Solution:

Let $r$ be the common radius of the six circles. Adjacent circles are externally tangent, so their centers are $2r$ apart, and the six centers form a regular hexagon with side $2r.$ Since a regular hexagon's circumradius equals its side length, each center is at distance $2r$ from the center $O$ of $\mathcal{C}.$ Internal tangency to $\mathcal{C}$ means the distance from $O$ to each small center plus $r$ equals $30,$ so $3r = 30$ and $r = 10.$

Therefore $$K = \pi\left(30^2 - 6 \cdot 10^2\right) = 300\pi \approx 942.48,$$ and $\lfloor K\rfloor = 942.$

Solution:

The $n$th term of $S_k$ is $1 + (n-1)k,$ so $2005$ is a term exactly when $(n-1)k = 2004$ for some positive integer $n,$ that is, exactly when $k$ divides $2004.$ Every divisor works, since $n - 1 = \frac{2004}{k}$ is then a positive integer.

Since $2004 = 2^2 \cdot 3 \cdot 167,$ the number of divisors is $(2+1)(1+1)(1+1) = 12.$

Solution:

An integer with exactly three proper divisors has exactly four divisors in total, so it is either $n = pq$ with $p$ and $q$ distinct primes (proper divisors $1, p, q$) or $n = p^3$ with $p$ prime (proper divisors $1, p, p^2$).

In the first case we need $p$ and $q$ both less than $50.$ There are $15$ primes below $50,$ giving $\binom{15}{2} = 105$ such numbers. In the second case we need $p^2 \lt 50,$ which holds for $p = 2, 3, 5, 7,$ giving $4$ more.

The total is $105 + 4 = 109.$

Solution:

Let the band have $n$ members, with $n = s^2 + 5$ for the square formation and $n = x(x + 7)$ for the rectangular formation with $x$ columns. Multiplying $x^2 + 7x = s^2 + 5$ by $4$ and completing the square gives $$(2x + 7)^2 - (2s)^2 = 69,$$ so $(2x + 7 - 2s)(2x + 7 + 2s) = 69.$

Writing $69 = 1 \cdot 69 = 3 \cdot 23$ with the larger factor second: from $1 \cdot 69$ we get $2x + 7 = 35$ and $2s = 34,$ so $x = 14,$ $s = 17,$ and $n = 17^2 + 5 = 294.$ From $3 \cdot 23$ we get $2x + 7 = 13$ and $2s = 10,$ so $x = 3,$ $s = 5,$ and $n = 30.$

The maximum is $294,$ achieved by a $21 \times 14$ rectangle.

Solution:

Choose the coin orientations and the gold/silver positions independently. Record the orientations from bottom to top as a string of U (engraved face up) and D (engraved face down). Two adjacent coins are face to face exactly when the lower coin's engraved side faces up and the upper coin's engraved side faces down — that is, exactly when a U is immediately followed by a D.

A string of U's and D's avoids the pattern UD exactly when every D precedes every U, so the string is $\text{D}^i\text{U}^{8-i}$ for some $i = 0, 1, \ldots, 8:$ there are $9$ allowable orientation strings. Independently, the gold coins occupy $4$ of the $8$ positions in $\binom{8}{4} = 70$ ways.

The total is $9 \cdot 70 = 630.$

Solution:

Adding $1$ to both sides turns the left side into a perfect fourth power: $$(x - 1)^4 = x^4 - 4x^3 + 6x^2 - 4x + 1 = 2006.$$ So $x - 1$ is a fourth root of $2006:$ the four roots are $x = 1 \pm \sqrt[4]{2006}$ (real) and $x = 1 \pm i\sqrt[4]{2006}$ (nonreal).

The product of the conjugate pair of nonreal roots is $$\left(1 + i\sqrt[4]{2006}\right)\left(1 - i\sqrt[4]{2006}\right) = 1 + \sqrt{2006}.$$ Since $44^2 = 1936 \lt 2006 \lt 2025 = 45^2,$ we have $45 \lt P \lt 46,$ so $\lfloor P\rfloor = 45.$

Solution:

Extend rays $AD$ and $BC$ until they meet at $P.$ Triangle $ABP$ has $60^\circ$ angles at $A$ and $B,$ so it is equilateral: $PA = PB = AB.$ Writing $x = AB,$ we get $PD = PA - AD = x - 10$ and $PC = PB - BC = x - 8.$

The Law of Cosines in triangle $PDC,$ with $\angle P = 60^\circ$ and $DC = 12,$ gives $$144 = (x-10)^2 + (x-8)^2 - (x-10)(x-8) = x^2 - 18x + 84,$$ so $x^2 - 18x - 60 = 0$ and $x = 9 + \sqrt{81 + 60} = 9 + \sqrt{141}.$

Thus $p + q = 9 + 141 = 150.$

Solution:

Let $y = 2^{111x}.$ Then $2^{333x-2} = \frac{y^3}{4},$ $2^{111x+2} = 4y,$ and $2^{222x+1} = 2y^2,$ so the equation becomes $\frac{y^3}{4} + 4y = 2y^2 + 1,$ that is, $$y^3 - 8y^2 + 16y - 4 = 0.$$ Since the three roots $x_1, x_2, x_3$ are real and $y = 2^{111x}$ is strictly increasing, they correspond to three positive real roots $y_1, y_2, y_3$ of the cubic.

Each $x_i = \frac{1}{111}\log_2 y_i,$ so $$x_1 + x_2 + x_3 = \frac{1}{111}\log_2(y_1 y_2 y_3) = \frac{1}{111}\log_2 4 = \frac{2}{111},$$ using Vieta's formulas for the product of the roots. Then $m + n = 2 + 111 = 113.$

Solution:

Each unit cube has one "bad edge": the edge shared by its two unpainted faces. The larger cube's surface is entirely orange exactly when every unit cube's bad edge touches no visible face. A uniformly random orientation places the bad edge uniformly among the cube's $12$ edge positions, so for each unit cube we count the edge positions both of whose faces are hidden.

A corner cube shows $3$ faces meeting at a vertex; the safe edges are those of the $3$ hidden faces meeting at the opposite vertex, so the probability is $\frac{3}{12} = \frac{1}{4}.$ An edge cube shows $2$ adjacent faces, which touch $4 + 4 - 1 = 7$ edges, leaving $5$ safe: probability $\frac{5}{12}.$ A face-center cube shows $1$ face touching $4$ edges, leaving $8$ safe: probability $\frac{8}{12} = \frac{2}{3}.$ The center cube is always fine.

With $8$ corner, $12$ edge, and $6$ face-center cubes, the probability is $$\left(\frac{1}{4}\right)^{8}\left(\frac{5}{12}\right)^{12}\left(\frac{2}{3}\right)^{6} = \frac{5^{12}}{2^{34} \cdot 3^{18}},$$ so $a + b + c + p + q + r = 12 + 34 + 18 + 5 + 2 + 3 = 74.$

Solution:

The median to $\overline{BC}$ passes through the midpoint $M = \left(\frac{35}{2}, \frac{39}{2}\right)$ of $\overline{BC}.$ The line through $M$ with slope $-5$ is $y = -5x + 107,$ and $A$ lies on this line, so $A = (p, -5p + 107)$ and $q = -5p + 107.$

By the shoelace formula with $B = (12, 19)$ and $C = (23, 20),$ $$[ABC] = \frac{1}{2}\left|{-p} + 12\bigl(20 - q\bigr) + 23\bigl(q - 19\bigr)\right| = \frac{1}{2}\left|980 - 56p\right| = 70,$$ so $|56p - 980| = 140,$ giving $p = 15$ or $p = 20.$

Since $p + q = p + (-5p + 107) = 107 - 4p,$ the smaller value $p = 15$ gives the larger sum $107 - 60 = 47.$

Solution:

Scale to a semicircle of radius $1$ and ask for the smallest square containing it when its diameter makes angle $\theta$ with one pair of sides, where $0 \le \theta \le 90^\circ.$ Squeeze the semicircle between two pairs of parallel lines in the square's two side directions: in each direction one line of the pair is tangent to the arc and the other passes through an endpoint of the diameter, and the distances between the pairs are $1 + \cos\theta$ and $1 + \sin\theta.$ So the smallest enclosing square in that orientation has side $\max\{1 + \cos\theta,\ 1 + \sin\theta\},$ which is minimized when $\theta = 45^\circ,$ giving side $1 + \frac{\sqrt{2}}{2} = \frac{2 + \sqrt{2}}{2}.$

Scaling this optimal configuration so the square has side $8,$ the radius becomes $r = \frac{8}{(2 + \sqrt{2})/2} = \frac{16}{2 + \sqrt{2}} = 8\left(2 - \sqrt{2}\right),$ so $$d = 2r = 16\left(2 - \sqrt{2}\right) = 32 - 16\sqrt{2} = 32 - \sqrt{512}.$$

Thus $m + n = 32 + 512 = 544.$

Solution:

Divisors of $n$ pair up as $d$ and $\frac{n}{d},$ so $\tau(n)$ is odd exactly when $n$ is a perfect square. Hence $S(n)$ changes parity exactly at the squares, which means $S(n)$ is odd exactly when the number of squares up to $n,$ namely $\lfloor\sqrt{n}\rfloor,$ is odd.

For each $k,$ there are $2k + 1$ integers $n$ with $\lfloor\sqrt{n}\rfloor = k,$ namely $k^2 \le n \le k^2 + 2k.$ Since $44^2 = 1936 \le 2005 \lt 2025 = 45^2,$ the odd values $k = 1, 3, \ldots, 43$ all have their full blocks within range, so $$a = \sum_{k \text{ odd},\, k \le 43} (2k + 1) = 2(1 + 3 + \cdots + 43) + 22 = 2 \cdot 484 + 22 = 990.$$

Then $b = 2005 - 990 = 1015,$ and $|a - b| = 25.$

Solution:

The forbidden right-angle turns say exactly that a rightward step may never immediately follow an upward step, and vice versa; a diagonal step may follow or precede anything. So at each lattice point $(x, y)$ track three counts $D(x,y),$ $R(x,y),$ $U(x,y):$ the numbers of legal paths from $(0,0)$ arriving there by a diagonal, rightward, or upward step. The rules give $$D(x,y) = D + R + U \text{ at } (x-1, y-1), \quad R(x,y) = D(x-1,y) + R(x-1,y), \quad U(x,y) = D(x,y-1) + U(x,y-1).$$

Starting from the single empty path at $(0,0)$ (which may begin with any step), fill in the grid up to $(5,5).$ Along the axes only all-rightward or all-upward paths survive, and the interior builds up quickly; at $(5, 5)$ the three counts come out to $27,$ $28,$ and $28.$

The total number of paths is $27 + 28 + 28 = 83.$

Solution:

Since segments $\overline{AC}$ and $\overline{BD}$ cross, $A$ and $C$ lie on opposite sides of the square, as do $B$ and $D.$ Let $m$ be the slope of the side through $B,$ so that side lies on $mx - y + 9 - 10m = 0,$ and the perpendicular side through $C$ lies on $x + my - 8 = 0.$ The side length of the square equals both the distance between the parallel sides through $B$ and $D$ and the distance between the sides through $A$ and $C:$ $$\frac{|{-4m} - 7 + 9 - 10m|}{\sqrt{m^2 + 1}} = \frac{|12m - 8|}{\sqrt{m^2 + 1}},$$ so $|2 - 14m| = |12m - 8|,$ giving $m = \frac{5}{13}$ or $m = -3.$

For $m = \frac{5}{13},$ the points $A$ and $C$ fall on opposite sides of the line through $B,$ which is impossible if that line contains a side of the square, so $m = -3.$ Then the side length is $\frac{|12(-3) - 8|}{\sqrt{9 + 1}} = \frac{44}{\sqrt{10}},$ so $$K = \frac{44^2}{10} = \frac{1936}{10} \qquad \text{and} \qquad 10K = 1936.$$

The remainder when $1936$ is divided by $1000$ is $936.$

Solution:

Let $M$ be the midpoint of $\overline{AB},$ and let the incircle cut median $\overline{CM}$ at $S$ and $N,$ with $CS = SN = NM = \frac{1}{3}CM.$ Let the incircle touch $\overline{AB}$ at $T$ and $\overline{AC}$ at $R.$ By Power of a Point, $$MT^2 = MN \cdot MS = \frac{CM}{3} \cdot \frac{2\,CM}{3} = \frac{2}{9}CM^2, \qquad CR^2 = CS \cdot CN = \frac{2}{9}CM^2,$$ so $MT = CR.$ Since $AR = AT$ (tangents from $A$), we get $AC = AR + RC = AT + TM = AM = 10.$

Write $a = BC$ and $s = \frac{20 + a + 10}{2} = 15 + \frac{a}{2}.$ The standard tangent length gives $AT = s - a,$ so $MT = AM - AT = 10 - \left(15 - \frac{a}{2}\right) = \frac{a - 10}{2},$ while the median length formula gives $CM^2 = \frac{2 \cdot 10^2 + 2a^2 - 20^2}{4} = \frac{a^2 - 100}{2}.$ Substituting into $MT^2 = \frac{2}{9}CM^2:$ $$\frac{(a - 10)^2}{4} = \frac{a^2 - 100}{9} \quad\Longrightarrow\quad 9(a - 10) = 4(a + 10) \quad\Longrightarrow\quad a = 26.$$

Then the sides are $20,$ $26,$ $10$ with $s = 28,$ and Heron's formula gives $$[ABC] = \sqrt{28 \cdot 8 \cdot 2 \cdot 18} = \sqrt{8064} = 24\sqrt{14},$$ so $m + n = 24 + 14 = 38.$