The six centers form a regular hexagon of side $2r,$ and a regular hexagon’s circumradius equals its side

Internal tangency to $\mathcal{C}$ gives $2r + r = 30;$ then subtract the six small disks from the big one

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 $$\begin{aligned} K &= \pi\left(30^2 - 6 \cdot 10^2\right) \\ &= 300\pi \approx 942.48, \end{aligned}$$ and $\lfloor K\rfloor = 942.$

The term $2005$ appears in $S_k$ exactly when $1 + (n-1)k = 2005$ for some positive integer $n$

So $k$ must be a divisor of $2004 = 2^2 \cdot 3 \cdot 167;$ count the divisors from the prime factorization

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 $(2+1)(1+1)(1+1) = 12.$

A number has exactly four divisors in all — hence three proper ones — exactly when it is $pq$ for distinct primes $p$ and $q,$ or a prime cube $p^3$

Count pairs of distinct primes less than $50,$ then add the primes $p$ with $p^2 \lt 50$

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 $105 + 4 = 109.$

With $x$ columns and $x + 7$ rows, the count satisfies $x(x+7) = s^2 + 5$ for some integer $s$

Multiply by $4$ and complete the square: $(2x+7)^2 - (2s)^2 = 69.$ Factor $69$ as a product of two positive integers.

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 $$\begin{aligned} &(2x + 7 - 2s) \\ &\quad {}\cdot (2x + 7 + 2s) = 69. \end{aligned}$$

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.

Handle orientations and colors separately. Record each coin’s orientation, bottom to top, as $U$ (face up) or $D$ (face down).

Face to face happens exactly when a $U$ sits directly below a $D,$ so all the $D$’s must come before all the $U$’s

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 $9 \cdot 70 = 630.$

The left side is one term short of the binomial expansion of $(x - 1)^4;$ add $1$ to both sides

From $(x-1)^4 = 2006,$ the nonreal roots are $1 \pm i\sqrt[4]{2006};$ multiply this conjugate pair

Adding $1$ to both sides turns the left side into a perfect fourth power: $$\begin{aligned} &(x - 1)^4 = x^4 - 4x^3 + 6x^2 \\ &\quad {}- 4x + 1 = 2006. \end{aligned}$$ 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 $$\begin{aligned} &\left(1 + i\sqrt[4]{2006}\right) \\ &\quad {}\cdot \left(1 - i\sqrt[4]{2006}\right) \\ &= 1 + \sqrt{2006}. \end{aligned}$$ Since $44^2 = 1936$ $\lt 2006$ $\lt 2025 = 45^2,$ we have $45 \lt P \lt 46,$ so $\lfloor P\rfloor = 45.$

Extend $\overline{AD}$ and $\overline{BC}$ to meet at $P;$ the two $60^\circ$ angles make triangle $ABP$ equilateral

Apply the Law of Cosines in triangle $PDC,$ where $PD = AB - 10,$ $PC = AB - 8,$ and $\angle P = 60^\circ$

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 $$\begin{aligned} 144 &= (x-10)^2 + (x-8)^2 \\ &\quad {}- (x-10)(x-8) \\ &= x^2 - 18x + 84, \end{aligned}$$ so $x^2 - 18x - 60 = 0$ and $x = 9 + \sqrt{81 + 60} = 9 + \sqrt{141}.$

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

Substitute $y = 2^{111x}$ to turn the equation into a cubic in $y$

Add the logarithms of the three $y$-values, then use Vieta’s formulas for their product

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 $$\begin{aligned} x_1 + x_2 + x_3 &= \frac{1}{111}\log_2(y_1 y_2 y_3) \\ &= \frac{1}{111}\log_2 4 \\ &= \frac{2}{111}, \end{aligned}$$ using Vieta’s formulas for the product of the roots. Then $m + n = 2 + 111 = 113.$

Each small cube’s two unpainted faces share one edge of that cube; the surface is all orange exactly when every such edge is buried

A random orientation puts that edge uniformly among the $12$ edge positions. A corner cube hides $3$ of them, an edge cube $5,$ a face cube $8.$

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 $$\begin{aligned} &\left(\frac{1}{4}\right)^{8}\left(\frac{5}{12}\right)^{12}\left(\frac{2}{3}\right)^{6} \\ &= \frac{5^{12}}{2^{34} \cdot 3^{18}}, \end{aligned}$$ so $$\begin{aligned} &a + b + c + p + q + r \\ &\quad {}= 12 + 34 + 18 \\ &\qquad {}+ 5 + 2 + 3 \\ &\quad {}= 74. \end{aligned}$$