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

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.

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.

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 $3a = 4b.$ The digits have to fit, with $a \le 6$ and $b \le 8,$ and the only pair that works is $a = 4, b = 3.$ So $a + b = 7.$ Therefore, the answer is A.

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.

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.

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.

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.

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.

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.

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.

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

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 $\dfrac{x}{x + y} = \dfrac{3\sqrt3/14}{\sqrt3/2} = \dfrac{3}{7}.$ Thus, A is the correct answer.

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.

Factor $k^3 + 6k^2 + 8k = k(k + 2)(k + 4),$ then split into partial fractions: $\dfrac{1}{k(k + 2)(k + 4)} = \dfrac{1/8}{k}$ $- \dfrac{1/4}{k + 2}$ $+ \dfrac{1/8}{k + 4}.$ 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.

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. A $6$-cycle has $2^6 + 2 = 66$ proper $3$-colorings altogether. Of these, $6$ use only two colors (type $(3,3,0)$) and $36$ use one color three times (type $(3,2,1)$). That leaves $66 - 6 - 36 = 24.$ Therefore, the answer is D.