Solution:

If the leading digit is $a,$ the digits are $a,$ $a - 1,$ $a - 2,$ $a - 3$ with $a = 3, 4, \ldots, 9,$ so $$n = 1000a + 100(a-1) + 10(a-2) + (a-3) = 1111a - 123.$$

Since $1111 = 30 \cdot 37 + 1$ and $123 = 3 \cdot 37 + 12,$ we get $n \equiv a - 12 \equiv a + 25 \pmod{37}.$ For $a = 3, \ldots, 9$ the values $a + 25$ run through $28, 29, \ldots, 34,$ each already less than $37,$ so these are exactly the seven possible remainders.

Their sum is $28 + 29 + \cdots + 34 = 7 \cdot 31 = 217.$

Solution:

The $m$ integers of $\mathcal{A}$ have mean $\frac{2m}{m} = 2,$ so they are centered at $2;$ since the mean of consecutive integers is an integer only when there are an odd number of them, $m$ is odd and the greatest element of $\mathcal{A}$ is $2 + \frac{m-1}{2}.$ The $2m$ integers of $\mathcal{B}$ have mean $\frac{1}{2},$ so they are $1 - m, \ldots, 0, 1, \ldots, m,$ with greatest element $m.$

The condition is $$\left| 2 + \frac{m-1}{2} - m \right| = \left| \frac{3 - m}{2} \right| = 99,$$ so $|3 - m| = 198,$ giving $m = 201$ (since $m \gt 0$). Indeed $201$ is odd, as required, so $m = 201.$

Solution:

Every pair of vertices determines exactly one of three things: an edge, a diagonal of a face, or a space diagonal. There are $\binom{26}{2} = 325$ pairs of vertices in all.

Of these, $60$ are edges. The $24$ triangular faces have no diagonals, while each of the $12$ quadrilateral faces has $2,$ for $24$ face diagonals (no two faces share a diagonal, since the polyhedron is convex).

The number of space diagonals is $325 - 60 - 24 = 241.$

Solution:

Let a segment $\overline{PQ}$ in $\mathcal{S}$ have endpoints on two sides meeting at corner $A,$ and let $M$ be its midpoint. Triangle $PAQ$ is right-angled at $A$ with hypotenuse $PQ = 2,$ and the median to the hypotenuse of a right triangle is half the hypotenuse, so $AM = 1.$ Conversely every point at distance $1$ from a corner (between the two adjacent sides) is such a midpoint, so the midpoints form four quarter-circle arcs of radius $1$ centered at the corners of the square.

The region these arcs enclose is the square with the four quarter-disks removed, of area $$4 - 4 \cdot \frac{\pi}{4} = 4 - \pi \approx 0.86.$$ Therefore $100k = 86.$

Solution:

Alpha's daily ratios were $\frac{160}{300} = \frac{8}{15}$ and $\frac{140}{200} = \frac{7}{10}.$ Since $\frac{8}{15} \lt \frac{7}{10},$ Beta's score was less than $\frac{7}{10}$ of the points attempted on each day, so Beta's total score was less than $\frac{7}{10} \cdot 500 = 350,$ hence at most $349.$

A total of $349$ is achievable: Beta can score $1$ out of $2$ points attempted on day one (and $\frac{1}{2} \lt \frac{8}{15}$) and $348$ out of $498$ on day two (and $\frac{348}{498} \lt \frac{7}{10}$ because $3480 \lt 3486$).

So Beta's largest possible two-day ratio is $\frac{349}{500},$ which is in lowest terms since $349$ is prime, and $m + n = 349 + 500 = 849.$

Solution:

A four-digit snakelike number satisfies $a_1 \lt a_2 \gt a_3 \lt a_4.$ First count the arrangements of any four distinct digits $w \lt x \lt y \lt z$ into this pattern. The largest digit $z$ must sit in position $2$ or $4.$ If $z$ is in position $4,$ the other three form $a_1 \lt a_2 \gt a_3,$ so the largest of them takes position $2$ and the remaining two can go in either order: $2$ ways. If $z$ is in position $2,$ any of the other three digits can be $a_1,$ and then $a_3 \lt a_4$ fixes the rest: $3$ ways. So each set of four digits admits exactly $5$ snakelike orders.

If $0$ is not among the digits, all $5$ orders give valid numbers: $\binom{9}{4} \cdot 5 = 630.$ If $0$ is among them, note $0$ must occupy position $1$ or $3$ (positions $2$ and $4$ must exceed a neighbor), and position $1$ is forbidden. With $0$ in position $3,$ any of the other three digits can be $a_4,$ and $a_1 \lt a_2$ fixes the rest, so $3$ of the $5$ orders survive: $\binom{9}{3} \cdot 3 = 252.$

The total is $630 + 252 = 882.$

Solution:

Write the product as $\prod_{k=1}^{15} (1 + a_k x)$ with $a_k = (-1)^k k.$ An $x^2$ term arises by choosing the $x$-term from two factors, so $C = \sum_{i \lt j} a_i a_j,$ and $$C = \frac{\left(\sum a_k\right)^2 - \sum a_k^2}{2}.$$

The alternating sum is $(-1 + 2) + (-3 + 4) + \cdots + (-13 + 14) - 15 = 7 - 15 = -8,$ and $\sum a_k^2 = 1^2 + 2^2 + \cdots + 15^2 = \frac{15 \cdot 16 \cdot 31}{6} = 1240.$

Thus $C = \frac{64 - 1240}{2} = -588,$ so $|C| = 588.$

Solution:

The congruent angles and congruent segments force the vertices of a regular star to be equally spaced on a circle, visited by taking a constant step: number $n$ equally spaced points $0, 1, \ldots, n - 1$ and connect every $d$th point. The path visits all $n$ points exactly when $\gcd(d, n) = 1,$ and the segments actually cross (making a star rather than a convex polygon) exactly when $2 \le d \le n - 2.$ Steps $d$ and $n - d$ trace the same figure in opposite directions, while different values otherwise give non-similar stars, since a dilation matching the circles would have to match the turning angles.

For $n = 1000 = 2^3 \cdot 5^3,$ the number of $d$ with $\gcd(d, 1000) = 1$ is $1000\left(1 - \frac{1}{2}\right)\left(1 - \frac{1}{5}\right) = 400.$ Removing $d = 1$ and $d = 999$ leaves $398$ values, which pair up as $\{d, 1000 - d\},$ so the number of non-similar regular $1000$-pointed stars is $\frac{398}{2} = 199.$

Solution:

A segment cuts the rectangle into a triangle and a trapezoid only if it runs from a vertex to a point on a nonadjacent side, so $U_2$ is a right triangle whose legs lie along two sides of the rectangle, one leg being a full side ($6$ or $7$). Since $U_1 \sim U_2,$ the cut in the $3$-$4$-$5$ right triangle $ABC$ must also produce a right triangle, so it is parallel to a leg, and then $U_1 \sim ABC.$ Hence $U_2$ is a $3$-$4$-$5$ triangle too: its legs are $6$ and $\frac{9}{2}$ (full side $6$) or $7$ and $\frac{21}{4}$ (full side $7$); the other orientations need legs $8$ or $\frac{28}{3},$ which do not fit.

In both cases the trapezoid $V_2$ has two right angles and an acute angle between the cut and its longer base with tangent $\frac{6}{9/2} = \frac{7}{21/4} = \frac{4}{3}.$ In triangle $ABC,$ a cut parallel to the leg of length $3$ gives $V_1$ an acute angle with tangent $\frac{4}{3},$ matching, while a cut parallel to the leg of length $4$ gives tangent $\frac{3}{4},$ which cannot match. So the cut is parallel to the side of length $3,$ and the parallel bases of $V_1$ are the cut segment $s$ and the side of length $3.$

Similarity of the trapezoids forces $s : 3$ to equal the ratio of the bases of $V_2,$ which is $\frac{7 - 9/2}{7} = \frac{5}{14}$ in the first case and $\frac{6 - 21/4}{6} = \frac{1}{8}$ in the second. Then $[U_1] = \left(\frac{s}{3}\right)^2 [ABC],$ giving $\left(\frac{5}{14}\right)^2 \cdot 6 = \frac{75}{98}$ or $\left(\frac{1}{8}\right)^2 \cdot 6 = \frac{3}{32}.$ The minimum is $\frac{3}{32},$ so $m + n = 3 + 32 = 35.$

Solution:

Place $A = (0, 0),$ $B = (36, 0),$ $C = (36, 15).$ For the circle to lie in the rectangle, its center must lie in the rectangle $[1, 35] \times [1, 14],$ of area $34 \cdot 13 = 442,$ and the center is uniformly distributed there. The diagonal $\overline{AC}$ lies on the line $5x - 12y = 0,$ and the circle misses it exactly when the center's distance $\frac{|5x - 12y|}{13}$ exceeds $1,$ that is, $|5x - 12y| \gt 13.$

The line $5x - 12y = 13$ meets $y = 1$ at $x = 5$ and $x = 35$ at $y = \frac{27}{2},$ so below the diagonal the favorable region is the right triangle with vertices $(5, 1),$ $(35, 1),$ $(35, \tfrac{27}{2}),$ with legs $30$ and $\frac{25}{2}$ and area $\frac{1}{2} \cdot 30 \cdot \frac{25}{2} = \frac{375}{2}.$ Rotating $180^\circ$ about the rectangle's center $(18, \tfrac{15}{2}),$ which lies on the diagonal, maps the inner rectangle and the diagonal to themselves, so the region above the diagonal has the same area.

The probability is $\frac{375}{442},$ and since $442 = 2 \cdot 13 \cdot 17$ shares no factor with $375 = 3 \cdot 5^3,$ we get $m + n = 375 + 442 = 817.$

Solution:

The cone has radius $3,$ height $4,$ and slant height $5,$ so its painted surface consists of lateral area $\pi \cdot 3 \cdot 5 = 15\pi$ and base area $9\pi,$ totaling $24\pi.$ Suppose the cut is at similarity ratio $t,$ so $\mathcal{C}$ is a cone with radius $3t$ and slant height $5t.$ Then $\mathcal{C}$'s painted surface is only its lateral area $15\pi t^2,$ and $\mathcal{F}$'s painted surface is the rest, $24\pi - 15\pi t^2.$ The volumes are in ratio $t^3$ to $1 - t^3.$

Setting the two ratios equal, $$\frac{15 t^2}{24 - 15 t^2} = \frac{t^3}{1 - t^3},$$ so $15 t^2 (1 - t^3) = t^3 (24 - 15 t^2),$ which simplifies to $15 t^2 = 24 t^3,$ giving $t = \frac{5}{8}.$

Then $k = \frac{t^3}{1 - t^3} = \frac{125/512}{387/512} = \frac{125}{387},$ which is in lowest terms since $387 = 3^2 \cdot 43,$ so $m + n = 125 + 387 = 512.$

Solution:

For $0 \lt x \le 1,$ the condition $\lfloor \log_2(1/x) \rfloor = 2k$ (for an integer $k \ge 0$) means $2k \le \log_2(1/x) \lt 2k + 1,$ i.e. $x \in \left(2^{-2k-1}, 2^{-2k}\right].$ These intervals have total length $$\sum_{k \ge 0} 2^{-2k-1} = \frac{1/2}{1 - 1/4} = \frac{2}{3}.$$ Similarly, $\lfloor \log_5(1/y) \rfloor$ is even for $y \in \left(5^{-2k-1}, 5^{-2k}\right],$ intervals of total length $\sum_{k \ge 0} \frac{4}{5} \cdot 25^{-k} = \frac{4/5}{1 - 1/25} = \frac{5}{6}.$

The graph of $\mathcal{S}$ is the product of these two sets, so its area is $\frac{2}{3} \cdot \frac{5}{6} = \frac{5}{9},$ and $m + n = 5 + 9 = 14.$

Solution:

For $x \ne 1,$ write $1 + x + \cdots + x^{17} = \frac{x^{18} - 1}{x - 1},$ so $$(x - 1)^2 P(x) = (x^{18} - 1)^2 - x^{17}(x - 1)^2 = x^{36} - x^{19} - x^{17} + 1 = (x^{19} - 1)(x^{17} - 1).$$ Hence the zeros of $P$ are the $34$ complex numbers other than $1$ satisfying $x^{17} = 1$ or $x^{19} = 1;$ all lie on the unit circle, with angles $\alpha = \frac{k}{17}$ $(k = 1, \ldots, 16)$ and $\alpha = \frac{k}{19}$ $(k = 1, \ldots, 18).$

The five smallest of these angles are $$\frac{1}{19} \lt \frac{1}{17} \lt \frac{2}{19} \lt \frac{2}{17} \lt \frac{3}{19},$$ whose sum is $\frac{6}{19} + \frac{3}{17} = \frac{102 + 57}{323} = \frac{159}{323}.$ Since $159 = 3 \cdot 53$ and $323 = 17 \cdot 19,$ this is in lowest terms, and $m + n = 159 + 323 = 482.$

Solution:

The rope runs from its anchor $A$ at the base of the tower, hugs the wall up to a point $P,$ then goes straight to its end $Q,$ which is at height $4$ and at distance $8 + 4 = 12$ from the tower's axis. Unroll the cylinder's wall into a plane: a taut rope becomes a single straight segment of length $20$ rising $4$ feet, so its horizontal projection has length $\sqrt{20^2 - 4^2} = 8\sqrt{6},$ and every piece of the rope has the same ratio $\frac{20}{8\sqrt{6}} = \frac{5}{2\sqrt{6}}$ of length to horizontal projection.

Viewed from above, the free portion $PQ$ is tangent to the circle of radius $8$ at $P$ from a point at distance $12,$ so its horizontal projection has length $\sqrt{12^2 - 8^2} = 4\sqrt{5}.$ Therefore $$PQ = \frac{5}{2\sqrt{6}} \cdot 4\sqrt{5} = \frac{10\sqrt{5}}{\sqrt{6}} = \frac{5\sqrt{30}}{3}.$$

The rope touching the tower has length $20 - \frac{5\sqrt{30}}{3} = \frac{60 - \sqrt{750}}{3},$ and $c = 3$ is prime, so $a + b + c = 60 + 750 + 3 = 813.$

Solution:

Work backwards: $f(z') = z$ for $z' = 10z$ (always) and for $z' = z - 1$ (provided $z - 1$ is not a multiple of $10$ and $z - 1 \ne 1,$ i.e. $z$ does not end in $1$ and $z \ne 2$). So the integers with $d(x) = n$ form column $n$ of a tree rooted at $1:$ the columns begin $\{1\},$ $\{10\},$ $\{9, 100\},$ $\{8, 90, 99, 1000\},$ and every vertex has two children except $2$ and the vertices ending in $1,$ which have only the child $10z.$

Locate those one-child vertices. Since $2 \to 3 \to \cdots \to 10 \to 1,$ the vertex $2$ sits in column $10.$ A vertex ending in $1$ is reached by subtracting $1$ nine times from a vertex ending in $0,$ so such vertices sit $9$ columns after the multiples of $10$ in the tree. Columns $2$ through $10$ double perfectly (no one-child vertices occur that early), so column $j$ has $2^{j-2}$ vertices, of which the multiples of $10$ — the children $10z$ of column $j - 1$ — number $2^{j-3}.$ Hence for $12 \le k \le 19,$ column $k$ contains $2^{k-12}$ vertices ending in $1$ (column $11$ has none, because the multiple of $10$ in column $2$ is $10$ itself, whose descendant $1$ is excluded — that exclusion is exactly the missing child of $2$).

A one-child vertex in column $k$ removes $2^{19-k}$ of the potential $2^{18}$ vertices from column $20.$ Therefore $$m = 2^{18} - 2^{9} - \sum_{k=12}^{19} 2^{k-12} \cdot 2^{19-k} = 2^{18} - 2^9 - 8 \cdot 2^7 = 2^9(2^9 - 1 - 2) = 2^9 \cdot 509.$$ Since $509$ is prime, the sum of the distinct prime factors of $m$ is $2 + 509 = 511.$