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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.