What is the product of all the solutions to the equation \begingathered \log_7x2023×\log_289x2023\ =\log_2023x2023? \endgathered

(A) (\log_20237×\log_2023289)^2

(E) (\log_7 2023×\log_2892023)^2

2023 AMC 12A — Problems & Online Timed Exam

Timed

1:15:00

1.

Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B.$ Alicia bikes towards $B$ at $18$ miles per hour. Leaving at the same time, Beth bikes toward $A$ at $12$ miles per hour. How many miles from City $A$ will they be when they meet?

$20$

$24$

$25$

$26$

$27$

Answer: E

Concepts:distance rate and time

Difficulty rating: 890

Small Hint:

Their closing speed is $18+12$ miles per hour

Big Hint:

Find the time until they meet, then multiply by Alicia’s speed

Solution:

The gap between them closes at $18+12=30$ miles per hour, so they meet after $\dfrac{45}{30}=1.5$ hours.

In that time Alicia has ridden $18\cdot 1.5=27$ miles from City $A.$

Thus, the correct answer is E.

2.

The weight of $\tfrac13$ of a large pizza together with $3\tfrac12$ cups of orange slices is the same as the weight of $\tfrac34$ of a large pizza together with $\tfrac12$ cup of orange slices. A cup of orange slices weighs $\tfrac14$ of a pound. What is the weight, in pounds, of a large pizza?

$1\tfrac45$

$2$

$2\tfrac25$

$3$

$3\tfrac35$

Answer: A

Concepts:linear equation fraction

Difficulty rating: 1020

Small Hint:

Let the pizza weigh $P$ pounds; each cup of orange slices weighs $\tfrac14$

Big Hint:

Set the two weights equal and gather the $P$ terms on one side

Solution:

Let the pizza weigh $P$ pounds. Then $3\tfrac12$ cups weigh $\tfrac72\cdot\tfrac14=\tfrac78$ and $\tfrac12$ cup weighs $\tfrac12\cdot\tfrac14=\tfrac18.$

The equation is $\tfrac13 P+\tfrac78=\tfrac34 P+\tfrac18.$

Subtracting gives $\tfrac78-\tfrac18=\left(\tfrac34-\tfrac13\right)P,$ so $\tfrac34=\tfrac{5}{12}P$ and $P=\dfrac{9}{5}=1\tfrac45.$

Thus, the correct answer is A.

3.

How many positive perfect squares less than $2023$ are divisible by $5?$

$8$

$9$

$10$

$11$

$12$

Answer: A

Concepts:perfect square divisibility

Difficulty rating: 1130

Small Hint:

A perfect square is divisible by $5$ exactly when its root is a multiple of $5$

Big Hint:

The squares are $(5k)^2$ ; find how many satisfy $(5k)^2\lt 2023$

Solution:

A perfect square is divisible by $5$ only if its root is, so the squares are $(5k)^2=25k^2.$

Since $44^2=1936\lt 2023$ $\lt 2025=45^2,$ the root can be $5,10,\ldots,40,$ which is $k=1$ through $8.$

Thus, the correct answer is A.

4.

How many digits are in the base-ten representation of $8^5\cdot 5^{10}\cdot 15^5?$

$14$

$15$

$16$

$17$

$18$

Answer: E

Concepts:prime factorization digits

Difficulty rating: 1200

Small Hint:

Factor each base into primes: $8=2^3,$ $15=3\cdot 5$

Big Hint:

Pair every $2$ with a $5$ to form a power of $10,$ leaving a small factor

Solution:

Writing everything in primes, $\begin{gathered} 8^5\cdot 5^{10}\cdot 15^5\\ {}=2^{15}\cdot 5^{10}\cdot 3^5\cdot 5^5\\ {}=3^5\cdot 2^{15}\cdot 5^{15}. \end{gathered}$

This equals $3^5\cdot 10^{15}=243\cdot 10^{15},$ which is $243$ followed by $15$ zeros, for a total of $18$ digits.

Thus, the correct answer is E.

5.

Janet rolls a standard $6$ -sided die $4$ times and keeps a running total of the numbers she rolls. What is the probability that at some point, her running total will equal $3?$

$\dfrac{2}{9}$

$\dfrac{49}{216}$

$\dfrac{25}{108}$

$\dfrac{17}{72}$

$\dfrac{13}{54}$

Answer: B

Concepts:dice (probability)casework

Difficulty rating: 1270

Small Hint:

Once the running total passes $3$ it can never return, so only the first few rolls matter

Big Hint:

Add the probabilities of the disjoint openings $3;$ $1,2;$ $2,1;$ and $1,1,1$

Solution:

The running total is increasing, so it hits $3$ exactly when one of these disjoint openings occurs: a first roll of $3;$ rolls $1,2;$ rolls $2,1;$ or rolls $1,1,1.$

Their probabilities are $\begin{gathered} \dfrac16+\dfrac1{36}+\dfrac1{36}+\dfrac1{216}\\ {}=\dfrac{36+6+6+1}{216}\\ {}=\dfrac{49}{216}. \end{gathered}$

Thus, the correct answer is B.

6.

Points $A$ and $B$ lie on the graph of $y=\log_2 x.$ The midpoint of $\overline{AB}$ is $(6,2).$ What is the positive difference between the $x$ -coordinates of $A$ and $B?$

$2\sqrt{11}$

$4\sqrt{3}$

$8$

$4\sqrt{5}$

$9$

Answer: D

Concepts:logarithm midpoint algebraic manipulation

Difficulty rating: 1350

Small Hint:

The $x$ -coordinates sum to $2\cdot 6=12$

Big Hint:

The $y$ -coordinates average to $2,$ so $\log_2 x_1+\log_2 x_2=4,$ giving $x_1x_2=16$

Solution:

Let the $x$ -coordinates be $x_1$ and $x_2.$ The midpoint gives $x_1+x_2=12,$ and the average of the $y$ -values gives $\log_2 x_1+\log_2 x_2=4,$ so $x_1x_2=2^4=16.$

Then $\begin{gathered} |x_1-x_2|\\ {}=\sqrt{(x_1+x_2)^2-4x_1x_2}\\ {}=\sqrt{144-64}\\ {}=\sqrt{80}\\ {}=4\sqrt5. \end{gathered}$

Thus, the correct answer is D.

7.

A digital display shows the current date as an $8$ -digit integer consisting of a $4$ -digit year, followed by a $2$ -digit month, followed by a $2$ -digit date within the month. For example, Arbor Day this year is displayed as $20230428.$ For how many dates in $2023$ will each digit appear an even number of times in the $8$ -digit display for that date?

$5$

$6$

$7$

$8$

$9$

Answer: E

Concepts:digits parity casework

Difficulty rating: 1380

Small Hint:

The fixed year $2023$ already contributes two $2$ ’s, one $0,$ and one $3$

Big Hint:

The four month-day digits must make the counts of $0$ and $3$ odd, the count of $2$ even, and every other digit even

Solution:

The year contributes the digits $2,0,2,3,$ so $2$ appears twice while $0$ and $3$ each appear once. For every digit to end up with an even count, the four digits of the month and day must supply an odd number of $0$ ’s, an odd number of $3$ ’s, and an even number of every other digit.

With only four digits available, the month-day string must use exactly one $0,$ one $3,$ and a repeated pair of some digit. Checking valid months and days leaves nine dates: $01\text{-}13,$ $01\text{-}31,$ $02\text{-}23,$ $03\text{-}11,$ $03\text{-}22,$ $10\text{-}13,$ $10\text{-}31,$ $11\text{-}03,$ and $11\text{-}30.$

Thus, the correct answer is E.

8.

Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1.$ If she scores an $11$ on each of the next three quizzes, her mean will increase by $2.$ What is the mean of her quiz scores currently?

$4$

$5$

$6$

$7$

$8$

Answer: D

Concepts:mean system of equations

Difficulty rating: 1440

Small Hint:

Let her current mean be $m$ over $n$ quizzes, with total $S=mn$

Big Hint:

One extra $11$ gives $\dfrac{S+11}{n+1}=m+1;$ simplify to $m+n=10$

Solution:

Let the current mean be $m$ over $n$ quizzes, so the total is $S=mn.$ Adding one $11$ gives $\dfrac{mn+11}{n+1}=m+1,$ which simplifies to $m+n=10.$

Adding three $11$ ’s gives $\dfrac{mn+33}{n+3}=m+2,$ which simplifies to $3m+2n=27.$

Solving $m+n=10$ and $3m+2n=27$ gives $m=7.$

Thus, the correct answer is D.

9.

A square of area $2$ is inscribed in a square of area $3,$ creating four congruent triangles, as shown below. What is the ratio of the shorter leg to the longer leg in the shaded right triangle?

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$2-\sqrt{3}$

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

$\sqrt{2}-1$

Answer: C

Concepts:square (geometry)Pythagorean Theorem system of equations

Difficulty rating: 1500

Small Hint:

The outer square has side $\sqrt3$ and the inner square has side $\sqrt2$

Big Hint:

If the legs are $p$ and $q,$ then $p+q=\sqrt3$ and $p^2+q^2=2$

Solution:

The outer square has side $\sqrt3$ and the inner square has side $\sqrt2.$ Each triangle is right, with legs $p$ and $q$ along an outer side, so $p+q=\sqrt3,$ and with hypotenuse an inner side, so $p^2+q^2=2.$

Then $(p+q)^2=3$ gives $2pq=1,$ so $p$ and $q$ are the roots of $t^2-\sqrt3\,t+\tfrac12=0,$ namely $\dfrac{\sqrt3\pm 1}{2}.$

The ratio of shorter to longer leg is $\begin{gathered} \dfrac{\sqrt3-1}{\sqrt3+1}\\ {}=\dfrac{(\sqrt3-1)^2}{2}\\ {}=2-\sqrt3. \end{gathered}$

Thus, the correct answer is C.

10.

Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2.$ What is $x+y?$

$12$

$18$

$24$

$36$

$42$

Answer: D

Concepts:system of equations substitution

Difficulty rating: 1560

Small Hint:

Taking square roots, $y-x=\pm 2y$

Big Hint:

Positivity rules out one sign and forces $x=3y;$ substitute into $y^3=x^2$

Solution:

From $(y-x)^2=4y^2$ we get $y-x=\pm 2y.$ The choice $y-x=2y$ gives $x=-y\lt 0,$ impossible, so $y-x=-2y,$ meaning $x=3y.$

Substituting into $y^3=x^2=9y^2$ gives $y=9,$ hence $x=27$ and $x+y=36.$

Thus, the correct answer is D.

11.

What is the degree measure of the acute angle formed by lines with slopes $2$ and $\tfrac13?$

$30$

$37.5$

$45$

$52.5$

$60$

Answer: C

Concepts:slope trigonometry

Difficulty rating: 1570

Small Hint:

Use $\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$

Big Hint:

Both the numerator and denominator work out to $\tfrac53$

Solution:

The tangent of the angle between the lines is $\left|\dfrac{2-\tfrac13}{1+2\cdot\tfrac13}\right| =\left|\dfrac{\tfrac53}{\tfrac53}\right|=1.$

The acute angle with tangent $1$ is $45^\circ.$

Thus, the correct answer is C.

12.

What is the value of $\begin{gathered} 2^3-1^3+4^3-3^3+6^3-5^3\\ {}+\cdots+18^3-17^3? \end{gathered}$

$2023$

$2679$

$2941$

$3159$

$3235$

Answer: D

Concepts:sum and difference of cubes sum of first n squares

Difficulty rating: 1630

Small Hint:

Group the terms into pairs $(2k)^3-(2k-1)^3$ for $k=1$ to $9$

Big Hint:

Each pair equals $12k^2-6k+1;$ sum using $\sum k^2$ and $\sum k$

Solution:

Group into pairs $(2k)^3-(2k-1)^3$ for $k=1,\ldots,9.$ Expanding, $\begin{gathered} (2k)^3-(2k-1)^3\\ {}=12k^2-6k+1. \end{gathered}$

Summing for $k=1$ to $9,$ with $\sum k^2=285$ and $\sum k=45,$ gives $\begin{gathered} 12\cdot 285-6\cdot 45+9\\ {}=3420-270+9\\ {}=3159. \end{gathered}$

Thus, the correct answer is D.

13.

In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?

$15$

$36$

$45$

$48$

$66$

Answer: B

Concepts:combinations divisibility

Difficulty rating: 1660

Small Hint:

Let there be $L$ left-handed and $2L$ right-handed players, so $3L$ players total

Big Hint:

If right-handers win $R$ games, the total is $2.4R,$ so the total must be a multiple of $12$

Solution:

Let there be $L$ left-handed and $2L$ right-handed players, for $3L$ players and $\binom{3L}{2}$ games total.

If right-handers win $R$ games, left-handers win $1.4R,$ so the total is $2.4R=\tfrac{12}{5}R.$ For this to be an integer count, the total number of games must be a multiple of $12.$

The left-handers can win at most every game involving at least one left-hander, namely $\binom{L}{2}+2L^2=\tfrac{5L^2-L}{2}.$ On the other hand, they must win $\tfrac{7}{12}\binom{3L}{2}=\tfrac{7L(3L-1)}8$ games. Comparing these quantities gives $21L-7\le 20L-4,$ so $L\le 3.$

Testing $L=1,2,3$ gives totals $3,15,36;$ only $36$ is a multiple of $12.$ It is achievable: the $3$ left-handers can win all $18$ mixed games and their $3$ internal games, giving $21=1.4\cdot 15$ wins.

Thus, the correct answer is B.

14.

How many complex numbers satisfy the equation $z^5=\overline{z},$ where $\overline{z}$ is the conjugate of the complex number $z?$

$2$

$3$

$5$

$6$

$7$

Answer: E

Concepts:complex number roots of unity

Difficulty rating: 1730

Small Hint:

Take absolute values: $|z|^5=|z|,$ so $|z|=0$ or $|z|=1$

Big Hint:

Multiplying by $z$ gives $z^6=z\overline{z}=|z|^2;$ handle $z=0$ separately

Solution:

Taking magnitudes gives $|z|^5=|z|,$ so $|z|=0$ or $|z|=1.$ The value $z=0$ works, giving one solution.

If $|z|=1,$ multiply the equation by $z$ to get $z^6=z\overline{z}=|z|^2=1.$ This has $6$ distinct roots, all of modulus $1.$

Altogether there are $1+6=7$ solutions.

Thus, the correct answer is E.

15.

Usain is walking for exercise by zigzagging across a $100$ -meter by $30$ -meter rectangular field, beginning at point $A$ and ending on the segment $\overline{BC}.$ He wants to increase the distance walked by zigzagging as shown in the figure below ( $APQRS$ ). What angle $\theta=\angle PAB$ $=\angle QPC$ $=\angle RQB=\cdots$ will produce a length that is $120$ meters? (Do not assume the zigzag path has exactly four segments as shown; there could be more or fewer.)

$\arccos\tfrac{5}{6}$

$\arccos\tfrac{4}{5}$

$\arccos\tfrac{3}{10}$

$\arcsin\tfrac{4}{5}$

$\arcsin\tfrac{5}{6}$

Answer: A

Concepts:trigonometry right triangle

Difficulty rating: 1800

Small Hint:

The horizontal projection of each zigzag segment is its length multiplied by $\cos\theta$

Big Hint:

Add the horizontal projections of all the segments and compare the $100$ -meter progress with the $120$ -meter path

Solution:

Every segment of the zigzag makes angle $\theta$ with a horizontal side of the field. Therefore a segment of length $s$ advances $s\cos\theta$ meters horizontally. This remains true for the last segment even if it ends before crossing the full width of the field.

Adding the horizontal projections over the entire $120$ -meter path gives $120\cos\theta=100.$

Therefore $\cos\theta=\dfrac56,$ so $\theta=\arccos\dfrac56.$

Thus, the correct answer is A.

16.

Consider the set of complex numbers $z$ satisfying $|1+z+z^2|=4.$ The maximum value of the imaginary part of $z$ can be written in the form $\dfrac{\sqrt{m}}{n},$ where $m$ and $n$ are relatively prime positive integers. What is $m+n?$

$20$

$21$

$22$

$23$

$24$

Answer: B

Concepts:complex number optimization

Difficulty rating: 1840

Small Hint:

Write $z=x+yi$ and expand $|1+z+z^2|^2=16$

Big Hint:

Maximizing $y$ forces the symmetric case $x=-\tfrac12;$ solve the resulting equation for $y$

Solution:

Write $z=x+yi.$ Then $1+z+z^2$ $=(1+x+x^2-y^2)$ $+y(1+2x)i,$ and the constraint is $\begin{gathered} (1+x+x^2-y^2)^2\\ {}+y^2(1+2x)^2=16. \end{gathered}$

At a point where $y$ is maximal on this closed, bounded curve, implicit differentiation (or Lagrange multipliers) gives $(1+2x)\bigl(P+2y^2\bigr)=0,$ where $P=1+x+x^2-y^2.$ But $P+2y^2=x^2+x+1+y^2\gt 0,$ so $x=-\tfrac12.$

Then $1+2x=0,$ so the constraint reduces to $\left(\tfrac34-y^2\right)^2=16.$ Taking $\tfrac34-y^2=-4$ gives $y^2=\tfrac{19}{4},$ so the maximum is $y=\dfrac{\sqrt{19}}{2}.$

Here $m=19$ and $n=2,$ so $m+n=21.$

Thus, the correct answer is B.

17.

Flora the frog starts at $0$ on the number line and makes a sequence of jumps to the right. In any one jump, independent of previous jumps, Flora leaps a positive integer distance $m$ with probability $\dfrac{1}{2^m}.$ What is the probability that Flora will eventually land at $10?$

$\dfrac{5}{512}$

$\dfrac{45}{1024}$

$\dfrac{127}{1024}$

$\dfrac{511}{1024}$

$\dfrac{1}{2}$

Answer: E

Concepts:recursive probability recursion

Difficulty rating: 1910

Small Hint:

Let $a_n$ be the probability Flora ever lands on $n;$ condition on the size of the first jump

Big Hint:

The recursion $a_n=\sum_{k=1}^{n}\dfrac{a_{n-k}}{2^k}$ gives the same value for every $n\ge 1$

Solution:

Let $a_n$ be the probability that Flora ever lands exactly on $n,$ with $a_0=1.$ Conditioning on the first jump, $a_n=\sum_{k=1}^{n}\dfrac{1}{2^k}\,a_{n-k}.$

We prove by induction that $a_n=\tfrac12$ for every $n\ge 1.$ The case $n=1$ is immediate. If the claim holds through $n-1,$ then $\begin{aligned} a_n&=\frac{1}{2^n}a_0 +\sum_{k=1}^{n-1}\frac{1}{2^k}\cdot\frac12\\ &=\frac{1}{2^n} +\frac12\left(1-\frac{1}{2^{n-1}}\right)\\ &=\frac12. \end{aligned}$

Hence the probability of landing on $10$ is $\dfrac12.$

Thus, the correct answer is E.

18.

Circle $C_1$ and $C_2$ each have radius $1,$ and the distance between their centers is $\tfrac12.$ Circle $C_3$ is the largest circle internally tangent to both $C_1$ and $C_2.$ Circle $C_4$ is internally tangent to both $C_1$ and $C_2$ and externally tangent to $C_3.$ What is the radius of $C_4?$

$\dfrac{1}{14}$

$\dfrac{1}{12}$

$\dfrac{1}{10}$

$\dfrac{3}{28}$

$\dfrac{1}{9}$

Answer: D

Concepts:tangent circles coordinate geometry

Difficulty rating: 1990

Small Hint:

By symmetry $C_3$ is centered at the midpoint of $C_1C_2;$ its radius is $1-\tfrac14=\tfrac34$

Big Hint:

Place $C_4$ on the perpendicular bisector; its center is $1-r$ from a unit circle’s center and $\tfrac34+r$ from the center of $C_3$

Solution:

Put the centers at $O_1=\left(-\tfrac14,0\right)$ and $O_2=\left(\tfrac14,0\right).$ By symmetry $C_3$ is centered at the origin, and internal tangency to $C_1$ gives radius $1-\tfrac14=\tfrac34.$

Let $C_4$ have radius $r,$ centered at $(0,k)$ on the axis of symmetry. External tangency to $C_3$ gives $k=\tfrac34+r,$ and internal tangency to $C_1$ gives $\sqrt{\tfrac{1}{16}+k^2}=1-r.$

Substituting, $\tfrac{1}{16}+\left(\tfrac34+r\right)^2=(1-r)^2,$ which simplifies to $\tfrac72 r=\tfrac38,$ so $r=\dfrac{3}{28}.$

Thus, the correct answer is D.

19.

What is the product of all the solutions to the equation $\begin{gathered} \log_{7x}2023\cdot\log_{289x}2023\\ {}=\log_{2023x}2023? \end{gathered}$

$(\log_{2023}7\cdot\log_{2023}289)^2$

$\log_{2023}7\cdot\log_{2023}289$

$1$

$\log_7 2023\cdot\log_{289}2023$

$(\log_7 2023\cdot\log_{289}2023)^2$

Answer: C

Concepts:logarithm Vieta’s Formulas

Difficulty rating: 2040

Small Hint:

Convert to base $2023$ and set $t=\log_{2023}x;$ note $2023=7\cdot 17^2,$ so $\log_{2023}7+\log_{2023}289=1$

Big Hint:

The equation becomes a quadratic in $t$ whose two roots sum to $0$

Solution:

Let $a=\log_{2023}7$ and $b=\log_{2023}289.$ Since $2023=7\cdot 289,$ we have $a+b=1.$ Writing $t=\log_{2023}x,$ each logarithm becomes a reciprocal, and the equation turns into $(1+t)=(a+t)(b+t).$

Expanding and using $a+b=1,$ the linear terms cancel, leaving $t^2+(ab-1)=0.$ Its two roots satisfy $t_1+t_2=0.$

The corresponding solutions multiply to $x_1x_2=2023^{t_1}\cdot 2023^{t_2}$ $=2023^{\,t_1+t_2}$ $=2023^0=1.$

Thus, the correct answer is C.

20.

Rows $1,$ $2,$ $3,$ $4,$ and $5$ of a triangular array of integers are shown below.

$\begin{array}{ccccccccc} &&&&1&&&&\\ &&&1&&1&&&\\ &&1&&3&&1&&\\ &1&&5&&5&&1&\\ 1&&7&&11&&7&&1 \end{array}$

Each row after the first row is formed by placing a $1$ at each end of the row, and each interior entry is $1$ greater than the sum of the two numbers diagonally above it in the previous row. What is the units digit of the sum of the $2023$ numbers in the $2023$ rd row?

$1$

$3$

$5$

$7$

$9$

Answer: C

Concepts:recursion units digit

Difficulty rating: 2110

Small Hint:

Let $S_n$ be the sum of row $n;$ each interior entry adds two copies of the row above plus a $1$

Big Hint:

This yields $S_n=2S_{n-1}+(n-2),$ whose closed form is $S_n=2^n-n$

Solution:

Let $S_n$ be the sum of row $n.$ Each interior entry is $1$ more than the sum of the two entries above it, and summing over the row gives the recurrence $S_n=2S_{n-1}+(n-2).$

With $S_1=1,$ this solves to $S_n=2^n-n$ (check: $S_5=32-5=27$ $=1+7+11+7+1$ ).

So $S_{2023}=2^{2023}-2023.$ Since powers of $2$ cycle with units digits $2,4,8,6$ and $2023\equiv 3\pmod 4,$ $2^{2023}$ ends in $8.$ Then $8-3=5$ gives units digit $5.$

Thus, the correct answer is C.

21.

If $A$ and $B$ are vertices of a polyhedron, define the distance $d(A,B)$ to be the minimum number of edges of the polyhedron one must traverse in order to connect $A$ and $B.$ For example, if $\overline{AB}$ is an edge of the polyhedron, then $d(A,B)=1,$ but if $\overline{AC}$ and $\overline{CB}$ are edges and $\overline{AB}$ is not an edge, then $d(A,B)=2.$ Let $Q,$ $R,$ and $S$ be randomly chosen distinct vertices of a regular icosahedron (regular polyhedron made up of $20$ equilateral triangles). What is the probability that $d(Q,R)\gt d(R,S)?$

$\dfrac{7}{22}$

$\dfrac{1}{3}$

$\dfrac{3}{8}$

$\dfrac{5}{12}$

$\dfrac{1}{2}$

Answer: A

Concepts:graph theory basic probability symmetry

Difficulty rating: 2170

Small Hint:

Fix $R.$ The other $11$ vertices split into $5$ at distance $1,$ $5$ at distance $2,$ and $1$ at distance $3$

Big Hint:

By symmetry $P(d(Q,R)\gt d(R,S))$ $=\dfrac{1-P(\text{equal})}{2}$

Solution:

Fix $R.$ Among the other $11$ vertices of the icosahedron, $5$ are at distance $1,$ $5$ are at distance $2,$ and $1$ (the antipode) is at distance $3.$

Choosing ordered distinct $Q,S,$ the probability that $d(Q,R)=d(R,S)$ is $\dfrac{5\cdot 4+5\cdot 4}{11\cdot 10}=\dfrac{40}{110}=\dfrac{4}{11}.$

By the symmetry between $Q$ and $S,$ $\begin{gathered} P(d(Q,R)\gt d(R,S))\\ {}=\dfrac{1-\tfrac{4}{11}}{2}\\ {}=\dfrac{7}{22}. \end{gathered}$

Thus, the correct answer is A.

22.

Let $f$ be the unique function defined on the positive integers such that $\sum_{d\mid n} d\cdot f\left(\frac{n}{d}\right)=1$ for all positive integers $n,$ where the sum is taken over all positive divisors of $n.$ What is $f(2023)?$

$-1536$

$96$

$108$

$116$

$144$

Answer: B

Concepts:prime factorization recursion

Difficulty rating: 2270

Small Hint:

The relation determines $f$ recursively; $f$ turns out to be multiplicative

Big Hint:

Compute $f(1)=1,$ $f(p)=1-p,$ and $f(p^2)=1-p;$ here $2023=7\cdot 17^2$

Solution:

Setting $n=1$ gives $f(1)=1.$ For a prime $p,$ $n=p$ gives $f(p)+p\cdot f(1)=1,$ so $f(p)=1-p.$ For $n=p^2,$ $f(p^2)+p\,f(p)+p^2 f(1)=1$ gives $f(p^2)=1-p.$

Since the defining relation is a Dirichlet convolution of multiplicative functions, $f$ is multiplicative. With $2023=7\cdot 17^2,$ $\begin{gathered} f(2023)=f(7)\cdot f(17^2)\\ {}=(1-7)(1-17)\\ {}=(-6)(-16)\\ {}=96. \end{gathered}$

Thus, the correct answer is B.

23.

How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation $\begin{gathered} (1+2a)(2+2b)(2a+b)\\ {}=32ab? \end{gathered}$

$0$

$1$

$2$

$3$

an infinite number

Answer: B

Concepts:AM-GM Inequality

Difficulty rating: 2380

Small Hint:

Apply AM-GM separately to $1+2a,$ $2+2b,$ and $2a+b$

Big Hint:

The product is at least $32ab,$ so equality must hold in all three inequalities at once

Solution:

By AM-GM, $1+2a\ge 2\sqrt{2a},$ $2+2b\ge 4\sqrt{b},$ and $2a+b\ge 2\sqrt{2ab}.$ Multiplying, $\begin{gathered} (1+2a)(2+2b)(2a+b)\\ {}\ge 16\sqrt{2a}\cdot\sqrt{b}\cdot\sqrt{2ab}\\ {}=32ab. \end{gathered}$

Equality requires $1=2a,$ $2=2b,$ and $2a=b$ simultaneously. These give $a=\tfrac12,$ $b=1,$ which are consistent, so there is exactly one solution.

Thus, the correct answer is B.

24.

Let $K$ be the number of sequences $A_1,$ $A_2,$ $\ldots,$ $A_n$ such that $n$ is a positive integer less than or equal to $10,$ each $A_i$ is a subset of $\{1,2,3,\ldots,10\},$ and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n,$ inclusive. For example, $\{\},$ $\{5,7\},$ $\{2,5,7\},$ $\{2,5,7\},$ $\{2,5,6,7,9\}$ is one such sequence, with $n=5.$ What is the remainder when $K$ is divided by $10?$

$1$

$3$

$5$

$7$

$9$

Answer: C

Concepts:subsets multiplication principle modular arithmetic

Difficulty rating: 2520

Small Hint:

For a fixed $n,$ each of the $10$ elements independently chooses the first set it enters, or never enters

Big Hint:

That gives $(n+1)^{10}$ chains, so $K=\sum_{k=2}^{11}k^{10};$ reduce each term modulo $10$

Solution:

For a fixed length $n,$ each element of $\{1,\ldots,10\}$ independently either never appears or first appears in one of $A_1,\ldots,A_n,$ giving $n+1$ choices. Hence there are $(n+1)^{10}$ chains of length $n.$

Summing, $K=\sum_{n=1}^{10}(n+1)^{10}=\sum_{k=2}^{11}k^{10}.$ Modulo $10,$ the terms $k=2,\ldots,11$ reduce to $4,9,6,5,6,9,4,1,0,1,$ which sum to $45\equiv 5.$

Thus, the correct answer is C.

25.

There is a unique sequence of integers $a_1,$ $a_2,$ $\cdots a_{2023}$ such that $\begin{gathered} \tan 2023x\\ {}=\tiny\dfrac{a_1\tan x+a_3\tan^3 x+a_5\tan^5 x+\cdots+a_{2023}\tan^{2023}x}{1+a_2\tan^2 x+a_4\tan^4 x\cdots+a_{2022}\tan^{2022}x} \end{gathered}$ whenever $\tan 2023x$ is defined. What is $a_{2023}?$

$-2023$

$-2022$

$-1$

$1$

$2023$

Answer: C

Concepts:De Moivre’s Theorem binomial theorem

Difficulty rating: 2650

Small Hint:

Expand $(\cos x+i\sin x)^{2023}$ and use $\tan 2023x=\dfrac{\operatorname{Im}}{\operatorname{Re}}$

Big Hint:

Dividing by $\cos^{2023}x,$ the top coefficient of $\tan^{2023}x$ is $(-1)^{\frac{2023-1}{2}}\binom{2023}{2023}$

Solution:

By De Moivre, $(\cos x+i\sin x)^{2023}$ $=\cos 2023x+i\sin 2023x.$ Expanding the left side and taking the ratio of imaginary to real parts gives $\tan 2023x$ as the stated rational function of $\tan x$ after dividing numerator and denominator by $\cos^{2023}x.$

The coefficient $a_{2023}$ is the coefficient of $\tan^{2023}x$ in the numerator, which comes from the $k=2023$ term: $\begin{gathered} a_{2023}=(-1)^{\frac{2023-1}{2}}\binom{2023}{2023}\\ {}=(-1)^{1011}\\ {}=-1. \end{gathered}$

Thus, the correct answer is C.