Solution:

Dividing by $4$ gives $5m + 3n = 503.$ Reducing modulo $3,$ we need $2m \equiv 503 \equiv 2 \pmod{3},$ so $m \equiv 1 \pmod{3}.$ Write $m = 3k + 1$ with $k \ge 0;$ then $3n = 503 - 5(3k + 1) = 498 - 15k,$ so $n = 166 - 5k.$

This is positive exactly when $5k \le 165,$ that is $k \le 33.$ So $k = 0, 1, \ldots, 33$ all work, giving $34$ ordered pairs.

Solution:

Let $r$ be the shared common ratio. Then $a_{15} = 27r^{14}$ and $b_{11} = 99r^{10},$ so $27r^{14} = 99r^{10}$ gives $r^4 = \frac{99}{27} = \frac{11}{3}.$

Therefore $$a_9 = 27r^8 = 27\left(\frac{11}{3}\right)^2 = 27 \cdot \frac{121}{9} = 3 \cdot 121 = 363.$$

Solution:

Each department contributes exactly two committee members. If every department sends one man and one woman, there are $2 \cdot 2 = 4$ choices per department, for $4^3 = 64$ committees.

Otherwise some department sends two men. To keep three of each gender, another department must then send its two women, and the remaining department sends one man and one woman. There are $3$ ways to pick the all-male department, $2$ ways to pick the all-female department, and $2 \cdot 2 = 4$ choices in the mixed department (the two-man and two-woman selections are forced), for $3 \cdot 2 \cdot 4 = 24$ committees.

The total is $64 + 24 = 88.$

Solution:

Let the field have length $L$ (west) and width $W$ (south) with $L \gt W,$ and let $x$ be the distance from $D$ to the southeast corner. Ana rides around the perimeter a distance $2L + W - x,$ Bob rides $W + x,$ and Cao rides $\sqrt{W^2 + x^2},$ all in the same time: $$\frac{2L + W - x}{8.6} = \frac{W + x}{6.2} = \frac{\sqrt{W^2 + x^2}}{5}.$$

The first equality gives $L = \frac{6W + 37x}{31}.$ Squaring the second, $25(W + x)^2 = 38.44\,(W^2 + x^2),$ which simplifies to $168W^2 - 625Wx + 168x^2 = 0,$ factoring as $(24W - 7x)(7W - 24x) = 0.$

The root $x = \frac{7W}{24}$ gives $L = \frac{13W}{24} \lt W,$ which is impossible, so $x = \frac{24W}{7}$ and then $L = \frac{30W}{7}.$ The ratio is $L : W : x = 30 : 7 : 24,$ and $p + q + r = 30 + 7 + 24 = 61.$

Solution:

Folding the star along the sides of $S'$ lifts the four triangular points so that their tips (the midpoints of the sides of $S$) meet at a single apex $V.$ Let $M$ be the center of $S'$ and $P$ the midpoint of one of its sides. In the flat figure, the distance from $P$ to the tip of its triangle is $20 - \frac{15}{2} = \frac{25}{2},$ and this becomes the slant $PV$ after folding.

Triangle $PMV$ has a right angle at $M,$ with $PM = \frac{15}{2},$ so the height is $$VM = \sqrt{\left(\tfrac{25}{2}\right)^2 - \left(\tfrac{15}{2}\right)^2} = \sqrt{100} = 10.$$ The volume is $\frac{1}{3} \cdot 15^2 \cdot 10 = 750.$

Solution:

The distance is $|(1 + 2i)z^3 - z^5| = |z|^3 \cdot |1 + 2i - z^2| = 125\,|1 + 2i - z^2|.$ As $z$ runs over the circle $|z| = 5$ with $b \gt 0,$ the square $z^2$ attains every point of the circle $|w| = 25$ (the condition $b \gt 0$ merely selects one of the two square roots). The point of that circle farthest from $1 + 2i$ is diametrically opposite in direction: $$z^2 = -25 \cdot \frac{1 + 2i}{|1 + 2i|} = -5\sqrt{5}\,(1 + 2i).$$

Squaring, $z^4 = 125\,(1 + 2i)^2 = 125\,(-3 + 4i) = -375 + 500i,$ so $c + d = -375 + 500 = 125.$

Solution:

There are $\binom{12}{8} = 495$ members of $S$ below $2^{12}$ and $\binom{13}{8} = 1287$ below $2^{13},$ so $N$ has $13$ binary digits, and exactly $1287 - 1000 = 287$ members below $2^{13}$ exceed it. The $\binom{11}{6} = 462$ members whose binary representations begin $11$ are the largest ones, so $N$ begins with $11$ and is the $462 - 287 = 175$th smallest of them.

Among these, $\binom{9}{6} = 84$ begin $1100,$ the next $\binom{8}{5} = 56$ begin $11010,$ and the next $\binom{7}{4} = 35$ begin $110110.$ Since $84 + 56 + 35 = 175,$ the number $N$ is the largest member beginning $110110,$ namely $1101101111000$ in binary.

Its value is $2^{12} + 2^{11} + 2^9 + 2^8 + 2^6 + 2^5 + 2^4 + 2^3 = 7032,$ so the remainder upon division by $1000$ is $32.$

Solution:

Multiplying the two equations gives $$zw + 12i + 20i - \frac{240}{zw} = (5 + i)(-4 + 10i) = -30 + 46i,$$ so $zw - \frac{240}{zw} = -30 + 14i.$ Setting $v = zw$ yields $v^2 + (30 - 14i)v - 240 = 0.$

By the quadratic formula, $v = -15 + 7i \pm \sqrt{(15 - 7i)^2 + 240} = -15 + 7i \pm \sqrt{416 - 210i}.$ Writing $(a + bi)^2 = 416 - 210i$ requires $a^2 - b^2 = 416$ and $ab = -105,$ which gives $a + bi = \pm(21 - 5i).$ Hence $v = 6 + 2i$ or $v = -36 + 12i,$ with $|v|^2 = 40$ or $1440.$

The smaller value is attained: $z = 1 - i,$ $w = 2 + 4i$ satisfies both equations with $zw = 6 + 2i.$ So the smallest possible value of $|zw|^2$ is $40.$

Solution:

From the double-angle formula, $$\frac{\sin 2x}{\sin 2y} = \frac{2\sin x \cos x}{2\sin y \cos y} = 3 \cdot \frac{1}{2} = \frac{3}{2}.$$

Squaring the given equations, $\sin^2 x = 9\sin^2 y$ and $\cos^2 x = \frac{1}{4}\cos^2 y.$ Adding, $1 = 9(1 - \cos^2 y) + \frac{1}{4}\cos^2 y,$ so $\frac{35}{4}\cos^2 y = 8$ and $\cos^2 y = \frac{32}{35}.$ Then $\cos 2y = 2\cos^2 y - 1 = \frac{29}{35}$ and $\cos 2x = 2\cos^2 x - 1 = \frac{1}{2}\cos^2 y - 1 = -\frac{19}{35},$ so $\frac{\cos 2x}{\cos 2y} = -\frac{19}{29}.$

The requested value is $\frac{3}{2} - \frac{19}{29} = \frac{87 - 38}{58} = \frac{49}{58},$ and $p + q = 49 + 58 = 107.$

Solution:

Fix $\lfloor x \rfloor = a \ge 1.$ For $a \le x \lt a + 1$ the product $n = ax$ ranges over $[a^2, a^2 + a),$ and every integer $n$ in that interval is achieved by $x = \frac{n}{a}$ (which indeed has floor $a$). So each $a$ contributes exactly $a$ values of $n,$ namely $a^2, a^2 + 1, \ldots, a^2 + a - 1.$

For $a = 31$ the largest value is $31^2 + 30 = 991 \lt 1000,$ so all values through $a = 31$ qualify, while $a = 32$ already starts at $1024.$ The count is $1 + 2 + \cdots + 31 = \frac{31 \cdot 32}{2} = 496.$

Solution:

Combining fractions, $f_1(x) = \frac{2(3x + 1) - 9}{3(3x + 1)} = \frac{6x - 7}{9x + 3}.$ Composing once, $f_2(x) = \frac{6 f_1(x) - 7}{9 f_1(x) + 3} = \frac{-3x - 7}{9x - 6},$ and composing again gives $f_3(x) = x.$

So the iteration is periodic with period $3.$ Since $1001 \equiv 2 \pmod{3},$ we have $f_{1001} = f_2,$ and the equation becomes $$\frac{-3x - 7}{9x - 6} = x - 3,$$ that is $9x^2 - 33x + 18 = -3x - 7,$ or $9x^2 - 30x + 25 = (3x - 5)^2 = 0.$

The unique solution is $x = \frac{5}{3},$ so $m + n = 5 + 3 = 8.$

Solution:

Being $p$-safe depends only on $n \bmod p:$ it requires $3 \le n \bmod p \le p - 3.$ That allows $2$ residues modulo $7,$ $6$ residues modulo $11,$ and $8$ residues modulo $13.$ Since $7 \cdot 11 \cdot 13 = 1001,$ the Chinese remainder theorem gives exactly $2 \cdot 6 \cdot 8 = 96$ safe residues modulo $1001,$ so each block of $1001$ consecutive integers contains $96$ safe numbers.

The integers $1$ through $10010$ form ten such blocks, containing $960$ safe numbers. It remains to discard the safe numbers among $10001, \ldots, 10010.$ Their residues modulo $7$ run $5, 6, 0, 1, 2, 3, 4, 5, 6, 0,$ so only $10006$ and $10007$ are $7$-safe; both are also $11$-safe (residues $7$ and $8$) and $13$-safe (residues $9$ and $10$).

Therefore the count is $960 - 2 = 958.$

Solution:

Write $s = \sqrt{111}$ and $r = \sqrt{11}.$ Since each triangle $AD_iE_k$ is congruent to $\triangle ABC,$ we have $AD_i = AE_k = s,$ so $D_1$ and $D_2$ are the two intersections of the circle of radius $s$ about $A$ with the circle of radius $r$ about $B;$ they are mirror images across line $AB,$ so $\angle BAD_1 = \angle BAD_2 = \theta$ with $D_1$ and $D_2$ on opposite sides of $AB.$ Each $E_k$ is the image of its $D_i$ rotated $\pm 60^\circ$ about $A.$ Measuring signed angles from ray $AB,$ with $C$ at $+60^\circ,$ the rays $AD_i$ sit at $\pm\theta$ and the rays $AE_k$ at $\pm\theta \pm 60^\circ,$ so the four angles $\angle CAE_k$ are $\theta,$ $\theta,$ $120^\circ - \theta,$ and $120^\circ + \theta.$

Since $AC = AE_k = s,$ the law of cosines gives $(CE_k)^2 = 2s^2(1 - \cos\angle CAE_k).$ Using $\cos(120^\circ - \theta) + \cos(120^\circ + \theta) = 2\cos 120^\circ \cos\theta = -\cos\theta,$ the four angles' cosines sum to $2\cos\theta - \cos\theta = \cos\theta,$ so $$\sum_{k=1}^{4}(CE_k)^2 = 2s^2(4 - \cos\theta).$$

Applying the law of cosines in triangle $ABD_1$ (with $AB = AD_1 = s$) gives $r^2 = 2s^2(1 - \cos\theta),$ so $2s^2\cos\theta = 2s^2 - r^2.$ Therefore the sum equals $8s^2 - (2s^2 - r^2) = 6s^2 + r^2 = 6 \cdot 111 + 11 = 677.$

Solution:

An arrangement in which everyone shakes exactly two hands is a disjoint union of cycles of length at least $3$ covering all nine people. The possible cycle-length partitions of $9$ are $3+3+3,$ $3+6,$ $4+5,$ and $9.$ On $k$ given people, the number of distinct cycles is $\frac{(k-1)!}{2}.$

For $3+3+3:$ split into three unordered triples in $\frac{1}{3!}\binom{9}{3}\binom{6}{3} = 280$ ways, one cycle each: $280.$ For $3+6:$ $\binom{9}{3} \cdot 1 \cdot \frac{5!}{2} = 84 \cdot 60 = 5040.$ For $4+5:$ $\binom{9}{4} \cdot \frac{3!}{2} \cdot \frac{4!}{2} = 126 \cdot 3 \cdot 12 = 4536.$ For a single $9$-cycle: $\frac{8!}{2} = 20160.$

In total $N = 280 + 5040 + 4536 + 20160 = 30016,$ so the remainder modulo $1000$ is $16.$

Solution:

Let $E'$ be the point of $\omega$ diametrically opposite $E.$ Since $\overline{DE}$ is a diameter of $\gamma,$ the angle $\angle DFE = 90^\circ,$ and since $\overline{EE'}$ is a diameter of $\omega,$ also $\angle E'FE = 90^\circ.$ Both $FD$ and $FE'$ are perpendicular to $FE,$ so $D$ lies on line $E'F:$ the point $F$ is the second intersection of line $E'D$ with $\omega.$

Set $B = (0, 0)$ and $C = (7, 0);$ then $A = \left(\frac{65}{14}, \frac{15\sqrt{3}}{14}\right).$ The bisector gives $\frac{BD}{DC} = \frac{AB}{AC} = \frac{5}{3},$ so $D = \left(\frac{35}{8}, 0\right).$ Since $E$ is the midpoint of arc $BC$ not containing $A,$ both $E$ and $E'$ lie on the vertical line $x = \frac{7}{2}$ through the center $O = \left(\frac{7}{2}, -\frac{7\sqrt{3}}{6}\right),$ which satisfies $|OB| = |OA|$ with $R^2 = \frac{49}{3}.$ Thus $E = \left(\frac{7}{2}, -\frac{7\sqrt{3}}{2}\right)$ and $E' = \left(\frac{7}{2}, \frac{7\sqrt{3}}{6}\right).$

The direction from $E'$ to $D$ is proportional to $(3, -4\sqrt{3}),$ and the point $E' + t\,(3, -4\sqrt{3})$ lies on $\omega$ when $57t^2 - 56t = 0,$ so $t = \frac{56}{57}$ gives $F = \left(\frac{245}{38}, -\frac{105\sqrt{3}}{38}\right).$ Then $$AF^2 = \left(\frac{240}{133}\right)^2 + 3\left(\frac{510}{133}\right)^2 = \frac{837900}{17689} = \frac{900}{19},$$ so $m + n = 900 + 19 = 919.$