry)

Difficulty rating: 2150

Small Hint:

Record the particle’s location at times $1,4,9,16,\ldots$

Big Hint:

Compare $1989$ with the consecutive squares $44^2$ and $45^2$

Solution:

At time $n^2,$ the particle is at $(n,0)$ for odd $n$ and at $(0,n)$ for even $n.$ Thus at time $44^2=1936$ it is at $(0,44).$ It then moves $44$ units right, reaching $(44,44)$ at time $1980,$ and moves downward for the next $9$ minutes. At time $1989$ it is therefore at $(44,35).$

Thus the correct answer is D.

24.

Five people are sitting at a round table. Let $f\ge0$ be the number of people sitting next to at least one female and $m\ge0$ be the number of people sitting next to at least one male. The number of possible ordered pairs $(f,m)$ is

$7$

$8$

$9$

$10$

$11$

Concepts:circular arrangements casework systematic listing

Difficulty rating: 2210

Small Hint:

Classify arrangements by the number of females

Big Hint:

With two females, separate the adjacent and nonadjacent cases; obtain three-female cases by symmetry

Solution:

For $0,1,2$ females, the possible pairs are respectively $\begin{aligned} &(0,5),\ (2,5),\\ &(4,5),\ (3,4), \end{aligned}$ where the two-female cases distinguish adjacent from nonadjacent females. Swapping the sexes gives $\begin{aligned} &(5,0),\ (5,2),\\ &(5,4),\ (4,3). \end{aligned}$ These are $8$ distinct ordered pairs.

Thus the correct answer is B.

25.

In a certain cross-country meet between two teams of five runners each, a runner who finishes in the $n$ th position contributes $n$ to his team’s score. The team with the lower score wins. If there are no ties among the runners, how many different winning scores are possible?

$10$

$13$

$27$

$120$

$126$

Concepts:subsets extremal argument systematic listing

Difficulty rating: 2270

Small Hint:

The two team scores add to $1+2+\cdots+10$

Big Hint:

Determine the minimum winning score and verify that every integer below half the total is attainable

Solution:

The two scores sum to $55,$ so the winning score is at most $27.$ Its minimum is $1+2+3+4+5=15.$ Every score from $15$ through $20$ is obtained by $\{1,2,3,4,k\}$ for $k=5,\ldots,10.$ Scores $21$ through $25$ use $\{1,2,3,k,10\}$ for $k=5,\ldots,9,$ and $26,27$ use $\{1,2,4,9,10\}$ and $\{1,2,5,9,10\}.$ Thus all $13$ integers from $15$ through $27$ are possible.

Thus the correct answer is B.

26.

A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is

$\frac{\sqrt3}{12}$

$\frac{\sqrt6}{16}$

$\frac16$

$\frac{\sqrt2}{8}$

$\frac14$

Concepts:cube geometry polyhedron volume

Difficulty rating: 1940

Small Hint:

Scale the cube to side length $2$ and place its center at the origin

Big Hint:

The octahedron has vertices $(\pm1,0,0),(0,\pm1,0),(0,0,\pm1)$

Solution:

Take a cube of side $2,$ so its volume is $8.$ The face centers are $(\pm1,0,0),(0,\pm1,0),(0,0,\pm1).$ In each octant, the octahedron cuts out a tetrahedron with three perpendicular unit edges and volume $\frac{1}{6}.$ Thus its volume is $8(\frac{1}{6})=\frac{4}{3},$ and the ratio is $\frac{\frac{4}{3}}{8}=\frac{1}{6}.$

Thus the correct answer is C.

27.

Let $n$ be a positive integer. If the equation $2x+2y+z=n$ has $28$ solutions in positive integers $x,$ $y$ and $z,$ then $n$ must be either

$14$ or $15$

$15$ or $16$

$16$ or $17$

$17$ or $18$

$18$ or $19$

Concepts:Diophantine Equation summation counting pairs

Difficulty rating: 2340

Small Hint:

Group solutions according to $s=x+y$

Big Hint:

For fixed $s,$ count the positive ordered pairs $(x,y)$ and require $n-2s\ge1$

Solution:

For $s=x+y\ge2,$ there are $s-1$ positive ordered pairs $(x,y),$ and $z=n-2s$ is positive when $s\le\lfloor\frac{n-1}{2}\rfloor.$ Writing $m=\lfloor\frac{n-1}{2}\rfloor,$ the number of solutions is $\sum_{s=2}^{m}(s-1)=\frac{m(m-1)}2.$ Setting this equal to $28$ gives $m=8.$ Hence $\lfloor\frac{n-1}{2}\rfloor=8,$ so $n=17$ or $18.$

Thus the correct answer is D.

28.

Find the sum of the roots of $\tan^2x-9\tan x+1=0$ that are between $x=0$ and $x=2\pi$ radians.

$\frac\pi2$

$\pi$

$\frac{3\pi}2$

$3\pi$

$4\pi$

Concepts:quadratic trigonometry algebraic manipulation

Difficulty rating: 2260

Small Hint:

Let the two positive roots in $\tan x$ be $r$ and $s$

Big Hint:

Use $rs=1$ to relate $\arctan r+\arctan s,$ then include the period of tangent

Solution:

The two roots $r,s$ of $t^2-9t+1=0$ are positive and satisfy $rs=1.$ Therefore their acute arctangents add to $\frac{\pi}{2}.$ Each tangent value occurs twice between $0$ and $2\pi,$ with the second occurrence shifted by $\pi.$ The sum of all four roots is $2\left(\frac\pi2\right)+2\pi=3\pi.$

Thus the correct answer is D.

29.

Find $\sum_{k=0}^{49}(-1)^k\binom{99}{2k},$ where $\binom{n}{j}=\frac{n!}{j!(n-j)!}.$

$-2^{50}$

$-2^{49}$

$0$

$2^{49}$

$2^{50}$

Concepts:binomial theorem complex number roots of unity

Difficulty rating: 2760

Small Hint:

Recognize $(-1)^k$ as $i^{2k}$

Big Hint:

Take the real part of the binomial expansion of $(1+i)^{99}$

Solution:

The sum is the real part of $(1+i)^{99}=2^{\frac{99}{2}}e^{\frac{99\pi i}{4}}.$ Since $99\equiv3\pmod8,$ its real part is $\begin{aligned} 2^{\frac{99}{2}}\cos\frac{3\pi}{4} &=2^{\frac{99}{2}}\left(-\frac{\sqrt2}{2}\right)\\ &=-2^{49}. \end{aligned}$

Thus the correct answer is B.

30.

Suppose that $7$ boys and $13$ girls line up in a row. Let $S$ be the number of places in the row where a boy and a girl are standing next to each other. For example, for the row $GBBGGGBGBG$ $GGBGBGGBGG$ we have $S=12.$ The average value of $S$ (if all possible orders of these $20$ people are considered) is closest to

$9$

$10$

$11$

$12$

$13$

Concepts:expected value double counting permutations

Difficulty rating: 2550

Small Hint:

Use an indicator for each of the $19$ adjacent pairs

Big Hint:

For a fixed adjacent pair, compute the probability of seeing $BG$ or $GB$

Solution:

For each of the $19$ adjacent position pairs, the probability of mixed sexes is $\frac7{20}\cdot\frac{13}{19} +\frac{13}{20}\cdot\frac7{19}.$ By linearity of expectation, $\begin{aligned} \mathbb E[S] &=19\cdot\frac{2\cdot7\cdot13}{20\cdot19}\\ &=\frac{91}{10}=9.1, \end{aligned}$ which is closest to $9.$

Thus the correct answer is A.