Scale so the radius is $2.$ The chord lies at distance $1$ from the center, so each radius to an endpoint of the chord makes a $60^\circ$ angle with the bisected radius, and the two endpoint radii form a central angle of $120^\circ.$ The isosceles triangle they cut off has area $\frac{1}{2} \cdot 2 \cdot 2 \sin 120^\circ = \sqrt{3},$ and the whole disk has area $4\pi.$

The smaller region is the $120^\circ$ sector minus the triangle, $\frac{4\pi}{3} - \sqrt{3},$ and the larger region is the rest, $\frac{8\pi}{3} + \sqrt{3}.$ The ratio is $$\frac{\frac{8\pi}{3} + \sqrt{3}}{\frac{4\pi}{3} - \sqrt{3}} = \frac{8\pi + 3\sqrt{3}}{4\pi - 3\sqrt{3}},$$ which has the required form with $(a, b, c, d, e, f) = (8, 3, 3, 4, 3, 3).$

The product is $8 \cdot 3 \cdot 3 \cdot 4 \cdot 3 \cdot 3 = 2592,$ whose remainder upon division by $1000$ is $592.$

The combinations match exactly when both draw two reds, both draw two blues, or both draw one candy of each color. The probability that Terry draws two reds is $\frac{\binom{10}{2}}{\binom{20}{2}} = \frac{45}{190} = \frac{9}{38},$ after which $8$ reds and $10$ blues remain, so Mary draws two reds with probability $\frac{\binom{8}{2}}{\binom{18}{2}} = \frac{28}{153}.$ That case has probability $\frac{9}{38} \cdot \frac{28}{153} = \frac{14}{323},$ and by symmetry two blues each is also $\frac{14}{323}.$

For mixed draws, Terry succeeds with probability $\frac{10 \cdot 10}{\binom{20}{2}} = \frac{10}{19},$ leaving $9$ of each color, and Mary with probability $\frac{9 \cdot 9}{\binom{18}{2}} = \frac{9}{17},$ for $\frac{10}{19} \cdot \frac{9}{17} = \frac{90}{323}.$

The total is $\frac{14 + 14 + 90}{323} = \frac{118}{323}.$ Since $118 = 2 \cdot 59$ and $323 = 17 \cdot 19,$ the fraction is in lowest terms, and $m + n = 118 + 323 = 441.$

Let the block measure $p \times q \times r.$ A cube is hidden exactly when it touches none of the three visible faces, so the hidden cubes form a $(p-1) \times (q-1) \times (r-1)$ block, giving $(p-1)(q-1)(r-1) = 231$ $= 3 \cdot 7 \cdot 11.$

The ways to write $231$ as a product of three positive integers are $3 \cdot 7 \cdot 11,$ $1 \cdot 3 \cdot 77,$ $1 \cdot 7 \cdot 33,$ $1 \cdot 11 \cdot 21,$ and $1 \cdot 1 \cdot 231,$ giving blocks $4 \times 8 \times 12,$ $2 \times 4 \times 78,$ $2 \times 8 \times 34,$ $2 \times 12 \times 22,$ and $2 \times 2 \times 232,$ with volumes $384,$ $624,$ $544,$ $528,$ and $928.$

The smallest is $N = 384.$

All $99$ positive integers below $100$ qualify. A qualifying $3$-digit number is either a repdigit ($9$ of them) or uses a leading digit $a \ge 1$ together with a second value $b \ne a$ in some of the last two positions: $2^2 - 1 = 3$ patterns, each realized in $9 \cdot 9$ ways ($9$ choices for $a,$ then $9$ for $b$), for $9 + 3 \cdot 81 = 252$ numbers.

Similarly a qualifying $4$-digit number is a repdigit ($9$) or has $b \ne a$ appearing in a nonempty subset of the last three positions: $2^3 - 1 = 7$ patterns, each in $9 \cdot 9$ ways, for $9 + 7 \cdot 81 = 576$ numbers.

The total is $99 + 252 + 576 = 927.$

Measure time so that $1000$ workers complete a quarter of the job in $1$ unit; the schedule allows $4$ units in all. With $900$ workers the second quarter takes $\frac{10}{9}$ units, and with $800$ workers the third quarter takes $\frac{10}{8} = \frac{5}{4}$ units. The time used so far is $$1 + \frac{10}{9} + \frac{5}{4} = \frac{121}{36},$$ leaving $4 - \frac{121}{36} = \frac{23}{36}$ of a unit for the last quarter.

The last quarter requires $1000$ worker-units of labor, so the workforce $w$ must satisfy $w \cdot \frac{23}{36} \ge 1000,$ that is $w \ge \frac{36000}{23} \approx 1565.2,$ so at least $1566$ workers are needed.

Since $800$ remain on the job, at least $1566 - 800 = 766$ additional workers must be hired.

Say the first monkey takes $8x$ bananas, keeping $6x$ and giving $x$ to each of the others; the second takes $8y,$ keeping $2y$ and giving $3y$ to each; the third takes $24z,$ keeping $2z$ and giving $11z$ to each. All divisions are whole numbers exactly when $x,$ $y,$ $z$ are positive integers. The final amounts are $6x + 3y + 11z,$ $x + 2y + 11z,$ and $x + 3y + 2z.$

The ratio $3 : 2 : 1$ says the first amount is triple the third and the second is double the third: $$\begin{aligned} &6x + 3y + 11z = 3(x + 3y + 2z) \\ &\quad \Longrightarrow\quad 3x + 5z = 6y, \end{aligned}$$ $$\begin{aligned} &x + 2y + 11z = 2(x + 3y + 2z) \\ &\quad \Longrightarrow\quad x + 4y = 7z. \end{aligned}$$ Substituting $x = 7z - 4y$ into the first equation gives $26z = 18y,$ so $9y = 13z.$ Thus $y = 13n$ and $z = 9n$ for a positive integer $n,$ and then $x = 63n - 52n = 11n.$

The total is $8x + 8y + 24z$ $= (88 + 104 + 216)n$ $= 408n,$ least when $n = 1:$ the answer is $408.$

Folding reflects $B$ to $B'$ across the crease, so $B'E = BE = 17.$ In right triangle $AEB',$ $AB' = \sqrt{17^2 - 8^2} = 15,$ and $AB = AE + EB = 25.$ Place $A = (0, 0),$ $B = (25, 0),$ $B' = (0, 15).$

Points on the crease are equidistant from $B$ and $B',$ so $\overline{EF}$ is perpendicular to $\overline{BB'}.$ Since $BB'$ has slope $-\frac{3}{5},$ the crease through $E = (8, 0)$ has slope $\frac{5}{3},$ and it meets the line $CD$ (at height $h = BC$) at $x = 8 + \frac{3h}{5}.$ The condition $CF = 3$ gives $$25 - \left(8 + \frac{3h}{5}\right) = 3,$$ so $$h = \frac{70}{3}.$$

The perimeter is $2\left(25 + \frac{70}{3}\right) = \frac{290}{3},$ so $m + n = 290 + 3 = 293.$

Since $2004 = 2^2 \cdot 3 \cdot 167,$ we have $2004^{2004} = 2^{4008} \cdot 3^{2004} \cdot 167^{2004},$ so its divisors are $N = 2^i 3^j 167^k$ with $i \le 4008$ and $j, k \le 2004.$ Such an $N$ has $(i+1)(j+1)(k+1)$ divisors, so we need $(i+1)(j+1)(k+1) = 2004.$

Every ordered triple of positive integers with product $2004$ yields admissible exponents, since each factor is at most $2004.$ Counting prime by prime: the exponent $2$ of the prime $2$ is split among the three factors in $\binom{2+2}{2} = 6$ ways by stars and bars, and each of the primes $3$ and $167$ goes to one of the $3$ factors.

The count is $6 \cdot 3 \cdot 3 = 54.$

Let $a_2 = r.$ The geometric condition gives $a_3 = r^2,$ the arithmetic condition gives $a_4 = 2r^2 - r = r(2r-1),$ then $a_5 = (2r-1)^2,$ and so on: inductively $$\begin{aligned} a_{2k+1} &= \bigl(kr - (k-1)\bigr)^2, \\ a_{2k+2} &= \bigl(kr - (k-1)\bigr) \\ &\quad {}\cdot \bigl((k+1)r - k\bigr). \end{aligned}$$ In particular $a_9 = (4r-3)^2$ and $a_{10} = (4r-3)(5r-4),$ so $a_9 + a_{10} = (4r-3)(9r-7)$ $= 646.$ Expanding gives $36r^2 - 55r - 625 = 0,$ which factors as $(r - 5)(36r + 125) = 0,$ so $r = 5.$

With $r = 5$ we get $kr - (k-1) = 4k + 1,$ so $a_{2k+1} = (4k+1)^2$ and $a_{2k+2} = (4k+1)(4k+5);$ the sequence is increasing. Since $a_{17} = 33^2 = 1089 \gt 1000$ while $a_{16} = 29 \cdot 33 = 957,$ the greatest term below $1000$ is $a_{16} = 957.$

Therefore $n + a_n = 16 + 957 = 973.$

The set $\mathcal{S}$ consists of the $\binom{40}{2} = 780$ numbers $2^a + 2^b$ with $0 \le a \lt b \le 39.$ Since $2^a$ is coprime to $9,$ we have $9 \mid 2^a(2^{b-a} + 1)$ exactly when $2^{b-a} \equiv -1 \pmod{9}.$ The powers of $2$ modulo $9$ cycle through $2, 4, 8, 7, 5, 1$ with period $6,$ so $2^d \equiv 8 \equiv -1$ exactly when $d \equiv 3 \pmod{6}.$

For each difference $d = b - a$ there are $40 - d$ pairs, so the number of multiples of $9$ in $\mathcal{S}$ is $$\begin{aligned} &\sum_{d = 3, 9, \ldots, 39} (40 - d) \\ &= 37 + 31 + 25 \\ &\quad {}+ 19 + 13 + 7 + 1 \\ &= 133. \end{aligned}$$

The probability is $\frac{133}{780},$ and since $133 = 7 \cdot 19$ while $780 = 2^2 \cdot 3 \cdot 5 \cdot 13,$ it is in lowest terms. Thus $p + q = 133 + 780 = 913.$

The slant height is $\sqrt{600^2 + (200\sqrt{7})^2}$ $= \sqrt{360000 + 280000}$ $= 800.$ Cutting the cone along the ruling through the starting point and unrolling gives a sector of radius $800$ whose arc has the base circumference $2\pi \cdot 600 = 1200\pi;$ since a full circle of radius $800$ has circumference $1600\pi,$ the central angle is $\frac{3}{4} \cdot 360^\circ = 270^\circ.$ A point on the exact opposite side of the cone is halfway around, which in the unrolled sector is $135^\circ$ away.

The shortest crawl is the straight segment between the two points, at radii $125$ and $375\sqrt{2}$ with a $135^\circ$ angle between them. By the law of cosines, $$\begin{aligned} d^2 &= 125^2 + (375\sqrt{2})^2 \\ &\quad {}- 2 \cdot 125 \cdot 375\sqrt{2} \cos 135^\circ \\ &= 15625 + 281250 + 93750 \\ &= 390625. \end{aligned}$$

Thus $d = 625.$