Solution:

Since $\frac{1}{\log_b a} = \log_a b,$ the two terms equal $2\log_{2000^6} 4 = \log_{2000^6} 16$ and $3\log_{2000^6} 5 = \log_{2000^6} 125.$ Their sum is $$\log_{2000^6}(16 \cdot 125) = \log_{2000^6} 2000 = \frac{1}{6}.$$

Since $\gcd(1, 6) = 1,$ the answer is $m + n = 1 + 6 = 7.$

Solution:

Factor $(x - y)(x + y) = 2000^2 = 2^8 \cdot 5^6.$ The factors $x - y$ and $x + y$ have the same parity, and their product is even, so both must be even. Writing $x - y = 2a$ and $x + y = 2b$ gives $ab = 2^6 \cdot 5^6 = 10^6.$

Each ordered pair of positive integers $(a, b)$ with $ab = 10^6$ yields exactly one solution $x = a + b,$ $y = b - a$ with $x \gt 0,$ and $10^6$ has $7 \cdot 7 = 49$ divisors, hence $49$ such pairs. Replacing $(a, b)$ by $(-a, -b)$ gives the $49$ solutions with $x \lt 0,$ and $x = 0$ is impossible since $-y^2 \lt 2000^2.$

In total there are $49 + 49 = 98$ lattice points.

Solution:

After the matching pair is removed, $38$ cards remain: nine numbers with four cards each and one number with only two cards. The number of ways to draw a pair is $9\binom{4}{2} + \binom{2}{2} = 54 + 1 = 55,$ out of $\binom{38}{2} = 703$ equally likely draws.

Since $703 = 19 \cdot 37$ shares no factor with $55 = 5 \cdot 11,$ the probability $\frac{55}{703}$ is in lowest terms, and $m + n = 55 + 703 = 758.$

Solution:

Write $N = 2^a m$ with $m$ odd. The odd divisors of $N$ are exactly the divisors of $m,$ so $d(m) = 6.$ Every even divisor is $2^k$ (for $1 \le k \le a$) times an odd divisor, so there are $a \cdot d(m) = 6a$ of them, and $6a = 12$ gives $a = 2.$

So $N = 4m$ where $m$ is the smallest odd number with exactly $6$ divisors. The shapes are $p^5$ (smallest $3^5 = 243$) and $p^2 q$ (smallest $3^2 \cdot 5 = 45$), so $m = 45$ and $N = 4 \cdot 45 = 180.$

Solution:

Choose which five rings to use in $\binom{8}{5} = 56$ ways, and order them (reading down the first finger, then the second, and so on) in $5! = 120$ ways. It remains to split the ordered list into four possibly empty consecutive blocks, one per finger: the number of compositions of $5$ into $4$ nonnegative parts, which by stars and bars is $\binom{8}{3} = 56.$

Therefore $n = 56 \cdot 120 \cdot 56 = 376320,$ whose leftmost three nonzero digits are $376.$

Solution:

Let the bases be $b$ and $b + 100.$ The midsegment has length $b + 50$ and splits the trapezoid into two trapezoids of equal height, whose areas are proportional to the sums of their parallel sides, $b + (b + 50)$ and $(b + 50) + (b + 100).$ Setting $\frac{2b + 50}{2b + 150} = \frac{2}{3}$ gives $b = 75,$ so the bases are $75$ and $175.$

Extend the legs to meet at an apex, creating similar triangles: a segment parallel to the bases with length $\ell$ cuts off a triangle of area $c\ell^2$ for a fixed constant $c.$ The segment of length $x$ bisects the trapezoid's area exactly when $cx^2 - c \cdot 75^2 = c \cdot 175^2 - cx^2,$ so $$x^2 = \frac{75^2 + 175^2}{2} = 18125.$$

Then $x^2/100 = 181.25,$ and the greatest integer not exceeding it is $181.$

Solution:

Multiply both sides by $19!.$ Each term on the left becomes $\frac{19!}{k!\,(19 - k)!} = \binom{19}{k}$ for $k = 2, \ldots, 9,$ while the right side becomes $19N.$

Since $\binom{19}{k} = \binom{19}{19 - k},$ the first half of the binomial row sums to $\sum_{k=0}^{9} \binom{19}{k} = \frac{2^{19}}{2} = 2^{18},$ so $$\sum_{k=2}^{9}\binom{19}{k} = 2^{18} - \binom{19}{0} - \binom{19}{1} = 262144 - 20 = 262124.$$

Hence $N = \frac{262124}{19} = 13796,$ so $\frac{N}{100} = 137.96,$ and the greatest integer less than this is $137.$

Solution:

Place $B = (0, 0),$ $A = (\sqrt{11}, 0),$ $C = (0, h),$ and $D = (d, h),$ so that $\overline{BC}$ is vertical and $BC^2 = h^2.$ The diagonals give vectors $\overrightarrow{AC} = (-\sqrt{11}, h)$ and $\overrightarrow{BD} = (d, h),$ and perpendicularity means $-\sqrt{11}\,d + h^2 = 0,$ so $d = \frac{h^2}{\sqrt{11}}.$

Then $AD^2 = (d - \sqrt{11})^2 + h^2 = 1001.$ Setting $u = h^2,$ this becomes $\frac{(u - 11)^2}{11} + u = 1001,$ that is, $$u^2 - 11u - 10890 = 0.$$ The positive root is $u = \frac{11 + \sqrt{121 + 43560}}{2} = \frac{11 + 209}{2} = 110,$ so $BC^2 = 110.$

Solution:

From $z + \frac{1}{z} = 2\cos 3^\circ$ we get $z^2 - (2\cos 3^\circ) z + 1 = 0,$ so $z = \cos 3^\circ \pm i \sin 3^\circ,$ a point on the unit circle. By de Moivre's theorem, $$z^{2000} + \frac{1}{z^{2000}} = 2\cos(2000 \cdot 3^\circ) = 2\cos 6000^\circ.$$

Since $6000 = 16 \cdot 360 + 240,$ this equals $2\cos 240^\circ = -1.$ The least integer greater than $-1$ is $0.$

Solution:

Let the incircle have center $I$ and radius $r.$ The tangent lengths from $A,$ $B,$ $C,$ $D$ are $19,$ $26,$ $37,$ $23,$ and $I$ lies on each angle bisector, so the half-angles $\alpha, \beta, \gamma, \delta$ at the four vertices satisfy $\tan\alpha = \frac{r}{19},$ $\tan\beta = \frac{r}{26},$ $\tan\gamma = \frac{r}{37},$ $\tan\delta = \frac{r}{23},$ with $\alpha + \beta + \gamma + \delta = 180^\circ.$

Then $\tan(\alpha + \gamma) = -\tan(\beta + \delta),$ and the tangent addition formula turns this into $$\frac{\frac{r}{19} + \frac{r}{37}}{1 - \frac{r^2}{19 \cdot 37}} = -\frac{\frac{r}{26} + \frac{r}{23}}{1 - \frac{r^2}{26 \cdot 23}}, \qquad\text{i.e.}\qquad \frac{56r}{703 - r^2} = \frac{49r}{r^2 - 598}.$$

Cross-multiplying gives $56r^2 - 56 \cdot 598 = 49 \cdot 703 - 49r^2,$ so $105r^2 = 33488 + 34447 = 67935$ and $r^2 = 647.$

Solution:

Since all vertices are lattice points, $w = \overrightarrow{BC}$ is an integer vector with $|w| = |\overrightarrow{AD}| = \sqrt{50},$ so $w$ is one of $(\pm 1, \pm 7),$ $(\pm 7, \pm 1),$ $(\pm 5, \pm 5).$ Write $\overrightarrow{AD} = (1, 7) = s\,\hat{u} + h\,\hat{v}$ where $\hat{u}$ points along $\overline{AB}$ and $\hat{v}$ is perpendicular. Because $\overline{AB} \parallel \overline{CD},$ the vector $w$ has the same perpendicular component $h,$ and the equal leg lengths force its $\hat{u}$-component to be $-s$ (the value $+s$ gives a parallelogram). Hence $(1, 7) - w = 2s\,\hat{u}$ is parallel to $\overline{AB}.$

Discard $w = (1, 7)$ (parallelogram) and $w = (-1, -7)$ (then $h = 0,$ degenerate). The choices $w = (1, -7)$ and $w = (-1, 7)$ make $(1, 7) - w$ vertical or horizontal, which is forbidden. The remaining eight choices give $(1, 7) - w$ equal to $(-6, 6),$ $(-6, 8),$ $(8, 6),$ $(8, 8),$ $(-4, 2),$ $(-4, 12),$ $(6, 2),$ $(6, 12),$ with slopes $-1,$ $-\frac{4}{3},$ $\frac{3}{4},$ $1,$ $-\frac{1}{2},$ $-3,$ $\frac{1}{3},$ $2;$ each is realizable by placing $B$ suitably far along $\hat{u}.$

The sum of the absolute values is $1 + \frac{4}{3} + \frac{3}{4} + 1 + \frac{1}{2} + 3 + \frac{1}{3} + 2 = \frac{119}{12},$ so $m + n = 119 + 12 = 131.$

Solution:

The foot of the perpendicular from $O$ to the plane of $ABC$ is equidistant from $A,$ $B,$ and $C$ (the slant segments to the vertices all have length $20$), so it is the circumcenter of triangle $ABC.$

By Heron's formula with $s = 21,$ the area is $K = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84,$ so the circumradius is $R = \frac{abc}{4K} = \frac{13 \cdot 14 \cdot 15}{336} = \frac{65}{8}.$ The distance from $O$ to the plane is $$\sqrt{20^2 - \left(\tfrac{65}{8}\right)^2} = \sqrt{\frac{25600 - 4225}{64}} = \frac{\sqrt{21375}}{8} = \frac{15\sqrt{95}}{8}.$$

Here $\gcd(15, 8) = 1$ and $95 = 5 \cdot 19$ is squarefree, so $m + n + k = 15 + 95 + 8 = 118.$

Solution:

Group the equation as $$2(1000x^6 - 1) + x(100x^4 + 10x^2 + 1) = 0.$$ Since $1000x^6 - 1 = (10x^2)^3 - 1 = (10x^2 - 1)(100x^4 + 10x^2 + 1),$ the left side factors as $$(100x^4 + 10x^2 + 1)\big(2(10x^2 - 1) + x\big) = (100x^4 + 10x^2 + 1)(20x^2 + x - 2).$$

The quartic factor is always positive, so the two real roots are the roots of $20x^2 + x - 2 = 0,$ namely $x = \frac{-1 \pm \sqrt{161}}{40}.$ The root of the form $\frac{m + \sqrt{n}}{r}$ is $\frac{-1 + \sqrt{161}}{40},$ with $m = -1,$ $n = 161,$ $r = 40,$ and $\gcd(-1, 40) = 1.$ Thus $m + n + r = -1 + 161 + 40 = 200.$

Solution:

Since $(i+1)! - i! = i \cdot i!,$ telescoping gives $a! - b! = \sum_{i=b}^{a-1} i \cdot i!$ for $a \gt b.$ Group the given number as $$16! + (48! - 32!) + (80! - 64!) + \cdots + (2000! - 1984!),$$ with $62$ parenthesized groups $(32j + 16)! - (32j)!$ for $j = 1, \ldots, 62.$

The group for $j$ contributes factorial-base digits $f_i = i$ for $32j \le i \le 32j + 15,$ and the lone $16!$ contributes $f_{16} = 1;$ all other digits are $0.$ Every digit satisfies $0 \le f_i \le i,$ so by uniqueness this is the factorial base expansion.

In the alternating sum, $f_{16} = 1$ sits at an even index and contributes $-1.$ Each group's range starts at an even index and has length $16,$ so it splits into $8$ consecutive pairs, each contributing $-i + (i + 1) = 1,$ for $+8$ per group. The total is $62 \cdot 8 - 1 = 495.$

Solution:

Since $\sin 1^\circ = \sin\big((k+1)^\circ - k^\circ\big) = \sin(k+1)^\circ \cos k^\circ - \cos(k+1)^\circ \sin k^\circ,$ dividing by $\sin k^\circ \sin(k+1)^\circ$ gives $$\frac{1}{\sin k^\circ \sin(k+1)^\circ} = \frac{\cot k^\circ - \cot(k+1)^\circ}{\sin 1^\circ}.$$ So the sum times $\sin 1^\circ$ equals $\cot 45^\circ - \cot 46^\circ + \cot 47^\circ - \cot 48^\circ + \cdots + \cot 133^\circ - \cot 134^\circ,$ with $+$ signs on odd arguments and $-$ signs on even arguments.

Because $\cot(180^\circ - x) = -\cot x$ and supplementary arguments here have the same parity, the terms cancel in supplementary pairs: $+\cot 133^\circ$ cancels $+\cot 47^\circ,$ $-\cot 134^\circ$ cancels $-\cot 46^\circ,$ and so on for every pair of arguments summing to $180^\circ.$ The only survivors are $\cot 45^\circ = 1$ (its partner $135^\circ$ is out of range) and $-\cot 90^\circ = 0.$

Hence the sum equals $\frac{\cot 45^\circ}{\sin 1^\circ} = \frac{1}{\sin 1^\circ},$ so the least such $n$ is $1.$