Solution:

The condition says $\binom{n}{6} = 6\binom{n}{3}.$ Dividing the binomial coefficients, $$\frac{\binom{n}{6}}{\binom{n}{3}} = \frac{(n-3)(n-4)(n-5)}{6 \cdot 5 \cdot 4} = 6,$$ so $(n-3)(n-4)(n-5) = 720 = 10 \cdot 9 \cdot 8.$

Since the product $(n-3)(n-4)(n-5)$ is increasing in $n,$ the only solution is $n - 3 = 10,$ that is, $n = 13.$

Solution:

Fill the first guest's bag one roll at a time. The first roll can be anything; the second must avoid the $2$ remaining rolls of the first roll's type, succeeding with probability $\frac{6}{8};$ and the third must be one of the $3$ rolls of the missing type among the remaining $7.$ So the first bag has one roll of each type with probability $\frac{6}{8} \cdot \frac{3}{7} = \frac{9}{28}.$

Given that, six rolls remain, two of each type, and the same argument gives $\frac{4}{5} \cdot \frac{2}{4} = \frac{2}{5}$ for the second bag. The third bag is then automatically one of each type. The probability is $\frac{9}{28} \cdot \frac{2}{5} = \frac{9}{70},$ so $m + n = 9 + 70 = 79.$

Solution:

Let the original series have first term $a$ and ratio $r,$ so $\frac{a}{1-r} = 2005.$ The squared series is geometric with first term $a^2$ and ratio $r^2,$ so $$\frac{a^2}{1-r^2} = \frac{a}{1-r} \cdot \frac{a}{1+r} = 2005 \cdot \frac{a}{1+r} = 10 \cdot 2005,$$ which gives $\frac{a}{1+r} = 10.$

Dividing the two equations, $\frac{1+r}{1-r} = \frac{2005}{10},$ so $2(1+r) = 401(1-r),$ giving $403r = 399$ and $r = \frac{399}{403}.$ Since $399 = 3 \cdot 7 \cdot 19$ and $403 = 13 \cdot 31,$ the fraction is in lowest terms, and $m + n = 399 + 403 = 802.$

Solution:

From the factorizations $10^{10} = 2^{10} 5^{10},$ $15^7 = 3^7 5^7,$ and $18^{11} = 2^{11} 3^{22},$ the divisor counts are $11 \cdot 11 = 121,$ $8 \cdot 8 = 64,$ and $12 \cdot 23 = 276.$

The divisors common to two of the numbers are exactly the divisors of their gcd: $\gcd(10^{10}, 15^7) = 5^7$ has $8$ divisors, $\gcd(10^{10}, 18^{11}) = 2^{10}$ has $11,$ and $\gcd(15^7, 18^{11}) = 3^7$ has $8.$ Only $1$ divides all three numbers.

By inclusion-exclusion, the count is $121 + 64 + 276 - 8 - 11 - 8 + 1 = 435.$

Solution:

Let $x = \log_a b.$ Since $\log_b a = \frac{1}{x},$ the equation becomes $x + \frac{6}{x} = 5,$ i.e. $x^2 - 5x + 6 = 0,$ so $x = 2$ or $x = 3.$ That means $b = a^2$ or $b = a^3.$

For $b = a^2 \le 2005$ we need $2 \le a \le 44$ (since $44^2 = 1936$ and $45^2 = 2025$), giving $43$ pairs. For $b = a^3 \le 2005$ we need $2 \le a \le 12$ (since $12^3 = 1728$ and $13^3 = 2197$), giving $11$ pairs.

In total there are $43 + 11 = 54$ ordered pairs.

Solution:

The new stack, read from the bottom up, is $n+1,\ 1,\ n+2,\ 2,\ \ldots,\ 2n,\ n.$ So pile $B$'s cards occupy the even positions from the top in reverse order, and pile $A$'s cards occupy the odd positions in reverse order: a card at original position $i \le n$ (pile $A$) moves to position $2(n - i) + 1,$ while a card at position $i \gt n$ (pile $B$) moves to position $2(2n - i) + 2.$

Since $131$ is odd, card $131$ can keep its position only if it comes from pile $A,$ so $131 = 2(n - 131) + 1,$ which gives $n = 196.$ Indeed $131 \le 196,$ and the stack is magical because card $262$ from pile $B$ also stays fixed: $2(2n - 262) + 2 = 262.$ The stack has $2n = 392$ cards.

Solution:

Let $y = \sqrt[16]{5}.$ Multiplying the numerator and denominator by $y - 1$ telescopes the denominator by repeated difference of squares: $$x = \frac{4(y-1)}{(y^8+1)(y^4+1)(y^2+1)(y+1)(y-1)} = \frac{4(y-1)}{y^{16} - 1} = \frac{4(y-1)}{4} = y - 1.$$

Hence $x + 1 = y = 5^{1/16},$ and $(x+1)^{48} = 5^{48/16} = 5^3 = 125.$

Solution:

Let $P_1,$ $P_2,$ $P_3$ be the centers of the circles and $R$ the radius of $\mathcal{C}_3.$ External tangency gives $P_1P_2 = 4 + 10 = 14,$ and internal tangency gives $P_3P_1 = R - 4$ and $P_3P_2 = R - 10.$ Since the centers are collinear, $(R - 4) + (R - 10) = 14,$ so $R = 14$ and $P_3$ lies on $\overline{P_1P_2}$ with $P_3P_1 = 10$ and $P_3P_2 = 4.$

Drop perpendiculars $P_1X,$ $P_2Y,$ $P_3Z$ to the chord, so $P_1X = 4,$ $P_2Y = 10,$ and $Z$ is the midpoint of the chord. The distance from a point moving along line $P_1P_2$ to the tangent line changes linearly, and $P_3$ is $\frac{10}{14}$ of the way from $P_1$ to $P_2,$ so $$P_3Z = 4 + \frac{10}{14}(10 - 4) = \frac{58}{7}.$$

The half-chord is $\sqrt{14^2 - \left(\frac{58}{7}\right)^2} = \frac{\sqrt{9604 - 3364}}{7} = \frac{4\sqrt{390}}{7},$ so the chord has length $\frac{8\sqrt{390}}{7}.$ Since $390 = 2 \cdot 3 \cdot 5 \cdot 13$ is squarefree and $\gcd(8, 7) = 1,$ the answer is $m + n + p = 8 + 390 + 7 = 405.$

Solution:

Since $\sin t + i\cos t = i(\cos t - i \sin t)$ and $\sin nt + i\cos nt = i(\cos nt - i\sin nt),$ de Moivre's theorem (applied to angle $-t$) gives $$(\sin t + i \cos t)^n = i^n(\cos t - i\sin t)^n = i^n(\cos nt - i \sin nt).$$

So the equation holds for all real $t$ exactly when $i^n = i,$ that is, when $n \equiv 1 \pmod 4.$ The values $n = 1, 5, 9, \ldots, 997$ give exactly $250$ positive integers up to $1000.$

Solution:

Place the octahedron's vertices at $(\pm 1, 0, 0),$ $(0, \pm 1, 0),$ $(0, 0, \pm 1).$ It is two square pyramids glued along the square with vertices $(\pm 1, 0, 0)$ and $(0, \pm 1, 0),$ which has area $2,$ and each pyramid has height $1,$ so $$V_{\mathcal{O}} = 2 \cdot \frac{1}{3} \cdot 2 \cdot 1 = \frac{4}{3}.$$

Each face centroid is the average of that face's three vertices, e.g. $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right),$ so the cube has vertices $\left(\pm\frac{1}{3}, \pm\frac{1}{3}, \pm\frac{1}{3}\right).$ Its edge is $\frac{2}{3}$ and its volume is $\frac{8}{27}.$

The ratio is $\frac{4/3}{8/27} = \frac{9}{2},$ so $m + n = 9 + 2 = 11.$

Solution:

Multiplying the recurrence by $a_k$ gives $a_{k+1} a_k = a_k a_{k-1} - 3,$ so the products $b_k = a_k a_{k-1}$ form an arithmetic sequence with common difference $-3.$ Since $b_1 = 72 \cdot 37 = 2664 = 3 \cdot 888,$ we get $$b_k = 2664 - 3(k - 1) = 3(889 - k).$$

Thus $b_k \gt 0$ for $k \le 888,$ so no term before $a_{889}$ can vanish (and the recurrence never divides by zero), while $b_{889} = a_{889} a_{888} = 0$ with $a_{888} \ne 0.$ Hence $a_{889} = 0,$ and $m = 889.$

Solution:

Let $G$ be the midpoint of $\overline{AB},$ so $OG \perp AB$ and $OG = 450.$ With $\alpha = \angle EOG$ and $\beta = \angle FOG$ on either side of ray $OG,$ we have $EG = 450\tan\alpha,$ $FG = 450\tan\beta,$ and $\alpha + \beta = 45^\circ.$ From $EG + FG = EF = 400,$ we get $\tan\alpha + \tan\beta = \frac{8}{9}.$

The tangent addition formula gives $$1 = \tan 45^\circ = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta},$$ so $\tan\alpha\tan\beta = 1 - \frac{8}{9} = \frac{1}{9}.$ Hence $\tan\alpha$ and $\tan\beta$ are the roots of $9t^2 - 8t + 1 = 0,$ namely $\frac{4 \pm \sqrt{7}}{9}.$

Since $AE = 450 - EG$ and $BF = 450 - FG,$ the condition $AE \lt BF$ means $EG \gt FG,$ so $\tan\beta = \frac{4 - \sqrt{7}}{9}.$ Then $BF = 450 - 450 \cdot \frac{4 - \sqrt{7}}{9} = 250 + 50\sqrt{7},$ and $p + q + r = 250 + 50 + 7 = 307.$

Solution:

Let $S(x) = P(x) - x - 3,$ so $S(17) = S(24) = -10.$ Since $S(x) + 10$ has integer coefficients and vanishes at $17$ and $24,$ $$S(x) = -10 + (x - 17)(x - 24)Q(x)$$ for some polynomial $Q$ with integer coefficients.

If $P(n) = n + 3$ for an integer $n,$ then $(n-17)(n-24)Q(n) = 10,$ so $(n-17)(n-24)$ divides $10.$ The factors $n - 17$ and $n - 24$ are integers differing by $7$ whose product divides $10,$ so they are $\{2, -5\}$ or $\{5, -2\},$ giving $n = 19$ and $n = 22.$ Both occur, for example, for $P(x) = x - 7 - (x-17)(x-24).$

Hence $n_1 \cdot n_2 = 19 \cdot 22 = 418.$

Solution:

A cevian $AD$ splits the opposite side in the ratio $$\frac{BD}{DC} = \frac{[ABD]}{[ACD]} = \frac{\frac{1}{2} AB \cdot AD \sin\angle BAD}{\frac{1}{2} AC \cdot AD \sin\angle CAD} = \frac{AB \sin\angle BAD}{AC \sin\angle CAD},$$ and similarly $\frac{BE}{EC} = \frac{AB \sin\angle BAE}{AC \sin\angle CAE}.$

Since $\angle BAE = \angle CAD,$ we also have $\angle BAD = \angle CAE$ (each is that common angle plus $\angle EAD$), so multiplying the two ratios cancels all the sines: $\frac{BD}{DC} \cdot \frac{BE}{EC} = \frac{AB^2}{AC^2}.$ With $BD = 9,$ $DC = 6,$ $AB = 13,$ and $AC = 14,$ this gives $$\frac{BE}{EC} = \frac{13^2}{14^2} \cdot \frac{6}{9} = \frac{169}{294}.$$

Hence $BE = 15 \cdot \frac{169}{169 + 294} = \frac{2535}{463}.$ Since $463$ is prime and does not divide $2535 = 3 \cdot 5 \cdot 13^2,$ the fraction is in lowest terms, and $q = 463.$

Solution:

Completing the square gives $\omega_1\colon (x+5)^2 + (y-12)^2 = 256$ and $\omega_2\colon (x-5)^2 + (y-12)^2 = 16,$ with centers $F_1 = (-5, 12)$ and $F_2 = (5, 12)$ and radii $16$ and $4.$ If a circle with center $P$ and radius $r$ is internally tangent to $\omega_1$ and externally tangent to $\omega_2,$ then $PF_1 = 16 - r$ and $PF_2 = 4 + r,$ so $PF_1 + PF_2 = 20.$

Thus $P$ lies on the ellipse with foci $F_1$ and $F_2$ and major axis $20:$ the semimajor axis is $10,$ the center-to-focus distance is $5,$ so the semiminor axis squared is $100 - 25 = 75,$ giving $$\frac{x^2}{100} + \frac{(y - 12)^2}{75} = 1,$$ i.e. $3x^2 + 4y^2 - 96y + 576 = 300.$ Substituting $y = ax$ yields $(3 + 4a^2)x^2 - 96ax + 276 = 0.$

The line $y = ax$ contains such a center exactly when this quadratic has a real root, i.e. $(96a)^2 - 4 \cdot 276\,(3 + 4a^2) \ge 0,$ which simplifies to $4800a^2 \ge 3312,$ so $a^2 \ge \frac{69}{100}.$ The smallest positive such $a$ has $m^2 = \frac{69}{100},$ and $p + q = 69 + 100 = 169.$