2025 AMC 10B — 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: 860

Small Hint:

Each mug uses $20$ grams, so divide the total mass by $20$

Big Hint:

A partial mug does not count, so round $\frac{350}{20}$ down to a whole number

Solution:

Each mug needs $20$ grams, so we divide: $\frac{350}{20} = 17.5.$ A half mug isn’t a mug, so round down. That leaves $17.$ Thus, B is the correct answer.

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 perfect square pattern recognition

Difficulty rating: 990

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:

One full block $1, 4, 9, 6, 5, 6, 9, 4, 1, 0$ sums to $45;$ there are $202$ full blocks and $5$ leftover terms

Solution:

The ones digit of $n^2$ depends only on the ones digit of $n,$ so the list repeats every $10$ terms: $1, 4, 9, 6, 5, 6, 9, 4, 1, 0.$ One block sums to $45.$ Now $2025 = 202 \cdot 10 + 5,$ so we get $202$ full blocks plus the first five terms: $202 \cdot 45$ $+ (1 + 4 + 9 + 6 + 5)$ $= 9090$ $+ 25 = 9115.$ Therefore, the answer is D.

3.

A Pascal-like triangle has $10$ as the top row and $10$ followed by $1$ as the second row. In each subsequent row the first number is $10,$ the last number is $1,$ and, as in the standard Pascal’s Triangle, each other number in the row is the sum of the two numbers directly above it. The first four rows are shown below.

What is the sum of the digits of the sum of the numbers in the $11$ th row?

$11$

$13$

$14$

$16$

$17$

Answer: D

Concepts:Pascal’s Triangle recursion geometric sequence

Difficulty rating: 1270

Small Hint:

Track only the row sums. Each interior entry is counted by both of the two entries below it

Big Hint:

Show $S_n = 2 S_{n-1}$ for $n \ge 3$ with $S_2 = 11,$ then compute $S_{11}$

Solution:

Let $S_n$ be the sum of row $n.$ Each entry in row $n-1$ feeds the two entries just below it, and the fixed border numbers $10$ and $1$ exactly make up for the terms lost at the edges. So $S_n = 2 S_{n-1}$ for $n \ge 3.$ With $S_2 = 11,$ this gives $S_n = 11 \cdot 2^{n-2},$ so $S_{11} = 11 \cdot 2^9 = 5632.$ Its digits sum to $5 + 6 + 3 + 2 = 16.$ Thus, D is the correct answer.

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 place value Diophantine Equation

Difficulty rating: 1130

Small Hint:

By place value, $\underline{a}\,\underline{b}$ in base seven is $7a + b,$ and $\underline{b}\,\underline{a}$ in base nine is $9b + a$

Big Hint:

The equation reduces to $3a = 4b;$ find digits with $a$ a valid base-seven digit

Solution:

By place value, $\underline{a}\,\underline{b}$ in base seven is $7a + b,$ and $\underline{b}\,\underline{a}$ in base nine is $9b + a.$ Set them equal: $7a + b = 9b + a,$ so $6a = 8b,$ that is $3a = 4b.$ Thus $a = 4t$ and $b = 3t.$ Because these are two-digit numerals, $t \ne 0;$ because $a$ is a base-seven digit, $t \le 1.$ Hence $a = 4$ and $b = 3,$ so $a + b = 7.$ Therefore, the answer is A.

5.

In $\triangle ABC,$ $AB = 10,$ $AC = 18,$ and $\angle B = 130^\circ.$ Let $O$ be the center of the circle containing points $A,$ $B,$ and $C.$ What is the degree measure of $\angle CAO?$

$20$

$30$

$40$

$50$

$60$

Answer: C

Concepts:circumcircle, circumcenter, and circumradius inscribed angle isosceles triangle

Difficulty rating: 1310

Small Hint:

$O$ is the circumcenter, so $OA = OB = OC.$ Relate the central angle $\angle AOC$ to the inscribed angle $\angle B$

Big Hint:

Because $\angle B$ is obtuse, $\angle AOC = 360^\circ - 2\angle B;$ then use the isosceles triangle $OAC$

Solution:

Since $O$ is the circumcenter, $OA = OB = OC.$ The inscribed angle $\angle B = 130^\circ$ subtends arc $AC,$ and because $B$ is obtuse, the central angle is $\angle AOC = 360^\circ$ $- 2 \cdot 130^\circ = 100^\circ.$ Triangle $OAC$ is isosceles, so $\angle CAO = \tfrac{180^\circ - 100^\circ}{2} = 40^\circ.$ (The lengths $AB$ and $AC$ never enter.) Thus, C is the correct answer.

6.

The line $y = \tfrac{1}{3}x + 1$ divides the square region defined by $0 \le x \le 2$ and $0 \le y \le 2$ into an upper region and a lower region. The line $x = a$ divides the lower region into two regions of equal area. Then $a$ can be written as $\sqrt{s} - t,$ where $s$ and $t$ are positive integers. What is $s + t?$

$18$

$19$

$20$

$21$

$22$

Answer: C

Concepts:trapezoid area quadratic

Difficulty rating: 1410

Small Hint:

The lower region left of $x = a$ is a trapezoid with vertical sides $1$ and $\tfrac{a}{3} + 1$ and width $a$

Big Hint:

The total lower area is $\tfrac{8}{3},$ so set $a + \tfrac{a^2}{6}$ equal to half of it, $\tfrac{4}{3},$ and solve the quadratic

Solution:

The lower region has area $\int_0^2\left(\tfrac{x}{3} + 1\right)dx = \tfrac{2}{3} + 2 = \tfrac{8}{3}.$ The slice with $0 \le x \le a$ is a trapezoid of area $a + \tfrac{a^2}{6}.$ We want that to be half the total, namely $\tfrac{4}{3},$ so $a^2 + 6a - 8 = 0$ and $a = -3 + \sqrt{17}.$ Then $s = 17,$ $t = 3,$ and $s + t = 20.$ Therefore, the answer is C.

7.

Frances stands $15$ meters directly south of a locked gate in a fence that runs east-west. Immediately behind the fence is a box of chocolates, located $x$ meters east of the locked gate. An unlocked gate lies $9$ meters east of the box, and another unlocked gate lies $8$ meters west of the locked gate. Frances can reach the box by walking toward an unlocked gate, passing through it, and walking toward the box. It happens that the total distance Frances would travel would be the same via either unlocked gate. What is the value of $x?$

$3\tfrac{2}{7}$

$3\tfrac{3}{7}$

$3\tfrac{4}{7}$

$3\tfrac{5}{7}$

$3\tfrac{6}{7}$

Answer: C

Concepts:coordinate geometry Pythagorean Theorem radical

Difficulty rating: 1500

Small Hint:

Put the locked gate at the origin and Frances at $(0, -15);$ each route is a straight walk to a gate plus a walk along the fence to the box

Big Hint:

The west route totals $17 + (x + 8);$ set it equal to $\sqrt{(x + 9)^2 + 225} + 9$ and square

Solution:

Put the fence on the $x$ -axis, the locked gate at the origin, and Frances at $(0, -15).$ Then the box is at $(x, 0),$ the east gate at $(x + 9, 0),$ and the west gate at $(-8, 0).$ The east route is $\sqrt{(x + 9)^2 + 15^2} + 9;$ the west route is $\sqrt{8^2 + 15^2}$ $+ (x + 8) = 17 + x + 8.$ Set them equal: $\sqrt{(x + 9)^2 + 225} = x + 16.$ Square and simplify to get $18x + 306 = 32x + 256,$ so $x = \tfrac{50}{14} = \tfrac{25}{7} = 3\tfrac{4}{7}.$ Thus, C is the correct answer.

8.

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 $A \ne 0.$ ” After a pause, Max says, “That was a good price.” What is $A + B?$

$7$

$8$

$11$

$14$

$15$

Answer: C

Concepts:divisibility modular arithmetic place value

Difficulty rating: 1350

Small Hint:

Write the total in cents: $\underline{A}\,\underline{B}\,\underline{B}.\underline{B}\,\underline{A}$ is $10001A + 1110B$ cents, and it must be divisible by $36$

Big Hint:

Reduce the condition modulo $36$ to $7A + 6B \equiv 0;$ test digits to find the unique $A, B$

Solution:

In cents the total is $10000A + 1000B + 100B$ $+ 10B + A$ $= 10001A + 1110B.$ Split evenly among $36$ shirts, so it’s divisible by $36.$ Now $10001 \equiv 29$ and $1110 \equiv 30 \pmod{36},$ so we need $29A + 30B \equiv 0,$ which reduces to $7A + 6B \equiv 0 \pmod{36}.$ The only digit solution with $A \ne 0$ is $A = 6, B = 5,$ since $7 \cdot 6 + 6 \cdot 5 = 72.$ That’s $\$655.56,$ or $\$18.21$ a shirt, so $A + B = 11.$ Therefore, the answer is C.

9.

How many ordered triples of integers $(x, y, z)$ satisfy the following system of inequalities?

$-x - y - z \le -2$ $-x + y + z \le 2$ $x - y + z \le 2$ $x + y - z \le 2$

$4$

$8$

$11$

$15$

$17$

Answer: C

Concepts:substitution parity systematic listing

Difficulty rating: 1560

Small Hint:

The last three inequalities are symmetric in $x, y, z;$ let $p = -x + y + z,$ $q = x - y + z,$ $r = x + y - z$

Big Hint:

Then $p, q, r \le 2$ with equal parity and $p + q + r = x + y + z \ge 2;$ count such triples

Solution:

Let $p = -x + y + z,$ $q = x - y + z,$ $r = x + y - z.$ The last three inequalities say $p, q, r \le 2,$ the first says $x + y + z \ge 2,$ and $p + q + r = x + y + z.$ Since $x = \tfrac{q + r}{2}$ and so on, $p, q, r$ must all share the same parity. Now count triples with each part $\le 2,$ equal parity, and sum in $[2, 6].$ The even ones are $(2,2,2),$ the permutations of $(2,2,0),$ of $(2,0,0),$ and of $(2,2,-2),$ giving $10.$ The only odd one is $(1,1,1).$ That’s $11$ in all, and each yields a unique $(x, y, z).$ Thus, C is the correct answer.

10.

Let $f(n) = n^3 - 5n^2 + 2n + 8,$ and let $g(n) = n^3 - 6n^2 + 5n + 12.$ What is the sum of all integer values of $n$ for which $\dfrac{f(n)}{g(n)}$ is also an integer?

$2$

$3$

$4$

$5$

$6$

Answer: A

Concepts:polynomial factoring divisibility

Difficulty rating: 1510

Small Hint:

Both cubics have $n = -1$ as a root; factor each and cancel common factors

Big Hint:

The ratio simplifies to $\dfrac{n - 2}{n - 3} = 1 + \dfrac{1}{n - 3};$ it is an integer only when $n - 3$ divides $1$

Solution:

Factor both cubics: $f(n) = (n + 1)(n - 2)(n - 4)$ and $g(n) = (n + 1)(n - 3)(n - 4).$ Away from $n \in \{-1, 3, 4\},$ where $g$ vanishes or the ratio is $\tfrac{0}{0},$ the common factors cancel and $\dfrac{f(n)}{g(n)} = \dfrac{n - 2}{n - 3} = 1 + \dfrac{1}{n - 3}.$ That’s an integer only when $n - 3 = \pm 1,$ so $n = 2$ or $n = 4.$ But $n = 4$ kills $g,$ so only $n = 2$ survives, and the sum is $2.$ Therefore, the answer is A.

11.

On Monday, $6$ students went to the tutoring center at the same time, and each one was randomly assigned to one of the $6$ tutors on duty. On Tuesday, the same $6$ students showed up, the same $6$ tutors were on duty, and the students were again randomly assigned to the tutors. What is the probability that exactly $2$ students met with the same tutor both Monday and Tuesday?

$\dfrac{1}{16}$

$\dfrac{3}{16}$

$\dfrac{1}{4}$

$\dfrac{3}{8}$

$\dfrac{1}{2}$

Answer: B

Concepts:derangements permutations basic probability

Difficulty rating: 1590

Small Hint:

Each day’s assignment is a permutation; comparing the two days is like a single random permutation, and “same tutor” means a fixed point

Big Hint:

Choose the $2$ matched students in $\binom{6}{2}$ ways and derange the other $4$ (there are $D_4 = 9$ derangements); divide by $6!$

Solution:

Each day’s assignment is a permutation of the $6$ students among the $6$ tutors. Comparing the two days, the number who keep the same tutor is the number of fixed points of $\tau = \pi_{\text{Tue}}^{-1}\pi_{\text{Mon}},$ itself a uniformly random permutation of $6$ elements. We want exactly $2$ fixed points, so choose those $2$ in $\binom{6}{2}$ ways and derange the other $4,$ where $D_4 = 9.$ The probability is $\dfrac{\binom{6}{2} D_4}{6!} = \dfrac{15 \cdot 9}{720} = \dfrac{3}{16}.$ Thus, B is the correct answer.

12.

The figure below shows an equilateral triangle, a rhombus with a $60^\circ$ angle, and a regular hexagon, each of them containing some mutually tangent congruent disks. Let $T,$ $R,$ and $H,$ respectively, denote the ratio in each case of the total area of the disks to the area of the enclosing polygon.

Which of the following is true?

$T = H = R$

$H \lt R = T$

$H = R \lt T$

$H \lt R \lt T$

$H \lt T \lt R$

Answer: C

Concepts:tangent circles area ratio regular polygon

Difficulty rating: 1710

Small Hint:

For a disk tangent to two sides meeting at angle $\theta,$ its center is at distance $\dfrac{r}{\sin(\frac{\theta}{2})}$ from the vertex

Big Hint:

Use the tangency conditions to write each disk radius $r$ in terms of the polygon’s size, then compute (disk area) / (polygon area) in each case

Solution:

Take the triangle with side $s.$ Three disks of radius $r$ give $s = 2r(1 + \sqrt3),$ so $T = \dfrac{3\pi r^2}{(\frac{\sqrt3}{4})s^2}$ $= \dfrac{(2\sqrt3 - 3)\pi}{2} \approx 0.73.$ For the rhombus with side $a,$ the two disks sit on the long diagonal $a\sqrt3 = 6r,$ so $r = \tfrac{a}{2\sqrt3}$ and $R = \dfrac{2\pi r^2}{(\frac{a^2\sqrt3}{2})} = \dfrac{\pi\sqrt3}{9} \approx 0.60.$ For the hexagon with side $a,$ each of the six disks touches a side at its midpoint, again giving $r = \tfrac{a}{2\sqrt3}$ and $H = \dfrac{6\pi r^2}{(\frac{3\sqrt3}{2})a^2}$ $= \dfrac{\pi\sqrt3}{9} \approx 0.60.$ So $H = R \lt T.$ Therefore, the answer is C.

13.

The altitude to the hypotenuse of a $30$ - $60$ - $90^\circ$ 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 $\dfrac{x}{x + y}?$

$\dfrac{3}{7}$

$\dfrac{\sqrt3}{4}$

$\dfrac{4}{9}$

$\dfrac{5}{11}$

$\dfrac{4\sqrt3}{15}$

Answer: A

Concepts:special right triangle altitude median (geometry)

Difficulty rating: 1660

Small Hint:

Set the right angle at the origin with legs along the axes (short leg $1,$ long leg $\sqrt3$ ); find the foot of the altitude and the median line

Big Hint:

Intersect the altitude with the median to the short side, then take the ratio of the near segment to the whole altitude

Solution:

Place the right angle at $C = (0,0),$ the short leg $CB = 1$ with $B = (1, 0),$ and the long leg $CA = \sqrt3$ with $A = (0, \sqrt3).$ The altitude from $C$ to hypotenuse $AB$ has foot $H = \left(\tfrac34, \tfrac{\sqrt3}{4}\right)$ and runs along $x = \sqrt3\,y.$ The median from $A$ to the midpoint $\left(\tfrac12, 0\right)$ of $CB$ meets that altitude at $\left(\tfrac37, \tfrac{\sqrt3}{7}\right).$ This cuts $CH$ (length $\tfrac{\sqrt3}{2}$ ) into $\tfrac{4\sqrt3}{14}$ and $\tfrac{3\sqrt3}{14},$ so $x = \tfrac{3\sqrt3}{14}$ and $\dfrac{x}{x + y} = \dfrac{\frac{3\sqrt3}{14}}{\frac{\sqrt3}{2}} = \dfrac{3}{7}.$ Thus, A is the correct answer.

14.

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 conditional probability

Difficulty rating: 1500

Small Hint:

The captains are the three tallest exactly when the three tallest end up in three different groups

Big Hint:

Assign the top three to slots one at a time: the probabilities that each new one lands in a fresh group are $\tfrac{6}{8}$ then $\tfrac{3}{7}$

Solution:

The captains are the three tallest exactly when those three land in three different groups, since each is then the tallest of its own group. Drop them into the $9$ slots one at a time ( $3$ per group). The second tallest misses the first’s group with probability $\tfrac{6}{8},$ and the third misses both with probability $\tfrac{3}{7}.$ So the probability is $\tfrac{6}{8} \cdot \tfrac{3}{7} = \tfrac{9}{28}.$ Therefore, the answer is C.

15.

The sum

$\sum_{k=1}^{\infty} \frac{1}{k^3 + 6k^2 + 8k}$

can be expressed as $\dfrac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a + b?$

$89$

$97$

$102$

$107$

$129$

Answer: D

Concepts:partial fractions telescoping

Difficulty rating: 1600

Small Hint:

Factor the denominator as $k(k + 2)(k + 4)$ and use partial fractions

Big Hint:

The pieces $\tfrac{\frac{1}{8}}{k} - \tfrac{\frac{1}{4}}{k + 2} + \tfrac{\frac{1}{8}}{k + 4}$ telescope; only $k = 1, 2, 3, 4$ leave a remainder

Solution:

Factor $k^3 + 6k^2 + 8k = k(k + 2)(k + 4),$ then split into partial fractions: $\dfrac{1}{k(k + 2)(k + 4)} = \dfrac{\frac{1}{8}}{k}$ $- \dfrac{\frac{1}{4}}{k + 2}$ $+ \dfrac{\frac{1}{8}}{k + 4}.$ Summing over all $k,$ the coefficient of $\tfrac1n$ cancels for $n \ge 5,$ so only the first few terms survive: $\tfrac18\left(1 + \tfrac12 - \tfrac13 - \tfrac14\right)$ $= \tfrac18 \cdot \tfrac{11}{12} = \tfrac{11}{96}.$ So $a + b = 11 + 96 = 107.$ Thus, D is the correct answer.

16.

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 complementary counting casework

Difficulty rating: 1800

Small Hint:

The unequal sectors form a fixed cycle of $6$ positions; you need proper colorings using each color exactly twice

Big Hint:

A $6$ -cycle has $2^6 + 2 = 66$ proper $3$ -colorings; remove those that do not use each color exactly twice

Solution:

The six unequal sectors form a fixed cycle of $6$ distinguishable positions, so we want proper $3$ -colorings of a $6$ -cycle that use each color exactly twice. There are $(3-1)^6 + (3-1) = 66$ proper $3$ -colorings altogether. Of these, $6$ use only two colors: choose the missing color in $3$ ways, then alternate the other two colors in $2$ ways. Another $36$ have color counts $(3,2,1)$ : choose the color used three times in $3$ ways, choose one of the $2$ alternating sets of positions for it, choose the color used twice in $2$ ways, and choose its $2$ positions among the other $3$ in $3$ ways. The remaining $66 - 6 - 36 = 24$ use each color exactly twice. Therefore, the answer is D.

17.

Consider a decreasing sequence of $n$ positive integers $x_1 \gt x_2 \gt x_3 \gt \cdots \gt x_n$ that satisfies the following conditions. The average 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 summation bounding to limit cases

Difficulty rating: 1910

Small Hint:

Let $A_k$ be the average of the first $k$ terms; the conditions give $A_k = 2028 - k$ for $k \ge 3$

Big Hint:

Then $x_k = k A_k$ $- (k-1)A_{k-1} = 2029$ $- 2k$ for $k \ge 4;$ require it to stay a positive integer

Solution:

Let $A_k$ be the average of the first $k$ terms. Then $A_3 = 2025$ and $A_k = A_{k-1} - 1$ for $k \ge 4,$ so $A_k = 2028 - k.$ The partial sum is $S_k = k(2028 - k),$ and for $k \ge 4$ the terms are $x_k = S_k - S_{k-1} = 2029 - 2k,$ namely $x_4 = 2021, x_5 = 2019, \ldots.$ These stay positive as long as $2029 - 2k \gt 0,$ that is $k \le 1014,$ with $x_{1014} = 1.$ This bound is attainable: take $(x_1,x_2,x_3) = (2030,2023,2022),$ which is decreasing, sums to $6075,$ and lies above $x_4 = 2021.$ Thus the greatest possible value is $n = 1014,$ so B is the correct answer.

18.

What is the ones digit of the sum $\lfloor\sqrt{1}\rfloor + \lfloor\sqrt{2}\rfloor + \lfloor\sqrt{3}\rfloor$ $+ \cdots + \lfloor\sqrt{2024}\rfloor$ $+ \lfloor\sqrt{2025}\rfloor?$ (Recall that $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x.$ )

$1$

$2$

$3$

$5$

$8$

Answer: D

Concepts:floor and ceiling functions sum of first n squares

Difficulty rating: 1730

Small Hint:

$\lfloor\sqrt{n}\rfloor = m$ for the $2m + 1$ values $m^2 \le n \le (m + 1)^2 - 1$

Big Hint:

Since $\sqrt{2025} = 45,$ sum $m(2m + 1)$ for $m = 1$ to $44$ and add the final term $45$

Solution:

For each $m,$ $\lfloor\sqrt{n}\rfloor = m$ on the $2m + 1$ integers $m^2 \le n \le (m + 1)^2 - 1.$ Since $\sqrt{2025} = 45,$ the terms with $1 \le m \le 44$ contribute $\sum_{m=1}^{44} m(2m + 1),$ and $n = 2025$ tacks on $45.$ That sum is $\sum_{m=1}^{44}(2m^2 + m)$ $= 2 \cdot \tfrac{44 \cdot 45 \cdot 89}{6}$ $+ \tfrac{44 \cdot 45}{2} = 58740$ $+ 990 = 59730,$ so the total is $59775.$ Its ones digit is $5.$ Therefore, the answer is D.

19.

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:power scaling of length, area, and volume volume similarity

Difficulty rating: 1660

Small Hint:

The container is a piece of a pyramid; the volume from the apex up to side length $w$ scales as $w^3$

Big Hint:

Volumes split as $2^3 - 1^3 = 7$ (up to the midline, side $2$ ) and $3^3 - 2^3 = 19$ (remainder); scale the $35$ minutes

Solution:

The container is a square frustum: a horizontal slice at height fraction $t$ has side $1 + 2t.$ Extend the sides up to their apex, and the volume out to where the side length is $w$ scales as $w^3.$ So the whole container is $3^3 - 1^3 = 26$ parts, the piece up to the midline (side $2$ ) is $2^3 - 1^3 = 7$ parts, and the rest is $3^3 - 2^3 = 19$ parts. Those $7$ parts take $35$ minutes, so each part is $5$ minutes. The remaining $19$ parts take $95$ minutes. Thus, D is the correct answer.

20.

Four congruent semicircles are inscribed in a square of side length $1$ so that their diameters are on the sides of the square, one endpoint of each diameter is at a vertex of the square, and adjacent semicircles are tangent to each other. A small circle centered at the center of the square is tangent to each of the four semicircles, as shown below.

The diameter of the small circle can be written as $(\sqrt{a} + b)(\sqrt{c} + d),$ where $a,$ $b,$ $c,$ and $d$ are integers. What is $a + b + c + d?$

$3$

$5$

$8$

$9$

$11$

Answer: A

Concepts:tangent circles coordinate geometry radical

Difficulty rating: 1930

Small Hint:

First find the semicircle radius $\rho$ from the tangency of two adjacent semicircles whose diameters start at neighboring vertices

Big Hint:

The small radius equals (distance from the square’s center to a semicircle’s center) minus $\rho;$ then factor $2t$ into two binomials

Solution:

Let each semicircle have radius $\rho,$ with centers like $(\rho, 0)$ and $(1, \rho).$ Adjacent semicircles are tangent, so these centers are $2\rho$ apart: $(1 - \rho)^2 + \rho^2 = 4\rho^2.$ This gives $2\rho^2 + 2\rho - 1 = 0,$ so $\rho = \tfrac{\sqrt3 - 1}{2}.$ The small circle of radius $t$ sits at $\left(\tfrac12, \tfrac12\right),$ and it’s tangent to a semicircle when its distance to that center equals $\rho + t.$ That distance is $\sqrt{2 - \sqrt3},$ so $t = \sqrt{2 - \sqrt3} - \rho,$ and the diameter is $2t = \sqrt6 - \sqrt2$ $- \sqrt3 + 1 = (\sqrt3 - 1)(\sqrt2 - 1).$ So $a + b + c + d = 3$ $+ (-1) + 2 + (-1) = 3.$ Therefore, the answer is A.

21.

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: 2100

Small Hint:

The rules form a directed cycle: red needs a blue neighbor, blue needs a yellow, yellow needs a red

Big Hint:

Count all valid colorings of the fixed labeled grid, then group them by the $8$ rotations and reflections of the square

Solution:

First count colorings of a grid whose positions are distinguished. Fix the center square as red and list the four edge-middle colors cyclically. Up to a rotation or reflection, the only possible edge patterns are $YRBR$ and $YRBY.$ The first has $4$ placements and $3$ possible cyclic corner strings, $BRYB, BYRB,$ and $BYYB;$ the second has $8$ placements and $2$ possible corner strings, $BYBB$ and $BYRB.$ Thus there are $4 \cdot 3 + 8 \cdot 2 = 28$ colorings with a red center. The center has $3$ possible colors, so there are $84$ labeled colorings.

Now apply Burnside’s lemma. The identity fixes all $84$ colorings. No nonidentity rotation fixes a valid coloring. Each of the two reflections across a horizontal or vertical axis fixes $6$ colorings, while each diagonal reflection fixes none. Therefore, the number of colorings up to rotations and reflections is $\dfrac{84 + 6 + 6}{8} = 12.$ Thus, C is the correct answer.

22.

A seven-digit positive integer is chosen at random. What is the probability that the number is divisible by $11,$ given that the sum of its digits is $61?$

$\dfrac{3}{14}$

$\dfrac{3}{11}$

$\dfrac{2}{7}$

$\dfrac{4}{11}$

$\dfrac{3}{7}$

Answer: A

Concepts:conditional probability divisibility stars and bars

Difficulty rating: 2040

Small Hint:

A digit sum of $61$ is $2$ below the maximum $63,$ so the digits are all $9$ except a total shortfall of $2$

Big Hint:

For divisibility by $11$ use the alternating digit sum; express it through the shortfalls on the odd- versus even-position digits

Solution:

A digit sum of $61 = 63 - 2$ means all seven digits are $9$ except for a total deficit of $2,$ which gives $\binom{2 + 6}{6} = 28$ numbers. For divisibility by $11$ we need $O - E \equiv 0 \pmod{11},$ where $O$ sums the $4$ odd-position digits and $E$ the $3$ even ones. Write the deficits as $d_O + d_E = 2.$ Then $O - E = 9 - d_O$ $+ d_E = 11 - 2 d_O,$ a multiple of $11$ only when $d_O = 0.$ So all of the deficit $2$ falls on the $3$ even positions, giving $\binom{2 + 2}{2} = 6$ ways. The probability is $\tfrac{6}{28} = \tfrac{3}{14}.$ Therefore, the answer is A.

23.

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

Difficulty rating: 2300

Small Hint:

At row $i,$ column $j,$ Horace writes $91(i - 1) + j$ and Vera writes $141(j - 1) + i;$ set them equal

Big Hint:

The equation reduces to $9i - 14j = -5;$ find the columns $j$ that make $i$ an integer in range

Solution:

At row $i,$ column $j,$ Horace writes $91(i - 1) + j$ and Vera writes $141(j - 1) + i.$ Set them equal and simplify to get $9i - 14j = -5,$ so $i = \tfrac{14j - 5}{9},$ an integer exactly when $j \equiv 1 \pmod 9.$ For $j = 1, 10, 19, \ldots, 91,$ that’s $11$ values, and $i$ runs $1, 15, 29, \ldots, 141,$ all within range. So $11$ squares match. Thus, C is the correct answer.

24.

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 $\tfrac12$ and it disappears with probability $\tfrac12.$ For $n = 1,$ $2,$ or $3,$ if it is at $n,$ then it moves to $n + 1$ with probability $\tfrac14,$ it moves to $n - 1$ with probability $\tfrac14,$ and it disappears with probability $\tfrac12.$

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: 2170

Small Hint:

Let $p_n$ be the probability of ever reaching $4$ starting from $n,$ with $p_4 = 1;$ set up one equation per state

Big Hint:

Use $p_0 = \tfrac12 p_1$ and $p_n = \tfrac14 p_{n-1} + \tfrac14 p_{n+1}$ for $n = 1, 2, 3,$ then solve the system

Solution:

Let $p_n$ be the probability of reaching $4$ from position $n,$ with $p_4 = 1.$ The rules give $p_0 = \tfrac12 p_1,$ $p_1 = \tfrac14 p_0 + \tfrac14 p_2,$ $p_2 = \tfrac14 p_1 + \tfrac14 p_3,$ and $p_3 = \tfrac14 p_2 + \tfrac14.$ Work upward: $p_1 = \tfrac27 p_2$ and $p_2 = \tfrac{7}{26} p_3.$ These unwind to $p_3 = \tfrac{26}{97},$ $p_2 = \tfrac{7}{97},$ $p_1 = \tfrac{2}{97},$ and finally $p_0 = \tfrac{1}{97}.$ Therefore, the answer is E.

25.

Square $ABCD$ has sides of length $4.$ Points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{CD},$ respectively, with $AP = \tfrac{8}{5}$ and $DQ = \tfrac{10}{3}.$ A path begins along the line segment from $P$ to $Q$ and continues by reflecting against the sides of $ABCD$ (with congruent incoming and outgoing angles), as shown in the figure. If the path hits a vertex of the square, then it terminates there; otherwise it continues forever.

At which vertex does the path terminate?

$A$

$B$

$C$

$D$

The path continues forever.

Answer: B

Concepts:reflection (geometry)parity greatest common divisor coordinate geometry

Difficulty rating: 2520

Small Hint:

Reflecting billiard paths is easiest by “unfolding”: follow a straight line through a grid of reflected copies of the square

Big Hint:

Find the first grid corner $(4a, 4b)$ the line hits; the parities of $a$ and $b$ determine which actual vertex it is

Solution:

Place $A = (0,0),$ $B = (4,0),$ $C = (4,4),$ $D = (0,4),$ so $P = \left(0, \tfrac85\right)$ and $Q = \left(\tfrac{10}{3}, 4\right).$ The initial direction is $\left(\tfrac{10}{3}, \tfrac{12}{5}\right) \parallel (25, 18).$ Unfold the billiard into a grid of reflected copies and follow the straight line from $P.$ At a grid corner $(4a,4b),$ the horizontal coordinate gives $t = \tfrac{4a}{25}.$ The vertical coordinate then requires $10 + 18a = 25b,$ so $a \equiv 5 \pmod{25}.$ The first positive possibility is $a = 5,$ giving $b = 4$ and the unfolded corner $(20,16).$ Crossing $5$ cells across (odd) puts it on the side $x = 4,$ and $4$ cells up (even) puts it on $y = 0.$ That’s vertex $(4,0) = B.$ Thus, B is the correct answer.