2012 AMC 12B — Problems & Online Timed Exam

Timed

1:15:00

1.

Each third-grade classroom at Pearl Creek Elementary has $18$ students and $2$ pet rabbits. How many more students than rabbits are there in all $4$ of the third-grade classrooms?

$48$

$56$

$64$

$72$

$80$

Answer: C

Concepts:whole number operations

Difficulty rating: 560

Small Hint:

Each classroom has $18-2$ more students than rabbits

Big Hint:

Multiply the per-classroom difference by $4$

Solution:

Each classroom has $18-2=16$ more students than rabbits.

Across all $4$ classrooms there are $4\cdot16=64$ more students than rabbits.

Thus, the correct answer is C.

2.

A circle of radius $5$ is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is $2:1.$ What is the area of the rectangle?

$50$

$100$

$125$

$150$

$200$

Answer: E

Concepts:circle rectangle

Difficulty rating: 730

Small Hint:

The width of the rectangle equals the diameter of the circle

Big Hint:

The length is twice the width

Solution:

The circle is inscribed, so the width of the rectangle equals the diameter, $2\cdot5=10.$

The length is then $2\cdot10=20,$ so the area is $10\cdot20=200.$

Thus, the correct answer is E.

3.

For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid $3$ acorns in each of the holes it dug. The squirrel hid $4$ acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed $4$ fewer holes. How many acorns did the chipmunk hide?

$30$

$36$

$42$

$48$

$54$

Answer: D

Concepts:linear equation

Difficulty rating: 860

Small Hint:

Let $h$ be the number of holes the chipmunk dug, so the squirrel dug $h-4$

Big Hint:

The equal totals give $3h=4(h-4)$

Solution:

Let $h$ be the number of holes the chipmunk dug. The chipmunk hid $3h$ acorns and the squirrel hid $4(h-4)$ acorns.

Since they hid the same number, $3h=4(h-4),$ which gives $h=16.$

The chipmunk hid $3\cdot16=48$ acorns.

Thus, the correct answer is D.

4.

Suppose that the euro is worth $1.30$ dollars. If Diana has $500$ dollars and Étienne has $400$ euros, by what percent is the value of Étienne’s money greater than the value of Diana’s money?

$2$

$4$

$6.5$

$8$

$13$

Answer: B

Concepts:percentage unit conversion

Difficulty rating: 1050

Small Hint:

Convert Étienne’s euros to dollars first

Big Hint:

The percent increase is $\dfrac{\text{Étienne} - \text{Diana}}{\text{Diana}}\cdot100\%$

Solution:

Étienne’s money is worth $400\cdot1.30=520$ dollars, while Diana has $500$ dollars.

The percent by which Étienne’s value exceeds Diana’s is $\frac{520-500}{500}\cdot100\%=4\%.$

Thus, the correct answer is B.

5.

Two integers have a sum of $26.$ When two more integers are added to the first two integers the sum is $41.$ Finally when two more integers are added to the sum of the previous four integers the sum is $57.$ What is the minimum number of even integers among the $6$ integers?

$1$

$2$

$3$

$4$

$5$

Answer: A

Concepts:parity

Difficulty rating: 1200

Small Hint:

Look at the sum of each new pair: $26,$ $41-26,$ and $57-41$

Big Hint:

A pair has an odd sum only when exactly one of its two integers is even

Solution:

The three successive pairs have sums $26,$ $41-26=15,$ and $57-41=16.$

A pair sums to an even number when its two integers share parity, and to an odd number when exactly one is even. Only the middle pair sums to an odd number, so it must contain at least one even integer.

The other two pairs can be all odd, so as few as $1$ even integer is possible, for example $1,25,1,14,1,15.$

Thus, the correct answer is A.

6.

In order to estimate the value of $x-y$ where $x$ and $y$ are real numbers with $x \gt y \gt 0,$ Xiaoli rounded $x$ up by a small amount, rounded $y$ down by the same amount, and then subtracted her rounded values. Which of the following statements is necessarily correct?

Her estimate is larger than $x-y.$

Her estimate is smaller than $x-y.$

Her estimate equals $x-y.$

Her estimate equals $y-x.$

Her estimate is $0.$

Answer: A

Concepts:estimation inequality

Difficulty rating: 1130

Small Hint:

Write the rounded values as $x+d$ and $y-d$ for a small $d \gt 0$

Big Hint:

Subtract to compare $(x+d)-(y-d)$ with $x-y$

Solution:

Let $d \gt 0$ be the small amount. Xiaoli computes $(x+d)-(y-d)$ $=(x-y)+2d.$

Since $2d \gt 0,$ her estimate exceeds the true value $x-y.$

Thus, the correct answer is A.

7.

Small lights are hung on a string $6$ inches apart in the order red, red, green, green, green, red, red, green, green, green, and so on continuing this pattern of $2$ red lights followed by $3$ green lights. How many feet separate the $3$ rd red light and the $21$ st red light?

Note: $1$ foot is equal to $12$ inches.

$18$

$18.5$

$20$

$20.5$

$22.5$

Answer: E

Concepts:pattern recognition unit conversion

Difficulty rating: 1270

Small Hint:

Group the lights into repeating blocks of $5$ (two red, three green)

Big Hint:

The $3$ rd red light starts block $2$ and the $21$ st red light starts block $11$

Solution:

The lights repeat in blocks of $5,$ so consecutive blocks start $5\cdot6=30$ inches, or $2.5$ feet, apart.

Each block has one odd-numbered red light beginning it. The $3$ rd red light begins the $2$ nd block and the $21$ st red light begins the $11$ th block.

The distance between them is $(11-2)\cdot2.5=22.5$ feet.

Thus, the correct answer is E.

8.

A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?

$729$

$972$

$1024$

$2187$

$2304$

Answer: A

Concepts:multiplication principle arrangements with restrictions

Difficulty rating: 1380

Small Hint:

Fix Friday as cake, then count the other six days

Big Hint:

Each remaining day has $3$ choices, anything except the dessert served the following day

Solution:

Friday is fixed as cake. Work outward from Friday.

Each of the other six days (Saturday, then Thursday, Wednesday, Tuesday, Monday, Sunday) can be any dessert except the one served on the neighboring already-chosen day, giving $3$ choices each.

The number of menus is $3^6=729.$

Thus, the correct answer is A.

9.

It takes Clea $60$ seconds to walk down an escalator when it is not operating, and only $24$ seconds to walk down the escalator when it is operating. How many seconds does it take Clea to ride down the operating escalator when she just stands on it?

$36$

$40$

$42$

$48$

$52$

Answer: B

Concepts:rate system of equations

Difficulty rating: 1440

Small Hint:

Let the escalator length be fixed and write Clea’s walking rate and the escalator’s rate

Big Hint:

From $24(x+r)=60x$ find $r$ in terms of $x,$ then use $rt=60x$

Solution:

Let $x$ be Clea’s walking rate and $r$ the escalator’s rate, with the length equal to $60x.$ Walking on the moving escalator gives $24(x+r)=60x,$ so $r=\tfrac32x.$

Standing takes time $t$ with $rt=60x,$ so $\tfrac32xt=60x$ and $t=40$ seconds.

Thus, the correct answer is B.

10.

What is the area of the polygon whose vertices are the points of intersection of the curves $x^2 + y^2 = 25$ and $(x - 4)^2 + 9y^2 = 81?$

$24$

$27$

$36$

$37.5$

$42$

Answer: B

Concepts:system of equations substitution triangle area

Difficulty rating: 1500

Small Hint:

Solve the first equation for $y^2$ and substitute into the second

Big Hint:

The three intersection points form a triangle; use one vertical side as the base

Solution:

From $x^2+y^2=25$ we get $y^2=25-x^2.$ Substituting into $(x-4)^2+9y^2=81$ gives $x^2+x-20=0,$ so $x=4$ or $x=-5.$

The intersection points are $(-5,0),$ $(4,3),$ and $(4,-3).$

The vertical side from $(4,3)$ to $(4,-3)$ has length $6,$ and the horizontal distance to $(-5,0)$ is $9,$ so the area is $\tfrac12\cdot6\cdot9=27.$

Thus, the correct answer is B.

11.

In the equation below, $A$ and $B$ are consecutive positive integers, and $A,$ $B,$ and $A + B$ represent number bases: $132_A + 43_B = 69_{A+B}.$ What is $A + B?$

$9$

$11$

$13$

$15$

$17$

Answer: C

Concepts:number base quadratic

Difficulty rating: 1560

Small Hint:

Rewrite each numeral in terms of its base, e.g. $132_A = A^2+3A+2$

Big Hint:

Try both $B=A-1$ and $B=A+1$ and solve the resulting quadratic in $A$

Solution:

Writing the numerals out, $132_A=A^2+3A+2,$ $43_B=4B+3,$ and $69_{A+B}=6(A+B)+9.$

With $B=A+1,$ the equation becomes $A^2+3A+2$ $+4(A+1)+3$ $=6(2A+1)+9,$ which simplifies to $(A-6)(A+1)=0.$ The positive solution is $A=6,$ so $B=7.$

(The case $B=A-1$ gives $A^2-5A-2=0,$ which has no integer solution.)

Therefore $A+B=13.$

Thus, the correct answer is C.

12.

How many sequences of zeros and/or ones of length $20$ have all the zeros consecutive, or all the ones consecutive, or both?

$190$

$192$

$211$

$380$

$382$

Answer: E

Concepts:inclusion-exclusion combinations

Difficulty rating: 1660

Small Hint:

Let $A$ be the sequences with all zeros together and $B$ those with all ones together, then use $|A\cup B|=|A|+|B|-|A\cap B|$

Big Hint:

A sequence with the zeros together is fixed by choosing the first and last zero positions, giving $1+20+\binom{20}{2}$ of them

Solution:

Let $A$ be the sequences in which all zeros are consecutive and $B$ those in which all ones are consecutive.

For $A,$ there is one all-ones sequence, $20$ sequences with exactly one zero, and $\binom{20}{2}=190$ sequences with two or more zeros (choose the first and last zero position). So $|A|=1+20+190=211,$ and by symmetry $|B|=211.$

A sequence in $A\cap B$ is a block of zeros followed by a block of ones, or the reverse; there are $|A\cap B|=40$ of these.

Therefore $|A\cup B|=211+211-40$ $=382.$

Thus, the correct answer is E.

13.

Two parabolas have equations $y = x^2 + ax + b$ and $y = x^2 + cx + d,$ where $a,$ $b,$ $c,$ and $d$ are integers (not necessarily different), each chosen independently by rolling a fair six-sided die. What is the probability that the parabolas have at least one point in common?

$\dfrac{1}{2}$

$\dfrac{25}{36}$

$\dfrac{5}{6}$

$\dfrac{31}{36}$

$1$

Answer: D

Concepts:basic probability complementary counting dice (probability)

Difficulty rating: 1590

Small Hint:

Setting the two right sides equal, the $x^2$ terms cancel, leaving $ax+b=cx+d$

Big Hint:

There is no common point exactly when $a=c$ and $b\neq d;$ find that probability and subtract from $1$

Solution:

The parabolas meet where $x^2+ax+b=x^2+cx+d,$ i.e. $ax+b=cx+d.$ This has no solution exactly when the lines are parallel and distinct: $a=c$ and $b\neq d.$

The probability that $a=c$ is $\tfrac16,$ and the probability that $b\neq d$ is $\tfrac56,$ so the probability of no common point is $\tfrac16\cdot\tfrac56=\tfrac5{36}.$

The probability of at least one common point is $1-\tfrac5{36}=\tfrac{31}{36}.$

Thus, the correct answer is D.

14.

Bernardo and Silvia play the following game. An integer between $0$ and $999,$ inclusive, is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds $50$ to it and passes the result to Bernardo. The winner is the last person who produces a number less than $1000.$ Let $N$ be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of $N?$

$7$

$8$

$9$

$10$

$11$

Answer: A

Concepts:work backwards inequality

Difficulty rating: 1730

Small Hint:

Bernardo wins on a turn when his doubled result reaches $1000,$ forcing Silvia’s next number to be at least $1000$

Big Hint:

Work backwards: find the smallest start that first reaches $\ge1000$ after Bernardo’s move, over one, two, three, or four rounds

Solution:

Bernardo wins after a round when his doubled number $2n+50\ge1000$ but the previous numbers stayed below $1000.$ The smallest $n$ with $2n+50\ge1000$ is $475.$

Working backwards, the smallest starting values that lead to a win after two, three, and four rounds are the smallest integers with $2n+50\ge475,$ $\ge213,$ and $\ge82,$ namely $213,$ $82,$ and $16.$ No start wins after more than four rounds.

So $N=16,$ and the sum of its digits is $1+6=7.$

Thus, the correct answer is A.

15.

Jesse cuts a circular paper disk of radius $12$ along two radii to form two sectors, the smaller having a central angle of $120$ degrees. He makes two circular cones, using each sector to form the lateral surface of a cone. What is the ratio of the volume of the smaller cone to that of the larger?

$\dfrac{1}{8}$

$\dfrac{1}{4}$

$\dfrac{\sqrt{10}}{10}$

$\dfrac{\sqrt{5}}{6}$

$\dfrac{\sqrt{10}}{5}$

Answer: C

Concepts:cone volume Pythagorean Theorem

Difficulty rating: 1800

Small Hint:

The arc length of each sector becomes the circumference of that cone’s base, giving its base radius

Big Hint:

Each cone has slant height $12,$ so find each height with the Pythagorean theorem before comparing $\tfrac13\pi r^2 h$

Solution:

Each sector forms a cone with slant height $12.$ The smaller sector’s arc length is $\tfrac{120}{360}\cdot2\pi\cdot12=8\pi,$ so its base radius is $4$ and its height is $\sqrt{12^2-4^2}=8\sqrt2.$

The larger sector (central angle $240^\circ$ ) has arc length $16\pi,$ base radius $8,$ and height $\sqrt{12^2-8^2}=4\sqrt5.$

The ratio of volumes is $\frac{\tfrac13\pi\cdot4^2\cdot8\sqrt2}{\tfrac13\pi\cdot8^2\cdot4\sqrt5} =\frac{\sqrt{10}}{10}.$

Thus, the correct answer is C.

16.

Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?

$108$

$132$

$671$

$846$

$1105$

Answer: B

Concepts:casework multiplication principle

Difficulty rating: 1840

Small Hint:

Each song is liked by one of the pairs $\{AB\},\{AC\},\{BC\},$ by a single girl, or by no one; all three pairs must appear

Big Hint:

Split into the case where all four songs go to pairs and the case where exactly three songs cover the three pairs

Solution:

Each song is liked by exactly one of the three pairs, by a single girl, or by no one. Every pair must be represented.

Case $1:$ every song is liked by a pair. One pair gets two of the four songs ( $\binom42=6$ ways, and $3$ choices for which pair), and the other two pairs get one song each ( $2$ ways). This gives $3\cdot6\cdot2=36.$

Case $2:$ three songs go to the three pairs (one each) and the fourth song is liked by a single girl or no one. Assigning the four songs to these four roles gives $4!=24$ ways, and the leftover role has $4$ options (Amy, Beth, Jo, or no one): $24\cdot4=96.$

The total is $36+96=132.$

Thus, the correct answer is B.

17.

Square $PQRS$ lies in the first quadrant. Points $(3, 0),$ $(5, 0),$ $(7, 0),$ and $(13, 0)$ lie on lines $SP,$ $RQ,$ $PQ,$ and $SR,$ respectively. What is the sum of the coordinates of the center of the square $PQRS?$

$6$

$6.2$

$6.4$

$6.6$

$6.8$

Answer: C

Concepts:coordinate geometry slope system of equations

Difficulty rating: 1910

Small Hint:

Let $\theta$ be the angle line $PQ$ makes with the $x$ -axis; the square condition relates the projections of the given segments

Big Hint:

From $2\cos\theta=6\sin\theta$ get $\tan\theta=\tfrac13,$ then intersect the two lines through the midpoints $(4,0)$ and $(10,0)$

Solution:

Let $\theta$ be the acute angle line $PQ$ makes with the $x$ -axis. Sides $SR=PQ$ span the segment from $(3,0)$ to $(5,0)$ as $2\cos\theta,$ while $SP=QR$ span the segment from $(7,0)$ to $(13,0)$ as $6\sin\theta.$

Since the square has equal sides, $2\cos\theta=6\sin\theta,$ so $\tan\theta=\tfrac13.$ Thus lines $SP,RQ$ have slope $3$ and lines $SR,PQ$ have slope $-\tfrac13.$

The center lies on the line through $(4,0)$ with slope $3$ and the line through $(10,0)$ with slope $-\tfrac13:$ $y=3(x-4),$ $y=-\tfrac13(x-10).$ These meet at $(4.6,1.8).$

The sum of the coordinates is $4.6+1.8=6.4.$

Thus, the correct answer is C.

18.

Let $(a_1, a_2, \ldots, a_{10})$ be a list of the first $10$ positive integers such that for each $2 \le i \le 10$ either $a_i + 1$ or $a_i - 1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?

$120$

$512$

$1024$

$181{,}440$

$362{,}880$

Answer: B

Concepts:bijection combinations

Difficulty rating: 1990

Small Hint:

The numbers greater than or equal to $a_1$ must appear in increasing order, and those less than $a_1$ in decreasing order

Big Hint:

Choosing which positions hold the small numbers determines the list, giving $\sum_k \binom{9}{k}$

Solution:

Once $a_1=k$ is fixed, the numbers $k,k+1,\ldots,10$ must appear left to right in increasing order, and the numbers $1,\ldots,k-1$ must appear from right to left in increasing order (so each new small number has its successor already placed).

For each $k,$ the list is determined by choosing which of the $9$ positions after the first hold the numbers below $k,$ giving $\binom{9}{k-1}$ lists.

Summing, $\begin{aligned} \sum_{k=1}^{10}\binom{9}{k-1} &= \sum_{j=0}^{9}\binom{9}{j} \\ &= 2^9=512. \end{aligned}$

Thus, the correct answer is B.

19.

A unit cube has vertices $P_1,$ $P_2,$ $P_3,$ $P_4,$ $P_1',$ $P_2',$ $P_3',$ and $P_4'.$ Vertices $P_2,$ $P_3,$ and $P_4$ are adjacent to $P_1,$ and for $1 \le i \le 4,$ vertices $P_i$ and $P_i'$ are opposite to each other. A regular octahedron has one vertex in each of the segments $P_1P_2,$ $P_1P_3,$ $P_1P_4,$ $P_1'P_2',$ $P_1'P_3',$ and $P_1'P_4'.$ What is the octahedron’s side length?

$\dfrac{3\sqrt{2}}{4}$

$\dfrac{7\sqrt{6}}{16}$

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

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

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

Answer: A

Concepts:cube geometry Pythagorean Theorem symmetry

Difficulty rating: 2110

Small Hint:

By symmetry each octahedron vertex is the same distance $t$ from $P_1$ (or $P_1'$ ) along its edge

Big Hint:

Set the squared distance between two vertices sharing $P_1$ equal to the squared distance across, and solve for $t$

Solution:

Place $P_1$ at the origin with edges along the axes, and let each of the three octahedron vertices near $P_1$ be a distance $t$ from $P_1;$ by symmetry the three near $P_1'$ are also a distance $t$ from $P_1'.$

Two vertices sharing $P_1,$ such as $(t,0,0)$ and $(0,t,0),$ are a distance $t\sqrt2$ apart. A vertex near $P_1,$ say $(t,0,0),$ and the appropriate vertex near $P_1',$ say $(1,1-t,1),$ must be the same distance apart.

Setting the two squared side lengths equal and using the cube’s unit edges yields $t=\tfrac34,$ so the side length is $t\sqrt2=\dfrac{3\sqrt2}{4}.$

Thus, the correct answer is A.

20.

A trapezoid has side lengths $3,$ $5,$ $7,$ and $11.$ The sum of all the possible areas of the trapezoid can be written in the form of $r_1\sqrt{n_1} + r_2\sqrt{n_2} + r_3,$ where $r_1,$ $r_2,$ and $r_3$ are rational numbers and $n_1$ and $n_2$ are positive integers not divisible by the square of a prime. What is the greatest integer less than or equal to $r_1 + r_2 + r_3 + n_1 + n_2?$

$57$

$59$

$61$

$63$

$65$

Answer: D

Concepts:trapezoid Heron’s Formula casework

Difficulty rating: 2150

Small Hint:

Slide one parallel side over: a trapezoid with parallel sides $a\lt c$ and legs $b,d$ gives a triangle with sides $b,d,c-a$

Big Hint:

The triangle inequality forces the longer parallel side to be $11;$ apply Heron’s formula in the cases $a=3,5,7$

Solution:

For a trapezoid with parallel sides $a\lt c$ and legs $b,d,$ translating a leg forms a triangle with sides $b,$ $d,$ and $c-a.$ The triangle inequality forces the longer parallel side to be $c=11.$

If $a=3,$ the triangle has sides $5,7,8$ with area $10\sqrt3,$ and the trapezoid has area $\tfrac{35}{2}\sqrt3.$ If $a=5,$ the triangle has sides $3,6,7$ with area $4\sqrt5,$ giving trapezoid area $\tfrac{32}{3}\sqrt5.$ If $a=7,$ the triangle has sides $3,4,5,$ a right triangle, giving trapezoid area $27.$

The total is $\tfrac{35}{2}\sqrt3+\tfrac{32}{3}\sqrt5+27,$ so $\begin{gathered} r_1+r_2+r_3+n_1+n_2 \\ = \tfrac{35}{2}+\tfrac{32}{3}+27+3+5 \\ = 63+\tfrac16. \end{gathered}$

The greatest integer at most this value is $63.$

Thus, the correct answer is D.

21.

Square $AXYZ$ is inscribed in equiangular hexagon $ABCDEF$ with $X$ on $\overline{BC},$ $Y$ on $\overline{DE},$ and $Z$ on $\overline{EF}.$ Suppose that $AB = 40$ and $EF = 41(\sqrt{3} - 1).$ What is the side-length of the square?

$29\sqrt{3}$

$\dfrac{21}{2}\sqrt{2} + \dfrac{41}{2}\sqrt{3}$

$20\sqrt{3} + 16$

$20\sqrt{2} + 13\sqrt{3}$

$21\sqrt{6}$

Answer: A

Concepts:equiangular polygon congruence (geometry)Pythagorean Theorem

Difficulty rating: 2170

Small Hint:

Drop the square’s vertex $A$ onto a line through $A$ perpendicular to the parallel sides $EF$ and $CB$

Big Hint:

Let $u=BX;$ the equiangular $60^\circ$ angles give congruent corner triangles, so express $EF$ in terms of $u$ and solve for the side $s^2=(20+u)^2+(20\sqrt3)^2$

Solution:

Extend $EF$ and $CB$ to a line through $A$ perpendicular to both, meeting them at $H$ and $J.$ Since $\angle ABJ=60^\circ,$ we have $BJ=20$ and $AJ=20\sqrt3.$ With $u=BX,$ the Pythagorean theorem gives $s^2=(20+u)^2+(20\sqrt3)^2.$

The equiangular angles make the four corner triangles congruent, and chasing the equal segments along $EF$ yields $\begin{aligned} u+20\sqrt3 &= 41(\sqrt3-1) \\ &\quad {}+\frac{20+u}{\sqrt3}, \end{aligned}$ so $u=21\sqrt3-20.$

Since $20+u=21\sqrt3,$ we get $\begin{aligned} s^2 &= (21\sqrt3)^2+(20\sqrt3)^2 \\ &= 3(441+400)=3\cdot29^2, \end{aligned}$ giving $s=29\sqrt3.$

Thus, the correct answer is A.

22.

A bug travels from $A$ to $B$ along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?

$2112$

$2304$

$2368$

$2384$

$2400$

Answer: E

Concepts:casework multiplication principle

Difficulty rating: 2270

Small Hint:

There are three left-pointing “back” segments; split into cases by which subset of them the path uses

Big Hint:

With no back segment used, the path picks one forward segment per column, giving $2\cdot2\cdot4\cdot4\cdot4\cdot2\cdot2=2^{10}$

Solution:

Label the seven columns of forward (rightward) segments; a path with no back segment simply chooses one forward segment in each column. The numbers of choices are $2,2,4,4,4,2,2,$ giving $2^{10}$ paths.

Let $s_1,s_2,s_3$ be the three left-pointing back segments (in columns $2,4,6$ ). Analyzing which columns become forced once a back segment is traversed gives $2^8$ paths for each of $\{s_1\}$ and $\{s_3\},$ $2^6$ for $\{s_1,s_3\},$ $2^9$ for $\{s_2\},$ $2^7$ for each of $\{s_1,s_2\}$ and $\{s_2,s_3\},$ and $2^5$ for $\{s_1,s_2,s_3\}.$

Adding, $\begin{aligned} &2^{10}+2\cdot2^8+2^6 \\ &\quad {}+2^9+2\cdot2^7+2^5=2400. \end{aligned}$

Thus, the correct answer is E.

23.

Consider all polynomials of a complex variable, $P(z) = 4z^4 + az^3$ $+ bz^2 + cz + d,$ where $a,$ $b,$ $c,$ and $d$ are integers, $0 \le d \le c \le b \le a \le 4,$ and the polynomial has a zero $z_0$ with $|z_0| = 1.$ What is the sum of all values $P(1)$ over all the polynomials with these properties?

$84$

$92$

$100$

$108$

$120$

Answer: B

Concepts:roots of unity complex number triangle inequality

Difficulty rating: 2380

Small Hint:

Apply the triangle inequality to $4z_0^5=(z_0-1)P(z_0)+4z_0^5;$ equality forces the coefficient jumps to concentrate

Big Hint:

First handle the cases where $z_0^k=1$ for some $1\le k\le4;$ otherwise equality in the triangle inequality permits only one nonzero coefficient jump

Solution:

Because $|z_0|=1,$ applying the triangle inequality to the identity $\begin{aligned} &4z_0^5-(z_0-1)P(z_0) \\ &\quad =z_0^4(4-a)+z_0^3(a-b) \\ &\quad {}+z_0^2(b-c)+z_0(c-d)+d \end{aligned}$ has left side of absolute value $4.$ The nonnegative coefficient jumps on the right sum to $4,$ so the triangle inequality is an equality and all its nonzero complex summands point in the same direction.

If two jumps are nonzero, their quotient shows that $z_0^k$ is a positive real for some $1\le k\le4.$ Since $|z_0|=1,$ this means $z_0^k=1.$ The cases $k=2,3,4$ give, respectively, $a=4,b=c,d=0;$ $a=b=4,c=d=0;$ and a polynomial already in the first family. If no such power exists, equality forces exactly one nonzero jump. The constant jump gives $a=b=c=d=4;$ any other jump again forces $z_0^{5-j}=1$ and returns to the cases just listed. Hence the polynomials are exactly $4z^4+4z^3+bz^2+bz$ for $0\le b\le4,$ together with $4z^4+4z^3+4z^2$ and $4z^4+4z^3+4z^2+4z+4.$

Their values at $1$ are $20,$ $12,$ and $8+2b;$ summing gives $\begin{gathered} 20+12+\sum_{b=0}^{4}(8+2b) \\ = 32+40+20 \\ = 92. \end{gathered}$

Thus, the correct answer is B.

24.

Define the function $f_1$ on the positive integers by setting $f_1(1) = 1$ and if $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$ is the prime factorization of $n \gt 1,$ then $\begin{aligned} &f_1(n) = (p_1 + 1)^{e_1 - 1}(p_2 + 1)^{e_2 - 1} \\ &\quad \cdots (p_k + 1)^{e_k - 1}. \end{aligned}$ For every $m \ge 2,$ let $f_m(n) = f_1(f_{m-1}(n)).$ For how many $N$ in the range $1 \le N \le 400$ is the sequence $(f_1(N), f_2(N), f_3(N), \ldots)$ unbounded? Note: a sequence of positive numbers is unbounded if for every integer $B,$ there is a member of the sequence greater than $B.$

$15$

$16$

$17$

$18$

$19$

Answer: D

Concepts:prime factorization divisibility recursion

Difficulty rating: 2520

Small Hint:

If $N_2$ is a multiple of $N_1$ then $f_1(N_2)$ is a multiple of $f_1(N_1),$ so unboundedness is inherited by multiples; find the minimal “essential” $N$

Big Hint:

An essential $N$ has every exponent at least $2$ and at most two prime factors; analyze powers of $2,3$ first, then the few remaining prime powers and prime pairs below $400$

Solution:

If $N_2$ is a multiple of $N_1$ then $f_1(N_2)$ is a multiple of $f_1(N_1),$ so if $S_{N_1}$ is unbounded so is $S_{N_2}.$ Call $N$ essential if it is unbounded but no proper divisor is. An essential $N$ must have all exponents at least $2,$ and $(p_1\cdots p_k)^2\le400$ forces at most two primes.

For $n=2^a3^b,$ two iterations send the exponent pair to $(2a-4,2b-3).$ Thus the orbit is unbounded exactly when $a\ge5$ or $b\ge4,$ producing the essential values $2^5=32$ and $3^4=81.$ For a single other prime, the bound leaves only $5^2,5^3,7^2,7^3,11^2,13^2,17^2,19^2;$ direct application of $f_1$ leaves only $7^3=343$ essential. With two primes, $p_1p_2\le20$ leaves the pairs $(2,5),(2,7),(3,5)$ besides $(2,3).$ Applying $f_1$ to their possible squareful products leaves only $2^4\cdot5^2=400.$

Their multiples up to $400$ number $\lfloor\frac{400}{32}\rfloor=12,$ $\lfloor\frac{400}{81}\rfloor=4,$ $\lfloor\frac{400}{343}\rfloor=1,$ and $\lfloor\frac{400}{400}\rfloor=1,$ with no overlaps, for a total of $12+4+1+1=18.$

Thus, the correct answer is D.

25.

Let $S = \{(x, y) : x \in \{0, 1, 2, 3, 4\},$ $y \in \{0, 1, 2, 3, 4, 5\},$ and $(x, y) \ne (0, 0)\}.$ Let $T$ be the set of all right triangles whose vertices are in $S.$ For every right triangle $t = \triangle ABC$ with vertices $A,$ $B,$ and $C$ in counter-clockwise order and right angle at $A,$ let $f(t) = \tan(\angle CBA).$ What is $\prod_{t \in T} f(t)?$

$1$

$\dfrac{625}{144}$

$\dfrac{125}{24}$

$6$

$\dfrac{625}{24}$

Answer: B

Concepts:trigonometry telescoping symmetry

Difficulty rating: 2650

Small Hint:

Pairing a triangle with its reflection swaps $\tan(\angle CBA)$ with $\tan(\angle ACB),$ whose product is $1$

Big Hint:

Reflections collapse the whole product down to a handful of triangles through the origin, and only six survive

Solution:

Isosceles right triangles contribute $f(t)=1.$ For a scalene right triangle, reflecting across a suitable line pairs it with a triangle $t_1$ so that $f(t)f(t_1)$ $=\tan(\angle CBA)\tan(\angle ACB)$ $=1.$

Successive reflections (across $x=2,$ then $x=y,$ then $y=\tfrac52$ ) reduce the product to the reciprocal of the product over just six triangles of the form $OYZ$ with $Y$ on the top row.

Those six give $\begin{aligned} &\frac15\cdot\frac25\cdot\frac35\cdot\frac45\cdot\frac{3\sqrt2}{\sqrt2} \\ &\quad {}\cdot\frac{4\sqrt2}{2\sqrt2}=\frac{144}{625}, \end{aligned}$ so the required product is its reciprocal, $\dfrac{625}{144}.$

Thus, the correct answer is B.