Solution:

Factoring gives $$\dfrac{2(1-2+3-4+5-6+7)}{3(1-2+3-4+5-6+7)}.$$ The identical alternating sums cancel, leaving $\dfrac{2}{3}.$

Thus, the correct answer is C.

Solution:

Each day's pills cost $546 \div 14 = 39$ dollars. If $x$ is the cost of a green pill, then the pink pill costs $x-1,$ so $x+(x-1)=39.$ Solving gives $x=20.$

Thus, the correct answer is D.

Solution:

The first $8$ odd counting numbers sum to $1+3+\cdots+15=64.$

Letting $n$ be the smallest even integer, $$n+(n+2)+(n+4)+(n+6)+(n+8)=5n+20=60,$$ so $n=8.$

Thus, the correct answer is B.

Solution:

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

Cost is minimized by placing the most expensive flower in the smallest region, so the total is $$(3)(4)+(2.5)(6)+(2)(15)+(1.5)(20)+(1)(21)=108.$$

Thus, the correct answer is A.

Solution:

The lawn has area $90 \cdot 150 = 13{,}500$ square feet.

Each foot Moe walks mows an effective strip $2$ feet wide, so he mows $2 \cdot 5000 = 10{,}000$ square feet per hour. The time needed is $$\dfrac{13{,}500}{10{,}000}=1.35 \text{ hours}.$$

Thus, the correct answer is C.

Solution:

Since the length and height are in ratio $4:3,$ the length, height, and diagonal form a $4:3:5$ right triangle. The diagonal is $27,$ so the horizontal length is $$\dfrac{4}{5}(27)=21.6,$$ which is closest to $21.5.$

Thus, the correct answer is D.

Solution:

The value is $1$ for $n=1,2,3;$ it is $2$ for $n=4,\ldots,8;$ it is $3$ for $n=9,\ldots,15;$ and it is $4$ for $n=16.$ The sum is $$3\cdot 1 + 5\cdot 2 + 7\cdot 3 + 1\cdot 4 = 38.$$

Thus, the correct answer is B.

Solution:

Let the terms be $a, ar, ar^2, ar^3,\dots$ with $ar=2$ and $ar^3=6.$ Dividing gives $r^2=3,$ so $r=\pm\sqrt3.$

Then $a=\dfrac{2}{r}=\pm\dfrac{2}{\sqrt3}=\pm\dfrac{2\sqrt3}{3}.$ The choice $-\dfrac{2\sqrt3}{3}$ matches the negative case.

Thus, the correct answer is B.

Solution:

Writing everything base $5,$ the left side is $5^{-4}$ and the right side is $$5^{(48-26-34)/x}=5^{-12/x}.$$ Setting exponents equal, $-4=-\dfrac{12}{x},$ so $x=3.$

Thus, the correct answer is B.

Solution:

The old scheme allows $26 \cdot 10^4$ plates and the new scheme allows $26^3 \cdot 10^3$ plates. The increase factor is $$\dfrac{26^3 \cdot 10^3}{26 \cdot 10^4}=\dfrac{26^2}{10}.$$

Thus, the correct answer is C.

Solution:

The lines are $y-15=3(x-10)$ and $y-15=5(x-10).$

Setting $y=0$ gives $x$-intercepts $x=5$ and $x=7.$ The distance between $(5,0)$ and $(7,0)$ is $2.$

Thus, the correct answer is A.

Solution:

Let $a, b, c$ be the original portions. Then $a+b+c=1000$ and $(a-100)+2(b+c)=1500.$

Substituting $b+c=1000-a$ into the second equation, $$a-100+2(1000-a)=1500,$$ so $a=400.$

Thus, the correct answer is C.

Solution:

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

The two-digit numbers with digit sum $3$ are $12, 21, 30$ $(3$ of them$).$ Those with digit sum $12$ are $39, 48, 57, 66, 75, 84, 93$ $(7$ of them$).$ In all there are $10.$

Thus, the correct answer is E.

Solution:

Because $a$ must be divisible by $5,$ and $3^8 \cdot 5^2$ is divisible by $5^2$ but not $5^3,$ we need $b\le 2.$

Taking $b=2$ gives $a=\sqrt{3^8 \cdot 5^2}=3^4 \cdot 5=405,$ so $a+b=407.$ This beats $b=1,$ which gives $a+b=164{,}026.$

Thus, the correct answer is D.

Solution:

Each match eliminates exactly one player. Since $100$ players start and all but the champion are eliminated, there are $99$ matches.

Because $99=9 \cdot 11,$ it is divisible by $11$ but satisfies none of the other options.

Thus, the correct answer is E.

Solution:

With $m$ main courses, the number of dinners is $3 \cdot m \cdot 2m = 6m^2.$ This must be at least $365.$

So $m^2 \ge \dfrac{365}{6}\approx 60.8.$ Since $7^2=49$ is too small but $8^2=64$ works, $m=8.$

Thus, the correct answer is E.

Solution:

The melted volume equals the cone's volume, so $$\dfrac34 \cdot \dfrac43 \pi r^3 = \dfrac13 \pi r^2 h.$$

Simplifying gives $\pi r^3 = \dfrac13 \pi r^2 h,$ so $h=3r.$ The ratio of height to radius is $3:1.$

Thus, the correct answer is B.

Solution:

For even $n,$ the factors are five 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 fixed divisor works: $n=2$ gives $3 \cdot 5 \cdot 7 \cdot 9 \cdot 11,$ whose greatest common divisor with other cases such as $n=10$ is exactly $15.$

Thus, the correct answer is D.

Solution:

The large semicircle has area $\dfrac12 \pi (2)^2 = 2\pi.$

Removing the small semicircles deletes a region equal to five congruent $60^\circ$ sectors of radius $1$ plus two equilateral triangles of side $1.$ Each sector has area $\dfrac{\pi}{6}$ and each triangle has area $\dfrac{\sqrt3}{4}.$

The shaded area is $$2\pi - 5 \cdot \dfrac{\pi}{6} - 2 \cdot \dfrac{\sqrt3}{4} = \dfrac{7}{6}\pi - \dfrac{\sqrt3}{2}.$$

Thus, the correct answer is E.

Solution:

Here $FG = CD - DF - GC = 5 - 1 - 2 = 2.$ Let $h$ be the distance from $E$ down to line $CD.$ Since $\triangle FEG \sim \triangle AEB$ with ratio $\dfrac{FG}{AB}=\dfrac25,$ we have $$\dfrac{h}{h+3}=\dfrac25,$$ so $h=2.$

The height of $\triangle AEB$ from $E$ to $AB$ is $h+3=5,$ giving area $\dfrac12 \cdot 5 \cdot 5 = \dfrac{25}{2}.$

Thus, the correct answer is D.

Solution:

The bag always holds $4$ beads. All are red at the end precisely when both greens are drawn.

Drawing green then green has probability $\dfrac24 \cdot \dfrac14 = \dfrac18.$ Green, red, green has probability $\dfrac24 \cdot \dfrac34 \cdot \dfrac14 = \dfrac{3}{32}.$ Red, green, green has probability $\dfrac24 \cdot \dfrac24 \cdot \dfrac14 = \dfrac{1}{16}.$

The total is $$\dfrac18 + \dfrac{3}{32} + \dfrac{1}{16} = \dfrac{9}{32}.$$

Thus, the correct answer is C.

Solution:

Each $12$-hour period has $12 + 78 = 90$ chimes, so each full day has $180.$

After 11:15 AM on February 26, the rest of that day has $91$ chimes. Adding $180$ for each following full day, the count reaches $1891$ by the end of March 8.

The remaining chimes fall on March 9: the 2003rd chime is the $112$th chime of that day, occurring at 3:30 PM on March 9.

Thus, the correct answer is B.

Solution:

Let $O$ be the center. The octagon splits into $8$ congruent triangles from $O,$ so $\triangle AOB$ has area $\dfrac18.$

Since $O$ is the midpoint of $AE,$ triangles $AOB$ and $BOE$ have equal area, so $\triangle ABE$ has area $\dfrac14.$ The rectangle $ABEF$ is twice this, namely $\dfrac12.$

Thus, the correct answer is D.

Solution:

The common difference is $-2y,$ so the third and fourth terms must be $x-3y$ and $x-5y.$ Thus $xy=x-3y$ and $\dfrac{x}{y}=x-5y.$

From $\dfrac{x}{y}=x-5y$ we get $x=xy-5y^2,$ and substituting $xy=x-3y$ gives $-3y-5y^2=0.$ Since $y\ne 0,$ $y=-\dfrac35$ and then $x=-\dfrac98.$

The fifth term is $x-7y=-\dfrac98 + \dfrac{21}{5}=\dfrac{123}{40}.$

Thus, the correct answer is E.

Solution:

Write the number as $\overline{ab23}.$ It is divisible by $3$ when $a+b+2+3=a+b+5$ is divisible by $3,$ that is, when $a+b\equiv 1 \pmod 3.$

The two-digit prefix $\overline{ab}$ ranges over the $90$ values from $10$ to $99,$ and exactly one third of them satisfy this, giving $\dfrac{90}{3}=30.$

Thus, the correct answer is B.