Solution:

Since $\$20$ is $2000$ cents, the tax is $$0.0145 \times 2000 = 29$$ cents per hour.

Thus, the correct answer is E.

Solution:

The $8$ unanswered problems are worth $2.5 \times 8 = 20$ points, so Charlyn needs at least $100 - 20 = 80$ more points from correct answers.

Each correct answer is worth $6$ points, and the smallest multiple of $6$ that is at least $80$ is $84 = 6 \times 14.$ So she needs at least $14$ correct answers.

Thus, the correct answer is C.

Solution:

Writing $x = 100 - 2y,$ the value of $x$ is a positive integer precisely when $y$ is a positive integer with $1 \le y \le 49.$

This gives $49$ valid ordered pairs.

Thus, the correct answer is B.

Solution:

Bertha has $30 - 6 = 24$ granddaughters, none of whom have daughters.

These granddaughters are the children of $24 / 6 = 4$ of Bertha's daughters, so exactly $4$ women have daughters.

Therefore the number of women with no daughters is $30 - 4 = 26.$

Thus, the correct answer is E.

Solution:

The $y$-intercept of the line is between $0$ and $1,$ so $0 \lt b \lt 1.$

The slope is negative and shallow, between $-1$ and $0,$ so $-1 \lt m \lt 0.$

The product $mb$ is therefore negative with absolute value less than $1,$ giving $-1 \lt mb \lt 0.$

Thus, the correct answer is B.

Solution:

Compute each difference by factoring:

$U - V = 2004^{2005},$ $V - W = 2004^{2004},$ $W - X = 2001 \cdot 2004^{2004},$ $X - Y = 2004^{2004},$ and $Y - Z = 2003 \cdot 2004^{2003}.$

Since $2004^{2005} = 2004 \cdot 2004^{2004}$ exceeds each of the others, none of which reaches $2004^{2005},$ the difference $U - V$ is the largest.

Thus, the correct answer is A.

Solution:

After three rounds the players $A,$ $B,$ and $C$ have $14,$ $13,$ and $12$ tokens, respectively. Every subsequent three rounds reduces each player's supply by one token.

After $36$ rounds they have $3,$ $2,$ and $1$ tokens. In the $37$th round player $A,$ who has the most, gives away all three of their tokens and runs out, ending the game.

Thus, the correct answer is B.

Solution:

Let $x,$ $y,$ and $z$ be the areas of $\triangle ADE,$ $\triangle BDC,$ and $\triangle ABD,$ respectively.

Then $\triangle ABE$ has area $\tfrac12 \cdot 4 \cdot 8 = 16 = x + z,$ and $\triangle ABC$ has area $\tfrac12 \cdot 4 \cdot 6 = 12 = y + z.$

The requested difference is $$x - y = (x + z) - (y + z) = 16 - 12 = 4.$$

Thus, the correct answer is B.

Solution:

Multiplying the diameter by $\tfrac54$ multiplies the base area by $\left(\tfrac54\right)^2 = \tfrac{25}{16}.$

To keep the volume fixed, the height must be multiplied by $\tfrac{16}{25} = 0.64.$ That is a decrease of $1 - 0.64 = 0.36,$ or $36\%.$

Thus, the correct answer is C.

Solution:

The sum of a set of consecutive integers equals the number of terms times their mean, and for consecutive integers the mean equals the median.

So the median is $$\dfrac{7^5}{49} = \dfrac{7^5}{7^2} = 7^3 = 343.$$

Thus, the correct answer is C.

Solution:

If Paula has $n$ coins, their total value is $20n$ cents. Adding a quarter gives $n + 1$ coins worth $20n + 25$ cents, which must also equal $21(n + 1)$ cents.

So $20n + 25 = 21(n + 1),$ giving $n = 4.$

Four coins totalling $80$ cents must be three quarters and one nickel, so the number of dimes is $0.$

Thus, the correct answer is A.

Solution:

Line $AC$ passes through $(0, 9)$ with slope $\tfrac{8 - 9}{2 - 0} = -\tfrac12,$ so its equation is $y = -\tfrac12 x + 9.$ Setting $y = x$ gives $A' = (6, 6).$

Line $BC$ passes through $(0, 12)$ with slope $-2,$ so $y = -2x + 12.$ Setting $y = x$ gives $B' = (4, 4).$

Then $$A'B' = \sqrt{(6 - 4)^2 + (6 - 4)^2} = 2\sqrt{2}.$$

Thus, the correct answer is B.

Solution:

There are $\binom{9}{2} = 36$ pairs of points, and each pair determines a line.

However, there are three horizontal, three vertical, and two diagonal lines that each pass through three collinear points of $S.$ Each such line is counted $3$ times, an overcount of $2$ per line.

With $8$ such lines, the number of distinct lines is $36 - 2 \cdot 8 = 20.$

Thus, the correct answer is B.

Solution:

The arithmetic progression is $9,$ $9 + d,$ $9 + 2d.$ After the additions, the geometric progression is $9,$ $11 + d,$ $29 + 2d.$

The geometric condition gives $(11 + d)^2 = 9(29 + 2d),$ which simplifies to $d^2 + 4d - 140 = 0,$ so $d = 10$ or $d = -14.$

The corresponding third terms $29 + 2d$ are $49$ and $1,$ so the smallest possible value is $1.$

Thus, the correct answer is A.

Solution:

Starting at opposite ends, when they first meet they have together run half the track. Between the first and second meetings, they together run a full track length.

Since Brenda runs at a constant speed and covered $100$ meters before the first meeting, she covers $2 \cdot 100 = 200$ meters between the two meetings.

Adding Sally's $150$ meters over that same interval gives a track length of $200 + 150 = 350$ meters.

Thus, the correct answer is C.

Solution:

The expression is defined if and only if $\log_{2003}(\log_{2002}(\log_{2001} x)) \gt 0,$ that is, $\log_{2002}(\log_{2001} x) \gt 1.$

This holds if and only if $\log_{2001} x \gt 2002,$ which is equivalent to $x \gt 2001^{2002}.$

Therefore $c = 2001^{2002}.$

Thus, the correct answer is B.

Solution:

Applying $f(2n) = n \cdot f(n)$ with $n = 2^{k},$ we get $f(2^{k+1}) = 2^{k} \cdot f(2^{k}).$

Unwinding from $f(2^1) = f(2) = 1 \cdot f(1) = 2^0,$ the exponents accumulate: $$f(2^n) = 2^{0 + 1 + 2 + \cdots + (n-1)} = 2^{n(n-1)/2}.$$

Therefore $f(2^{100}) = 2^{100 \cdot 99 / 2} = 2^{4950}.$

Thus, the correct answer is D.

Solution:

Let $F$ be the point where $CE$ touches the semicircle. Since $CB$ and $CF$ are both tangents from $C,$ we have $CF = CB = 2.$ Similarly, with $x = AE,$ the tangents from $E$ give $EF = EA = x.$

Thus $CE = CF + FE = 2 + x.$ In right triangle $CDE,$ where $CD = 2$ and $DE = 2 - x,$ $$(2 - x)^2 + 2^2 = (2 + x)^2.$$

Expanding gives $8x = 4,$ so $x = \tfrac12$ and $CE = 2 + \tfrac12 = \tfrac52.$

Thus, the correct answer is D.

Solution:

Circle $A$ has radius $1$ and passes through the center of $D$ while being internally tangent to $D,$ so $D$ has radius $2.$

Place the center of $D$ at the origin, with $A$ centered at $(-1, 0).$ By symmetry, $B$ has center $(x, y)$ and radius $r,$ with $C$ its mirror image across the horizontal axis, so the two congruent circles touch on that axis and $y = r.$

Internal tangency to $D$ gives $x^2 + y^2 = (2 - r)^2,$ and external tangency to $A$ gives $(x + 1)^2 + y^2 = (1 + r)^2.$

Subtracting and using $y = r$ yields $x = \tfrac23$ and $r = \tfrac89.$ The radius of circle $B$ is $\tfrac89.$

Thus, the correct answer is D.

Solution:

Represent the choices as a point $(a, b)$ in the unit square. Each of $a$ and $b$ rounds to $0$ if below $\tfrac12$ and to $1$ otherwise, while $c = a + b$ rounds based on $\tfrac12$ and $\tfrac32.$

The equation $A + B = C$ holds in these regions:

if $a + b \lt \tfrac12$ then $A = B = C = 0;$ if exactly one of $a, b$ is at least $\tfrac12$ and $a + b \lt \tfrac32$ then that variable rounds to $1 = C;$ and if $a + b \ge \tfrac32$ then $A = B = 1$ and $C = 2.$

These regions consist of two corner triangles of area $\tfrac18$ each and two central strips, with combined area $\tfrac34.$ Since the square has area $1,$ the probability is $\tfrac34.$

Thus, the correct answer is E.

Solution:

The series is geometric with first term $1$ and ratio $\cos^2 \theta,$ so its sum is $\dfrac{1}{1 - \cos^2 \theta} = \dfrac{1}{\sin^2 \theta} = 5.$

Thus $\sin^2 \theta = \tfrac15,$ and $$\cos 2\theta = 1 - 2\sin^2 \theta = 1 - \tfrac25 = \tfrac35.$$

Thus, the correct answer is D.

Solution:

Let the centers of the three unit spheres be $A,$ $B,$ $C,$ forming an equilateral triangle of side $2$ at height $1$ above the plane, and let $E$ be the center of the large sphere directly above the centroid $D$ of $\triangle ABC.$

The distance from a vertex to the centroid is $AD = \tfrac{2\sqrt3}{3},$ and $AE = 1 + 2 = 3,$ so $$DE = \sqrt{3^2 - \left(\tfrac{2\sqrt3}{3}\right)^2} = \sqrt{9 - \tfrac{4}{3}} = \dfrac{\sqrt{69}}{3}.$$

Since $D$ is $1$ unit above the plane and the top of the large sphere is $2$ units above $E,$ the total height is $$1 + \dfrac{\sqrt{69}}{3} + 2 = 3 + \dfrac{\sqrt{69}}{3}.$$

Thus, the correct answer is B.

Solution:

Since $z_1 = a_1 + b_1 i = 0$ is a root, $c_0 = P(0) = 0.$

The nonreal zeros occur in conjugate pairs, so $\sum b_k = 0,$ and the hypothesis then forces $\sum a_k = 0.$ The coefficient $c_{2003}$ equals $-c_{2004}$ times the sum of the roots $\sum a_k + i \sum b_k = 0,$ so $c_{2003} = 0.$

Because the degree is even, at least one of $z_2, \ldots, z_{2004}$ is real, making one $b_k = 0,$ so $b_2 b_3 \cdots b_{2004} = 0.$ Thus (A) through (D) all must be $0.$

On the other hand, $\sum_{k=1}^{2004} c_k = P(1),$ and a valid polynomial such as $P(x) = x(x - 2)(x - 3) \cdots (x - 2003)\left(x + \sum_{k=2}^{2003} k\right)$ has $P(1) \ne 0.$ So only $\sum c_k$ can be nonzero.

Thus, the correct answer is E.

Solution:

A radius-$1$ disk covers segment $\overline{AB}$ exactly when its center is within $1$ of both $A$ and $B.$ That region $R$ is the lens where the two unit circles centered at $A$ and $B$ overlap.

Each unit circle passes through the other's center, so the lens is bounded by two $120^\circ$ arcs. Two $120^\circ$ sectors of area $\tfrac{\pi}{3}$ overlap in two equilateral triangles of total area $\tfrac{\sqrt3}{2},$ giving $R$ area $\tfrac{2\pi}{3} - \tfrac{\sqrt3}{2}.$

The set $S$ consists of all points within $1$ of $R.$ Beyond $R$ itself, this adds two $60^\circ$ sectors of radius $1$ (each area $\tfrac{\pi}{6}$) and two $120^\circ$ annuli of outer radius $2$ and inner radius $1$ (each area $\pi$).

Therefore the area of $S$ is $$\left(\tfrac{2\pi}{3} - \tfrac{\sqrt3}{2}\right) + 2 \cdot \tfrac{\pi}{6} + 2\pi = 3\pi - \tfrac{\sqrt3}{2}.$$

Thus, the correct answer is C.

Solution:

Since $n^3 \cdot a_n = 133.\overline{133}_n = a_n + n^2 + 3n + 3,$ we get $$a_n = \dfrac{n^2 + 3n + 3}{n^3 - 1} = \dfrac{(n+1)^3 - 1}{n(n^3 - 1)}.$$

Writing $n^3 - 1 = (n - 1)(n^2 + n + 1)$ and $(n+1)^3 - 1 = n\big((n+1)^2 + (n+1) + 1\big),$ the product $a_4 a_5 \cdots a_{99}$ telescopes to $$\dfrac{3!}{99!} \cdot \dfrac{100^3 - 1}{6^3} = \dfrac{3!}{99!} \cdot \dfrac{99(100^2 + 100 + 1)}{63}.$$

This simplifies to $\dfrac{(2)(10101)}{(21)(98!)} = \dfrac{962}{98!},$ so $m = 962$ (with the smallest possible $n = 98$).

Thus, the correct answer is E.