2017 AMC 12B — Problems & Online Timed Exam

Timed

1:15:00

1.

Kymbrea’s comic book collection currently has $30$ comic books in it, and she is adding to her collection at the rate of $2$ comic books per month. LaShawn’s collection currently has $10$ comic books in it, and he is adding to his collection at the rate of $6$ comic books per month. After how many months will LaShawn’s collection have twice as many comic books as Kymbrea’s?

$1$

$4$

$5$

$20$

$25$

Answer: E

Concepts:linear equation

Difficulty rating: 920

Small Hint:

After $m$ months the collections hold $30 + 2m$ and $10 + 6m$ comic books

Big Hint:

Set $10 + 6m = 2(30 + 2m)$ and solve for $m$

Solution:

After $m$ months, Kymbrea has $30 + 2m$ comic books and LaShawn has $10 + 6m.$ Setting $10 + 6m = 2(30 + 2m)$ gives $10 + 6m = 60 + 4m,$ so $2m = 50$ and $m = 25.$

Thus, the correct answer is E.

2.

Real numbers $x,$ $y,$ and $z$ satisfy the inequalities

$0 \lt x \lt 1, \quad -1 \lt y \lt 0,$ and $1 \lt z \lt 2.$

Which of the following numbers is necessarily positive?

$y + x^2$

$y + xz$

$y + y^2$

$y + 2y^2$

$y + z$

Answer: E

Concepts:inequality counterexample

Difficulty rating: 1020

Small Hint:

Since $y \gt -1$ and $z \gt 1,$ what can you say about $y + z?$

Big Hint:

For the other four, try $x = \tfrac18,$ $y = -\tfrac14,$ $z = \tfrac32$ to force a negative value

Solution:

Adding $y \gt -1$ and $z \gt 1$ gives $y + z \gt 0,$ so $y + z$ is always positive. Each of the other four choices can be made negative: with $x = \tfrac18,$ $y = -\tfrac14,$ $z = \tfrac32,$ every one of $y + x^2,$ $y + xz,$ $y + y^2,$ and $y + 2y^2$ is negative.

Thus, the correct answer is E.

3.

Suppose that $x$ and $y$ are nonzero real numbers such that

$\frac{3x + y}{x - 3y} = -2.$

What is the value of $\frac{x + 3y}{3x - y}?$

$-3$

$-1$

$1$

$2$

$3$

Answer: D

Concepts:algebraic manipulation substitution

Difficulty rating: 1130

Small Hint:

Clear the denominator: $3x + y = -2(x - 3y)$

Big Hint:

This simplifies to $x = y;$ substitute into the target expression

Solution:

The equation gives $3x + y = -2(x - 3y)$ $= -2x + 6y,$ so $5x = 5y,$ meaning $x = y.$ Then $\frac{x + 3y}{3x - y} = \frac{y + 3y}{3y - y} = \frac{4y}{2y} = 2.$

Thus, the correct answer is D.

4.

Samia set off on her bicycle to visit her friend, traveling at an average speed of $17$ kilometers per hour. When she had gone half the distance to her friend’s house, a tire went flat, and she walked the rest of the way at $5$ kilometers per hour. In all it took her $44$ minutes to reach her friend’s house. In kilometers rounded to the nearest tenth, how far did Samia walk?

$2.0$

$2.2$

$2.8$

$3.4$

$4.4$

Answer: C

Concepts:distance rate and time linear equation

Difficulty rating: 1270

Small Hint:

Let $d$ be the one-way distance she biked (and also walked); use time $= \dfrac{\text{distance}}{\text{rate}}$

Big Hint:

Solve $\dfrac{d}{17} + \dfrac{d}{5} = \dfrac{44}{60}$ for $d$

Solution:

Let $2d$ be the total distance, so she biked $d$ at $17$ km/h and walked $d$ at $5$ km/h. The total time in hours is $\frac{d}{17} + \frac{d}{5} = \frac{44}{60}.$ Combining the left side gives $\dfrac{22d}{85} = \dfrac{11}{15},$ so $d = \dfrac{11}{15} \cdot \dfrac{85}{22} = \dfrac{17}{6} = 2.833\ldots$ She walked about $2.8$ kilometers.

Thus, the correct answer is C.

5.

The data set $[6, 19, 33, 33, 39,$ $41,$ $41,$ $43,$ $51,$ $57]$ has median $Q_2 = 40,$ first quartile $Q_1 = 33,$ and third quartile $Q_3 = 43.$ An outlier in a data set is a value that is more than $1.5$ times the interquartile range below the first quartile $(Q_1)$ or more than $1.5$ times the interquartile range above the third quartile $(Q_3),$ where the interquartile range is defined as $Q_3 - Q_1.$ How many outliers does this data set have?

$0$

$1$

$2$

$3$

$4$

Answer: B

Concepts:median (data)

Difficulty rating: 1130

Small Hint:

The interquartile range is $Q_3 - Q_1 = 43 - 33 = 10$

Big Hint:

Count values below $33 - 1.5(10)$ or above $43 + 1.5(10)$

Solution:

The interquartile range is $43 - 33 = 10,$ so $1.5$ times it is $15.$ Outliers are values less than $33 - 15 = 18$ or greater than $43 + 15 = 58.$ Only $6$ falls below $18,$ and nothing exceeds $58,$ so there is exactly $1$ outlier.

Thus, the correct answer is B.

6.

The circle having $(0, 0)$ and $(8, 6)$ as the endpoints of a diameter intersects the $x$ -axis at a second point. What is the $x$ -coordinate of this point?

$4\sqrt{2}$

$6$

$5\sqrt{2}$

$8$

$6\sqrt{2}$

Answer: D

Concepts:coordinate geometry circle midpoint

Difficulty rating: 1350

Small Hint:

The center is the midpoint $(4, 3)$ and the radius is $\sqrt{4^2 + 3^2} = 5$

Big Hint:

Set $y = 0$ in $(x - 4)^2 + (y - 3)^2 = 25$

Solution:

The center is the midpoint of the diameter, $(4, 3),$ and the radius is $\sqrt{4^2 + 3^2} = 5.$ The circle is $(x - 4)^2 + (y - 3)^2 = 25.$ Setting $y = 0$ gives $(x - 4)^2 = 16,$ so $x = 0$ or $x = 8.$ The second intersection with the $x$ -axis is at $x = 8.$

Thus, the correct answer is D.

7.

The functions $\sin(x)$ and $\cos(x)$ are periodic with least period $2\pi.$ What is the least period of the function $\cos(\sin(x))?$

$\dfrac{\pi}{2}$

$\pi$

$2\pi$

$4\pi$

It’s not periodic.

Answer: B

Concepts:trigonometric identity function

Difficulty rating: 1500

Small Hint:

Use that $\sin(x + \pi) = -\sin(x)$ and $\cos$ is even

Big Hint:

To rule out a smaller period, note $\cos(\sin(x)) = 1$ only when $\sin(x) = 0,$ i.e. at multiples of $\pi$

Solution:

Since $\cos(\sin(x + \pi))$ $= \cos(-\sin(x))$ $= \cos(\sin(x)),$ the function has period $\pi.$ It cannot be smaller: $\cos(\sin(x)) = 1$ exactly when $\sin(x) = 0,$ which happens only at integer multiples of $\pi,$ so the maxima are spaced $\pi$ apart. The least period is $\pi.$

Thus, the correct answer is B.

8.

The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?

$\dfrac{\sqrt{3} - 1}{2}$

$\dfrac{1}{2}$

$\dfrac{\sqrt{5} - 1}{2}$

$\dfrac{\sqrt{2}}{2}$

$\dfrac{\sqrt{6} - 1}{2}$

Answer: C

Concepts:quadratic Pythagorean Theorem ratio and proportion

Difficulty rating: 1440

Small Hint:

With short side $x$ and long side $y,$ the diagonal is $\sqrt{x^2 + y^2};$ write $\dfrac{x}{y} = \dfrac{y}{\sqrt{x^2+y^2}}$

Big Hint:

Let $r = \dfrac{x^2}{y^2}.$ Squaring the ratio equation gives $\dfrac{1}{r} = r + 1$

Solution:

Let $x$ and $y$ be the short and long sides, so the diagonal is $\sqrt{x^2 + y^2}$ and $\dfrac{x^2}{y^2} = \dfrac{y^2}{x^2 + y^2}.$ Writing $r = \dfrac{x^2}{y^2},$ the right side is $\dfrac{y^2}{x^2 + y^2} = \dfrac{1}{r + 1},$ so $r = \dfrac{1}{r+1},$ giving $r^2 + r - 1 = 0.$ The positive root is $r = \dfrac{\sqrt{5} - 1}{2}.$

Thus, the correct answer is C.

9.

A circle has center $(-10, -4)$ and radius $13.$ Another circle has center $(3, 9)$ and radius $\sqrt{65}.$ The line passing through the two points of intersection of the two circles has equation $x + y = c.$ What is $c?$

$3$

$3\sqrt{3}$

$4\sqrt{2}$

$6$

$\dfrac{13}{2}$

Answer: A

Concepts:radical axis circle coordinate geometry

Difficulty rating: 1410

Small Hint:

Write both circles as $(x+10)^2 + (y+4)^2 = 169$ and $(x-3)^2 + (y-9)^2 = 65$

Big Hint:

Subtract one expanded equation from the other; the quadratic terms cancel

Solution:

The circles are $(x+10)^2 + (y+4)^2 = 169$ and $(x-3)^2 + (y-9)^2 = 65.$ Expanding and subtracting the second from the first cancels the $x^2$ and $y^2$ terms and simplifies to $x + y = 3.$ Any intersection point satisfies this, so it is the line through both, and $c = 3.$

Thus, the correct answer is A.

10.

At Typico High School, $60\%$ of the students like dancing, and the rest dislike it. Of those who like dancing, $80\%$ say that they like it, and the rest say that they dislike it. Of those who dislike dancing, $90\%$ say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?

$10\%$

$12\%$

$20\%$

$25\%$

$33\tfrac{1}{3}\%$

Answer: D

Concepts:conditional probability Bayes’ Theorem

Difficulty rating: 1440

Small Hint:

Students who like dancing but say they dislike it are $60\% \cdot 20\%$ of all students

Big Hint:

Divide that group by the total who say they dislike dancing

Solution:

Students who like dancing but say they dislike it make up $60\% \cdot 20\% = 12\%$ of all students. Students who dislike dancing and say so make up $40\% \cdot 90\% = 36\%.$ Among everyone who says they dislike dancing, the fraction who actually like it is $\frac{12}{12 + 36} = \frac{12}{48} = \frac{1}{4} = 25\%.$

Thus, the correct answer is D.

11.

Call a positive integer monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, $3,$ $23578,$ and $987620$ are monotonous, but $88,$ $7434,$ and $23557$ are not. How many monotonous positive integers are there?

$1024$

$1524$

$1533$

$1536$

$2048$

Answer: B

Concepts:subsets bijection inclusion-exclusion

Difficulty rating: 1590

Small Hint:

Each nonempty subset of $\{1, 2, \ldots, 9\}$ gives exactly one strictly increasing number

Big Hint:

Decreasing numbers correspond to subsets of $\{0, 1, \ldots, 9\}$ except $\varnothing$ and $\{0\};$ single digits are counted twice

Solution:

Strictly increasing monotonous numbers correspond to nonempty subsets of $\{1, \ldots, 9\},$ giving $2^9 - 1 = 511.$ Strictly decreasing ones correspond to subsets of $\{0, 1, \ldots, 9\}$ other than $\varnothing$ and $\{0\}$ (a leading $0$ is not allowed), giving $2^{10} - 2 = 1022.$ The nine single-digit numbers are counted in both, so the total is $511 + 1022 - 9 = 1524.$

Thus, the correct answer is B.

12.

What is the sum of the roots of $z^{12} = 64$ that have a positive real part?

$2$

$4$

$2 + 2\sqrt{3}$

$2\sqrt{2} + \sqrt{6}$

$(1 + \sqrt{3}) + (1 + \sqrt{3})i$

Answer: D

Concepts:roots of unity complex number symmetry

Difficulty rating: 1630

Small Hint:

The twelve roots lie on a circle of radius $\sqrt{2},$ spaced $30^\circ$ apart

Big Hint:

The roots with positive real part are at angles $0, \pm 30^\circ, \pm 60^\circ;$ imaginary parts cancel in pairs

Solution:

The roots of $z^{12} = 64$ lie on the circle of radius $64^{\frac{1}{12}} = \sqrt{2},$ at angles that are multiples of $30^\circ.$ Those with positive real part are at angles $0, \pm 30^\circ, \pm 60^\circ.$ Their imaginary parts cancel, so the sum is $\begin{aligned} &\sqrt{2} + 2\sqrt{2}\cos 30^\circ \\ &\quad {}+ 2\sqrt{2}\cos 60^\circ \\ &\quad {}= \sqrt{2}\bigl(1 + \sqrt{3} + 1\bigr) \\ &\quad {}= 2\sqrt{2} + \sqrt{6}. \end{aligned}$

Thus, the correct answer is D.

13.

In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?

$6$

$8$

$9$

$12$

$15$

Answer: D

Concepts:casework symmetry

Difficulty rating: 1660

Small Hint:

Count the labeled colorings first, then apply Burnside’s Lemma to the $6$ symmetries of the triangle

Big Hint:

A nonidentity rotation fixes none; under a reflection, the odd green and blue counts must use the two fixed disks

Solution:

Before accounting for symmetry, there are $\dfrac{6!}{3!2!}=60$ paintings. The two nonidentity rotations partition the disks into two $3$ -cycles, so neither can fix a painting having color counts $3,2,1.$

Each of the $3$ reflections fixes $2$ disks and swaps the other $4$ in $2$ pairs. For a painting to be fixed, the lone green disk and one of the $3$ blue disks must occupy the two fixed positions, in $2$ orders. Of the two swapped pairs, either one can be the red pair, giving $2\cdot2=4$ fixed paintings per reflection. Burnside’s Lemma therefore gives $\dfrac{60+3(4)}{6}=12$ distinct paintings.

Thus, the correct answer is D.

14.

An ice-cream novelty item consists of a cup in the shape of a $4$ -inch-tall frustum of a right circular cone, with a $2$ -inch-diameter base at the bottom and a $4$ -inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height $4$ inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches?

$8\pi$

$\dfrac{28\pi}{3}$

$12\pi$

$14\pi$

$\dfrac{44\pi}{3}$

Answer: E

Concepts:cone volume similarity

Difficulty rating: 1530

Small Hint:

The frustum is a full cone (radius $2,$ height $8$ ) minus a small cone (radius $1,$ height $4$ )

Big Hint:

Add the top solid cone of radius $2$ and height $4;$ use $V = \tfrac13 \pi r^2 h$

Solution:

Extending the frustum’s sides to a point, similar triangles show the frustum equals a cone of radius $2$ and height $8$ minus a cone of radius $1$ and height $4:$ $\begin{aligned} &\tfrac13 \pi (2^2)(8) \\ &\quad {}- \tfrac13 \pi (1^2)(4) \\ &\quad {}= \tfrac{32}{3}\pi - \tfrac{4}{3}\pi \\ &\quad {}= \tfrac{28}{3}\pi. \end{aligned}$ The top cone of radius $2$ and height $4$ adds $\tfrac13 \pi (2^2)(4) = \tfrac{16}{3}\pi.$ The total is $\tfrac{28}{3}\pi + \tfrac{16}{3}\pi = \tfrac{44}{3}\pi.$

Thus, the correct answer is E.

15.

Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB' = 3 \cdot AB.$ Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC' = 3 \cdot BC,$ and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA' = 3 \cdot CA.$ What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC?$

$9 : 1$

$16 : 1$

$25 : 1$

$36 : 1$

$37 : 1$

Answer: E

Concepts:area ratio area decomposition equilateral triangle

Difficulty rating: 1660

Small Hint:

Let $X$ be the area of $\triangle ABC.$ Draw $CB',$ $AC',$ $BA'$ to split $\triangle A'B'C'$ into pieces

Big Hint:

A triangle with $k$ times the base and the same height has $k$ times the area; find each piece as a multiple of $X$

Solution:

Let $X = [\triangle ABC],$ and draw segments $CB',$ $AC',$ and $BA'.$ Triangle $BB'C$ has base $BB' = 3 \cdot AB$ and the same altitude as $\triangle ABC$ from $C$ to line $AB,$ so its area is $3X;$ likewise $\triangle CC'A$ and $\triangle AA'B$ each have area $3X.$ Next, $\triangle AA'C'$ has $3$ times the base and the same height as $\triangle ACC',$ so its area is $9X;$ similarly $\triangle CC'B'$ and $\triangle BB'A'$ each have area $9X.$ Thus $\begin{aligned} &[\triangle A'B'C'] = X \\ &\quad {}+ 3(3X) + 3(9X) \\ &\quad {}= 37X, \end{aligned}$ so the ratio is $37 : 1.$

Thus, the correct answer is E.

16.

The number $21!$ $= 51{,}090{,}942{,}171{,}709{,}440{,}000$ has over $60{,}000$ positive integer divisors. One of them is chosen at random. What is the probability that it is odd?

$\dfrac{1}{21}$

$\dfrac{1}{19}$

$\dfrac{1}{18}$

$\dfrac{1}{2}$

$\dfrac{11}{21}$

Answer: B

Concepts:Legendre’s Formula factor counting prime factorization

Difficulty rating: 1730

Small Hint:

Find the exponent of $2$ in $21!$ using $\lfloor \frac{21}{2} \rfloor + \lfloor \frac{21}{4} \rfloor + \cdots$

Big Hint:

A divisor is odd iff it uses $2^0.$ With exponent $e,$ the odd divisors are a fraction $\dfrac{1}{e+1}$ of all divisors

Solution:

The exponent of $2$ in $21!$ is $\lfloor \frac{21}{2} \rfloor$ $+ \lfloor \frac{21}{4} \rfloor$ $+ \lfloor \frac{21}{8} \rfloor$ $+ \lfloor \frac{21}{16} \rfloor$ $= 10 + 5 + 2 + 1$ $= 18.$ Every divisor has the form $2^i b$ with $0 \le i \le 18$ and $b$ odd; it is odd exactly when $i = 0.$ So the fraction of odd divisors is $\dfrac{1}{18 + 1} = \dfrac{1}{19}.$

Thus, the correct answer is B.

17.

A coin is biased in such a way that on each toss the probability of heads is $\dfrac{2}{3}$ and the probability of tails is $\dfrac{1}{3}.$ The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?

The probability of winning Game A is $\dfrac{4}{81}$ less than the probability of winning Game B.

The probability of winning Game A is $\dfrac{2}{81}$ less than the probability of winning Game B.

The probabilities are the same.

The probability of winning Game A is $\dfrac{2}{81}$ greater than the probability of winning Game B.

The probability of winning Game A is $\dfrac{4}{81}$ greater than the probability of winning Game B.

Answer: D

Concepts:independent events basic probability

Difficulty rating: 1800

Small Hint:

Game A wins with probability $p^3 + (1-p)^3;$ Game B with $\bigl(p^2 + (1-p)^2\bigr)^2$

Big Hint:

Substitute $p = \dfrac{2}{3}$ and subtract the two probabilities

Solution:

Let $p = \dfrac23.$ Game A is won when all three tosses match: $p^3 + (1-p)^3.$ Game B needs the first pair to match and the second pair to match, each with probability $p^2 + (1-p)^2,$ so the win probability is $\bigl(p^2 + (1-p)^2\bigr)^2.$ With $p = \tfrac23,$ Game A gives $\left(\tfrac23\right)^3 + \left(\tfrac13\right)^3 = \tfrac{9}{27} = \tfrac13,$ and Game B gives $\left(\tfrac49 + \tfrac19\right)^2 = \left(\tfrac59\right)^2 = \tfrac{25}{81}.$ The difference is $\tfrac{27}{81} - \tfrac{25}{81} = \tfrac{2}{81},$ so Game A is $\tfrac{2}{81}$ more likely.

Thus, the correct answer is D.

18.

The diameter $\overline{AB}$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD = 3.$ Point $E$ is chosen so that $ED = 5$ and line $ED$ is perpendicular to line $AD.$ Segment $\overline{AE}$ intersects the circle at a point $C$ between $A$ and $E.$ What is the area of $\triangle ABC?$

$\dfrac{120}{37}$

$\dfrac{140}{39}$

$\dfrac{145}{39}$

$\dfrac{140}{37}$

$\dfrac{120}{31}$

Answer: D

Concepts:inscribed angle similarity area ratio

Difficulty rating: 1860

Small Hint:

$\angle ACB$ is inscribed in a semicircle, so it is a right angle; hence $\triangle ABC \sim \triangle AED$

Big Hint:

Their areas are in ratio $AB^2 : AE^2;$ compute $AE^2 = AD^2 + ED^2$

Solution:

Since $\angle ACB$ is inscribed in a semicircle, it is a right angle, so $\triangle ABC \sim \triangle AED$ (both right-angled and sharing angle $A$ ). Their areas are in ratio $AB^2 : AE^2.$ Here $AB = 4,$ so $AB^2 = 16,$ and $AD = AB + BD = 7,$ so $AE^2 = AD^2 + ED^2$ $= 49 + 25$ $= 74.$ The area of $\triangle AED$ is $\tfrac12 \cdot 7 \cdot 5 = \tfrac{35}{2}.$ Thus $[\triangle ABC] = \frac{16}{74} \cdot \frac{35}{2} = \frac{140}{37}.$

Thus, the correct answer is D.

19.

Let $N = 123456789101112\ldots4344$ be the $79$ -digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45?$

$1$

$4$

$9$

$18$

$44$

Answer: C

Concepts:modular arithmetic divisibility digits

Difficulty rating: 1910

Small Hint:

Since $45 = 9 \cdot 5,$ find $N \bmod 5$ (the last digit) and $N \bmod 9$ (the digit sum)

Big Hint:

The digit sum is a multiple of $9,$ so $N \equiv 0 \pmod 9;$ combine with the residue mod $5$

Solution:

The last digit of $N$ is $4,$ so $N \equiv 4 \pmod 5.$ For mod $9,$ sum the digits: the numbers $1$ – $9$ contribute their digits, the tens digits of $10$ – $44$ and the units digits together sum to $270,$ which is a multiple of $9,$ so $N \equiv 0 \pmod 9.$ The number $N - 9$ is then a multiple of $9,$ and its last digit is $5,$ so it is a multiple of $5;$ hence $N - 9$ is a multiple of $45.$ Therefore $N \equiv 9 \pmod{45}.$

Thus, the correct answer is C.

20.

Real numbers $x$ and $y$ are chosen independently and uniformly at random from the interval $(0, 1).$ What is the probability that $\lfloor \log_2 x \rfloor = \lfloor \log_2 y \rfloor,$ where $\lfloor r \rfloor$ denotes the greatest integer less than or equal to the real number $r?$

$\dfrac{1}{8}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

Answer: D

Concepts:geometric probability floor and ceiling functions geometric sequence

Difficulty rating: 1990

Small Hint:

$\lfloor \log_2 x \rfloor = -n$ exactly when $\dfrac{1}{2^n} \le x \lt \dfrac{1}{2^{n-1}}$

Big Hint:

For each $n,$ both $x$ and $y$ must land in that interval of length $\dfrac{1}{2^n};$ sum the resulting areas

Solution:

For each positive integer $n,$ $\lfloor \log_2 x \rfloor = -n$ exactly when $\dfrac{1}{2^n} \le x \lt \dfrac{1}{2^{n-1}},$ an interval of length $\dfrac{1}{2^n}.$ The event that both floors equal $-n$ is a square of area $\dfrac{1}{4^n}.$ Summing over all $n,$ the probability is $\sum_{n=1}^{\infty} \frac{1}{4^n} = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{1}{3}.$

Thus, the correct answer is D.

21.

Last year Isabella took $7$ math tests and received $7$ different scores, each an integer between $91$ and $100,$ inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was $95.$ What was her score on the sixth test?

$92$

$94$

$96$

$98$

$100$

Answer: E

Concepts:divisibility mean bounding to limit cases

Difficulty rating: 2040

Small Hint:

The sum $S$ of all $7$ scores is a multiple of $7$ and lies between $658$ and $679$

Big Hint:

Also $S - 95$ is a multiple of $6;$ then the first $5$ scores sum to a multiple of $5,$ forcing the sixth score’s residue

Solution:

Let $S$ be the sum of all seven scores. Then $S$ is a multiple of $7$ with $658 \le S \le 679,$ so $S \in \{658, 665, 672, 679\}.$ Since the average after six tests is an integer, $S - 95$ is a multiple of $6,$ which forces $S = 665.$ Then the first six scores sum to $570,$ a multiple of $5;$ the average after five tests is an integer, so the first five scores also sum to a multiple of $5,$ making the sixth score a multiple of $5.$ Since all scores differ and the seventh is $95,$ the sixth must be $100.$

Thus, the correct answer is E.

22.

Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn—one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?

$\dfrac{7}{576}$

$\dfrac{5}{192}$

$\dfrac{1}{36}$

$\dfrac{5}{144}$

$\dfrac{7}{48}$

Answer: B

Concepts:basic probability casework

Difficulty rating: 2330

Small Hint:

Each round has $4 \cdot 3 = 12$ equally likely (giver, receiver) outcomes, for $12^4$ total

Big Hint:

Count outcome sequences that return everyone to $4$ coins: each player’s net transfers must cancel; split into cases by the pattern of exchanges

Solution:

Each round has $4 \cdot 3 = 12$ equally likely (giver, receiver) pairs, so there are $12^4$ outcome sequences. Everyone ends with four coins exactly when the four transfers cancel. The favorable patterns are: a $4$ -cycle of gifts ( $24 \cdot 6 = 144$ ways), two disjoint mutual exchanges ( $24 \cdot 3 = 72$ ), one pair exchanging twice ( $6 \cdot 6 = 36$ ), and one player both giving to and receiving from each of two others ( $4 \cdot 3 \cdot 24 = 288$ ). These total $144 + 72 + 36 + 288 = 540.$ The probability is $\dfrac{540}{12^4} = \dfrac{540}{20736} = \dfrac{5}{192}.$

Thus, the correct answer is B.

23.

The graph of $y = f(x),$ where $f(x)$ is a polynomial of degree $3,$ contains points $A(2, 4),$ $B(3, 9),$ and $C(4, 16).$ Lines $AB,$ $AC,$ and $BC$ intersect the graph again at points $D,$ $E,$ and $F,$ respectively, and the sum of the $x$ -coordinates of $D,$ $E,$ and $F$ is $24.$ What is $f(0)?$

$-2$

$0$

$2$

$\dfrac{24}{5}$

$8$

Answer: D

Concepts:Vieta’s Formulas polynomial

Difficulty rating: 2370

Small Hint:

Since $A, B, C$ lie on $y = x^2,$ let $g(x) = f(x) - x^2,$ so $g(x) = a(x-2)(x-3)(x-4)$

Big Hint:

A line meets the cubic where $f(x) - L(x) = 0;$ by Vieta the three $x$ -coordinates of each such triple sum to $9 - \tfrac1a$

Solution:

The points $A, B, C$ lie on $y = x^2,$ so $g(x) = f(x) - x^2$ has roots $2, 3, 4:$ $g(x) = a(x-2)(x-3)(x-4)$ for some $a \ne 0.$ The coefficients of $x^3$ and $x^2$ in $f$ are $a$ and $1 - 9a,$ so by Vieta the three roots of $f(x) - L(x)$ (for any linear $L$ ) sum to $9 - \tfrac1a.$ The lines $AB, AC, BC$ meet the cubic in triples $\{2, 3, x_D\},$ $\{2, 4, x_E\},$ $\{3, 4, x_F\},$ so $\begin{aligned} &x_D + x_E + x_F \\ &\quad {}= 3\left(9 - \tfrac1a\right) \\ &\quad {}- 2(2 + 3 + 4) \\ &\quad {}= 9 - \tfrac3a = 24, \end{aligned}$ giving $a = -\tfrac15.$ Then $f(x) = x^2$ $- \tfrac15(x-2)(x-3)(x-4),$ so $f(0) = 0 - \tfrac15(-2)(-3)(-4)$ $= \tfrac{24}{5}.$

Thus, the correct answer is D.

24.

Quadrilateral $ABCD$ has right angles at $B$ and $C,$ $\triangle ABC \sim \triangle BCD,$ and $AB \gt BC.$ There is a point $E$ in the interior of $ABCD$ such that $\triangle ABC \sim \triangle CEB$ and the area of $\triangle AED$ is $17$ times the area of $\triangle CEB.$ What is $\dfrac{AB}{BC}?$

$1 + \sqrt{2}$

$2 + \sqrt{2}$

$\sqrt{17}$

$2 + \sqrt{5}$

$1 + 2\sqrt{3}$

Answer: D

Difficulty rating: 2550

Small Hint:

Set $BC = 1,$ $AB = r,$ and place $C = (0,0),$ $B = (0,1),$ $A = (r,1),$ $D = (\tfrac1r, 0)$

Big Hint:

Find $E$ from $\triangle ABC \sim \triangle CEB,$ compute both areas, and set $[\triangle AED] = 17[\triangle CEB]$

Solution:

Set $BC = 1$ and $AB = r \gt 1.$ The similarity $\triangle ABC \sim \triangle BCD$ with the right angles places the figure at $C = (0,0),$ $B = (0,1),$ $A = (r,1),$ $D = \bigl(\tfrac1r, 0\bigr).$ Let $E = (x, y)$ with $x, y \gt 0.$ From $\triangle ABC \sim \triangle CEB$ we get $\dfrac{x}{y} = \tan(\angle ECB)$ $= \tan(\angle BAC)$ $= \dfrac1r$ and $x^2 + y^2 = \dfrac{r^2}{1 + r^2},$ so $x = \dfrac{r}{1+r^2},$ $y = \dfrac{r^2}{1+r^2}.$ The two relevant areas are $\begin{aligned} [\triangle CEB]&=\dfrac{r}{2(1+r^2)},\\ [\triangle AED]&=\dfrac{r^4-r^2+1}{2r(1+r^2)}. \end{aligned}$ Setting the second equal to $17$ times the first gives $r^4 - 18r^2 + 1 = 0.$ Then $r^2 = 9 + 4\sqrt5 = (2 + \sqrt5)^2,$ so $r = 2 + \sqrt5.$

Thus, the correct answer is D.

25.

A set of $n$ people participate in an online video basketball tournament. Each person may be a member of any number of $5$ -player teams, but no two teams may have exactly the same $5$ members. The site statistics show a curious fact: The average, over all subsets of size $9$ of the set of $n$ participants, of the number of complete teams whose members are among those $9$ people is equal to the reciprocal of the average, over all subsets of size $8$ of the set of $n$ participants, of the number of complete teams whose members are among those $8$ people. How many values $n,$ $9 \le n \le 2017,$ can be the number of participants?

$477$

$482$

$487$

$557$

$562$

Answer: D

Concepts:double counting divisibility Chinese Remainder Theorem

Difficulty rating: 2650

Small Hint:

Let $T$ be the number of teams. Each team is counted $\binom{n-5}{4}$ times in the size- $9$ sum and $\binom{n-5}{3}$ times in the size- $8$ sum

Big Hint:

The condition becomes $T =$ $\dfrac{n(n-1)(n-2)(n-3)(n-4)}{2^5 \cdot 3^2 \cdot 5 \cdot 7};$ count $n$ making this an integer

Solution:

Let $T$ be the number of teams. Summing over size- $9$ subsets counts each team $\binom{n-5}{4}$ times and over size- $8$ subsets $\binom{n-5}{3}$ times. The averages are $\dfrac{\binom{n-5}{4}T}{\binom n9}$ and $\dfrac{\binom{n-5}{3}T}{\binom n8};$ setting the first equal to the reciprocal of the second and simplifying gives $\begin{aligned} &T \\ &\quad {}= \scriptsize \frac{n(n-1)(n-2)(n-3)(n-4)}{2^5 \cdot 3^2 \cdot 5 \cdot 7}. \end{aligned}$ We need this to be a positive integer with $n \ge 9.$ Let $N =$ $n(n-1)(n-2)(n-3)(n-4);$ as a product of five consecutive integers, $N$ is always divisible by $5.$ The condition that $N$ is divisible by $7$ holds for $5$ residues modulo $7;$ divisibility by $9$ holds for $7$ residues modulo $9;$ and divisibility by $32$ holds for $8$ residues modulo $16.$ The Chinese Remainder Theorem therefore gives $5\cdot7\cdot8=280$ solutions modulo $\operatorname{lcm}(7,9,16)=1008.$ So there are $560$ values in $1 \le n \le 2016;$ removing $n = 1, 2, 3, 4$ (which are below $9$ ) and adding $n = 2017$ (since $2017 \equiv 1 \pmod{1008}$ ) gives $560 - 4 + 1 = 557$ valid values.

Thus, the correct answer is D.