Solution:

Five sandwiches cost $5\cdot 3 = 15$ dollars and eight sodas cost $8\cdot 2 = 16$ dollars. Together they cost $15 + 16 = 31$ dollars.

Thus, the correct answer is A.

Solution:

By the definition, $h \otimes h = h^3 - h.$ Then $$h \otimes (h^3 - h) = h^3 - (h^3 - h) = h.$$

Thus, the correct answer is C.

Solution:

Mary's age is $\tfrac{3}{5}$ of Alice's, so Mary is $\tfrac{3}{5}\cdot 30 = 18$ years old.

Thus, the correct answer is B.

Solution:

The two minutes digits sum to at most $5 + 9 = 14,$ at $59$ minutes past the hour. For the hour, a single digit $9$ gives digit sum $9,$ which beats any two-digit hour $(10, 11, 12$ give at most $1 + 2 = 3).$

The largest total is $14 + 9 = 23,$ occurring at $9\!:\!59.$

Thus, the correct answer is E.

Solution:

Each plain slice costs $\$1.$ The $\$2$ anchovy charge is spread over the $4$ anchovy slices, adding $\$0.50$ each, so an anchovy slice costs $\$1.50.$

Dave ate $4$ anchovy slices and $1$ plain slice: $4\cdot 1.5 + 1 = \$7.$ Doug ate the $3$ remaining plain slices: $\$3.$ Dave paid $7 - 3 = \$4$ more.

Thus, the correct answer is D.

Solution:

The two hexagons form a square of area $8 \cdot 18 = 144,$ so the square has side $12.$

The staircase cut splits the width into three equal horizontal pieces of length $y,$ which together span the full width: $y + y + y = 18,$ so $y = 6.$ (The two vertical steps each rise $12 - 8 = 4,$ building the extra height of the square.)

Thus, the correct answer is A.

Solution:

Let Danielle be $x$ years old. Then Sally is $0.6x$ and Mary is $1.2(0.6x) = 0.72x.$

The sum $x + 0.6x + 0.72x = 2.32x = 23.2$ gives $x = 10.$ So Mary is $0.72(10) = 7.2$ years old, and on her next birthday she will be $8.$

Thus, the correct answer is B.

Solution:

The sum of $n$ consecutive integers equals $n$ times their median. For a sum of $15:$ $n = 2$ gives $7 + 8,$ $n = 3$ gives $4 + 5 + 6,$ and $n = 5$ gives $1 + 2 + 3 + 4 + 5.$

A run of four consecutive integers sums to an even number, and more than five terms already exceed $1 + 2 + 3 + 4 + 5 = 15.$ So there are $3$ sets.

Thus, the correct answer is C.

Solution:

Let $p$ be a pencil's cost and $s$ the cost of one pencil plus one eraser, in cents. Then $$13p + 3e = 3s + 10p = 100,$$ so $3s$ is a multiple of $10$ less than $100.$ Hence $s \in \{10, 20, 30\},$ with $p = 7, 4, 1$ respectively.

Since a pencil costs more than an eraser, $p \gt \tfrac{s}{2},$ which holds only for $s = 10$ (pencil $7,$ eraser $3).$ So one pencil and one eraser cost $10$ cents.

Thus, the correct answer is A.

Solution:

Let $k = \sqrt{120 - \sqrt{x}}$ be an integer. Then $k \ge 0$ and $k^2 = 120 - \sqrt{x} \le 120,$ so $0 \le k \le 10.$

Each such $k$ gives $\sqrt{x} = 120 - k^2 \ge 0,$ hence a distinct value $x = (120 - k^2)^2.$ That is $11$ values.

Thus, the correct answer is E.

Solution:

Expanding, $x^2 + 2xy + y^2 = x^2 + y^2,$ so $2xy = 0,$ i.e. $xy = 0.$

This is the union of the two coordinate axes, a pair of lines.

Thus, the correct answer is C.

Solution:

The top ring spans $20$ cm. Each ring below overlaps the ring above by $2$ cm (twice the $1$-cm thickness), so it adds its outside diameter minus $2.$

The lower rings have outside diameters $19, 18, \ldots, 3,$ contributing $17, 16, \ldots, 1.$ Thus the total distance is $$20 + (17 + 16 + \cdots + 1) = 20 + \frac{17 \cdot 18}{2} = 20 + 153 = 173 \text{ cm}.$$

Thus, the correct answer is B.

Solution:

If $r, s, t$ are the radii at the vertices, then $r + s = 3,\ r + t = 4,\ s + t = 5.$ Adding all three gives $r + s + t = 6,$ so $r = 1,\ s = 2,\ t = 3.$

The sum of the areas is $\pi(1^2 + 2^2 + 3^2) = 14\pi.$

Thus, the correct answer is E.

Solution:

A debt $D$ is resolvable if and only if $D = 300p + 210g = 30(10p + 7g)$ for integers $p, g.$ Thus $D$ is a multiple of $\gcd(300, 210) = 30,$ so no smaller positive debt works.

A debt of $\$30$ is achievable since $30 = 300(-2) + 210(3),$ i.e. give $3$ goats and receive $2$ pigs in change.

Thus, the correct answer is C.

Solution:

Since $\cos x = 0,$ we have $x = \tfrac{\pi}{2} + k\pi.$ Since $\cos(x + z) = \tfrac{1}{2},$ we have $x + z = 2n\pi \pm \tfrac{\pi}{3}.$

Taking $x = -\tfrac{\pi}{2}$ and $x + z = -\tfrac{\pi}{3}$ gives $z = -\tfrac{\pi}{3} + \tfrac{\pi}{2} = \tfrac{\pi}{6},$ the smallest positive value.

Thus, the correct answer is A.

Solution:

The radii satisfy $AC \perp CD$ and $BD \perp CD.$ By the Pythagorean theorem, $CE = \sqrt{5^2 - 3^2} = 4.$

Since $\triangle ACE \sim \triangle BDE,$ we get $\tfrac{DE}{CE} = \tfrac{BD}{AC} = \tfrac{8}{3},$ so $DE = 4 \cdot \tfrac{8}{3} = \tfrac{32}{3}.$ Then $$CD = CE + DE = 4 + \frac{32}{3} = \frac{44}{3}.$$

Thus, the correct answer is B.

Solution:

Set $B = (0, 0),$ $C = (s, 0),$ $A = (0, s),$ $D = (s, s),$ so that $E = \left(s + \tfrac{r}{\sqrt{2}},\ s + \tfrac{r}{\sqrt{2}}\right)$ lies on ray $BD.$

Since $AF$ is tangent to the circle, $AF^2 = AE^2 - r^2.$ Computing $AE^2$ and simplifying gives $9 + 5\sqrt{2} = s^2 + rs\sqrt{2}.$

Because $r$ and $s$ are rational, the rational and irrational parts match: $s^2 = 9$ and $rs = 5.$ Thus $s = 3,\ r = \tfrac{5}{3},$ and $r/s = \tfrac{5}{9}.$

Thus, the correct answer is B.

Solution:

Replacing $x$ by $1/x$ gives $f\!\left(\tfrac{1}{x}\right) + f(x) = \tfrac{1}{x}.$ Together with $f(x) + f\!\left(\tfrac{1}{x}\right) = x,$ this requires $x = \tfrac{1}{x},$ so $x = \pm 1.$

Both values are consistent, with $f(1) = \tfrac{1}{2}$ and $f(-1) = -\tfrac{1}{2}.$ So the largest possible domain is $\{-1, 1\}.$

Thus, the correct answer is E.

Solution:

Each circle's radius equals its center's $y$-coordinate, so both are tangent to the $x$-axis, which is a common external tangent. The two external tangents meet at the $x$-intercept of the line through the centers.

That line has slope $\tfrac{9 - 4}{14 - 2} = \tfrac{5}{12} = \tan\theta$ and passes through $(2, 4),$ meeting the $x$-axis at $\left(-\tfrac{38}{5}, 0\right).$

The other tangent makes angle $2\theta$ with the $x$-axis, so its slope is $$\tan 2\theta = \frac{2 \cdot \tfrac{5}{12}}{1 - \left(\tfrac{5}{12}\right)^2} = \frac{120}{119}.$$ Then $b = \tfrac{120}{119} \cdot \tfrac{38}{5} = \tfrac{912}{119}.$

Thus, the correct answer is E.

Solution:

From the start there are $3^7$ equally likely $7$-move walks. Consider a walk visiting all $8$ vertices: there are $3$ choices for the first move and $2$ for the second.

Labeling the first three vertices $A, B, C,$ the bug must next move to one of two vertices, and in each case the remaining moves are forced. This gives $3 \cdot 2 \cdot 3 = 18$ such walks.

The probability is $\dfrac{18}{3^7} = \dfrac{18}{2187} = \dfrac{2}{243}.$

Thus, the correct answer is C.

Solution:

For $j = 1, 2,$ the condition becomes $j + x^2 + y^2 \le 10^j(x + y),$ i.e. $$\left(x - \frac{10^j}{2}\right)^2 + \left(y - \frac{10^j}{2}\right)^2 \le \frac{10^{2j}}{2} - j.$$

These are disks with squared radii $\tfrac{100}{2} - 1 = 49$ for $S_1$ and $\tfrac{10000}{2} - 2 = 4998$ for $S_2.$ The area ratio is $\dfrac{4998}{49} = 102.$

Thus, the correct answer is E.

Solution:

Place the hexagon at the center of the circle. There are six congruent arcs from which three whole sides are visible; since the total probability is $\tfrac{1}{2},$ each arc measures $30^\circ.$

Take the arc centered at $(r, 0)$ with upper endpoint $P,$ so $\angle POA = 15^\circ.$ Then $P$ lies on the line containing a side whose distance from the center is the apothem $\sqrt{3}.$

Hence $\sqrt{3} = r\sin 15^\circ = r \cdot \dfrac{\sqrt{6} - \sqrt{2}}{4},$ giving $$r = \frac{4\sqrt{3}}{\sqrt{6} - \sqrt{2}} = 3\sqrt{2} + \sqrt{6}.$$

Thus, the correct answer is D.

Solution:

Each application of $A$ averages adjacent terms, so after $100$ steps the single remaining term is $$\frac{1}{2^{100}} \sum_{m=0}^{100} \binom{100}{m} x^m = \frac{(1 + x)^{100}}{2^{100}}.$$

Setting this equal to $\dfrac{1}{2^{50}}$ gives $(1 + x)^{100} = 2^{50},$ so $1 + x = 2^{1/2} = \sqrt{2}.$ Since $x \gt 0,$ we get $x = \sqrt{2} - 1.$

Thus, the correct answer is B.

Solution:

A term $x^a y^b z^c$ survives only when $a$ is even, since terms with odd $a$ cancel between the two expansions.

For each even $a$ with $0 \le a \le 2006,$ the exponent $b$ ranges over $2007 - a$ values and $c = 2006 - a - b$ is then determined. Summing over even $a:$ $$(2007 - 0) + (2007 - 2) + \cdots + (2007 - 2006) = 2007 + 2005 + \cdots + 1,$$ the sum of the first $1004$ odd positive integers, which is $1004^2 = 1{,}008{,}016.$

Thus, the correct answer is D.

Solution:

By property $(2),$ a valid $k$-element set is a $k$-subset of $\{k, k+1, \ldots, 15\}$ with no two consecutive elements.

Collapsing the gaps between chosen elements, these correspond bijectively to $k$-subsets of a $(17 - 2k)$-element set, counted by $\binom{17 - 2k}{k}.$ This is nonzero only for $k \le 5,$ so the total is $$\binom{15}{1} + \binom{13}{2} + \binom{11}{3} + \binom{9}{4} + \binom{7}{5} = 15 + 78 + 165 + 126 + 21 = 405.$$

Thus, the correct answer is E.