Solution:

Rows $12$ through $22$ inclusive make up $22 - 12 + 1 = 11$ rows.

Each row has $33$ seats, so the total is $33 \times 11 = 363.$

Thus, the correct answer is C.

Solution:

The numbers $70$ through $79$ give $10$ with a $7$ in the tens place.

The numbers $17, 27, \ldots, 97$ give $9$ with a $7$ in the units place.

Since $77$ is counted twice, the total is $10 + 9 - 1 = 18.$

Thus, the correct answer is B.

Solution:

Working backward from the fifth practice, the counts are $48, 24, 12, 6,$ and $3$ at the fourth, third, second, and first practices.

Thus, the correct answer is A.

Solution:

Since $6! = 720 = 2^4 \cdot 3^2 \cdot 5,$ the visible product uses only the primes $2, 3,$ and $5.$

Hiding $4$ leaves the fewest $2$'s, namely $2^2.$ Hiding $3$ or $6$ leaves the fewest $3$'s, namely one. Hiding $5$ leaves no factor of $5.$

Therefore $P$ is always divisible by $2^2 \cdot 3 = 12,$ but not necessarily by any larger number.

Thus, the correct answer is B.

Solution:

Setting $d = 0$ removes the subtraction, so we maximize $c \cdot a^b$ using $1, 2, 3.$

Taking $c = 1,$ $a = 3,$ $b = 2$ gives $3^2 = 9.$ The alternative $2^3 = 8$ is smaller, and any assignment with $c \gt 1$ forces a smaller power. The maximum is $9.$

Thus, the correct answer is D.

Solution:

For $m \lt n,$ we have $m! \cdot n! = (m!)^2 \cdot (m+1)(m+2)\cdots n,$ which is a perfect square precisely when $(m+1)\cdots n$ is a perfect square.

For the five choices this leftover factor is $99,$ $99 \cdot 100,$ $100,$ $100 \cdot 101,$ and $101.$ Only $100 = 10^2$ is a perfect square.

Therefore $99! \cdot 100! = (99! \cdot 10)^2$ is the perfect square.

Thus, the correct answer is C.

Solution:

Isabella received $\dfrac{10}{7}d$ Canadian dollars and spent $60,$ leaving $d.$ So $$\dfrac{10}{7}d - 60 = d.$$

Then $\dfrac{3}{7}d = 60,$ so $d = 140.$ The sum of its digits is $1 + 4 + 0 = 5.$

Thus, the correct answer is A.

Solution:

The two given directions are perpendicular, so the airport sits at the right angle of a right triangle with legs $8$ and $10.$

The distance between the downtowns is $\sqrt{8^2 + 10^2} = \sqrt{164} \approx 12.8,$ which is closest to $13.$

Thus, the correct answer is A.

Solution:

The square has area $10^2 = 100$ and the circle has area $\pi(10)^2 = 100\pi.$

Since the circle is centered at a vertex of the square, exactly one quarter of the circle, area $25\pi,$ lies inside the square.

The union has area $100 + 100\pi - 25\pi = 100 + 75\pi.$

Thus, the correct answer is B.

Solution:

The rows hold $1, 3, 5, \ldots$ cans, and the sum of the first $n$ odd numbers is $n^2.$

Setting $n^2 = 100$ gives $n = 10.$

Thus, the correct answer is D.

Solution:

There are $8 \cdot 8 = 64$ ordered pairs. The inequality $mn \gt m + n$ is equivalent to $(m-1)(n-1) \gt 1.$

This fails only when $m = 1,$ $n = 1,$ or $m = n = 2,$ which account for $8 + 8 - 1 + 1 = 16$ pairs.

The probability is $\dfrac{64 - 16}{64} = \dfrac{48}{64} = \dfrac{3}{4}.$

Thus, the correct answer is C.

Solution:

The annulus is the difference of the two circular areas, $\pi b^2 - \pi c^2.$

Because $\overline{XZ}$ is tangent to the small circle at $Z,$ it is perpendicular to the radius $\overline{OZ}.$ In right triangle $OZX$ with $OX = b,$ $OZ = c,$ and $XZ = a,$ we get $b^2 - c^2 = a^2.$

Therefore the area of the annulus is $\pi a^2.$

Thus, the correct answer is A.

Solution:

Each thickness ends with $5$ in the hundredths place. A stack of an odd number of coins keeps a $5$ there, and pairs give an odd digit in the tenths place, so a whole-number height requires the count to be a multiple of $4.$

A stack of $4$ coins is at most $4(1.95) = 7.8$ mm, and a stack of $12$ coins is at least $12(1.35) = 16.2$ mm, so only $8$ coins can total $14$ mm.

Indeed, $8$ quarters give $8(1.75) = 14$ mm.

Thus, the correct answer is B.

Solution:

Let there be $B$ blue marbles. After adding red marbles the total is $3B;$ after adding yellow marbles the total is $5B,$ still with $B$ blue.

Doubling the blue marbles gives $2B$ blue out of $6B$ total, which is $\dfrac{2B}{6B} = \dfrac{1}{3}.$

Thus, the correct answer is C.

Solution:

Swapping increases the value, so Patty has more nickels than dimes. Each swapped coin changes the total by $5$ cents, so she has $70 / 5 = 14$ more nickels than dimes.

With $n + d = 20$ and $n - d = 14,$ she has $17$ nickels and $3$ dimes.

Her coins are worth $17 \cdot 5 + 3 \cdot 10 = 115$ cents, or $\$1.15.$

Thus, the correct answer is A.

Solution:

The centers of the three unit circles form an equilateral triangle with side $2.$ Its center is the center of the large circle.

The distance from the center of an equilateral triangle to a vertex is $\dfrac{\text{side}}{\sqrt3} = \dfrac{2}{\sqrt3} = \dfrac{2\sqrt3}{3}.$

Adding the unit radius, the large radius is $1 + \dfrac{2\sqrt3}{3} = \dfrac{3 + 2\sqrt3}{3}.$

Thus, the correct answer is D.

Solution:

Let Jack's age be $10x + y$ and Bill's be $10y + x.$ In five years $10x + y + 5 = 2(10y + x + 5),$ which simplifies to $8x = 19y + 5.$

Since $x$ and $y$ are digits, the only solution is $y = 1,$ $x = 3.$

So Jack is $31$ and Bill is $13,$ a difference of $18.$

Thus, the correct answer is B.

Solution:

The area of $\triangle ACE$ is $\tfrac12(12)(16) = 96.$

Each corner triangle $\triangle ABF,$ $\triangle BCD,$ and $\triangle DEF$ has a base and an altitude that are $\tfrac34$ and $\tfrac14$ of a corresponding base and altitude of $\triangle ACE.$ So each has area $\tfrac14 \cdot \tfrac34 = \tfrac{3}{16}$ of $\triangle ACE.$

Hence $\dfrac{[BDF]}{[ACE]} = 1 - 3 \cdot \dfrac{3}{16} = 1 - \dfrac{9}{16} = \dfrac{7}{16}.$

Thus, the correct answer is E.

Solution:

The recurrence $a_{k+1} = a_{k-2} + a_{k-1} - a_k$ gives $a_{k+1} - a_{k-1} = -(a_k - a_{k-2}).$ The sequence begins $2001, 2002, 2003, 2000, 2005, 1998, \ldots$

So the even-position terms form the arithmetic sequence $2002, 2000, 1998, \ldots$ with common difference $-2.$ The $2004$th term is its $1002$nd term, $2002 + 1001(-2) = 0.$

Thus, the correct answer is C.

Solution:

Let $F$ be on $\overline{AC}$ with $DF \parallel BE,$ and write $ET = x,$ $BT = 4x.$

From $\triangle ATE \sim \triangle ADF,$ $\dfrac{DF}{x} = \dfrac{AD}{AT} = \dfrac{4}{3},$ so $DF = \dfrac{4x}{3}.$

From $\triangle BEC \sim \triangle DFC,$ $\dfrac{CD}{BC} = \dfrac{DF}{BE} = \dfrac{4x/3}{5x} = \dfrac{4}{15}.$

Therefore $\dfrac{CD}{BD} = \dfrac{CD/BC}{1 - CD/BC} = \dfrac{4/15}{11/15} = \dfrac{4}{11}.$

Thus, the correct answer is D.

Solution:

The first sequence is $1 + 3k$ with largest term $6010,$ and the second is $9 + 7j$ with a much larger last term, so the binding limit is $6010.$

A common value has the form $16 + 21m$ (the first shared term is $16,$ spaced by $\mathrm{lcm}(3, 7) = 21$). Requiring $16 + 21m \le 6010$ gives $0 \le m \le 285,$ that is $286$ common numbers.

The number of distinct values is $2004 + 2004 - 286 = 3722.$

Thus, the correct answer is A.

Solution:

Since $5^2 + 12^2 = 13^2,$ the triangle is right. Place it at $(0, 0),$ $(5, 0),$ $(0, 12).$ The circumcenter is the midpoint of the hypotenuse, $\left(\tfrac52, 6\right).$

The inradius satisfies $(12 - r) + (5 - r) = 13,$ so $r = 2$ and the incenter is $(2, 2).$

The distance is $\sqrt{\left(\tfrac52 - 2\right)^2 + (6 - 2)^2} = \sqrt{\tfrac14 + 16} = \dfrac{\sqrt{65}}{2}.$

Thus, the correct answer is D.

Solution:

Fixing the orientation, there are $2^6 = 64$ colorings.

A coloring works if all six faces match ($2$ ways), exactly five match ($\binom{6}{5} \cdot 2 = 12$ ways), or four faces share a color with the remaining pair being opposite faces of the other color ($3$ opposite pairs, $2$ colors, giving $6$ ways).

The total is $2 + 12 + 6 = 20,$ so the probability is $\dfrac{20}{64} = \dfrac{5}{16}.$

Thus, the correct answer is B.

Solution:

Let $\overline{AD}$ meet $\overline{BC}$ at $E.$ Since $\angle ABC$ and $\angle ADC$ subtend the same arc, they are equal, and $\angle EAB = \angle CAD,$ so $\triangle ABE \sim \triangle ADC.$

Hence $\dfrac{AD}{CD} = \dfrac{AB}{BE}.$

By the Angle Bisector Theorem, $\dfrac{BE}{EC} = \dfrac{AB}{AC},$ so $BE = \dfrac{AB \cdot BC}{AB + AC} = \dfrac{7 \cdot 9}{15}.$

Therefore $\dfrac{AD}{CD} = \dfrac{AB}{BE} = \dfrac{AB + AC}{BC} = \dfrac{15}{9} = \dfrac{5}{3}.$

Thus, the correct answer is B.

Solution:

Let the large circles have centers $A$ and $B,$ let $C$ be the center of the small circle, and let $D$ be a point where the two large circles meet.

Then $\triangle ACD$ is right with $AC = 1$ and $AD = 2,$ so $CD = \sqrt3,$ $\angle CAD = 60^\circ,$ and its area is $\dfrac{\sqrt3}{2}.$

One quarter of the shaded region equals the $60^\circ$ sector of the radius-$2$ circle (area $\dfrac{2\pi}{3}$) minus $\triangle ACD$ (area $\dfrac{\sqrt3}{2}$) minus a quarter of the small circle (area $\dfrac{\pi}{4}$), giving $\dfrac{2\pi}{3} - \dfrac{\sqrt3}{2} - \dfrac{\pi}{4} = \dfrac{5\pi}{12} - \dfrac{\sqrt3}{2}.$

Multiplying by $4,$ the shaded area is $\dfrac{5\pi}{3} - 2\sqrt3.$

Thus, the correct answer is B.