2025 AMC 12B 考试答案

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?

$16$

$17$

$18$

$19$

$20$

Concepts:floor and ceiling functions

Difficulty rating: 890

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?

$9025$

$9070$

$9090$

$9115$

$9160$

Concepts:units digit pattern recognition summation

Difficulty rating: 1020

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 $i = \sqrt{-1}?$

$6 - 5i$

$-10i$

$10i$

$-10$

$10$

Concepts:complex number

Difficulty rating: 1130

Solution:

$i(i-1) = i^2 - i = -1 - i,$ and $(i-2)(i-3) = i^2 - 5i + 6 = 5 - 5i.$ Then $(-1-i)(5-5i) = -5 + 5i - 5i + 5i^2 = -5 - 5 = -10.$

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 $a + b?$

$7$

$9$

$10$

$11$

$14$

Concepts:number base linear equation

Difficulty rating: 1200

Solution:

Setting $7a + b = 9b + a$ gives $6a = 8b,$ so $3a = 4b.$ The digits are $a = 4, b = 3,$ which check out since $43_7 = 31 = 34_9.$ Hence $a + b = 7.$

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 $x + y?$

$10$

$11$

$13$

$14$

$15$

Concepts:Diophantine Equation modular arithmetic

Difficulty rating: 1290

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 $A + B?$

$7$

$8$

$11$

$14$

$15$

Concepts:place value divisibility modular arithmetic

Difficulty rating: 1390

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. $7A + 6B \equiv 0 \pmod{36}.$ The only digit solution with $A \neq 0$ is $A = 6, B = 5$ ( $7 \cdot 6 + 6 \cdot 5 = 72$ ), giving $\$655.56 = 36 \times \$18.21.$ So $A + B = 11.$

Thus, the correct answer is C.

7.

What is the value of

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

$\dfrac{3}{4}$

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

$\dfrac{7}{8}$

$\dfrac{15}{16}$

$1$

Concepts:telescoping logarithm

Difficulty rating: 1420

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 $a + b?$

$13$

$17$

$20$

$30$

$68$

Concepts:Vieta’s Formulas polynomial radical

Difficulty rating: 1440

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 $6^{6^6}?$

$1$

$3$

$5$

$7$

$9$

Concepts:modular exponentiation digits

Difficulty rating: 1510

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 $\dfrac{x}{x+y}?$

$\dfrac{3}{7}$

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

$\dfrac{4}{9}$

$\dfrac{5}{11}$

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

Concepts:special right triangle coordinate geometry median (geometry)

Difficulty rating: 1580

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?

$\dfrac{2}{9}$

$\dfrac{2}{7}$

$\dfrac{9}{28}$

$\dfrac{1}{3}$

$\dfrac{3}{8}$

Concepts:basic probability symmetry

Difficulty rating: 1590

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.)

$9.68$

$10.14$

$10.50$

$11.32$

$12.00$

Concepts:area arc

Difficulty rating: 1690

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?

$12$

$16$

$18$

$24$

$28$

Concepts:graph theory multiplication principle

Difficulty rating: 1660

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 $n?$

$1013$

$1014$

$1016$

$2016$

$2025$

Concepts:mean arithmetic sequence bounding to limit cases

Difficulty rating: 1730

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?

$70$

$85$

$90$

$95$

$105$

Concepts:volume similarity ratio and proportion

Difficulty rating: 1800

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?

$0$

$\sqrt{2} - 1$

$2 - \sqrt{2}$

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

$3 - \sqrt{2}$

Concepts:clock trigonometric identity

Difficulty rating: 1840

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?

$3$

$9$

$12$

$18$

$27$

Concepts:Burnside’s Lemma casework

Difficulty rating: 1980

Solution:

Each red square needs a blue neighbor, each blue a yellow, and each yellow a red, forcing all three colors to appear in an interlocking pattern. A systematic check gives $84$ valid colorings of the labeled grid. Under the $8$ symmetries of the square, only two diagonal reflections fix any colorings — $6$ each — so Burnside's lemma gives $\tfrac{1}{8}(84 + 6 + 6) = 12$ distinct colorings.

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?

$\dfrac{5}{2}$

$3$

$\dfrac{16}{5}$

$\dfrac{7}{2}$

$\dfrac{15}{4}$

Concepts:expected value geometric distribution

Difficulty rating: 1770

Solution:

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

$7$

$10$

$11$

$12$

$19$

Concepts:Diophantine Equation modular arithmetic counting integers in a range

Difficulty rating: 2040

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 $4?$

$\dfrac{1}{101}$

$\dfrac{1}{100}$

$\dfrac{1}{99}$

$\dfrac{1}{98}$

$\dfrac{1}{97}$

Concepts:random walk recursive probability system of equations

Difficulty rating: 2110

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?

$20$

$22$

$24$

$26$

$28$

Concepts:law of cosines triangle area triangle inequality

Difficulty rating: 2170

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 $t_1^2 + t_2^2 = 290.$ The only integer values in the valid range $(1, 17)$ are $11$ and $13$ ( $121 + 169 = 290$ ), so 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 $|4z - 2| = 1?$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

$\dfrac{9}{16}$

$\dfrac{3}{4}$

$1$

Concepts:complex number triangle area optimization

Difficulty rating: 2270

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 $S?$

$3041$

$3542$

$3750$

$4044$

$4319$

Concepts:modular arithmetic divisibility sum of factors

Difficulty rating: 2380

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. $z \mid 2024.$ 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 $\sin(20\pi x) = \log_{20}(x)?$

$199$

$200$

$398$

$399$

$400$

Concepts:trigonometry logarithm counting intersections

Difficulty rating: 2520

Solution:

Since $|\sin| \le 1,$ solutions need $x \in \left[\tfrac{1}{20}, 20\right],$ where $\log_{20} x$ rises from $-1$ to $1.$ The curve $\sin(20\pi x)$ has period $\tfrac{1}{10},$ and on every full monotonic branch inside this interval it crosses the slowly increasing log exactly once. There are $398$ such full branches; the partial branch near $x = \tfrac{1}{20}$ adds one crossing, while the partial branch near $x = 20$ adds none (the sine cannot rise to the log's near- $1$ value there). This gives $399$ solutions.

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 $s^2?$

$6$

$\dfrac{25}{4}$

$\dfrac{13}{2}$

$\dfrac{27}{4}$

$7$

Concepts:equilateral triangle transformation circle

Difficulty rating: 2650

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.