2016 AMC 12A — Problems & Online Timed Exam

Timed

1:15:00

1.

What is the value of $\dfrac{11!-10!}{9!}?$

$99$

$100$

$110$

$121$

$132$

Answer: B

Concepts:factorial factoring

Difficulty rating: 920

Small Hint:

Factor $10!$ out of the numerator

Big Hint:

$11!-10!=10!(11-1)$ and $10!=10\cdot 9!$

Solution:

Factoring the numerator gives $\begin{gathered} \dfrac{11!-10!}{9!}\\ =\dfrac{10!(11-1)}{9!}\\ =\dfrac{10\cdot 9!\cdot 10}{9!}\\ =100. \end{gathered}$

Thus, the correct answer is B.

2.

For what value of $x$ does $10^x\cdot 100^{2x}=1000^5?$

$1$

$2$

$3$

$4$

$5$

Answer: C

Concepts:exponent

Difficulty rating: 1020

Small Hint:

Write every base as a power of $10$

Big Hint:

The left side becomes $10^{5x}$ and the right side $10^{15}$

Solution:

Since $100=10^2$ and $1000=10^3,$ the equation becomes $10^x\cdot 10^{4x}=10^{15},$ so $10^{5x}=10^{15}.$ Then $5x=15,$ giving $x=3.$

Thus, the correct answer is C.

3.

The remainder function can be defined for all real numbers $x$ and $y$ with $y\neq 0$ by $\text{rem}(x,y)=x-y\left\lfloor \dfrac{x}{y}\right\rfloor,$ where $\left\lfloor \dfrac{x}{y}\right\rfloor$ denotes the greatest integer less than or equal to $\dfrac{x}{y}.$ What is the value of $\text{rem}\left(\dfrac{3}{8},-\dfrac{2}{5}\right)?$

$-\dfrac{3}{8}$

$-\dfrac{1}{40}$

$0$

$\dfrac{3}{8}$

$\dfrac{31}{40}$

Answer: B

Concepts:custom operation floor and ceiling functions

Difficulty rating: 1200

Small Hint:

Compute $\dfrac{x}{y}=\dfrac{\frac{3}{8}}{-\frac{2}{5}}$ first

Big Hint:

$\left\lfloor -\dfrac{15}{16}\right\rfloor=-1$

Solution:

First, $\begin{gathered} \dfrac{x}{y}=\dfrac{\frac{3}{8}}{-\frac{2}{5}}\\ =\dfrac{3}{8}\cdot\left(-\dfrac{5}{2}\right)\\ =-\dfrac{15}{16}, \end{gathered}$ and $\left\lfloor -\dfrac{15}{16}\right\rfloor=-1.$

Therefore $\begin{gathered} \text{rem}\left(\dfrac{3}{8},-\dfrac{2}{5}\right)\\ =\dfrac{3}{8}-\left(-\dfrac{2}{5}\right)(-1)\\ =\dfrac{3}{8}-\dfrac{2}{5}\\ =\dfrac{15-16}{40}\\ =-\dfrac{1}{40}. \end{gathered}$

Thus, the correct answer is B.

4.

The mean, median, and mode of the $7$ data values $60,$ $100,$ $x,$ $40,$ $50,$ $200,$ $90$ are all equal to $x.$ What is the value of $x?$

$50$

$60$

$75$

$90$

$100$

Answer: D

Concepts:mean median (data)mode

Difficulty rating: 1100

Small Hint:

Set the mean equal to $x$

Big Hint:

$\dfrac{540+x}{7}=x,$ then confirm the median and mode

Solution:

The mean condition gives $\begin{gathered} \small \dfrac{60+100+x+40+50+200+90}{7}\\ =\dfrac{540+x}{7}\\ =x, \end{gathered}$ so $540+x=7x$ and $x=90.$

In nondecreasing order the data are $40, 50, 60, 90, 90, 100, 200,$ so the median is $90$ and the mode is $90,$ as required.

Thus, the correct answer is D.

5.

Goldbach’s conjecture states that every even integer greater than $2$ can be written as the sum of two prime numbers (for example, $2016=13+2003$ ). So far, no one has been able to prove that the conjecture is true, and no one has found a counterexample to show that the conjecture is false. What would a counterexample consist of?

an odd integer greater than $2$ that can be written as the sum of two prime numbers

an odd integer greater than $2$ that cannot be written as the sum of two prime numbers

an even integer greater than $2$ that can be written as the sum of two numbers that are not prime

an even integer greater than $2$ that can be written as the sum of two prime numbers

an even integer greater than $2$ that cannot be written as the sum of two prime numbers

Answer: E

Concepts:counterexample logical deduction

Difficulty rating: 1100

Small Hint:

A counterexample keeps the hypothesis but breaks the conclusion

Big Hint:

The hypothesis is “even integer greater than $2$ ”; the conclusion is “a sum of two primes”

Solution:

A counterexample must satisfy the hypothesis of being an even integer greater than $2$ while failing the conclusion that it can be written as the sum of two prime numbers.

Thus, the correct answer is E.

6.

A triangular array of $2016$ coins has $1$ coin in the first row, $2$ coins in the second row, $3$ coins in the third row, and so on up to $N$ coins in the $N$ th row. What is the sum of the digits of $N?$

$6$

$7$

$8$

$9$

$10$

Answer: D

Concepts:triangular number estimation

Difficulty rating: 1270

Small Hint:

$1+2+\cdots+N=\dfrac{N(N+1)}{2}$

Big Hint:

Solve $N(N+1)=4032$ by noting $N\approx\sqrt{4032}$

Solution:

The total number of coins is $\begin{gathered} 1+2+\cdots+N\\ =\dfrac{N(N+1)}{2}\\ =2016, \end{gathered}$ so $N(N+1)=4032.$ Since $63\cdot 64=4032,$ we have $N=63,$ and the sum of its digits is $6+3=9.$

Thus, the correct answer is D.

7.

Which of these describes the graph of $x^2(x+y+1)=y^2(x+y+1)?$

two parallel lines

two intersecting lines

three lines that all pass through a common point

three lines that do not all pass through a common point

a line and a parabola

Answer: D

Concepts:factoring difference of squares

Difficulty rating: 1410

Small Hint:

Move everything to one side and factor out $x+y+1$

Big Hint:

$(x-y)(x+y)(x+y+1)=0$ is the union of three lines

Solution:

Moving all terms to one side gives $(x^2-y^2)(x+y+1)=0,$ which factors as $(x-y)(x+y)(x+y+1)=0.$ The graph is therefore the union of the lines $x=y,$ $x=-y,$ and $x+y+1=0.$

The first two lines intersect at the origin, but the third line $x+y=-1$ is parallel to $x+y=0$ and does not pass through the origin. So the graph consists of three lines that do not all pass through a common point.

Thus, the correct answer is D.

8.

What is the area of the shaded region of the given $8\times 5$ rectangle?

$4\tfrac{3}{4}$

$5$

$5\tfrac{1}{4}$

$6\tfrac{1}{2}$

$8$

Answer: D

Concepts:triangle area symmetry

Difficulty rating: 1350

Small Hint:

Every drawn segment passes through the center of the rectangle

Big Hint:

The main diagonal splits the shaded region into four triangles, each with base $1$

Solution:

The diagonal of the rectangle from the upper-left corner to the lower-right corner divides the shaded region into four triangles, all meeting at the center of the rectangle.

Two of these triangles have a horizontal base of length $1$ and altitude $\frac12\cdot 5=\frac52,$ and the other two have a vertical base of length $1$ and altitude $\frac12\cdot 8=4.$ The total area is $\begin{gathered} 2\cdot\dfrac12\cdot 1\cdot\dfrac52\\ {}+2\cdot\dfrac12\cdot 1\cdot 4\\ =\dfrac52+4\\ =\dfrac{13}{2}. \end{gathered}$

Thus, the correct answer is D.

9.

The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is $\dfrac{a-\sqrt{2}}{b},$ where $a$ and $b$ are positive integers. What is $a+b?$

$7$

$8$

$9$

$10$

$11$

Answer: E

Concepts:diagonal rationalizing denominator

Difficulty rating: 1510

Small Hint:

Draw the diagonal of the unit square, which has length $\sqrt{2}$

Big Hint:

That diagonal equals two small-square diagonals plus one small-square side: $2x\sqrt2+x=\sqrt2$

Solution:

Let $x$ be the common side length. The diagonal of the unit square has length $\sqrt{2}$ and consists of two small-square diagonals (each $x\sqrt2$ ) plus one small-square side length $x,$ so $2x\sqrt2+x=\sqrt2.$

Solving, $\begin{gathered} x=\dfrac{\sqrt2}{2\sqrt2+1}\\ =\dfrac{\sqrt2\,(2\sqrt2-1)}{(2\sqrt2+1)(2\sqrt2-1)}\\ =\dfrac{4-\sqrt2}{7}. \end{gathered}$ Thus $a=4,$ $b=7,$ and $a+b=11.$

Thus, the correct answer is E.

10.

Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions “left” and “right” are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?

$1$

$2$

$3$

$4$

$5$

Answer: B

Concepts:invariant logical deduction

Difficulty rating: 1410

Small Hint:

The total rightward movement must equal the total leftward movement

Big Hint:

Bea moves $+2,$ Ceci moves $-1,$ and Dee and Edie cancel; find Ada’s shift

Solution:

The net displacement of all five friends is zero. Dee and Edie swapped seats, so their movements cancel. Bea moved $+2$ and Ceci moved $-1,$ a net of $+1,$ so Ada must move $-1$ to balance.

Ada returns to an end seat; since she moved one seat to the left, that seat must be seat $1,$ so she had been sitting in seat $2.$

Thus, the correct answer is B.

11.

Each of the $100$ students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are $42$ students who cannot sing, $65$ students who cannot dance, and $29$ students who cannot act. How many students have two of these talents?

$16$

$25$

$36$

$49$

$64$

Answer: E

Concepts:complementary counting inclusion-exclusion double counting

Difficulty rating: 1470

Small Hint:

Count how many students can do each activity

Big Hint:

Adding the three counts tallies two-talent students twice; subtract $100$

Solution:

The numbers who can sing, dance, and act are $100-42=58,$ $100-65=35,$ and $100-29=71,$ respectively, for a total of $58+35+71=164.$

Since no student has all three talents, each student has one or two talents, so single-talent students are counted once and two-talent students are counted twice. The number counted twice is $164-100=64.$

Thus, the correct answer is E.

12.

In $\triangle ABC,$ $AB=6,$ $BC=7,$ and $CA=8.$ Point $D$ lies on $\overline{BC},$ and $\overline{AD}$ bisects $\angle BAC.$ Point $E$ lies on $\overline{AC},$ and $\overline{BE}$ bisects $\angle ABC.$ The bisectors intersect at $F.$ What is the ratio $AF:FD?$

$3:2$

$5:3$

$2:1$

$7:3$

$5:2$

Answer: C

Concepts:angle bisector theorem ratio and proportion

Difficulty rating: 1500

Small Hint:

Apply the Angle Bisector Theorem in $\triangle ABC$ to find $BD$

Big Hint:

In $\triangle ABD,$ segment $BF$ bisects $\angle ABD,$ so $AF:FD=AB:BD$

Solution:

Applying the Angle Bisector Theorem to $\triangle ABC$ gives $BD:DC=AB:AC=6:8,$ so $BD=\dfrac{6}{6+8}\cdot 7=3.$

Now $BF$ lies along the bisector of $\angle ABD$ in $\triangle ABD,$ so by the Angle Bisector Theorem again, $\begin{gathered} AF:FD=AB:BD\\ =6:3=2:1. \end{gathered}$

Thus, the correct answer is C.

13.

Let $N$ be a positive multiple of $5.$ One red ball and $N$ green balls are arranged in a line in random order. Let $P(N)$ be the probability that at least $\dfrac{3}{5}$ of the green balls are on the same side of the red ball. Observe that $P(5)=1$ and that $P(N)$ approaches $\dfrac{4}{5}$ as $N$ grows large. What is the sum of the digits of the least value of $N$ such that $P(N)\lt\dfrac{321}{400}?$

$12$

$14$

$16$

$18$

$20$

Answer: A

Concepts:basic probability inequality

Difficulty rating: 1690

Small Hint:

Write $N=5k;$ the red ball has $5k+1$ equally likely positions

Big Hint:

$P(N)=\dfrac{4k+2}{5k+1};$ solve $\dfrac{4k+2}{5k+1}\lt\dfrac{321}{400}$

Solution:

Write $N=5k.$ Number the positions of the red ball $0,1,\ldots,5k$ from one end; there are $5k+1$ equally likely positions.

Fewer than $\dfrac{3}{5}$ of the green balls lie on each side exactly when the red ball is in one of the positions $2k+1,2k+2,\ldots,3k-1,$ which is $k-1$ positions. Hence $P(N)=1-\dfrac{k-1}{5k+1}=\dfrac{4k+2}{5k+1}.$

Solving $\dfrac{4k+2}{5k+1}\lt\dfrac{321}{400}$ gives $400(4k+2)\lt 321(5k+1),$ so $1600k+800\lt 1605k+321$ and $5k\gt 479,$ meaning $k\gt 95.8.$ Thus $k=96$ and $N=480,$ whose digit sum is $4+8+0=12.$

Thus, the correct answer is A.

14.

Each vertex of a cube is to be labeled with an integer from $1$ through $8,$ with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?

$1$

$3$

$6$

$12$

$24$

Answer: C

Concepts:cube geometry casework symmetry

Difficulty rating: 1730

Small Hint:

Each vertex lies on $3$ faces, so the six face-sums total $3(1+2+\cdots+8)$

Big Hint:

The face-sum is $18;$ show $1$ and $8$ must be on adjacent vertices

Solution:

Each vertex belongs to $3$ faces, so $6S=3(1+2+\cdots+8)=108,$ giving each face-sum $S=18.$

The four-element subsets containing $1$ with sum $18$ are $\{1,2,7,8\},$ $\{1,3,6,8\},$ $\{1,4,5,8\},$ and $\{1,4,6,7\}.$ Only one omits $8,$ so at least two of the three distinct faces through $1$ contain $8.$ Two vertices of a cube lie on two common faces exactly when they are adjacent; hence $1$ and $8$ are adjacent.

Rotate the cube so that $1$ is at the lower-left-front vertex and $8$ at the lower-right-front vertex. The unique face through $1$ that does not contain $8$ must use $4,6,7,$ and these can be placed in $3!=6$ orders. Each order forces $5,3,2$ at the three opposite vertices, and the remaining face sums are then $18.$ Hence there are $6$ arrangements.

Thus, the correct answer is C.

15.

Circles with centers $P,$ $Q,$ and $R,$ having radii $1,$ $2,$ and $3,$ respectively, lie on the same side of line $l$ and are tangent to $l$ at $P',$ $Q',$ and $R',$ respectively, with $Q'$ between $P'$ and $R'.$ The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of $\triangle PQR?$

$0$

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

$1$

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

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

Answer: D

Concepts:tangent circles Pythagorean Theorem shoelace formula

Difficulty rating: 1800

Small Hint:

The horizontal distance between two tangent points is $\sqrt{(r_1+r_2)^2-(r_2-r_1)^2}$

Big Hint:

Place $P,Q,R$ at heights $1,2,3$ and use the shoelace formula

Solution:

The centers lie at heights $1,$ $2,$ and $3$ above line $l.$ Since circle $Q$ is externally tangent to circle $P,$ we have $PQ=3,$ so the horizontal distance is $P'Q'=\sqrt{3^2-1^2}=\sqrt{8}.$ Since circle $Q$ is tangent to circle $R,$ we have $QR=5,$ so $Q'R'=\sqrt{5^2-1^2}=\sqrt{24}.$

Place $P=(0,1),$ $Q=(\sqrt8,2),$ and $R=(\sqrt8+\sqrt{24},3).$ By the shoelace formula, the area is $\begin{gathered} \small \dfrac12\left|\sqrt8(3-1)+(\sqrt8+\sqrt{24})(1-2)\right|\\ =\dfrac12\left(\sqrt{24}-\sqrt8\right)\\ =\sqrt6-\sqrt2. \end{gathered}$

Thus, the correct answer is D.

16.

The graphs of $y=\log_3 x,$ $y=\log_x 3,$ $y=\log_{\frac{1}{3}} x,$ and $y=\log_x\dfrac{1}{3}$ are plotted on the same set of axes. How many points in the plane with positive $x$ -coordinates lie on two or more of the graphs?

$2$

$3$

$4$

$5$

$6$

Answer: D

Concepts:logarithm substitution

Difficulty rating: 1860

Small Hint:

Let $u=\log_3 x$ and rewrite all four as $u,\ \dfrac1u,\ -u,\ -\dfrac1u$

Big Hint:

Find every $x$ where two of these four expressions are equal

Solution:

Let $u=\log_3 x.$ Then $\log_x 3=\dfrac1u,$ $\log_{\frac{1}{3}}x=-u,$ and $\log_x\dfrac13=-\dfrac1u.$ Two graphs meet where two of $u,\dfrac1u,-u,-\dfrac1u$ are equal for some valid $x\gt 0.$

Setting $u=\dfrac1u$ gives $u=\pm1,$ so $x=3$ or $x=\dfrac13;$ setting $-u=-\dfrac1u$ gives the same values. Setting $u=-u$ gives $u=0,$ i.e. $x=1,$ where $\log_3 x$ and $\log_{\frac{1}{3}}x$ are both $0.$ The remaining pairings have no real solution.

The distinct intersection points are $(1,0),$ $(3,1),$ $\left(\dfrac13,-1\right),$ $(3,-1),$ and $\left(\dfrac13,1\right),$ so there are $5.$

Thus, the correct answer is D.

17.

Let $ABCD$ be a square. Let $E,$ $F,$ $G,$ and $H$ be the centers, respectively, of equilateral triangles with bases $\overline{AB},$ $\overline{BC},$ $\overline{CD},$ and $\overline{DA},$ each exterior to the square. What is the ratio of the area of square $EFGH$ to the area of square $ABCD?$

$1$

$\dfrac{2+\sqrt{3}}{3}$

$\sqrt{2}$

$\dfrac{\sqrt{2}+\sqrt{3}}{2}$

$\sqrt{3}$

Answer: B

Concepts:equilateral triangle centroid area ratio

Difficulty rating: 1800

Small Hint:

The center of an equilateral triangle lies $\frac13$ of its height from the base

Big Hint:

Compare diagonals of the two squares; the area ratio is the square of the length ratio

Solution:

Let square $ABCD$ have side length $6.$ Each equilateral triangle has height $3\sqrt3,$ and its center lies $\frac13$ of that height, namely $\sqrt3,$ from the square’s side.

Square $ABCD$ has diagonal $6\sqrt2.$ Square $EFGH$ has diagonal equal to the side of $ABCD$ plus twice $\sqrt3,$ namely $6+2\sqrt3.$ The area ratio is the square of the ratio of diagonals: $\begin{gathered} \left(\dfrac{6+2\sqrt3}{6\sqrt2}\right)^2\\ =\left(\dfrac{3+\sqrt3}{3\sqrt2}\right)^2\\ =\dfrac{12+6\sqrt3}{18}\\ =\dfrac{2+\sqrt3}{3}. \end{gathered}$

Thus, the correct answer is B.

18.

For some positive integer $n,$ the number $110n^3$ has $110$ positive integer divisors, including $1$ and the number $110n^3.$ How many positive integer divisors does the number $81n^4$ have?

$110$

$191$

$261$

$325$

$425$

Answer: D

Concepts:factor counting prime factorization

Difficulty rating: 1910

Small Hint:

$110=2\cdot5\cdot11,$ so $110n^3$ has exactly three prime factors

Big Hint:

The exponents plus one multiply to $110=2\cdot5\cdot11;$ deduce $n=p\cdot q^3$

Solution:

The number $110n^3$ is divisible by the three distinct primes $2,5,11.$ If its prime exponents are $r_1,r_2,\ldots,$ then $(r_1+1)(r_2+1)\cdots=110,$ where $110=2\cdot5\cdot11.$ Because there are already at least three factors greater than $1,$ there are exactly three, so no other prime divides $n.$

Each exponent in $110n^3$ is $1\pmod3,$ so each divisor-count factor $r_i+1$ is $2\pmod3.$ The factors $2,5,11$ all have that form, so the exponents are $1,4,10$ in some order. After subtracting the exponent $1$ from $110$ and dividing by $3,$ the exponents in $n$ are $0,1,3$ in some order. Thus $n=pq^3$ for two distinct primes $p,q$ chosen from $2,5,11.$

Consequently $n^4$ has exponents $4$ and $12,$ while $81=3^4$ introduces a third prime because $n$ is not divisible by $3.$ Hence the number of divisors of $81n^4$ is $\begin{gathered} (4+1)(4+1)(12+1)\\ =5\cdot5\cdot13\\ =325. \end{gathered}$

Thus, the correct answer is D.

19.

Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process is $\dfrac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$ (For example, he succeeds if his sequence of tosses is HTHHHHHH.)

$69$

$151$

$257$

$293$

$313$

Answer: B

Concepts:random walk casework

Difficulty rating: 1990

Small Hint:

Count the $2^8$ toss sequences in which the running total reaches $4$

Big Hint:

With at most $2$ tails he always reaches $4;$ with $3$ or $4$ tails only some orders work

Solution:

Count the sequences of $8$ tosses whose running total reaches $4.$ With at most $2$ tails he certainly reaches $4,$ contributing $\begin{gathered} \binom80+\binom81+\binom82\\ =1+8+28\\ =37 \end{gathered}$ sequences.

With exactly $3$ tails, he can first reach $4$ on toss $4$ or toss $6.$ Reaching it on toss $4$ requires four initial heads, after which the remaining head has $4$ possible positions. Otherwise, tosses $5$ and $6$ are heads, one of the first four tosses is a tail, and the last two are tails, again giving $4$ possibilities. Thus this case contributes $8.$ With exactly $4$ tails, only HHHHTTTT works, giving $1.$ He cannot reach $4$ with fewer than $4$ heads.

So there are $37+8+1=46$ favorable sequences out of $2^8=256,$ a probability of $\dfrac{46}{256}=\dfrac{23}{128}.$ Then $a+b=23+128=151.$

Thus, the correct answer is B.

20.

A binary operation $\diamond$ has the properties that $a\diamond(b\diamond c)=(a\diamond b)\cdot c$ and that $a\diamond a=1$ for all nonzero real numbers $a,$ $b,$ and $c.$ (Here the dot $\cdot$ represents the usual multiplication operation.) The solution to the equation $2016\diamond(6\diamond x)=100$ can be written as $\dfrac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. What is $p+q?$

$109$

$201$

$301$

$3049$

$33{,}601$

Answer: A

Concepts:custom operation functional equation

Difficulty rating: 1910

Small Hint:

Use $a\diamond a=1$ to simplify $a\diamond(a\diamond a)$

Big Hint:

Show $a\diamond b=\dfrac{a}{b},$ so $2016\diamond(6\diamond x)=\dfrac{2016}{\frac{6}{x}}$

Solution:

Setting $b=c=a$ gives $a\diamond 1=a\diamond(a\diamond a)$ $=(a\diamond a)\cdot a=a.$ Then setting $c=b$ gives $a=a\diamond 1$ $=a\diamond(b\diamond b)=(a\diamond b)\cdot b,$ so $a\diamond b=\dfrac{a}{b}.$

Therefore $\begin{gathered} 2016\diamond(6\diamond x)\\ =2016\diamond\dfrac{6}{x}\\ =\dfrac{2016}{\frac{6}{x}}\\ =336x=100, \end{gathered}$ so $x=\dfrac{100}{336}=\dfrac{25}{84}$ and $p+q=25+84=109.$

Thus, the correct answer is A.

21.

A quadrilateral is inscribed in a circle of radius $200\sqrt{2}.$ Three of the sides of this quadrilateral have length $200.$ What is the length of its fourth side?

$200$

$200\sqrt{2}$

$200\sqrt{3}$

$300\sqrt{2}$

$500$

Answer: E

Concepts:cyclic quadrilateral law of cosines trigonometric identity

Difficulty rating: 2040

Small Hint:

Each side of length $200$ subtends a central angle $\theta$ with $\cos\theta=\frac34$

Big Hint:

The fourth chord has the same length as one subtending $3\theta;$ use $\cos 3\theta=4\cos^3\theta-3\cos\theta$

Solution:

Let $\theta$ be the central angle subtending a side of length $200,$ with radius $R=200\sqrt2.$ By the law of cosines on the isosceles triangle from the center, $\begin{gathered} 200^2=2R^2(1-\cos\theta)\\ =160000(1-\cos\theta), \end{gathered}$ so $\cos\theta=\dfrac34.$

The three equal sides use three consecutive arcs of angle $\theta,$ so the fourth arc has angle $2\pi-3\theta.$ Its chord has the same length as a chord with central angle $3\theta,$ and $\begin{gathered} \cos 3\theta=4\cos^3\theta-3\cos\theta\\ =4\cdot\dfrac{27}{64}-\dfrac94\\ =-\dfrac{9}{16}. \end{gathered}$ Its length squared is $\begin{gathered} 2R^2(1-\cos 3\theta)\\ =160000\left(1+\dfrac{9}{16}\right)\\ =160000\cdot\dfrac{25}{16}\\ =250000, \end{gathered}$ so the fourth side is $500.$

Thus, the correct answer is E.

22.

How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y)=72,$ $\text{lcm}(x,z)=600,$ and $\text{lcm}(y,z)=900?$

$15$

$16$

$24$

$27$

$64$

Answer: A

Concepts:least common multiple prime factorization casework

Difficulty rating: 2160

Small Hint:

Factor: $72=2^3\cdot3^2,$ $600=2^3\cdot3\cdot5^2,$ $900=2^2\cdot3^2\cdot5^2$

Big Hint:

Handle each prime $2,3,5$ separately, using $\max$ of the exponents

Solution:

Because $\text{lcm}(x,y)=2^3\cdot3^2$ and $\text{lcm}(x,z)=2^3\cdot3\cdot5^2,$ the factor $5^2$ divides $z$ while neither $x$ nor $y$ is divisible by $5.$ Also $3^2$ divides $y,$ while neither $x$ nor $z$ is divisible by $3^2,$ and $x$ must have the factor $2^3.$

Writing $x=2^3\cdot3^{\,j},$ $y=2^{\,k}\cdot3^2,$ and $z=2^{\,m}\cdot3^{\,n}\cdot5^2,$ the lcm conditions require $\max(j,n)=1$ and $\max(k,m)=2.$ The first pair can be $(1,0),(0,1),(1,1),$ and the second can be $(2,0),(2,1),(2,2),(0,2),(1,2).$ Thus there are $3\cdot5=15$ ordered triples.

Thus, the correct answer is A.

23.

Three numbers in the interval $[0,1]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?

$\dfrac{1}{6}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{5}{6}$

Answer: C

Concepts:geometric probability triangle inequality volume

Difficulty rating: 2160

Small Hint:

The triples $(x,y,z)$ fill a unit cube; failure means one number $\ge$ the sum of the other two

Big Hint:

Each region such as $z\ge x+y$ is a tetrahedron of volume $\frac16$

Solution:

The ordered triples $(x,y,z)$ fill the unit cube of volume $1.$ They fail to form a triangle exactly when one value is at least the sum of the other two.

The region $z\ge x+y$ is a tetrahedron with vertices $(0,0,0),(0,0,1),(0,1,1),(1,0,1)$ of volume $\frac16.$ The analogous regions $x\ge y+z$ and $y\ge x+z$ also have volume $\frac16$ and have disjoint interiors. So the failure probability is $3\cdot\frac16=\frac12,$ and the triangle probability is $1-\frac12=\frac12.$

Thus, the correct answer is C.

24.

There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b?$

$8$

$9$

$10$

$11$

$12$

Answer: B

Concepts:Vieta’s Formulas AM-GM Inequality

Difficulty rating: 2380

Small Hint:

If the roots are $r,s,t,$ then $r+s+t=a$ and $rst=a$

Big Hint:

Apply AM-GM to $r,s,t;$ the minimum $a$ forces $r=s=t$

Solution:

By Vieta’s formulas, the real roots $r,s,t$ satisfy $r+s+t=a,$ $rs+st+tr=b,$ and $rst=a,$ so $r+s+t=rst.$

The positive product means either all three roots are positive or two are negative. In the latter case write the roots as $-u,-v,w$ with $u,v,w\gt0.$ The positive sum forces $w\gt u+v,$ and then $b=uv-w(u+v)\lt0,$ a contradiction. Hence all three roots are positive.

By the AM-GM inequality, $27rst\le(r+s+t)^3=(rst)^3,$ so $a=rst\ge 3\sqrt3,$ with equality if and only if $r=s=t=\sqrt3.$ At this smallest $a,$ $\begin{gathered} b=rs+st+tr\\ =3r^2=3\cdot 3=9. \end{gathered}$

Thus, the correct answer is B.

25.

Let $k$ be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with $k+1$ digits. Every time Bernardo writes a number, Silvia erases the last $k$ digits of it. Bernardo then writes the next perfect square, Silvia erases the last $k$ digits of it, and this process continues until the last two numbers that remain on the board differ by at least $2.$ Let $f(k)$ be the smallest positive integer not written on the board. For example, if $k=1,$ then the numbers that Bernardo writes are $16,$ $25,$ $36,$ $49,$ and $64,$ and the numbers showing on the board after Silvia erases are $1,$ $2,$ $3,$ $4,$ and $6,$ and thus $f(1)=5.$ What is the sum of the digits of $f(2)+f(4)$ $+f(6)+\cdots+f(2016)?$

$7986$

$8002$

$8030$

$8048$

$8064$

Answer: E

Concepts:perfect square floor and ceiling functions digits

Difficulty rating: 2720

Small Hint:

For even $k=2j,$ the numbers shown are $\left\lfloor \frac{n^2}{10^{k}}\right\rfloor$ for $n\ge 10^{j}$

Big Hint:

Show $f(2j)=\dfrac{10^{2j}}{4}+10^{j},$ then add over $j$

Solution:

Take $k=2j.$ The smallest perfect square with $k+1$ digits is $10^{k}=(10^{j})^2,$ and after Silvia erases, the numbers shown are $\left\lfloor \frac{n^2}{10^{k}}\right\rfloor$ for $n=10^{j}, 10^{j}+1,\ldots$

Put $M=10^k.$ A jump of at least $2$ from $n$ to $n+1$ requires $(n+1)^2-n^2=2n+1\gt M,$ so write $n=\frac{M}{2}+m$ with $m\ge0.$ The case $m=0$ gives a jump of only $1,$ so the first larger jump has $m\ge1.$ Let $A=\left\lfloor \frac{n^2}{M}\right\rfloor$ and $B=\left\lfloor\frac{(n+1)^2}{M}\right\rfloor.$ Because $M$ is divisible by $4,$ $\begin{aligned} A&=\dfrac M4+m+\left\lfloor\dfrac{m^2}{M}\right\rfloor,\\ B&=\dfrac M4+m+1\\ &\quad{}+\left\lfloor\dfrac{(m+1)^2}{M}\right\rfloor. \end{aligned}$

Therefore the first jump of at least $2$ occurs at the first $m$ for which $m^2\lt M\le(m+1)^2.$ Since $\sqrt M=10^j,$ this is $m=10^j-1.$ The last displayed value before the gap is $\frac{M}{4}+10^j-1,$ so the smallest missing integer is $f(2j)=\dfrac{10^{2j}}4+10^j.$

Summing over $j=1,\ldots,1008,$ $\begin{gathered} \sum_{j=1}^{1008}f(2j)\\ =25\sum_{j=0}^{1007}10^{2j}\\ {}+10\sum_{j=0}^{1007}10^{j}\\ =\underbrace{2525\cdots25}_{2016\text{ digits}}\\ {}+\underbrace{111\cdots10}_{1009\text{ digits}}. \end{gathered}$ There are no carries, so the digit sum is $1008\cdot(2+5)$ $+1008\cdot 1=1008\cdot 8=8064.$

Thus, the correct answer is E.