Solution:

Write $x = 100h + 10t + u$ with digits $h,$ $t,$ $u.$ Then $y = 100u + 10t + h,$ so $$z = |x - y| = 99\,|h - u|.$$

Since both $x$ and $y$ are three-digit numbers, both $h$ and $u$ run from $1$ to $9,$ so $|h - u|$ can be any of $0, 1, \ldots, 8.$ Each choice gives a different multiple of $99,$ so there are $9$ distinct values of $z.$

Solution:

Compute the squared distances: $PQ^2 = 1^2 + 4^2 + 9^2 = 98,$ $QR^2 = 3^2 + 5^2 + 8^2 = 98,$ and $RP^2 = 4^2 + 9^2 + 1^2 = 98.$ So $P,$ $Q,$ and $R$ form an equilateral triangle with side $\sqrt{98} = 7\sqrt{2}.$

Three mutually equidistant vertices of a cube must be joined by face diagonals, and a face diagonal of a cube with edge $s$ has length $s\sqrt{2}.$ Thus $s = 7,$ and the surface area is $6 \cdot 7^2 = 294.$

Solution:

Adding the logs gives $\log_6(abc) = 6,$ so $abc = 6^6.$ In a geometric sequence $ac = b^2,$ hence $b^3 = 6^6,$ so $b = 36$ and $ac = 36^2 = 1296.$

Since the sequence is increasing, $b - a$ is a positive perfect square, so $a = 36 - k^2$ for some $k = 1, \ldots, 5,$ giving candidates $35, 32, 27, 20, 11.$ Also $a$ must divide $1296 = 2^4 \cdot 3^4,$ and of the candidates only $27$ does, with $c = 1296/27 = 48.$

Indeed $27, 36, 48$ is geometric with ratio $\frac{4}{3},$ and $a + b + c = 27 + 36 + 48 = 111.$

Solution:

The garden enclosed by the path is itself a hexagonal arrangement of unit hexagons with $n - 1$ on each side. Counting from the center outward in rings of $6, 12, \ldots$ hexagons, it contains $$1 + 6 + 12 + \cdots + 6(n-2) = 1 + 3(n-2)(n-1)$$ blocks, which for $n = 202$ is $1 + 3 \cdot 200 \cdot 201 = 120601.$

Each unit hexagon consists of $6$ equilateral triangles of side $1,$ so its area is $6 \cdot \frac{\sqrt{3}}{4} = 3 \cdot \frac{\sqrt{3}}{2}.$ The garden's area is therefore $3 \cdot 120601 = 361803$ times $\frac{\sqrt{3}}{2},$ so $m = 361803,$ and the remainder upon division by $1000$ is $803.$

Solution:

With $a = 2^n 3^m,$ $$\frac{6^a}{a^6} = \frac{2^a 3^a}{2^{6n} 3^{6m}},$$ which fails to be an integer exactly when $6n \gt a$ or $6m \gt a.$

If $m, n \ge 1,$ then $a \ge 3 \cdot 2^n \ge 6n$ (since $2^n \ge 2n$) and similarly $a \ge 2 \cdot 3^m \ge 6m,$ so no such $a$ works. If $m = 0,$ the condition is $2^n \lt 6n,$ which holds for $n = 1, 2, 3, 4,$ giving $a = 2, 4, 8, 16.$ If $n = 0,$ the condition is $3^m \lt 6m,$ which holds for $m = 1, 2,$ giving $a = 3, 9.$ (For $a = 1$ the condition fails.)

The sum is $2 + 4 + 8 + 16 + 3 + 9 = 42.$

Solution:

Since $\frac{1}{n^2 - 4} = \frac{1}{4}\left(\frac{1}{n-2} - \frac{1}{n+2}\right),$ the sum telescopes: $$1000 \sum_{n=3}^{10000} \frac{1}{n^2 - 4} = 250\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{9999} - \frac{1}{10000} - \frac{1}{10001} - \frac{1}{10002}\right).$$

The front part is $250 \cdot \frac{25}{12} = 520.8\overline{3},$ and the four tail terms subtract only about $250 \cdot \frac{4}{10000} = 0.1.$ The value is therefore about $520.73,$ so the closest integer is $521.$

Solution:

The sum is a multiple of $200$ exactly when $k(k+1)(2k+1)$ is a multiple of $1200 = 2^4 \cdot 3 \cdot 5^2.$ The factor $3$ always divides $k(k+1)(2k+1)$ (if $k \equiv 1 \pmod 3,$ then $3 \mid 2k+1$), so only $2^4$ and $5^2$ matter.

Since $2k+1$ is odd and $k,$ $k+1$ cannot both be even, $16$ must divide $k$ or $k+1,$ so $k \equiv 0$ or $15 \pmod{16}.$ Similarly $25$ must divide one of $k,$ $k+1,$ $2k+1,$ giving $k \equiv 0,$ $24,$ or $12 \pmod{25}.$ Combining each pair of congruences modulo $400,$ the smallest positive solutions are $112,$ $175,$ $224,$ $287,$ $399,$ and $400.$

The least is $k = 112:$ indeed $112 \cdot 113 \cdot 225 = (16 \cdot 7) \cdot 113 \cdot (9 \cdot 25)$ is a multiple of $1200.$

Solution:

The value $k$ is attained exactly when some integer $n$ satisfies $k \le \frac{2002}{n} \lt k + 1,$ that is, when the interval $\left(\frac{2002}{k+1}, \frac{2002}{k}\right]$ contains an integer. Its length is $\frac{2002}{k(k+1)},$ which is at least $1$ whenever $k(k+1) \le 2002$ — so every $k \le 44$ is attained.

For larger $k,$ check directly: $n = 44, 43, 42, 41, 40$ give $\left\lfloor \frac{2002}{n} \right\rfloor = 45, 46, 47, 48, 50.$ Since $\frac{2002}{41} \approx 48.8$ and $\frac{2002}{40} \gt 50,$ the value $49$ is never attained, so the least such $k$ is $49.$

Solution:

Count ordered pairs $(A, B)$ of disjoint subsets first: each of the $10$ elements goes in $A,$ in $B,$ or in neither, for $3^{10}$ pairs. Among these, $2^{10}$ have $A$ empty and $2^{10}$ have $B$ empty, with the pair $(\varnothing, \varnothing)$ counted in both, so $$3^{10} - 2 \cdot 2^{10} + 1 = 57002$$ ordered pairs have both subsets non-empty.

Disjoint non-empty subsets are never equal, so each set $\{A, B\}$ is counted twice, giving $n = \frac{57002}{2} = 28501.$ The remainder mod $1000$ is $501.$

Solution:

An angle of $x$ degrees is $\frac{\pi x}{180}$ radians, so we need $\sin \frac{\pi x}{180} = \sin x.$ Two angles have equal sines exactly when they differ by a multiple of $2\pi$ or sum to $\pi$ plus a multiple of $2\pi.$

The first case gives $x - \frac{\pi x}{180} = 2\pi j,$ so $x = \frac{360 j \pi}{180 - \pi},$ with least positive value $\frac{360\pi}{180 - \pi} \approx 6.4.$ The second gives $x + \frac{\pi x}{180} = (2k + 1)\pi,$ so $x = \frac{180(2k+1)\pi}{180 + \pi},$ with least positive value $\frac{180\pi}{180 + \pi} \approx 3.1.$ These are the two smallest solutions.

Matching $\frac{m\pi}{n - \pi}$ and $\frac{p\pi}{q + \pi}$ gives $m = 360,$ $n = 180,$ $p = 180,$ $q = 180,$ so $m + n + p + q = 900.$

Solution:

A geometric series with ratio $r$ and sum $1$ has first term $1 - r,$ so its second term is $r(1 - r).$ If the two ratios are $r$ and $s,$ then $r(1 - r) = s(1 - s)$ gives $r - s = r^2 - s^2,$ and since the series are distinct, $r \ne s,$ forcing $s = 1 - r.$

Say the series with ratio $r$ has third term $r^2(1 - r) = \frac{1}{8},$ i.e. $8r^3 - 8r^2 + 1 = 0.$ Substituting $t = 2r$ gives $t^3 - 2t^2 + 1 = (t - 1)(t^2 - t - 1) = 0.$ The root $t = 1$ makes $r = s = \frac{1}{2}$ (the series would coincide), and $r = \frac{1 - \sqrt{5}}{4}$ forces $s = 1 - r \gt 1,$ which diverges. So $r = \frac{1 + \sqrt{5}}{4}.$

The common second term is $$r(1 - r) = \frac{1 + \sqrt{5}}{4} \cdot \frac{3 - \sqrt{5}}{4} = \frac{2\sqrt{5} - 2}{16} = \frac{\sqrt{5} - 1}{8},$$ so $m = 5,$ $n = 1,$ $p = 8,$ and $100m + 10n + p = 518.$

Solution:

Record the player's progress as a path through points $(n, y),$ where $y$ is the number of shots made after $n$ attempts. The condition $a_n \le .4$ caps $y$ at $\lfloor 0.4n \rfloor,$ which for $n = 1, \ldots, 9$ is $0, 0, 1, 1, 2, 2, 2, 3, 3,$ and $a_{10} = .4$ means the path ends at $(10, 4).$

Count the allowed paths by adding, at each point, the counts of its two predecessors (a miss keeps $y,$ a make raises it by $1$). The counts at the maximum allowed heights for $n = 3, \ldots, 9$ come out to $1, 2, 2, 5, 9, 9, 23,$ and the tenth shot must be a make, so $23$ shot sequences qualify. Each consists of $4$ makes and $6$ misses, so the probability is $$23 \left(\tfrac{2}{5}\right)^4 \left(\tfrac{3}{5}\right)^6 = \frac{2^4 \, 3^6 \cdot 23}{5^{10}}.$$

Thus $\{p, q, r, s\} = \{2, 3, 23, 5\}$ and $(a, b, c) = (4, 6, 10),$ giving $(2 + 3 + 23 + 5)(4 + 6 + 10) = 33 \cdot 20 = 660.$

Solution:

Assign masses $5$ at $A,$ $6$ at $B,$ and $15$ at $C.$ Then $E$ balances $\overline{AC}$ ($5 \cdot 3 = 15 \cdot 1$) and $D$ balances $\overline{BC}$ ($6 \cdot 5 = 15 \cdot 2$), so the cevians $\overline{AD}$ and $\overline{BE}$ meet at the center of mass $P,$ of total mass $26.$ Extending $\overline{CP}$ to meet $\overline{AB}$ at $F,$ the mass at $F$ is $5 + 6 = 11,$ so on segment $CF$ we get $CP : PF = 11 : 15,$ that is, $\frac{FP}{FC} = \frac{15}{26}.$

The homothety centered at $F$ with ratio $\frac{15}{26}$ sends $C$ to $P$ and maps line $AB$ to itself; it carries line $CA$ to the parallel line through $P$ — which is line $PQ$ — and line $CB$ to line $PR.$ Hence it maps triangle $CAB$ onto triangle $PQR,$ and $$\frac{[PQR]}{[ABC]} = \left(\frac{15}{26}\right)^2 = \frac{225}{676}.$$

Since $\gcd(225, 676) = 1,$ the answer is $m + n = 225 + 676 = 901.$

Solution:

Let $T$ be the point where the circle touches $\overline{PM}.$ Since $\overline{AM} \perp \overline{AP}$ and $O$ lies on $\overline{AP}$ at distance $19$ from line $AM,$ the circle is tangent to $\overline{AM}$ at $A$ itself, so the two tangents from $M$ give $MT = MA.$ Right triangles $POT$ and $PMA$ (right angles at $T$ and $A$) share angle $P,$ so they are similar with ratio $\frac{OT}{MA} = \frac{19}{MA}.$

The small triangle's perimeter is $PO + OT + TP = (PA - 19) + 19 + TP = PA + PT,$ and since $MT = MA,$ $$PA + PT = PA + PM - MT = 152 - 2\,MA.$$ Perimeters of similar triangles are in the ratio of similarity, so $\frac{19}{MA} = \frac{152 - 2\,MA}{152},$ which simplifies to $MA^2 - 76\,MA + 1444 = (MA - 38)^2 = 0.$ Thus $MA = 38.$

The ratio of similarity is then $\frac{19}{38} = \frac{1}{2},$ so $PO = \frac{1}{2} PM.$ From the perimeter, $PA + PM = 152 - 38 = 114,$ and $PA = PO + 19,$ so $\frac{1}{2} PM + 19 + PM = 114,$ giving $PM = \frac{190}{3}$ and $OP = \frac{95}{3}.$ Hence $m + n = 95 + 3 = 98.$

Solution:

Both circles are tangent to the $x$-axis and to $y = mx,$ so both centers lie on the bisector of the first-quadrant angle between those lines. If the bisector makes angle $\alpha$ with the $x$-axis, then $m = \tan 2\alpha,$ and each center has the form $(x_i, \, x_i \tan\alpha)$ with radius $r_i = x_i \tan\alpha$ (its distance to the $x$-axis).

Since $(9, 6)$ lies on each circle, $(9 - x_i)^2 + (6 - x_i \tan\alpha)^2 = x_i^2 \tan^2\alpha,$ which expands to $$x_i^2 - (18 + 12\tan\alpha)\,x_i + 117 = 0.$$ Both $x_1$ and $x_2$ satisfy this one quadratic, so by Vieta's formulas $x_1 x_2 = 117.$ Then $r_1 r_2 = x_1 x_2 \tan^2\alpha = 117 \tan^2\alpha = 68,$ so $\tan^2\alpha = \frac{68}{117}$ and $\tan\alpha = \frac{2\sqrt{17}}{3\sqrt{13}}.$

Finally, $$m = \frac{2\tan\alpha}{1 - \tan^2\alpha} = \frac{2\tan\alpha}{49/117} = \frac{234}{49} \cdot \frac{2\sqrt{17}}{3\sqrt{13}} = \frac{156\sqrt{17}}{49\sqrt{13}} = \frac{12\sqrt{221}}{49},$$ so $a + b + c = 12 + 221 + 49 = 282.$