Solution:

Since $12^{12} = 2^{24} 3^{12},$ $6^6 = 2^6 3^6,$ and $8^8 = 2^{24},$ the number $k$ can involve no primes other than $2$ and $3,$ so write $k = 2^a 3^b.$ The least common multiple of the three numbers is then $2^{\max(24,\,a)} \, 3^{\max(6,\,b)}.$

Matching this to $2^{24} 3^{12}$ requires $\max(24, a) = 24,$ i.e. $0 \le a \le 24,$ and $\max(6, b) = 12,$ i.e. $b = 12.$ That gives $25$ choices for $a$ and one for $b,$ so there are $25$ values of $k.$

Solution:

The chains unpack into four conditions: $x \le 2y,$ $2y \le 60,$ $y \le 2x,$ and $2x \le 60.$ So $(x, y)$ lies in the square $1 \le x, y \le 30,$ and within it we must avoid $x \gt 2y$ and $y \gt 2x,$ which cannot both happen.

Pairs with $y \gt 2x:$ for each $x$ from $1$ to $14$ the values $y = 2x + 1, \ldots, 30$ work, giving $\sum_{x=1}^{14} (30 - 2x) = 420 - 210 = 210$ pairs. By the symmetry swapping $x$ and $y,$ there are also $210$ pairs with $x \gt 2y.$

The answer is $30 \cdot 30 - 210 - 210 = 480.$

Solution:

For $x \ge 0$ rewrite the equation as $2x(y + 20) = 400 - y^2 = (20 - y)(20 + y),$ so either $y = -20$ or $y = 20 - 2x.$ For $x \le 0$ it becomes $2x(y - 20) = -(y - 20)(y + 20),$ so either $y = 20$ or $y = -20 - 2x.$ The graph therefore consists of two horizontal rays and two rays of slope $-2.$

These rays bound a parallelogram: the top edge runs from $(-20, 20)$ to $(0, 20)$ along $y = 20,$ the bottom edge from $(0, -20)$ to $(20, -20)$ along $y = -20,$ and the two slanted edges of slope $-2$ connect them.

The parallelogram has horizontal base $20$ and height $40$ between the lines $y = 20$ and $y = -20,$ so its area is $20 \cdot 40 = 800.$

Solution:

A player's three tiles have an odd sum exactly when the player holds an odd number of odd tiles — one or three. The nine tiles include five odd and four even, and the only way to split five odd tiles into three groups of size one or three is $3 + 1 + 1.$

Count favorable deals: choose which player gets three odd tiles ($3$ ways), choose that player's odd tiles ($\binom{5}{3} = 10$ ways), give one of the two remaining odd tiles to each other player ($2$ ways), then split the four even tiles two and two between those players ($\binom{4}{2} = 6$ ways), for $3 \cdot 10 \cdot 2 \cdot 6 = 360$ deals. The total number of deals is $\binom{9}{3}\binom{6}{3} = 84 \cdot 20 = 1680.$

The probability is $\frac{360}{1680} = \frac{3}{14},$ so $m + n = 3 + 14 = 17.$

Solution:

Since $k(k-1)$ is even, $n_k = \frac{k(k-1)}{2}$ is an integer and $\cos\frac{k(k-1)\pi}{2} = \cos(n_k \pi) = (-1)^{n_k}.$ The parity of the triangular number $n_k$ depends only on $k \bmod 4:$ it is even for $k \equiv 0, 1 \pmod 4$ and odd for $k \equiv 2, 3 \pmod 4.$ So $A_k = n_k$ when $k \equiv 0, 1 \pmod 4$ and $A_k = -n_k$ when $k \equiv 2, 3 \pmod 4.$

Group the $80$ terms into $20$ consecutive blocks of four starting at $k = 19 \equiv 3 \pmod 4.$ Using $n_{j+1} - n_j = j,$ each block with $k \equiv 3 \pmod 4$ collapses: $$A_k + A_{k+1} + A_{k+2} + A_{k+3} = (n_{k+1} - n_k) - (n_{k+3} - n_{k+2}) = k - (k + 2) = -2.$$

The total is $20 \cdot (-2) = -40,$ so the requested absolute value is $40.$

Solution:

Let $a = \frac{PA}{AD}.$ Since $AQ \parallel DC,$ triangles $PAQ$ and $PDC$ are similar, so $\frac{PQ}{PC} = \frac{PA}{PD} = \frac{a}{a+1}.$ Since $BC \parallel AD,$ i.e. $BC \parallel PD,$ triangles $RBC$ and $RDP$ are similar, so $\frac{RC}{RP} = \frac{BC}{PD} = \frac{1}{a+1},$ which gives $\frac{RC}{PC} = \frac{1}{a+2}.$

Writing $PC = L,$ we get $PQ = \frac{a}{a+1}L,$ $RC = \frac{L}{a+2},$ and $$QR = L - PQ - RC = \frac{L}{(a+1)(a+2)}.$$ Hence $\frac{PQ}{QR} = a(a+2) = \frac{735}{112} = \frac{105}{16},$ so $16a^2 + 32a - 105 = 0,$ which factors as $(4a - 7)(4a + 15) = 0,$ giving $a = \frac{7}{4}.$

Finally $RC = (a + 1)\,QR = \frac{11}{4} \cdot 112 = 308.$

Solution:

Write $x_i = 2y_i - 1$ where each $y_i$ is a positive integer. Then $x_1 + x_2 + x_3 + x_4 = 98$ becomes $2(y_1 + y_2 + y_3 + y_4) - 4 = 98,$ so $y_1 + y_2 + y_3 + y_4 = 51.$

By stars and bars, the number of solutions in positive integers is $\binom{50}{3} = 19600.$ Therefore $\frac{n}{100} = 196.$

Solution:

Computing terms, $a_3 = 1000 - x,$ $a_4 = 2x - 1000,$ $a_5 = 2000 - 3x,$ $a_6 = 5x - 3000,$ and in general $$a_{2k+1} = 1000 F_{2k-1} - x F_{2k}, \qquad a_{2k+2} = x F_{2k+1} - 1000 F_{2k},$$ where $F_1 = F_2 = 1,$ $F_3 = 2, \ldots$ are the Fibonacci numbers. The sequence keeps going exactly as long as its terms stay nonnegative, so a long sequence requires $\frac{x}{1000}$ to be squeezed between the ratios $\frac{F_{2k}}{F_{2k+1}}$ and $\frac{F_{2k-1}}{F_{2k}}$ for larger and larger $k.$

For the first $13$ terms to be nonnegative we need $a_{12} = 89x - 55000 \ge 0$ and $a_{13} = 89000 - 144x \ge 0,$ i.e. $617.9\ldots \le x \le 618.05\ldots,$ so $x = 618.$ If $x \le 617$ the sequence turns negative by $a_{12},$ and if $x \ge 619$ it turns negative by $a_{13},$ so every other integer gives a shorter sequence.

Indeed $x = 618$ yields $1000, 618, 382, 236, 146, 90, 56, 34, 22, 12, 10, 2, 8, -6,$ a sequence of $14$ terms, the maximum possible. The answer is $618.$

Solution:

Let the arrival times be $x$ and $y$ minutes after 9 a.m., so $(x, y)$ is uniform in a $60 \times 60$ square. The two people meet exactly when $|x - y| \lt m.$

The non-meeting region $|x - y| \ge m$ consists of two right triangles with legs $60 - m,$ with total area $(60 - m)^2.$ Meeting with probability $40\%$ means $$(60 - m)^2 = 0.6 \cdot 3600 = 2160,$$ so $60 - m = \sqrt{2160} = 12\sqrt{15}.$

Thus $m = 60 - 12\sqrt{15},$ and $a + b + c = 60 + 12 + 15 = 87.$

Solution:

The eight centers are at height $100,$ at the vertices of a regular octagon of side $200$ (adjacent spheres are tangent). If the ninth sphere has radius $r,$ it rests on the surface with its center at height $r$ directly above the octagon's center, and tangency to each sphere gives $\sqrt{R^2 + (r - 100)^2} = r + 100,$ where $R$ is the octagon's circumradius. Hence $$R^2 = (r + 100)^2 - (r - 100)^2 = 400r.$$

A side of a regular octagon subtends $45^\circ$ at the center, so $200 = 2R \sin 22.5^\circ$ and, using $\sin^2 22.5^\circ = \frac{1 - \cos 45^\circ}{2} = \frac{2 - \sqrt{2}}{4},$ $$R^2 = \frac{10000}{\sin^2 22.5^\circ} = \frac{40000}{2 - \sqrt{2}} = 20000\,(2 + \sqrt{2}).$$

Then $r = \frac{R^2}{400} = 50\,(2 + \sqrt{2}) = 100 + 50\sqrt{2},$ so $a + b + c = 100 + 50 + 2 = 152.$

Solution:

The cube has side $20.$ Take $B = (0,0,0),$ $A = (20,0,0),$ $C = (0,20,0),$ and $D = (0,20,20),$ so $\overline{AD}$ is an interior diagonal. Then $P = (15, 0, 0),$ $Q = (0, 15, 0),$ $R = (0, 20, 10),$ and the plane through them is $2x + 2y - z = 30.$

Evaluating $2x + 2y - z$ at the cube's vertices and checking all twelve edges, the plane also crosses the edges at $(5, 20, 20),$ $(20, 5, 20),$ and $(20, 0, 10),$ so the cross-section is the hexagon with vertices $(15,0,0),$ $(0,15,0),$ $(0,20,10),$ $(5,20,20),$ $(20,5,20),$ $(20,0,10)$ in order. Its projection onto the $xy$-plane is the hexagon $(15,0),$ $(0,15),$ $(0,20),$ $(5,20),$ $(20,5),$ $(20,0),$ whose area by the shoelace formula is $175.$

The plane's unit normal $\frac{1}{3}(2, 2, -1)$ has vertical component of magnitude $\frac{1}{3},$ so projecting onto the $xy$-plane multiplies area by $\frac{1}{3}.$ The cross-section therefore has area $3 \cdot 175 = 525.$

Solution:

Place $A = (0, \sqrt{3}),$ $B = (-1, 0),$ $C = (1, 0),$ so $D = (0, 0),$ $E = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right),$ $F = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right).$ The $120^\circ$ rotation about the center $G$ sends $A \to B \to C$ and $D \to E \to F,$ so we may take the symmetric configuration $P = D + t(E - D),$ $Q = E + t(F - E),$ $R = F + t(D - F);$ the rotation then carries the condition "$P$ on $\overline{CQ}$" to the other two conditions, so it suffices to make $C,$ $P,$ $Q$ collinear.

With $P = \left(\frac{t}{2}, \frac{t\sqrt{3}}{2}\right)$ and $Q = \left(\frac{1}{2} - t, \frac{\sqrt{3}}{2}\right),$ the cross product of $\overrightarrow{CP}$ and $\overrightarrow{CQ}$ is a multiple of $1 - t - t^2,$ so $t^2 + t - 1 = 0$ and $t = \frac{\sqrt{5} - 1}{2}.$ Both triangles are equilateral with center $G = \left(0, \frac{\sqrt{3}}{3}\right),$ so the area ratio is $\frac{GA^2}{GP^2}.$ Using $t^2 = 1 - t,$ $$GP^2 = \frac{t^2}{4} + 3\left(\frac{t}{2} - \frac{1}{3}\right)^2 = t^2 - t + \frac{1}{3} = \frac{7}{3} - \sqrt{5}, \qquad GA^2 = \frac{4}{3}.$$

Hence $$\frac{[ABC]}{[PQR]} = \frac{4/3}{7/3 - \sqrt{5}} = \frac{4}{7 - 3\sqrt{5}} = 7 + 3\sqrt{5},$$ so $a = 7,$ $b = 3,$ $c = 5,$ and $a^2 + b^2 + c^2 = 49 + 9 + 25 = 83.$

Solution:

Split the nonempty subsets of $\{1, \ldots, 9\}$ by whether they contain $9.$ Those without $9$ contribute $S_8.$ A subset containing $9$ is $T \cup \{9\}$ for a (possibly empty) $T \subseteq \{1, \ldots, 8\},$ and since $9$ is its largest element, its complex power sum is the complex power sum of $T$ plus $9i^{|T| + 1}.$ Summing over all $T$ gives another $S_8$ plus $$\sum_{k=0}^{8} \binom{8}{k}\, 9\, i^{k+1} = 9i\,(1 + i)^8.$$

Since $(1 + i)^2 = 2i,$ we get $(1 + i)^8 = (2i)^4 = 16,$ so $$S_9 = 2S_8 + 144i = 2(-176 - 64i) + 144i = -352 + 16i.$$

Therefore $|p| + |q| = 352 + 16 = 368.$

Solution:

The condition $2mnp = (m + 2)(n + 2)(p + 2)$ rewrites as $$\left(1 + \frac{2}{m}\right)\left(1 + \frac{2}{n}\right)\left(1 + \frac{2}{p}\right) = 2.$$ If $m = 1$ the first factor alone is $3 \gt 2,$ and if $m = 2$ it equals $2$ while the other factors exceed $1;$ both are impossible. If $m \ge 5,$ then since $n \ge m$ the first two factors are at most $\left(\frac{7}{5}\right)^2 = \frac{49}{25},$ forcing $1 + \frac{2}{p} \ge \frac{50}{49},$ i.e. $p \le 98.$

For $m = 4$ the equation becomes $4np = 3(n + 2)(p + 2),$ i.e. $(n - 6)(p - 6) = 48,$ so $p \le 54.$ For $m = 3$ it becomes $6np = 5(n + 2)(p + 2),$ i.e. $np - 10n - 10p - 20 = 0,$ or $(n - 10)(p - 10) = 120.$ Both factors must be positive (if $n, p \lt 10$ the product $(10 - n)(10 - p)$ is at most $49$), so the largest $p$ comes from $n - 10 = 1:$ $n = 11$ and $p = 130.$ Indeed $2 \cdot 3 \cdot 11 \cdot 130 = 8580 = 5 \cdot 13 \cdot 132.$

Since every other case yields $p \le 98,$ the largest possible value is $p = 130.$

Solution:

A domino $(i, j)$ is an oriented edge of the complete graph on vertices $1, \ldots, 40,$ and the rule that $(i, j)$ and $(j, i)$ cannot both appear means each of the $\binom{40}{2} = 780$ edges is available at most once. A proper sequence is exactly a trail: a walk that repeats no edge. In any trail, every vertex other than the two endpoints is entered and left equally often, so it has even degree in the set of edges used.

In the complete graph every vertex has odd degree $39,$ so at least $38$ vertices must have odd degree in the set of unused edges, and a graph with $38$ odd-degree vertices has at least $\frac{38}{2} = 19$ edges. Hence at most $780 - 19 = 761$ dominos can be used.

Conversely, set aside the $19$ disjoint edges $(3,4),$ $(5,6),$ $\ldots,$ $(39,40).$ The remaining graph is connected and only vertices $1$ and $2$ have odd degree, so it has an Euler trail traversing all $761$ remaining edges; orienting each edge in the direction of travel gives a proper sequence of length $761.$