Answer: C

Solution:

Since $(-2)^{-2} = \dfrac{1}{(-2)^2} = \dfrac14,$ we get $2 - \dfrac14 = \dfrac74.$

Thus, the correct answer is C.

Answer: B

Solution:

The first two tasks together take $100$ minutes, so each task takes $50$ minutes.

The third task finishes $50$ minutes after 2:40 PM, at 3:30 PM.

Thus, the correct answer is B.

Answer: A

Solution:

Write $2x + 3y = 100.$ If $28$ were written twice, then $3y = 100 - 56 = 44,$ which is not a multiple of $3.$

So $28$ is written three times: $2x = 100 - 84 = 16,$ giving $x = 8.$

Thus, the correct answer is A.

Answer: B

Solution:

Marta is $6$th, so Jack is $5$th. Jack is $2$ behind Todd, so Todd is $3$rd. Todd is $1$ behind Rand, so Rand is $2$nd.

Rand is $6$ ahead of Hikmet, so Hikmet is $8$th. (David is $10$th.)

Thus, the correct answer is B.

Answer: B

Solution:

The Sharks won $1$ of the first $3$ games. To reach $95\%$ with the fewest extra games, they should win all $N$ additional games, giving a win fraction $\dfrac{1 + N}{3 + N}.$

Requiring $\dfrac{1 + N}{3 + N} \ge \dfrac{19}{20}$ gives $20 + 20N \ge 57 + 19N,$ so $N \ge 37.$

Thus, the correct answer is B.

Answer: A

Solution:

The body has $13 \times 13 = 169$ entries. A product is odd exactly when both factors are odd.

There are $6$ odd numbers among $0, 1, \ldots, 12,$ giving $6 \times 6 = 36$ odd entries. The fraction is $\dfrac{36}{169} = 0.213\ldots \approx 0.21.$

Thus, the correct answer is A.

Answer: D

Solution:

A regular $15$-gon has $L = 15$ lines of symmetry, and its smallest angle of rotational symmetry is $R = \dfrac{360}{15} = 24$ degrees.

Then $L + R = 15 + 24 = 39.$

Thus, the correct answer is D.

Answer: D

Solution:

Since $625 = 5^4,$ we have $625^{\log_5 2015} = 5^{4\log_5 2015} = \left(5^{\log_5 2015}\right)^4 = 2015^4.$

Taking the fourth root gives $\left(2015^4\right)^{1/4} = 2015.$

Thus, the correct answer is D.

Answer: C

Solution:

Let $x$ be the probability Larry wins. He wins right away with probability $\dfrac12,$ or both players miss (probability $\dfrac14$) and the game restarts.

So $x = \dfrac12 + \dfrac14 x,$ giving $\dfrac34 x = \dfrac12$ and $x = \dfrac23.$

Thus, the correct answer is C.

Answer: C

Solution:

Let the distinct sides be $a \lt b \lt c.$ Since $a + b \gt c,$ the perimeter exceeds $2c,$ so $2c \lt 15$ and $c \le 6.$

The scalene triples with perimeter less than $15$ are $(6,5,3),$ $(6,5,2),$ $(6,4,3),$ $(5,4,3),$ $(5,4,2),$ and $(4,3,2).$ Of these, only $(5,4,3)$ is a right triangle, leaving $5.$

Thus, the correct answer is C.

Answer: E

Solution:

The line meets the axes at $(5, 0)$ and $(0, 12),$ so the triangle is right with legs $5$ and $12$ and hypotenuse $13.$ Its area is $30.$

Two altitudes are the legs $5$ and $12;$ the altitude to the hypotenuse is $\dfrac{2 \cdot 30}{13} = \dfrac{60}{13}.$ The sum is $17 + \dfrac{60}{13} = \dfrac{281}{13}.$

Thus, the correct answer is E.

Answer: D

Solution:

Factoring gives $(x - b)\bigl(2x - (a + c)\bigr) = 0,$ so the roots are $b$ and $\dfrac{a + c}{2}.$ Their sum is $b + \dfrac{a + c}{2}.$

Using distinct digits, take $b = 9$ and $a + c = 8 + 7 = 15,$ giving $9 + 7.5 = 16.5.$

Thus, the correct answer is D.

Answer: B

Solution:

Angles $BAC$ and $BDC$ subtend arc $BC,$ so $\angle BDC = 70^\circ.$ Then $\angle ADC = \angle ADB + \angle BDC = 110^\circ.$

Since $ABCD$ is cyclic, $\angle ABC = 180^\circ - 110^\circ = 70^\circ = \angle BAC.$ Thus $\triangle ABC$ is isosceles with $AC = BC = 6.$

Thus, the correct answer is B.

Answer: D

Solution:

Let $z$ be the area shared by the circle and triangle. The requested difference is $(\text{circle} - z) - (\text{triangle} - z) = \text{circle} - \text{triangle}.$

The circle has area $\pi \cdot 2^2 = 4\pi,$ and the equilateral triangle has area $\dfrac{\sqrt3}{4}\cdot 4^2 = 4\sqrt3.$ The difference is $4\pi - 4\sqrt3 = 4(\pi - \sqrt3).$

Thus, the correct answer is D.

Answer: D

Solution:

Math and Science give $8$ points, so Rachelle needs at least $6$ more from English and History. The chance of a C is $1 - \dfrac16 - \dfrac14 = \dfrac{7}{12}$ in English and $1 - \dfrac14 - \dfrac13 = \dfrac{5}{12}$ in History.

Working over a denominator of $144:$ $8$ points has probability $\dfrac16\cdot\dfrac14 = \dfrac{6}{144};$ $7$ points has $\dfrac16\cdot\dfrac13 + \dfrac14\cdot\dfrac14 = \dfrac{17}{144};$ and $6$ points has $\dfrac16\cdot\dfrac{5}{12} + \dfrac14\cdot\dfrac13 + \dfrac{7}{12}\cdot\dfrac14 = \dfrac{43}{144}.$

The total is $\dfrac{6 + 17 + 43}{144} = \dfrac{66}{144} = \dfrac{11}{24}.$

Thus, the correct answer is D.

Answer: C

Solution:

The distance from the hexagon's center to a vertex is $6.$ A lateral edge has length $8,$ so the pyramid's height is $\sqrt{8^2 - 6^2} = \sqrt{28} = 2\sqrt7.$

The hexagon's area is $\dfrac{3\sqrt3}{2}\cdot 6^2 = 54\sqrt3.$ Thus the volume is $\dfrac13 \cdot 54\sqrt3 \cdot 2\sqrt7 = 36\sqrt{21}.$

Thus, the correct answer is C.

Answer: D

Solution:

Setting the two probabilities equal and cancelling the common powers of $\tfrac14$ and $\tfrac34$ gives $\binom{n}{2}\cdot\dfrac34 = \binom{n}{3}\cdot\dfrac14.$

This becomes $\dfrac{n(n-1)}{2}\cdot 3 = \dfrac{n(n-1)(n-2)}{6},$ so $\dfrac32 = \dfrac{n-2}{6},$ giving $n - 2 = 9$ and $n = 11.$

Thus, the correct answer is D.

Answer: D

Solution:

A composite number has at least two prime factors (with multiplicity), and the smallest prime is $2,$ so the least possible value is $2 + 2 = 4.$

Every integer greater than $3$ is attained: $r(2^k) = 2k$ covers the even values $\ge 4,$ and $r(2^k\cdot 3) = 2k + 3$ covers the odd values $\ge 5.$ So the range is the integers greater than $3.$

Thus, the correct answer is D.

Answer: C

Solution:

The center $O$ of the circle lies on the perpendicular bisectors of $XY$ and $ZW,$ which are the same as those of $AB$ and $AC.$ So $O$ is the circumcenter of $\triangle ABC,$ and since $\angle C = 90^\circ,$ $O$ is the midpoint of $AB.$

Let $a = \tfrac12 BC$ and $b = \tfrac12 CA.$ Then $a^2 + b^2 = 6^2,$ and computing $OX^2 = OW^2$ gives $12^2 + 6^2 = b^2 + (a + 2b)^2.$ Solving yields $a = b = 3\sqrt2,$ so $BC = CA = 6\sqrt2$ and the perimeter is $12 + 12\sqrt2.$

Thus, the correct answer is C.

Answer: B

Solution:

Computing $f(i, j)$ row by row from the definition, the column $j = 2$ stabilizes: $f(i, 2) = 1$ for all $i \ge 5.$

Since $2015 \ge 5,$ we get $f(2015, 2) = 1.$

Thus, the correct answer is B.

Answer: D

Solution:

A staircase of $t$ steps takes Cozy $\left\lceil \tfrac{t}{2} \right\rceil$ jumps and Dash $\left\lceil \tfrac{t}{5} \right\rceil$ jumps, and we need the difference to equal $19.$

Checking the possibilities, the valid values are $t = 63,$ $64,$ and $66,$ so $s = 63 + 64 + 66 = 193.$ Its digit sum is $1 + 9 + 3 = 13.$

Thus, the correct answer is D.

Answer: D

Solution:

First imagine everyone moves to the chair directly opposite. The condition becomes: each person must sit in the same chair or an adjacent one. The number of people who keep their seat must be even (otherwise an odd-length gap cannot be filled).

If $0$ keep their seat, everyone shifts left, shifts right, or swaps with a neighbor: $4$ ways. If $2$ keep their seats, those two are opposite or adjacent, giving $3 + 6 = 9$ ways, with the rest forced. If $4$ keep their seats, there are $6$ ways to choose them and the other two swap. If all $6$ stay, $1$ way. The total is $4 + 9 + 6 + 1 = 20.$

Thus, the correct answer is D.

Answer: B

Solution:

Numerically equal volume and surface area means $abc = 2(ab + bc + ca).$ Rearranging shows $a \le 6,$ and $a = 1$ or $a = 2$ give no solutions.

For each remaining $a,$ the equation factors: $a = 3$ gives $(b - 6)(c - 6) = 36$ with $5$ solutions; $a = 4$ gives $(b - 4)(c - 4) = 16$ with $3$ solutions; $a = 5$ gives $(3b - 10)(3c - 10) = 100$ with $1$ valid solution; and $a = 6$ gives $(b - 3)(c - 3) = 9$ with $1$ valid solution. That is $5 + 3 + 1 + 1 = 10$ triples.

Thus, the correct answer is B.

Answer: D

Solution:

Since every center is equidistant from $P$ and $Q,$ all four centers and $R$ lie on the perpendicular bisector of $PQ,$ with $PR = 24.$ Suppose $R$ lies between $A$ and $B.$ Let $y = AR$ and $x = \tfrac15$ of circle $A$'s radius. Then $y^2 + 24^2 = 25x^2$ and $(39 - y)^2 + 24^2 = 64x^2.$ Subtracting gives $x^2 = 39 - 2y,$ so $y^2 + 50y - 399 = 0$ and $y = 7.$ Thus $AR = 7$ and $BR = 32.$

Because the circles are noncongruent, $R$ does not lie between $C$ and $D.$ The analogous equations give $w^2 - 50w - 399 = 0$ with $w = CR = 57,$ so $DR = 96.$ The sum is $7 + 32 + 57 + 96 = 192.$

Thus, the correct answer is D.

Answer: B

Solution:

Place $P_0 = 0$ and let $z = e^{\pi i/6},$ so each step of length $k$ in direction $z^{k-1}$ gives $P_{2015} = \sum_{k=1}^{2015} k z^{k-1}.$ Summing this (a differentiated geometric series) leads to $P_{2015} = \dfrac{1}{(z - 1)^2}\bigl(2015 z^{2016} - 2016 z^{2015} + 1\bigr).$

Since $z^{12} = 1,$ we have $z^{2016} = 1$ and $z^{2015} = \tfrac1z,$ so $P_{2015} = \dfrac{2016}{z(z - 1)}.$ Using $|z - 1|^2 = 2 - \sqrt3 = \dfrac{(\sqrt3 - 1)^2}{2}$ and $|z| = 1,$ the distance is $\dfrac{2016}{|z - 1|} = 1008\sqrt6 + 1008\sqrt2.$

Hence $a + b + c + d = 1008 + 6 + 1008 + 2 = 2024.$

Thus, the correct answer is B.