2025 AMC 12B — Problems & Online Timed Exam

Timed

1:15:00

1.

The instructions on a $350$ -gram bag of coffee beans say that proper brewing of a large mug of pour-over coffee requires $20$ grams of coffee beans. What is the greatest number of properly brewed large mugs of coffee that can be made from the coffee beans in that bag?

$16$

$17$

$18$

$19$

$20$

Answer: B

Concepts:floor and ceiling functions

Difficulty rating: 890

Small Hint:

Divide the total grams by the grams needed per mug

Big Hint:

Only whole mugs count, so round $\dfrac{350}{20}$ down

Solution:

Each mug uses $20$ grams, and $\dfrac{350}{20} = 17.5.$ Only complete mugs can be brewed, so the greatest number is $17.$

Thus, the correct answer is B.

2.

Jerry wrote down the ones digit of each of the first $2025$ positive squares: $1,$ $4,$ $9,$ $6,$ $5,$ $6,$ $\ldots.$ What is the sum of all the numbers Jerry wrote down?

$9025$

$9070$

$9090$

$9115$

$9160$

Answer: D

Concepts:units digit pattern recognition summation

Difficulty rating: 1020

Small Hint:

The ones digit of $n^2$ depends only on the ones digit of $n,$ so the list repeats every $10$ terms

Big Hint:

Sum one full block of $10,$ multiply by the number of blocks, then add the leftover terms

Solution:

The ones digits of $1^2, 2^2, \ldots, 10^2$ are $1, 4, 9, 6, 5, 6, 9, 4, 1, 0,$ which sum to $45.$ The $2025$ terms contain $202$ full blocks ( $2020$ terms) plus $5$ more with digits $1, 4, 9, 6, 5$ summing to $25.$ The total is $202 \cdot 45 + 25 = 9090 + 25$ $= 9115.$

Thus, the correct answer is D.

3.

What is the value of $i(i-1)(i-2)(i-3),$ where $i = \sqrt{-1}?$

$6 - 5i$

$-10i$

$10i$

$-10$

$10$

Answer: D

Concepts:complex number

Difficulty rating: 1130

Small Hint:

Multiply in pairs: $i(i-1)$ and $(i-2)(i-3)$

Big Hint:

Use $i^2 = -1$ to reduce each product before multiplying the two results

Solution:

$i(i-1) = i^2 - i = -1 - i,$ and $(i-2)(i-3) = i^2 - 5i + 6$ $= 5 - 5i.$ Then $\begin{gathered} (-1-i)(5-5i) = -5 \\ {}+ 5i - 5i \\ {}+ 5i^2 \\ = -5 - 5 = -10. \end{gathered}$

Thus, the correct answer is D.

4.

The value of the two-digit number $\underline{a}\,\underline{b}$ in base seven equals the value of the two-digit number $\underline{b}\,\underline{a}$ in base nine. What is $a + b?$

$7$

$9$

$10$

$11$

$14$

Answer: A

Concepts:number base linear equation

Difficulty rating: 1200

Small Hint:

Write each number in terms of its base: $\underline{a}\,\underline{b}$ is $7a + b$ and $\underline{b}\,\underline{a}$ is $9b + a$

Big Hint:

Set $7a + b = 9b + a$ and simplify to a ratio between $a$ and $b$

Solution:

Setting $7a + b = 9b + a$ gives $6a = 8b,$ so $3a = 4b.$ Thus $a=4t$ and $b=3t.$ Because both are base-seven digits and $a\ne0,$ only $t=1$ works. Hence $a=4, b=3,$ and indeed $43_7 = 31 = 34_9.$ Therefore $a+b=7.$

Thus, the correct answer is A.

5.

Positive integers $x$ and $y$ satisfy the equation $57x + 22y = 400.$ What is the least possible value of $x + y?$

$10$

$11$

$13$

$14$

$15$

Answer: E

Concepts:Diophantine Equation modular arithmetic

Difficulty rating: 1290

Small Hint:

Reduce $57x + 22y = 400$ modulo $22$ to pin down $x$

Big Hint:

Since $x$ is positive and $57x \lt 400,$ only a few values of $x$ are possible

Solution:

Modulo $22,$ the equation gives $13x \equiv 4,$ so $x \equiv 2 \pmod{22}.$ With $57x \lt 400,$ the only option is $x = 2,$ which gives $22y = 286,$ so $y = 13.$ Then $x + y = 15.$

Thus, the correct answer is E.

6.

Emmy says to Max, “I ordered $36$ math club sweatshirts today.” Max asks, “How much did each shirt cost?” Emmy responds, “I’ll give you a hint. The total cost was $\$\underline{A}\,\underline{B}\,\underline{B}.\underline{B}\,\underline{A},$ where $A$ and $B$ are digits and $A \neq 0.$ ” After a pause, Max says, “That was a good price.” What is $A + B?$

$7$

$8$

$11$

$14$

$15$

Answer: C

Concepts:place value divisibility modular arithmetic

Difficulty rating: 1390

Small Hint:

In cents the total is $10001A + 1110B,$ and it must be divisible by $36$

Big Hint:

Reduce modulo $36$ to get $7A + 6B \equiv 0,$ then test digits with $A \neq 0$

Solution:

The total in cents is $10000A + 1110B + A$ $= 10001A + 1110B,$ which must be a multiple of $36.$ Since $10001 \equiv 29$ and $1110 \equiv 30 \pmod{36},$ the condition is $29A + 30B \equiv 0,$ i.e. $7A + 6B \equiv 0 \pmod{36}.$ Reducing once more modulo $6$ gives $A\equiv0\pmod6,$ so the only possible nonzero digit is $A=6.$ Then $42+6B$ is divisible by $36$ only for the digit $B=5.$ Indeed $\$655.56 = 36 \times \$18.21,$ so $A+B=11.$

Thus, the correct answer is C.

7.

What is the value of

$\sum_{n=2}^{255} \frac{\log_2\left(1 + \frac{1}{n}\right)}{(\log_2 n)(\log_2(n+1))}?$

$\dfrac{3}{4}$

$1 - \dfrac{1}{\log_2 255}$

$\dfrac{7}{8}$

$\dfrac{15}{16}$

$1$

Answer: C

Concepts:telescoping logarithm

Difficulty rating: 1420

Small Hint:

Write $\log_2\!\left(1 + \frac{1}{n}\right)$ $= \log_2(n+1) - \log_2 n$

Big Hint:

With $a_n = \log_2 n,$ each term becomes $\dfrac{1}{a_n} - \dfrac{1}{a_{n+1}}$

Solution:

Let $a_n = \log_2 n.$ The numerator equals $a_{n+1} - a_n,$ so each term is $\dfrac{a_{n+1} - a_n}{a_n a_{n+1}} = \dfrac{1}{a_n} - \dfrac{1}{a_{n+1}}.$ Telescoping from $n = 2$ to $255$ leaves $\dfrac{1}{\log_2 2} - \dfrac{1}{\log_2 256} = 1 - \dfrac{1}{8}$ $= \dfrac{7}{8}.$

Thus, the correct answer is C.

8.

There are integers $a$ and $b$ such that the polynomial $x^3 - 5x^2 + ax + b$ has $4 + \sqrt{5}$ as a root. What is $a + b?$

$13$

$17$

$20$

$30$

$68$

Answer: C

Concepts:Vieta’s Formulas polynomial radical

Difficulty rating: 1440

Small Hint:

Integer coefficients force the conjugate $4 - \sqrt{5}$ to be a root as well

Big Hint:

The three roots sum to $5,$ so find the third root, then expand to read off $a$ and $b$

Solution:

The conjugate $4 - \sqrt{5}$ is also a root, and these two are the roots of $x^2 - 8x + 11.$ The third root $r$ satisfies $8 + r = 5,$ so $r = -3.$ Then $(x^2 - 8x + 11)(x + 3)$ $= x^3 - 5x^2 - 13x + 33,$ giving $a = -13$ and $b = 33,$ so $a + b = 20.$

Thus, the correct answer is C.

9.

What is the tens digit of $6^{6^6}?$

$1$

$3$

$5$

$7$

$9$

Answer: C

Concepts:modular exponentiation digits

Difficulty rating: 1510

Small Hint:

The tens digit is determined by $6^{6^6} \bmod 100$

Big Hint:

For $n \ge 2,$ $6^n \bmod 100$ cycles with period $5;$ reduce the exponent $6^6$ modulo $5$

Solution:

Here $6^6 = 46656.$ For $n \ge 2,$ the last two digits of $6^n$ cycle with period $5$ through $36, 16, 96, 76, 56.$ Since $46656 \equiv 1 \pmod 5,$ $6^{46656}$ ends in $56,$ so the tens digit is $5.$

Thus, the correct answer is C.

10.

The altitude to the hypotenuse of a $30\text{-}60\text{-}90$ right triangle is divided into two segments of lengths $x \lt y$ by the median to the shortest side of the triangle. What is the ratio $\dfrac{x}{x+y}?$

$\dfrac{3}{7}$

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

$\dfrac{4}{9}$

$\dfrac{5}{11}$

$\dfrac{4\sqrt{3}}{15}$

Answer: A

Concepts:special right triangle coordinate geometry median (geometry)

Difficulty rating: 1580

Small Hint:

Place the right angle at the origin with the legs along the axes and coordinatize every point

Big Hint:

Find where the median to the shortest side crosses the altitude to the hypotenuse, then split the altitude at that point

Solution:

Take $C = (0,0),$ $A = (\sqrt{3}, 0),$ $B = (0, 1),$ so $AB$ is the hypotenuse and $BC$ is the shortest side. The altitude from $C$ meets $AB$ at $H = \left(\tfrac{\sqrt{3}}{4}, \tfrac{3}{4}\right).$ The median from $A$ to $M = \left(0, \tfrac{1}{2}\right)$ crosses the altitude at $P = \left(\tfrac{\sqrt{3}}{7}, \tfrac{3}{7}\right).$ This splits the altitude into $CP = \tfrac{2\sqrt{3}}{7} = \tfrac{4\sqrt{3}}{14}$ and $PH = \tfrac{3\sqrt{3}}{14},$ so $x = \tfrac{3\sqrt{3}}{14}$ and $\dfrac{x}{x+y} = \dfrac{3}{7}.$

Thus, the correct answer is A.

11.

Nine athletes, no two of whom are the same height, try out for the basketball team. One at a time, they draw a wristband at random, without replacement, from a bag containing $3$ blue bands, $3$ red bands, and $3$ green bands. They are divided into a blue group, a red group, and a green group. The tallest member of each group is named the group captain. What is the probability that the group captains are the three tallest athletes?

$\dfrac{2}{9}$

$\dfrac{2}{7}$

$\dfrac{9}{28}$

$\dfrac{1}{3}$

$\dfrac{3}{8}$

Answer: C

Concepts:basic probability symmetry

Difficulty rating: 1590

Small Hint:

The captains are the three tallest exactly when the three tallest land in three different groups

Big Hint:

Place the tallest three one at a time; each must avoid the groups already used, whose slots shrink the pool

Solution:

Each group has $3$ slots. The three tallest athletes are the captains precisely when they fall into three different groups. Placing them one at a time into the $9$ slots, the second must avoid the first’s group ( $6$ of the remaining $8$ slots) and the third must avoid both used groups ( $3$ of the remaining $7$ slots). The probability is $\dfrac{6}{8} \cdot \dfrac{3}{7} = \dfrac{9}{28}.$

Thus, the correct answer is C.

12.

The windshield wiper on the driver’s side of a large bus is depicted below.

Arm $\overline{AB}$ pivots back and forth around point $A,$ sweeping out an arc of $60^\circ,$ symmetric about the vertical line through $A.$ The wiper blade $\overline{CD}$ is attached to $B$ at its midpoint and stays vertical as the arm moves. The arm is $3$ feet long, and the wiper blade is $3.5$ feet tall. What is the area of the windshield cleaned by the wiper, in square feet, to the nearest hundredth? (Assume that the windshield is a flat vertical surface.)

$9.68$

$10.14$

$10.50$

$11.32$

$12.00$

Answer: C

Concepts:area arc

Difficulty rating: 1690

Small Hint:

As the arm turns, $B$ traces an arc of radius $3;$ the blade is always a vertical segment of height $3.5$ centered at $B$

Big Hint:

Each vertical line meets the cleaned region in a segment of constant height $3.5,$ so the area is that height times the horizontal width swept

Solution:

Put $A$ at the origin. Then $B = (3\sin\theta, 3\cos\theta)$ for $\theta \in [-30^\circ, 30^\circ],$ so the horizontal coordinate of $B$ ranges over $[-1.5, 1.5],$ a width of $3.$ At each horizontal position exactly one vertical blade of height $3.5$ passes through, so by Cavalieri’s principle the cleaned area is $3.5 \times 3 = 10.5$ square feet.

Thus, the correct answer is C.

13.

A circle has been divided into $6$ sectors of different sizes. Then $2$ of the sectors are painted red, $2$ painted green, and $2$ painted blue so that no two neighboring sectors are painted the same color. One such coloring is shown below.

How many different colorings are possible?

$12$

$16$

$18$

$24$

$28$

Answer: D

Concepts:graph theory multiplication principle

Difficulty rating: 1660

Small Hint:

Because same-colored sectors cannot touch, the two sectors of each color form a non-adjacent pair

Big Hint:

Count the ways to split the $6$ sectors into three non-adjacent pairs, then assign the three colors to the pairs

Solution:

The two sectors of each color must be a non-adjacent pair, so a coloring is a way to split the $6$ cyclic sectors into three non-adjacent pairs together with an assignment of the three colors. The non-adjacent pairs are the edges of the complement of the $6$ -cycle, the triangular prism, which has $4$ perfect matchings. Assigning the three colors in $3!$ ways gives $4 \times 6 = 24$ colorings.

Thus, the correct answer is D.

14.

Consider a decreasing sequence of $n$ positive integers $x_1 \gt x_2 \gt \cdots \gt x_n$ that satisfies the following two conditions:

• The average (arithmetic mean) of the first $3$ terms in the sequence is $2025.$

• For all $4 \le k \le n,$ the average of the first $k$ terms in the sequence is $1$ less than the average of the first $k-1$ terms in the sequence.

What is the greatest possible value of $n?$

$1013$

$1014$

$1016$

$2016$

$2025$

Answer: B

Concepts:mean arithmetic sequence bounding to limit cases

Difficulty rating: 1730

Small Hint:

Let $A_k$ be the average of the first $k$ terms; the second condition gives $A_k = 2028 - k$ for $k \ge 3$

Big Hint:

From $S_k = k A_k$ recover $x_k = 2029 - 2k$ for $k \ge 4,$ then require every term to stay positive

Solution:

The average of the first $k$ terms is $A_k = 2028 - k$ for $k \ge 3,$ so the partial sum is $S_k = k(2028 - k).$ For $k \ge 4,$ $x_k = S_k - S_{k-1} = 2029 - 2k,$ which is positive exactly when $k \le 1014.$ A valid start such as $x_1, x_2, x_3 = 2030, 2023, 2022$ keeps the whole sequence strictly decreasing, so the greatest possible $n$ is $1014.$

Thus, the correct answer is B.

15.

A container has a $1 \times 1$ square bottom, a $3 \times 3$ open square top, and four congruent trapezoidal sides, as shown. Starting when the container is empty, a hose that runs water at a constant rate takes $35$ minutes to fill the container up to the midline of the trapezoids.

How many more minutes will it take to fill the remainder of the container?

$70$

$85$

$90$

$95$

$105$

Answer: D

Concepts:volume similarity ratio and proportion

Difficulty rating: 1800

Small Hint:

At height fraction $t$ the cross-section is a square of side $1 + 2t,$ so its area is $(1 + 2t)^2$

Big Hint:

Compare the volume up to the midline ( $t = \tfrac{1}{2}$ ) with the total volume; fill time is proportional to volume

Solution:

At height fraction $t$ the square cross-section has side $1 + 2t,$ so the volume filled up to height $t$ is $\int_0^t (1 + 2u)^2\,du.$ Up to the midline $\left(t = \tfrac{1}{2}\right)$ this is $\tfrac{7}{6},$ and the full volume is $\tfrac{13}{3}.$ The remaining volume is $\tfrac{13}{3} - \tfrac{7}{6} = \tfrac{19}{6},$ which is $\tfrac{19}{7}$ times the first part. So the remainder takes $35 \cdot \tfrac{19}{7} = 95$ more minutes.

Thus, the correct answer is D.

16.

An analog clock starts at midnight and runs for $2025$ minutes before stopping. What is the tangent of the acute angle between the hour hand and the minute hand when the clock stops?

$0$

$\sqrt{2} - 1$

$2 - \sqrt{2}$

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

$3 - \sqrt{2}$

Answer: B

Concepts:clock trigonometric identity

Difficulty rating: 1840

Small Hint:

Reduce $2025$ minutes modulo $12$ hours to find where the hands point

Big Hint:

Find each hand’s angle, take the acute difference, then evaluate $\tan 22.5^\circ$

Solution:

$2025$ minutes is $33$ hours $45$ minutes, which modulo $12$ hours reads $9{:}45.$ The minute hand points at $270^\circ$ and the hour hand at $9.75 \times 30^\circ = 292.5^\circ,$ so the acute angle between them is $22.5^\circ.$ Using the half-angle value, $\tan 22.5^\circ = \sqrt{2} - 1.$

Thus, the correct answer is B.

17.

Each of the $9$ squares in a $3 \times 3$ grid is to be colored red, blue, or yellow in such a way that each red square shares an edge with at least one blue square, each blue square shares an edge with at least one yellow square, and each yellow square shares an edge with at least one red square. Colorings that can be obtained from one another by rotations and/or reflections are to be considered the same. How many different colorings are possible?

$3$

$9$

$12$

$18$

$27$

Answer: C

Concepts:Burnside’s Lemma casework

Difficulty rating: 1980

Small Hint:

Every color must appear, and the cyclic conditions red $\to$ blue $\to$ yellow $\to$ red tightly constrain the layout

Big Hint:

Count all valid colorings of the fixed (labeled) grid, then reduce by the $8$ rotations and reflections using Burnside’s lemma

Solution:

First count colorings of the grid with its positions labeled. Checking the $27$ possible rows in succession and rejecting a row as soon as a square whose neighbors are now known lacks its required next color gives the following complete count by the numbers of red, blue, and yellow squares: $\begin{array}{c|c} (\#R,\#B,\#Y)&\text{valid colorings}\\ \hline (2,4,3)&16\\ (3,2,4)&16\\ (3,3,3)&36\\ (4,3,2)&16 \end{array}$ Thus the identity symmetry fixes $16+16+36+16=84$ colorings.

For the other symmetries, the fixed-coloring counts are $0$ for each nontrivial rotation, $6$ for each reflection across a horizontal or vertical axis, and $0$ for each diagonal reflection. (For an axis reflection, a direct check of the three palindromic rows gives the $6$ possibilities.) Therefore Burnside’s lemma gives $\frac{84+2\cdot6}{8}=12.$

Thus, the correct answer is C.

18.

Awnik repeatedly plays a game that has a probability of winning of $\dfrac{1}{3}.$ The outcomes of the games are independent. What is the expected value of the number of games he will play until he has both won and lost at least once?

$\dfrac{5}{2}$

$3$

$\dfrac{16}{5}$

$\dfrac{7}{2}$

$\dfrac{15}{4}$

Answer: D

Concepts:expected value geometric distribution

Difficulty rating: 1770

Small Hint:

After the first game you already have one outcome; you then wait only for the opposite outcome

Big Hint:

The expected number of games to get an outcome of probability $p$ is $\dfrac{1}{p};$ condition on whether the first game is a win or a loss

Solution:

The first game produces one outcome. If it was a win (probability $\tfrac{1}{3}$ ), the expected wait for a loss is $\tfrac{1}{\frac{2}{3}} = \tfrac{3}{2};$ if it was a loss (probability $\tfrac{2}{3}$ ), the expected wait for a win is $\tfrac{1}{\frac{1}{3}} = 3.$ So the expected total is $1 + \tfrac{1}{3}\cdot\tfrac{3}{2} + \tfrac{2}{3}\cdot 3 = 1 + \tfrac{1}{2} + 2$ $= \tfrac{7}{2}.$

Thus, the correct answer is D.

19.

A rectangular grid of squares has $141$ rows and $91$ columns. Each square has room for two numbers. Horace and Vera each fill in the grid by putting the numbers from $1$ through $141 \times 91 = 12{,}831$ into the squares. Horace fills the grid horizontally: he puts $1$ through $91$ in order from left to right into row $1,$ puts $92$ through $182$ into row $2$ in order from left to right, and continues similarly through row $141.$ Vera fills the grid vertically: she puts $1$ through $141$ in order from top to bottom into column $1,$ then $142$ through $282$ into column $2$ in order from top to bottom, and continues similarly through column $91.$ How many squares get two copies of the same number?

$7$

$10$

$11$

$12$

$19$

Answer: C

Concepts:Diophantine Equation modular arithmetic counting integers in a range

Difficulty rating: 2040

Small Hint:

Write the Horace-number and the Vera-number of the square in row $r,$ column $c$ as formulas in $r$ and $c$

Big Hint:

Setting them equal gives $9r = 14c - 5;$ find which $c$ make $r$ an integer in range

Solution:

In row $r,$ column $c,$ Horace writes $(r-1)\cdot 91 + c$ and Vera writes $(c-1)\cdot 141 + r.$ Setting these equal gives $90r - 140c = -50,$ i.e. $9r = 14c - 5.$ This requires $c \equiv 1 \pmod 9,$ so $c = 1, 10, 19, \ldots, 91$ — that is $11$ values, and each yields a valid $r$ between $1$ and $141.$ So $11$ squares match.

Thus, the correct answer is C.

20.

A frog hops along the number line according to the following rules.

• It starts at $0.$

• If it is at $0,$ then it moves to $1$ with probability $\dfrac{1}{2}$ and it disappears with probability $\dfrac{1}{2}.$

• For $n = 1,$ $2,$ or $3,$ if it is at $n,$ then it moves to $n+1$ with probability $\dfrac{1}{4},$ it moves to $n-1$ with probability $\dfrac{1}{4},$ and it disappears with probability $\dfrac{1}{2}.$

What is the probability that the frog reaches $4?$

$\dfrac{1}{101}$

$\dfrac{1}{100}$

$\dfrac{1}{99}$

$\dfrac{1}{98}$

$\dfrac{1}{97}$

Answer: E

Concepts:random walk recursive probability system of equations

Difficulty rating: 2110

Small Hint:

Let $f(n)$ be the probability of eventually reaching $4$ starting from $n;$ write a first-step equation at each state

Big Hint:

Solve the linear system with $f(4) = 1$ and value $0$ whenever the frog disappears

Solution:

Let $f(n)$ be the probability of reaching $4$ from $n.$ Then $f(0) = \tfrac{1}{2} f(1),$ and for $n = 1, 2, 3,$ $f(n) = \tfrac{1}{4} f(n+1) + \tfrac{1}{4} f(n-1),$ with $f(4) = 1.$ Solving upward gives $f(2) = \tfrac{7}{2} f(1)$ and $f(3) = 13 f(1);$ then $4 \cdot 13 f(1) = 1 + \tfrac{7}{2} f(1)$ yields $f(1) = \tfrac{2}{97},$ so $f(0) = \tfrac{1}{97}.$

Thus, the correct answer is E.

21.

Two non-congruent triangles have the same area. Each triangle has sides of length $8$ and $9,$ and the third side of each triangle has integer length. What is the sum of the lengths of the third sides?

$20$

$22$

$24$

$26$

$28$

Answer: C

Concepts:law of cosines triangle area triangle inequality

Difficulty rating: 2170

Small Hint:

With sides $8$ and $9$ and included angle $\theta,$ the area is $36\sin\theta,$ so equal areas come from $\theta$ and $180^\circ - \theta$

Big Hint:

By the law of cosines the third sides satisfy $t^2 = 145 \mp 144\cos\theta,$ so $t_1^2 + t_2^2 = 290$

Solution:

The area with included angle $\theta$ is $36\sin\theta,$ so two triangles of equal area use angles $\theta$ and $180^\circ - \theta,$ with cosines $\pm\cos\theta.$ By the law of cosines the third sides satisfy $t^2 = 145 \mp 144\cos\theta,$ hence $t_1^2 + t_2^2 = 290.$ Since $290\equiv2\pmod8,$ both integer sides must be odd. Checking the odd squares in the triangle-inequality range $1\lt t\lt17$ leaves only $11^2+13^2=121+169=290.$ Therefore the sum is $24.$

Thus, the correct answer is C.

22.

What is the greatest possible area of the triangle in the complex plane with vertices $2z,$ $(1+i)z,$ and $(1-i)z,$ where $z$ is a complex number satisfying $|4z - 2| = 1?$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

$\dfrac{9}{16}$

$\dfrac{3}{4}$

$1$

Answer: C

Concepts:complex number triangle area optimization

Difficulty rating: 2270

Small Hint:

The vertices are $z$ times the fixed points $2, 1+i, 1-i,$ so the triangle is that fixed triangle scaled by $|z|$

Big Hint:

Its area is $|z|^2$ times a fixed area; the condition $|4z - 2| = 1$ is a circle, so maximize $|z|$ on it

Solution:

The vertices are $z \cdot 2,$ $z(1+i),$ and $z(1-i),$ so the triangle is the fixed triangle with vertices $2, 1+i, 1-i$ — which has area $1$ — scaled by $|z|,$ giving area $|z|^2.$ The condition $|4z - 2| = 1$ is the circle $\left|z - \tfrac{1}{2}\right| = \tfrac{1}{4},$ on which $|z|$ is at most $\tfrac{1}{2} + \tfrac{1}{4} = \tfrac{3}{4}.$ So the greatest area is $\left(\tfrac{3}{4}\right)^2 = \tfrac{9}{16}.$

Thus, the correct answer is C.

23.

Let $S$ be the set of all integers $z \gt 1$ such that for all pairs of nonnegative integers $(x, y)$ with $x \lt y \lt z,$ the remainder when $2025x$ is divided by $z$ is less than the remainder when $2025y$ is divided by $z.$ What is the sum of the elements of $S?$

$3041$

$3542$

$3750$

$4044$

$4319$

Answer: E

Concepts:modular arithmetic divisibility sum of factors

Difficulty rating: 2380

Small Hint:

The condition says the map $k \mapsto 2025k \bmod z$ is strictly increasing on $0, 1, \ldots, z-1$

Big Hint:

An increasing rearrangement of $0, 1, \ldots, z-1$ starting at $0$ must be the identity, forcing $2025 \equiv 1 \pmod z$

Solution:

The condition requires $k \mapsto 2025k \bmod z$ to be strictly increasing on $\{0, 1, \ldots, z-1\}.$ A strictly increasing list of $z$ distinct values in $[0, z-1]$ must be $0, 1, \ldots, z-1,$ so $2025 \equiv 1 \pmod z,$ i.e. $2024$ is divisible by $z.$ Since $2024 = 2^3 \cdot 11 \cdot 23,$ the sum of all its divisors is $(1+2+4+8)(1+11)(1+23)$ $= 15 \cdot 12 \cdot 24$ $= 4320.$ Excluding $z = 1$ leaves $4319.$

Thus, the correct answer is E.

24.

How many real numbers satisfy the equation $\sin(20\pi x) = \log_{20}(x)?$

$199$

$200$

$398$

$399$

$400$

Answer: D

Concepts:trigonometry logarithm counting intersections

Difficulty rating: 2520

Small Hint:

A solution needs $-1 \le \log_{20} x \le 1,$ so $x$ lies in $\left[\tfrac{1}{20}, 20\right]$

Big Hint:

On each monotonic half-wave of $\sin(20\pi x)$ the slowly rising log crosses once; count the branches carefully at both ends

Solution:

Since $|\sin(20\pi x)|\le1,$ every solution lies in $\left[\tfrac1{20},20\right].$ For $x\lt1$ the logarithm is negative, so only the $10$ negative sine lobes in $\left[\tfrac1{20},1\right]$ can contribute. Each has one crossing on each side of its minimum; on the last lobe, the second crossing is the endpoint $x=1.$ Hence this part contributes $20$ solutions.

For $x\gt1,$ only positive lobes contribute. There are $190$ of them: one in each interval from $1+\tfrac{k}{10}$ to $1+\tfrac{k}{10}+\tfrac1{20}$ for $0\le k\le189.$ Every one after the first has one crossing on each side of its maximum. The first has only one new crossing because its left endpoint is the already-counted solution $x=1.$ Thus $x\gt1$ contributes $1+2\cdot189=379$ more solutions.

For completeness, the claimed uniqueness on each half-lobe follows by differentiating the difference of the two sides. On a falling positive half-lobe it is strictly decreasing. On a rising positive half-lobe its derivative increases and then decreases once, so the difference crosses from negative to positive only once. Applying the same argument to $-\sin(20\pi x)+\log_{20}x$ handles a negative lobe. The total is $20+379=399.$

Thus, the correct answer is D.

25.

Three concentric circles have radii $1,$ $2,$ $3.$ An equilateral triangle with side length $s$ has one vertex on each circle. What is $s^2?$

$6$

$\dfrac{25}{4}$

$\dfrac{13}{2}$

$\dfrac{27}{4}$

$7$

Answer: E

Concepts:equilateral triangle transformation circle

Difficulty rating: 2650

Small Hint:

For a point at distances $p, q, r$ from the vertices of an equilateral triangle of side $a,$ there is a symmetric relation among $p^2, q^2, r^2, a^2$

Big Hint:

Use $3(p^4 + q^4 + r^4 + a^4)$ $= (p^2 + q^2 + r^2 + a^2)^2$ with $p, q, r = 1, 2, 3$

Solution:

For the common center at distances $1, 2, 3$ from the vertices of an equilateral triangle of side $s,$ the identity $3(1 + 16 + 81 + s^4)$ $= (1 + 4 + 9 + s^2)^2$ holds. This simplifies to $2s^4 - 28s^2 + 98 = 0,$ i.e. $(s^2 - 7)^2 = 0,$ so $s^2 = 7.$

Thus, the correct answer is E.