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.

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.

$i(i-1) = i^2 - i = -1 - i,$ 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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

$2025$ 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.

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.

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.

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.

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 $f(0) = \tfrac{1}{97}.$