Answer: C

Solution:

Factor $2$ from the numerator and $3$ from the denominator: $$\frac{2(1 - 2 + 3 - 4 + 5 - 6 + 7)}{3(1 - 2 + 3 - 4 + 5 - 6 + 7)}.$$

The equal parenthesized sums cancel, leaving $\dfrac{2}{3}.$

Thus, the correct answer is C.

Answer: D

Solution:

Over $14$ days the daily cost of the two pills is $$\frac{546}{14} = 39.$$

Let $g$ be the cost of a green pill. The pink pill costs $g - 1,$ so $$g + (g - 1) = 39,$$ giving $g = 20.$

Thus, the correct answer is D.

Answer: A

Solution:

The five regions have areas $4,$ $6,$ $15,$ $20,$ and $21$ square feet.

To minimize the cost, plant the most expensive flowers in the smallest regions. The least possible cost is $$3(4) + 2.5(6) + 2(15) + 1.5(20) + 1(21) = 108.$$

Thus, the correct answer is A.

Answer: C

Solution:

Because of the overlap, each pass adds a strip $28 - 4 = 24$ inches $= 2$ feet wide. So each foot Moe walks mows $2$ square feet, that is, $2 \cdot 5000 = 10000$ square feet per hour.

The lawn has area $90 \cdot 150 = 13500$ square feet, so the time is $$\frac{13500}{10000} = 1.35$$ hours.

Thus, the correct answer is C.

Answer: D

Solution:

A rectangle with side ratio $4 : 3$ has height, length, and diagonal in ratio $3 : 4 : 5.$ With diagonal $27,$ the horizontal length is $$\frac{4}{5}(27) = 21.6,$$ which is closest to $21.5.$

Thus, the correct answer is D.

Answer: B

Solution:

Let the first term be $a$ and the common ratio $r.$ Then $ar = 2$ and $ar^3 = 6,$ so $r^2 = 3$ and $r = \pm\sqrt{3}.$

The first term is $$a = \frac{2}{r} = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}.$$ The choice $-\dfrac{2\sqrt{3}}{3}$ appears among the options.

Thus, the correct answer is B.

Answer: D

Solution:

Let $n,$ $d,$ $q$ be the numbers of nickels, dimes, quarters. Then $n + d + q = 100$ and $n + 2d + 5q = 167$ (dividing the value equation by $5$).

Subtracting gives $d + 4q = 67,$ so $d = 67 - 4q.$

The largest $d$ is at $q = 0,$ giving $d = 67$ (with $n = 33$). The smallest occurs at $q = 16,$ giving $d = 3$ (with $n = 81$). The difference is $67 - 3 = 64.$

Thus, the correct answer is D.

Answer: E

Solution:

Let $y = \clubsuit(x).$ Since $x \le 99,$ we have $y \le 18.$ Then $\clubsuit(y) = 3$ requires $y = 3$ or $y = 12.$

The two-digit numbers with digit sum $3$ are $12, 21, 30$ ($3$ values), and those with digit sum $12$ are $39, 48, 57, 66, 75, 84, 93$ ($7$ values), for $10$ in all.

Thus, the correct answer is E.

Answer: D

Solution:

The slope of $f$ is $$\frac{f(6) - f(2)}{6 - 2} = \frac{12}{4} = 3.$$

Therefore $$f(12) - f(2) = 3(12 - 2) = 30.$$

Thus, the correct answer is D.

Answer: B

Solution:

Assume one triangle is attached to side $AB.$ The second triangle can be attached to a side that is one step away or two steps away from $AB.$

Attaching it to $BC$ or $CD$ gives two figures; attaching it to $AE$ or $DE$ gives figures that are mirror images of these across the pentagon's axis of symmetry.

So there are only $2$ non-congruent figures.

Thus, the correct answer is B.

Answer: C

Solution:

In $60$ real minutes the watch advances only $57$ minutes $36$ seconds $= 57.6$ minutes. So when the watch shows $t$ minutes past noon, the real elapsed time is $\dfrac{60}{57.6}t = \dfrac{25}{24}t$ minutes.

The watch reads 10:00 PM after $600$ recorded minutes, so the real elapsed time is $$\frac{25}{24}(600) = 625$$ minutes $= 10$ hours $25$ minutes past noon. The actual time is 10:25 PM.

Thus, the correct answer is C.

Answer: D

Solution:

For even $n,$ the five factors are consecutive odd numbers. Among any five consecutive odd numbers, at least one is divisible by $3$ and exactly one by $5,$ so the product is always divisible by $15.$

No larger divisor always works: the products for $n = 2$ and $n = 10$ are $3 \cdot 5 \cdot 7 \cdot 9 \cdot 11$ and $11 \cdot 13 \cdot 15 \cdot 17 \cdot 19,$ whose greatest common divisor is $15.$

Thus, the correct answer is D.

Answer: B

Solution:

Let $r$ be the common radius and $h$ the cone's height. The melted ice cream fills the cone, so $$\frac{3}{4}\cdot\frac{4}{3}\pi r^3 = \frac{1}{3}\pi r^2 h.$$

This simplifies to $\pi r^3 = \dfrac{1}{3}\pi r^2 h,$ so $h = 3r,$ a ratio of $3 : 1.$

Thus, the correct answer is B.

Answer: D

Solution:

Since $FG = 5 - 1 - 2 = 2$ and $\overline{FG} \parallel \overline{AB},$ triangles $FEG$ and $AEB$ are similar with ratio $\dfrac{FG}{AB} = \dfrac{2}{5}.$

Let the distance from $E$ to line $CD$ be $k.$ Then the distance from $E$ to $AB$ is $k + 3,$ and $$\frac{k}{k + 3} = \frac{2}{5},$$ giving $k = 2.$

The height of $\triangle AEB$ is $k + 3 = 5,$ so its area is $$\frac{1}{2}(5)(5) = \frac{25}{2}.$$

Thus, the correct answer is D.

Answer: D

Solution:

Let $O$ be the center of the octagon. Joining $O$ to the vertices splits the octagon into $8$ congruent triangles, so $\triangle AOB$ has area $\dfrac{1}{8}.$

Since $O$ is the midpoint of $\overline{AE},$ triangles $AOB$ and $BOE$ have equal areas, so $\triangle ABE$ has area $\dfrac{1}{4}.$

The rectangle $ABEF$ is split by diagonal $\overline{BE}$ into two congruent triangles, so $\triangle ABE$ is half of it. Hence $ABEF$ has area $\dfrac{1}{2}.$

Thus, the correct answer is D.

Answer: E

Solution:

The large semicircle has area $\dfrac{1}{2}\pi(2)^2 = 2\pi.$

Where the small semicircles overlap, adjacent ones meet at points a distance $1$ from two centers, forming equilateral triangles. The region removed from the large semicircle consists of five congruent $60^\circ$ sectors of radius $1,$ each of area $\dfrac{\pi}{6},$ together with two equilateral triangles of side $1,$ each of area $\dfrac{\sqrt{3}}{4}.$

The shaded area is $$2\pi - 5\cdot\frac{\pi}{6} - 2\cdot\frac{\sqrt{3}}{4} = \frac{7}{6}\pi - \frac{\sqrt{3}}{2}.$$

Thus, the correct answer is E.

Answer: D

Solution:

Let $X = \log x$ and $Y = \log y.$ Then $$X + 3Y = 1 \quad\text{and}\quad 2X + Y = 1.$$

Solving gives $X = \dfrac{2}{5}$ and $Y = \dfrac{1}{5},$ so $$\log(xy) = X + Y = \frac{3}{5}.$$

Thus, the correct answer is D.

Answer: B

Solution:

For the minimum $x,$ neither $x$ nor $y$ has prime factors other than $7$ and $11.$ Write $x = 7^c 11^d,$ so $7x^5 = 7^{5c+1} 11^{5d}.$ Writing $y = 7^m 11^n,$ we need $7^{5c+1}11^{5d} = 7^{13m}11^{13n+1}.$

Matching exponents: $5c + 1 \equiv 0 \pmod{13}$ gives the least $c = 5,$ and $5d \equiv 1 \pmod{13}$ gives the least $d = 8.$ So $a = 7,$ $b = 11,$ and $$a + b + c + d = 7 + 11 + 5 + 8 = 31.$$

Thus, the correct answer is B.

Answer: E

Solution:

The set $S$ contains $4 \cdot 4! = 96$ permutations, since the first term has $4$ choices and the remaining four terms can be arranged in $4!$ ways.

For the second term to be $2,$ the first term must be $3, 4,$ or $5$ (not $1,$ not $2$), giving $3$ choices, and the remaining three terms can be arranged in $3!$ ways: $3 \cdot 3! = 18.$

The probability is $\dfrac{18}{96} = \dfrac{3}{16},$ so $a + b = 3 + 16 = 19.$

Thus, the correct answer is E.

Answer: B

Solution:

The graph passes through $(-1, 0),$ $(1, 0),$ and $(0, 2).$ So $f(0) = d = 2.$

Adding $$f(1) + f(-1) = (a + b + c + d) + (-a + b - c + d) = 2b + 2d = 0,$$ so $b = -d = -2.$

Thus, the correct answer is B.

Answer: D

Solution:

Let $\beta = \pi - \alpha$ be the interior angle of $\triangle ABC$ at $B.$ By the Law of Cosines, $$AC^2 = 8^2 + 5^2 - 2(8)(5)\cos\beta = 89 - 80\cos\beta.$$

Then $AC \lt 7$ means $89 - 80\cos\beta \lt 49,$ i.e. $\cos\beta \gt \dfrac{1}{2},$ i.e. $\beta \lt \dfrac{\pi}{3}.$

As $\alpha$ is uniform on $(0, \pi),$ so is $\beta.$ The probability is $$\frac{\pi/3}{\pi} = \frac{1}{3}.$$

Thus, the correct answer is D.

Answer: C

Solution:

Let $O$ be the intersection of the diagonals. Then $\triangle AOB$ is right-angled at $O$ with legs $OA = 8$ and $OB = 15.$ Quadrilateral $OPNQ$ has right angles at $O,$ $P,$ and $Q,$ so it is a rectangle and $PQ = ON.$

The minimum of $ON$ is the altitude from $O$ to $\overline{AB}$ in $\triangle AOB.$ Since $AB = \sqrt{8^2 + 15^2} = 17,$ equating the two area expressions gives $$ON = \frac{OA \cdot OB}{AB} = \frac{8 \cdot 15}{17} = \frac{120}{17} \approx 7.06.$$

This is closest to $7.$

Thus, the correct answer is C.

Answer: A

Solution:

The intercepts occur where $1/x = k\pi,$ that is $x = \dfrac{1}{k\pi}$ for a nonzero integer $k.$

The condition $0.0001 \lt \dfrac{1}{k\pi} \lt 0.001$ becomes $$\frac{1000}{\pi} \lt k \lt \frac{10000}{\pi}.$$

The number of such integers is $$\left\lfloor \frac{10000}{\pi} \right\rfloor - \left\lfloor \frac{1000}{\pi} \right\rfloor = 3183 - 318 = 2865,$$ closest to $2900.$

Thus, the correct answer is A.

Answer: C

Solution:

The function $y = |x-a| + |x-b| + |x-c|$ is piecewise linear with slopes $-3, -1, 1, 3$ and corners at $x = a, b, c.$ The line $2x + y = 2003$ has slope $-2.$

A line of slope $-2$ meets this graph exactly once only if it passes through the leftmost corner $(a,\, b + c - 2a),$ where the graph's slope jumps from $-3$ to $-1.$ Substituting, $$2a + (b + c - 2a) = 2003,$$ so $b + c = 2003.$

Since $b \lt c,$ we need $c \gt \dfrac{2003}{2},$ so the minimum is $c = 1002$ (with $b = 1001$).

Thus, the correct answer is C.

Answer: D

Solution:

A chord has length less than the radius exactly when the arc it subtends is less than $60^\circ,$ since a chord of a $60^\circ$ arc equals the radius.

All three pairwise chords are shorter than the radius precisely when the three points all lie within some arc of $60^\circ.$

The probability that $n$ random points all lie within some arc of angle $L$ is $n\left(\dfrac{L}{2\pi}\right)^{n-1}.$ With $n = 3$ and $L = \dfrac{\pi}{3}$ (that is, $\dfrac{L}{2\pi} = \dfrac{1}{6}$), the probability is $$3\left(\frac{1}{6}\right)^2 = \frac{1}{12}.$$

Thus, the correct answer is D.