Solution:

The total ranges from $5 \cdot 2 = 10$ (all two-pointers) to $5 \cdot 3 = 15$ (all three-pointers). Swapping one two-pointer for a three-pointer raises the total by exactly $1,$ so every integer in between occurs.

The possible totals are $10, 11, 12, 13, 14, 15,$ which is $6$ values.

Thus, the correct answer is E.

Solution:

Reversing the second and fourth rows gives the array with rows $1\,2\,3\,4,$ $11\,10\,9\,8,$ $15\,16\,17\,18,$ and $25\,24\,23\,22.$

The main diagonal sums to $1 + 10 + 17 + 22 = 50,$ and the other diagonal sums to $4 + 9 + 16 + 25 = 54.$

The positive difference is $54 - 50 = 4.$

Thus, the correct answer is B.

Solution:

One player earns the most when the other $20$ players each receive the minimum salary of $\$15{,}000.$

Thus the maximum salary is $\$700{,}000 - 20 \cdot \$15{,}000 = \$700{,}000 - \$300{,}000 = \$400{,}000.$

Thus, the correct answer is C.

Solution:

Since $\angle AOC,$ $\angle COD,$ and $\angle DOB$ fill the straight angle over diameter $\overline{AB},$ $$\angle COD = 180^\circ - 30^\circ - 45^\circ = 105^\circ.$$

The sector's share of the circle is $\dfrac{105^\circ}{360^\circ} = \dfrac{7}{24}.$

Thus, the correct answer is D.

Solution:

Let $r$ be the number of roses and $c$ the number of carnations, so $3r + 2c = 50$ with $r, c \ge 0.$

Because $2c$ and $50$ are even, $3r$ must be even, forcing $r$ to be even. The largest possible $r$ is $16$ (since $3 \cdot 17 \gt 50$), so $r \in \{0, 2, 4, \ldots, 16\}.$

That gives $9$ values of $r,$ each determining a bouquet.

Thus, the correct answer is C.

Solution:

Each flip counts $100000$ steps, so the year's steps total $$44 \cdot 100000 + 50000 = 4{,}450{,}000.$$

At $1800$ steps per mile, the mileage is $\dfrac{4{,}450{,}000}{1800} \approx 2472,$ which is closest to $2500.$

Thus, the correct answer is A.

Solution:

Since $(y - x)^2 = (x - y)^2,$ both inputs to the operation are the same value $t = (x - y)^2.$

Therefore $(x - y)^2 \$ (y - x)^2 = t \$ t = (t - t)^2 = 0.$

Thus, the correct answer is A.

Solution:

Since $AB = 4\,BD$ and $AB + BD = AD,$ we have $5\,BD = AD,$ so $BD = \tfrac{1}{5}AD.$

Likewise $AC = 9\,CD$ with $AC + CD = AD$ gives $CD = \tfrac{1}{10}AD.$

Because $B$ and $C$ both measure from $A,$ $\;BC = BD - CD = \tfrac{1}{5}AD - \tfrac{1}{10}AD = \tfrac{1}{10}AD.$

Thus, the correct answer is C.

Solution:

Let $O$ be the center and $D$ the point where $\overline{OC}$ meets $\overline{AB}.$ Since $C$ is the midpoint of arc $AB,$ $\overline{OC}$ is the perpendicular bisector of the chord, so $AD = 3.$

In right triangle $ADO,$ $OD = \sqrt{5^2 - 3^2} = 4,$ so $DC = OC - OD = 5 - 4 = 1.$

Then in right triangle $ADC,$ $AC = \sqrt{AD^2 + DC^2} = \sqrt{3^2 + 1^2} = \sqrt{10}.$

Thus, the correct answer is A.

Solution:

Let $n$ be the number of bricks. Alone, Brenda lays $\tfrac{n}{9}$ bricks per hour and Brandon lays $\tfrac{n}{10}.$ Together, their rate is $\tfrac{n}{9} + \tfrac{n}{10} - 10.$

Working for $5$ hours completes the chimney: $$5\left(\tfrac{n}{9} + \tfrac{n}{10} - 10\right) = n.$$ Expanding, $\tfrac{5n}{9} + \tfrac{5n}{10} - 50 = n,$ so $\tfrac{95n}{90} - n = 50,$ giving $\tfrac{5n}{90} = 50.$

Hence $n = 900.$

Thus, the correct answer is B.

Solution:

The part above the water is a cone similar to the whole mountain, with volume $\tfrac{1}{8}$ of the total. Since volume scales as the cube of length, the above-water cone's height is $\sqrt[3]{\tfrac{1}{8}} = \tfrac{1}{2}$ of the full height.

So the above-water height is $8000 \cdot \tfrac{1}{2} = 4000$ feet.

The ocean depth at the base is the submerged height, $8000 - 4000 = 4000$ feet.

Thus, the correct answer is A.

Solution:

Since the mean of the first $n$ terms is $n,$ their sum is $n \cdot n = n^2.$

The $n$th term is the difference of consecutive sums, $n^2 - (n-1)^2 = 2n - 1.$

For $n = 2008,$ the term is $2 \cdot 2008 - 1 = 4015.$

Thus, the correct answer is B.

Solution:

Place $A = (0,0),$ $B = (1,0),$ $C = (1,1),$ $D = (0,1),$ so $\overline{AD}$ lies along the $y$-axis and distance from $\overline{AD}$ is the $x$-coordinate. The region lies in the strip $\tfrac13 \le x \le \tfrac23,$ which within the square has area $\tfrac13.$

Equilateral $\triangle ABE$ has $E = \left(\tfrac12, \tfrac{\sqrt3}{2}\right),$ with side $AE$ on $y = \sqrt3\,x$ and side $BE$ on $y = \sqrt3(1 - x).$ The area of the triangle inside the strip is $$\int_{1/3}^{1/2} \sqrt3\,x\,dx + \int_{1/2}^{2/3} \sqrt3(1 - x)\,dx = 2\int_{1/3}^{1/2}\sqrt3\,x\,dx = \frac{5\sqrt3}{36}.$$

Therefore $$[R] = \frac13 - \frac{5\sqrt3}{36} = \frac{12 - 5\sqrt3}{36}.$$

Thus, the correct answer is B.

Solution:

The circumference is $2\pi$ times the radius, so $$\log_{10}(b^4) = 2\pi \log_{10}(a^2).$$

Rewriting, $4\log_{10} b = 4\pi \log_{10} a,$ hence $\log_{10} b = \pi \log_{10} a.$

Therefore $\log_a b = \dfrac{\log_{10} b}{\log_{10} a} = \pi.$

Thus, the correct answer is C.

Solution:

The convex hull $S$ differs from $R$ only near the four corners of the square, where a small triangular gap forms. Each gap triangle has two sides of length $1$ (outer edges of adjacent triangles).

The angle between those two sides is $360^\circ - 90^\circ - 4 \cdot 60^\circ = 30^\circ,$ so each gap has area $$\tfrac12 \cdot 1 \cdot 1 \cdot \sin 30^\circ = \tfrac14.$$

The total area is $4 \cdot \tfrac14 = 1.$

Thus, the correct answer is C.

Solution:

The painted rectangle measures $(a - 2)$ by $(b - 2)$ and has half the area of the floor, so $$ab = 2(a - 2)(b - 2).$$

Expanding gives $0 = ab - 4a - 4b + 8,$ and adding $8$ yields $(a - 4)(b - 4) = 8.$

With $b \gt a \gt 0,$ the only valid factor pairs of $8$ are $(a - 4, b - 4) = (1, 8)$ and $(2, 4),$ giving $(a, b) = (5, 12)$ and $(6, 8).$

There are $2$ possibilities.

Thus, the correct answer is B.

Solution:

Since $AB$ is horizontal, take $A = (a, a^2)$ and $B = (-a, a^2),$ and let $C = (c, c^2).$ The right angle cannot be at $A$ or $B$ (that would need $c = \pm a$), so it is at $C.$

Then $CA \perp CB$ gives $(c + a)(c - a) = -1,$ so $a^2 - c^2 = 1.$ This value is the height of the triangle above $\overline{AB}.$

The area is $\tfrac12 \cdot AB \cdot \text{height} = \tfrac12 (2|a|)(1) = |a| = 2008,$ so $a^2 = 2008^2 = 4{,}032{,}064$ and the $y$-coordinate of $C$ is $c^2 = a^2 - 1 = 4{,}032{,}063.$

Its digit sum is $4 + 0 + 3 + 2 + 0 + 6 + 3 = 18.$

Thus, the correct answer is C.

Solution:

The square has side $\sqrt{196} = 14.$ Let $F$ and $G$ be the feet of the perpendiculars from $E$ to $AB$ and $CD.$ Then $FG = 14,$ $EF = \tfrac{2 \cdot 105}{14} = 15,$ and $EG = \tfrac{2 \cdot 91}{14} = 13.$

Triangle $EFG$ lies in a plane perpendicular to the base, so its altitude to $FG$ is the pyramid's height. By Heron's formula with $s = 21,$ its area is $\sqrt{21 \cdot 6 \cdot 8 \cdot 7} = 84,$ so the altitude to $FG$ is $\tfrac{2 \cdot 84}{14} = 12.$

The volume is $\tfrac13 \cdot 196 \cdot 12 = 784.$

Thus, the correct answer is E.

Solution:

Let $\alpha = a + bi$ and $\gamma = c + di.$ Then $f(1) = (4 + a + c) + (1 + b + d)i$ and $f(i) = (-4 - b + c) + (-1 + a + d)i.$

Both being real forces $1 + b + d = 0$ and $-1 + a + d = 0,$ i.e. $a = 1 - d$ and $b = -1 - d.$

Hence $$|\alpha| + |\gamma| = \sqrt{(1 - d)^2 + (1 + d)^2} + \sqrt{c^2 + d^2} = \sqrt{2 + 2d^2} + \sqrt{c^2 + d^2},$$ which is smallest when $c = d = 0,$ giving $\sqrt{2}.$

Thus, the correct answer is B.

Solution:

Number the pails so Michael is at pail $0$ and the truck at pail $1$ at time $0.$ Michael reaches pail $n$ at $40n$ seconds. The truck spends $20$ seconds between pails and $30$ stopped, so it leaves pail $n$ at $50(n - 1)$ seconds and (for $n \ge 2$) arrives at $50(n - 1) - 30.$

Michael is at pail $n$ while the truck is there exactly when $50(n-1) - 30 \le 40n \le 50(n-1),$ which simplifies to $5 \le n \le 8.$ So they meet at pail $5$ (at $t = 200,$ as the truck departs), pail $6$ ($t = 240$), pail $7$ ($t = 280$), and pail $8$ ($t = 320,$ as the truck arrives).

Between pails $6$ and $7$ the truck (moving at $10$ ft/s) pulls ahead of and is then overtaken by Michael once more, adding one crossing. In all, they meet $5$ times.

Thus, the correct answer is B.

Solution:

Let the centers be $(a, 0)$ and $(b, 1)$ with $a, b \in [0, 2].$ The circles (radius $1$ each) intersect iff the distance between centers is at most $2:$ $$\sqrt{(a - b)^2 + 1} \le 2 \iff |a - b| \le \sqrt{3}.$$

The pairs $(a, b)$ fill the square $[0, 2]^2$ of area $4.$ The failing region $|a - b| \gt \sqrt3$ is two right triangles, each with legs $2 - \sqrt3,$ of total area $(2 - \sqrt3)^2 = 7 - 4\sqrt3.$

So the favorable area is $4 - (7 - 4\sqrt3) = 4\sqrt3 - 3,$ and the probability is $$\frac{4\sqrt3 - 3}{4}.$$

Thus, the correct answer is E.

Solution:

After the $12$ cars park, $4$ spaces are empty, equally likely to be any $4$ of the $16,$ for $\binom{16}{4} = 1820$ equally likely sets.

Auntie Em fails exactly when no two empty spaces are adjacent. The number of ways to place $4$ non-adjacent empties among $16$ is $\binom{13}{4} = 715.$

So the probability she can park is $$1 - \frac{715}{1820} = \frac{1105}{1820} = \frac{17}{28}.$$

Thus, the correct answer is E.

Solution:

The sum of the base-$10$ logs of the divisors is the log of their product. A number $N$ with $d(N)$ divisors has divisor product $N^{d(N)/2}.$

Here $N = 10^n$ has $d(N) = (n + 1)^2$ divisors, so the product is $(10^n)^{(n+1)^2/2}$ and its log is $$\frac{n(n + 1)^2}{2} = 792.$$

Thus $n(n + 1)^2 = 1584 = 11 \cdot 144 = 11 \cdot 12^2,$ giving $n = 11.$

Thus, the correct answer is A.

Solution:

Let $a_n = A_0A_n$ and $c_n = a_n - a_{n-1}$ be the base of the $n$th equilateral triangle. Its apex $B_n$ lies above the midpoint at height $\tfrac{\sqrt3}{2}c_n,$ and being on $y = \sqrt{x}$ gives $$\left(\tfrac{\sqrt3}{2}c_n\right)^2 = a_{n-1} + \tfrac{c_n}{2}, \quad\text{i.e.}\quad \tfrac34 c_n^2 = a_{n-1} + \tfrac{c_n}{2}.$$

Writing the same relation for the previous triangle and subtracting gives $c_n = c_{n-1} + \tfrac23,$ and with $c_1 = \tfrac23$ we get $c_n = \tfrac{2n}{3}.$ Summing, $$a_n = \sum_{k=1}^n \frac{2k}{3} = \frac{n(n + 1)}{3}.$$

We need $\tfrac{n(n + 1)}{3} \ge 100,$ i.e. $n(n + 1) \ge 300.$ Since $16 \cdot 17 = 272$ and $17 \cdot 18 = 306,$ the least such $n$ is $17.$

Thus, the correct answer is C.

Solution:

Because $AB \parallel CD,$ $\angle A + \angle D = 180^\circ,$ so the bisectors of $\angle A$ and $\angle D$ meet at right angles, $\angle APD = 90^\circ.$ Then the midpoint $M$ of $\overline{AD}$ is the circumcenter of right triangle $APD,$ giving $MP = MA = MD$ and $MP \parallel AB.$ The same holds for the midpoint $N$ of $\overline{BC}$ with $Q,$ so $M, P, Q, N$ are collinear on the midline.

The midline has length $\tfrac{AB + CD}{2} = 15,$ while $MP = \tfrac{AD}{2} = \tfrac72$ and $QN = \tfrac{BC}{2} = \tfrac52.$ Hence $PQ = 15 - \tfrac72 - \tfrac52 = 9.$

Drawing $AE \parallel BC$ with $E$ on $\overline{CD}$ gives $AE = 5$ and $DE = CD - AB = 8.$ In $\triangle ADE,$ $\cos(\angle AED) = \tfrac{8^2 + 5^2 - 7^2}{2 \cdot 8 \cdot 5} = \tfrac12,$ so $\angle AED = 60^\circ$ and the trapezoid's height is $AF = 5\sin 60^\circ = \tfrac{5\sqrt3}{2}.$

The segment $PQ$ sits at half the height, so the hexagon splits into two trapezoids and $$[ABQCDP] = \frac{AF}{4}\bigl(AB + CD + 2\,PQ\bigr) = \frac{5\sqrt3/2}{4}(11 + 19 + 18) = 30\sqrt3.$$

Thus, the correct answer is B.