Small Hint:

Divide the equation by $4$ to get $5m + 3n = 503$

Big Hint:

Modulo $3$ the equation forces $m \equiv 1 \pmod{3};$ count the values $m = 3k + 1$ that keep $n$ positive

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.

Small Hint:

With common ratio $r,$ the condition $a_{15} = b_{11}$ says $27r^{14} = 99r^{10}$

Big Hint:

You only need $r^4,$ because $a_9 = 27r^8$

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

Small Hint:

Each department contributes exactly two members; classify by how the three men are spread across the departments

Big Hint:

Either every department sends one man and one woman, or one department sends two men, another sends two women, and the third sends one of each

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 $64 + 24 = 88.$

Small Hint:

Let the field be $L$ by $W$ and let $x$ be the distance from $D$ to the southeast corner; set the three riders’ travel times equal

Big Hint:

Equating Bob’s and Cao’s times gives $168W^2 - 625Wx + 168x^2 = 0,$ which factors into two linear factors

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: $$\begin{aligned} \frac{2L + W - x}{8.6} &= \frac{W + x}{6.2} \\ &= \frac{\sqrt{W^2 + x^2}}{5}. \end{aligned}$$

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 $p + q + r = 30 + 7 + 24 = 61.$

Small Hint:

The four triangular points fold up along the sides of $S',$ and their tips meet at the apex above the center of $S'$

Big Hint:

The slant from the midpoint of a side of $S'$ to the star tip has length $20 - \frac{15}{2};$ use a right triangle through the center to get the height

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 $$\begin{aligned} VM &= \sqrt{\left(\tfrac{25}{2}\right)^2 - \left(\tfrac{15}{2}\right)^2} \\ &= \sqrt{100} = 10. \end{aligned}$$ The volume is $\frac{1}{3} \cdot 15^2 \cdot 10 = 750.$

Small Hint:

Factor the distance as $|z|^3 \cdot |1 + 2\mathrm{i} - z^2|,$ so you need the point $z^2$ on the circle of radius $25$ farthest from $1 + 2\mathrm{i}$

Big Hint:

The farthest point lies in exactly the opposite direction: $z^2 = -5\sqrt{5}\,(1 + 2\mathrm{i});$ now square once more

Solution:

The distance is $|(1 + 2\mathrm{i})z^3 - z^5|$ $= |z|^3 \cdot |1 + 2\mathrm{i} - z^2|$ $= 125\,|1 + 2\mathrm{i} - z^2|.$ 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 $1 + 2\mathrm{i}$ is diametrically opposite in direction: $$\begin{aligned} z^2 &= -25 \cdot \frac{1 + 2\mathrm{i}}{|1 + 2\mathrm{i}|} \\ &= -5\sqrt{5}\,(1 + 2\mathrm{i}). \end{aligned}$$

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

Small Hint:

There are $\binom{12}{8} = 495$ members of $S$ below $2^{12},$ so $N$ has $13$ binary digits

Big Hint:

Fix the leading binary digits one at a time, counting members with each prefix, until the position pins down every remaining bit

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 $32.$

Small Hint:

Multiply the two equations: the cross terms become the constants $12\mathrm{i}$ and $20\mathrm{i},$ leaving an equation in the single unknown $zw$

Big Hint:

The quadratic in $zw$ needs $\sqrt{416 - 210\mathrm{i}};$ find it by setting $(a + b\mathrm{i})^2 = 416 - 210\mathrm{i}$ with integers $a$ and $b$

Solution:

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

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

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

Small Hint:

$\frac{\sin 2x}{\sin 2y}$ is just the product of the two given ratios

Big Hint:

Square both given equations and add, using $\sin^2 + \cos^2 = 1,$ to find $\cos^2 y;$ then apply $\cos 2\theta = 2\cos^2\theta - 1$

Solution:

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

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 $p + q = 49 + 58 = 107.$

Small Hint:

Fix $\lfloor x \rfloor = a;$ as $x$ runs from $a$ to $a + 1,$ the product $x\lfloor x \rfloor$ sweeps through the integers from $a^2$ to $a^2 + a - 1$

Big Hint:

Each $a$ contributes exactly $a$ values of $n;$ sum over all $a$ whose values stay below $1000$

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 $1 + 2 + \cdots + 31 = \frac{31 \cdot 32}{2} = 496.$

Small Hint:

Write $f_1$ as a single fraction and compute $f_2$ and $f_3$ by hand — a pattern appears

Big Hint:

$f_3(x) = x,$ so $f_{1001} = f_2;$ setting $f_2(x) = x - 3$ gives a quadratic with a double root

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 $m + n = 5 + 3 = 8.$

Small Hint:

$n$ is $p$-safe exactly when $3 \le n \bmod p \le p - 3$

Big Hint:

Since $7 \cdot 11 \cdot 13 = 1001,$ count the safe residues in each block of $1001$ consecutive integers, then adjust near the cutoff $10{,}000$

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 $960 - 2 = 958.$

Small Hint:

Each $E_k$ is $D_1$ or $D_2$ rotated $\pm 60^\circ$ about $A,$ so track the four angles $\angle CAE_k$

Big Hint:

With $\angle BAD_i = \theta,$ the four angles are $\theta,$ $\theta,$ $120^\circ - \theta,$ $120^\circ + \theta,$ and the last two cosines add up to $-\cos\theta$

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 $8s^2 - (2s^2 - r^2)$ $= 6s^2 + r^2$ $= 6 \cdot 111 + 11 = 677.$

Small Hint:

Everyone shaking exactly two hands splits the nine people into cycles of length at least $3$

Big Hint:

A cycle on $k$ chosen people can be formed in $\frac{(k-1)!}{2}$ ways; sum over the partitions $3+3+3,$ $3+6,$ $4+5,$ and $9$