Solution:

There are $\binom{15}{3} = 455$ ways to choose $3$ of the points. A choice fails to give a triangle of positive area exactly when the $3$ points are collinear, which happens only when all three lie on one side of the triangle. Including its endpoints, side $\overline{AB}$ contains $5$ points, $\overline{BC}$ contains $6,$ and $\overline{CA}$ contains $7,$ giving $\binom{5}{3} + \binom{6}{3} + \binom{7}{3} = 10 + 20 + 35 = 65$ collinear triples.

The number of triangles is $455 - 65 = 390.$

Solution:

Numbers leaving equal remainders upon division by $m$ differ by multiples of $m,$ so $m$ divides both $787 - 702 = 85$ and $855 - 787 = 68.$ Since $\gcd(85, 68) = 17$ and $m$ must exceed the positive remainder $r,$ we get $m = 17,$ and $r = 702 - 41 \cdot 17 = 5.$

Similarly $n$ divides both $722 - 412 = 310$ and $815 - 722 = 93,$ and $\gcd(310, 93) = 31,$ so $n = 31$ and $s = 412 - 13 \cdot 31 = 9,$ which indeed differs from $r.$

The requested sum is $17 + 31 + 5 + 9 = 62.$

Solution:

Here $d_n$ is the units digit of the triangular number $\frac{n(n+1)}{2}.$ Since $\frac{(n+20)(n+21)}{2} - \frac{n(n+1)}{2} = 20n + 210$ is a multiple of $10,$ the sequence $d_n$ is periodic with period $20.$ Computing one period, $$(d_1, d_2, \ldots, d_{20}) = (1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0),$$ which sums to $70.$

Since $2017 = 100 \cdot 20 + 17,$ the total is $100 \cdot 70$ plus the first $17$ terms of the period, which sum to $70 - (1 + 0 + 0) = 69.$ The sum is $7069,$ so the remainder is $69.$

Solution:

Since the apex is equidistant from all three base vertices, its foot is the circumcenter of the base. The base is isosceles with sides $20, 20, 24:$ its altitude to the side of length $24$ is $\sqrt{20^2 - 12^2} = 16,$ so its area is $K = \frac{1}{2} \cdot 24 \cdot 16 = 192,$ and its circumradius is $$R = \frac{abc}{4K} = \frac{20 \cdot 20 \cdot 24}{4 \cdot 192} = \frac{25}{2}.$$

The height of the pyramid is $\sqrt{25^2 - \left(\frac{25}{2}\right)^2} = \frac{25\sqrt{3}}{2},$ so the volume is $\frac{1}{3} \cdot 192 \cdot \frac{25\sqrt{3}}{2} = 800\sqrt{3}.$ Then $m + n = 800 + 3 = 803.$

Solution:

The integer parts must be equal: $8a + b = 12b + b,$ so $8a = 12b,$ that is $2a = 3b.$ Since $a$ and $b$ are nonzero base-eight digits (at most $7$), the only options are $(a, b) = (3, 2)$ and $(6, 4).$

The fractional parts must also match: $\frac{c}{8} + \frac{d}{64} = \frac{b}{12} + \frac{a}{144},$ i.e. $\frac{8c + d}{64} = \frac{12b + a}{144}.$ For $(a, b) = (6, 4)$ the right side is $\frac{54}{144} = \frac{24}{64},$ forcing $8c + d = 24,$ which has no solution with both digits nonzero and less than $8.$ For $(a, b) = (3, 2)$ the right side is $\frac{27}{144} = \frac{12}{64},$ so $8c + d = 12,$ giving $c = 1$ and $d = 4.$

Indeed $32.14_{\text{eight}} = 22.23_{\text{twelve}},$ and the requested number $\underline{a}\,\underline{b}\,\underline{c}$ is $321.$

Solution:

Each inscribed angle of the triangle subtends an arc of twice its measure, so the vertices split the circle into arcs of $2x,$ $2x,$ and $360 - 4x$ degrees. The chord fails to intersect the triangle exactly when both random points fall in the same arc, which has probability $$\left(\frac{2x}{360}\right)^2 + \left(\frac{2x}{360}\right)^2 + \left(\frac{360 - 4x}{360}\right)^2 = 1 - \frac{14}{25} = \frac{11}{25}.$$

Setting $y = \frac{x}{180},$ this reads $2y^2 + (1 - 2y)^2 = \frac{11}{25},$ which simplifies to $75y^2 - 50y + 7 = 0,$ with roots $y = \frac{1}{5}$ and $y = \frac{7}{15}.$ These give $x = 36$ and $x = 84,$ both legitimate base angles of an isosceles triangle.

The requested difference is $84 - 36 = 48.$

Solution:

By the symmetry $\binom{6}{a+b} = \binom{6}{6-(a+b)},$ substituting $c = 6 - a - b$ turns the sum into $$S = \sum_{a+b+c=6} \binom{6}{a}\binom{6}{b}\binom{6}{c}.$$

Each term counts the ways to choose $a$ elements from one $6$-element set, $b$ from a second, and $c$ from a third. Summed over all $a + b + c = 6,$ this counts every way to choose $6$ elements from the combined $18$-element set, so $S = \binom{18}{6} = 18564.$

The remainder upon division by $1000$ is $564.$

Solution:

Since $\angle OQP = \angle ORP = 90^\circ,$ both $Q$ and $R$ lie on the circle with diameter $\overline{OP},$ whose radius is $100.$ The angle $\angle QOR = |a - b|$ is an inscribed angle in this circle, so the chord satisfies $QR = 2 \cdot 100 \cdot \sin|a - b|.$ Because $|a - b| \lt 75^\circ,$ the condition $QR \le 100,$ i.e. $\sin|a - b| \le \frac{1}{2},$ is equivalent to $|a - b| \le 30.$

In the $75 \times 75$ square of equally likely pairs $(a, b),$ the region $|a - b| \gt 30$ consists of two right triangles with legs $75 - 30 = 45,$ so the probability is $$1 - \frac{45^2}{75^2} = 1 - \frac{9}{25} = \frac{16}{25}.$$

Therefore $m + n = 16 + 25 = 41.$

Solution:

Because $100 \equiv 1 \pmod{99},$ the recurrence gives $a_n \equiv a_{n-1} + n \pmod{99},$ so $$a_n \equiv 10 + 11 + \cdots + n = \frac{(n + 10)(n - 9)}{2} \pmod{99}.$$ We need $99 \mid \frac{(n+10)(n-9)}{2}.$ One of $n + 10$ and $n - 9$ is even, so this is the same as requiring $9$ and $11$ each to divide the product. Since the two factors differ by $19,$ they cannot both be multiples of $3.$

So $9$ must divide one factor entirely and $11$ the other (or one factor is divisible by $99$). Checking the cases: $99 \mid n - 9$ first at $n = 108;$ $99 \mid n + 10$ first at $n = 89;$ $9 \mid n + 10$ with $11 \mid n - 9$ first at $n = 53;$ and $11 \mid n + 10$ with $9 \mid n - 9$ first at $n = 45.$

The least is $n = 45,$ where $\frac{55 \cdot 36}{2} = 990$ is indeed a multiple of $99.$

Solution:

The argument of $\frac{z_3 - z_1}{z_2 - z_1}$ is the angle $\angle z_2 z_1 z_3,$ and the argument of $\frac{z - z_2}{z - z_3}$ is the angle between $\overline{zz_2}$ and $\overline{zz_3}.$ Their product is real exactly when these angles are equal or supplementary, which by the inscribed angle theorem happens exactly when $z_1,$ $z_2,$ $z_3,$ and $z$ are concyclic. So $z$ lies on the circumcircle of $z_1, z_2, z_3.$

The segment from $18 + 39i$ to $18 + 83i$ is vertical, so its perpendicular bisector is the horizontal line $y = 61.$ The segment from $z_2 = 18 + 39i$ to $z_3 = 78 + 99i$ has slope $1$ and midpoint $(48, 69),$ so its perpendicular bisector is $y - 69 = -(x - 48).$ Setting $y = 61$ gives $x = 56,$ so the center is $56 + 61i.$

The point of the circle with maximal imaginary part is directly above the center, so the real part of $z$ is $56.$

Solution:

Rename each of $1, 2, 3, 4$ as L and each of $6, 7, 8, 9$ as G. If $5$ is not a row median, then no row median equals $5,$ so $m \neq 5.$ Thus $5$'s row must contain one L and one G (reading L5G in some order), and the other two rows must supply one median below $5$ and one above. With the remaining three L's and three G's, those rows are either LLL and GGG, or LLG and LGG.

Count arrangements of letters: the three row types can be assigned to rows $1, 2, 3$ in $3! = 6$ ways, and the L5G row can be ordered in $3! = 6$ ways. In the first case LLL and GGG have $1$ ordering each, giving $6 \cdot 6 \cdot 1 = 36$ patterns; in the second, LLG and LGG each have $3$ orderings, giving $6 \cdot 6 \cdot 9 = 324$ patterns. That is $360$ letter patterns in all.

Finally the four L's can be filled with $1, 2, 3, 4$ in $4!$ ways and the four G's with $6, 7, 8, 9$ in $4!$ ways, so $Q = 360 \cdot 24^2 = 207360,$ whose remainder mod $1000$ is $360.$

Solution:

Since $1 \cdot 1 = 1,$ no product-free set contains $1.$ Split by the least element $t.$ If $t \ge 4,$ any product of two elements is at least $16 \gt 10,$ so every subset of $\{4, 5, \ldots, 10\}$ works: $2^7 = 128$ subsets, including the empty set.

If $t = 2:$ then $4 \notin S$ (as $2 \cdot 2 = 4$), while $7$ and $8$ are unrestricted ($2$ choices each). Among $\{3, 6, 9\},$ the constraints $2 \cdot 3 = 6$ and $3 \cdot 3 = 9$ leave exactly $\varnothing, \{3\}, \{6\}, \{9\}, \{6, 9\}$ — $5$ choices. Among $\{5, 10\},$ the constraint $2 \cdot 5 = 10$ leaves $3$ choices. That gives $2 \cdot 2 \cdot 5 \cdot 3 = 60$ sets. If $t = 3:$ then $9 \notin S$ (as $3 \cdot 3 = 9$), and any subset of $\{4, 5, 6, 7, 8, 10\}$ may be added since all other products exceed $10:$ $2^6 = 64$ sets.

In total there are $128 + 60 + 64 = 252$ product-free subsets.

Solution:

If $k^3 \le n \lt (k+1)^3,$ then the interval $(n, mn]$ contains the cube $(k+1)^3$ as long as $(k+1)^3 \le m k^3,$ i.e. $\left(1 + \frac{1}{k}\right)^3 \le m.$ Since $\left(1 + \frac{1}{k}\right)^3 \le 8$ for all $k \ge 1,$ every $m \ge 8$ has $Q(m) = 1.$

For $4 \le m \le 7:$ $n = 1$ fails since $(1, m]$ contains no cube, but for $n \ge 2$ the interval works: $8 \le 4n$ covers $2 \le n \le 7,$ and $\left(1 + \frac{1}{k}\right)^3 \le \frac{27}{8} \lt 4$ covers $k \ge 2.$ So $Q(4) = Q(5) = Q(6) = Q(7) = 2.$ For $m = 3:$ $n = 8$ fails (no cube in $(8, 24]$), while $27 \le 3n$ covers $9 \le n \le 26$ and $\left(\frac{4}{3}\right)^3 \lt 3$ covers $k \ge 3,$ so $Q(3) = 9.$ For $m = 2:$ $n = 31$ fails (no cube in $(31, 62]$), while $64 \le 2n$ covers $32 \le n \le 63$ and $\left(\frac{5}{4}\right)^3 \lt 2$ covers $k \ge 4,$ so $Q(2) = 32.$

Therefore $$\sum_{m=2}^{2017} Q(m) = 32 + 9 + 4 \cdot 2 + 2010 \cdot 1 = 2059,$$ and the remainder is $59.$

Solution:

Exponentiating the first equation twice: $\log_a(\log_a 2) + \log_a 24 - 128 = a^{128}$ becomes $\log_a(24 \log_a 2) = 128 + a^{128},$ so $24 \log_a 2 = a^{128} \cdot a^{a^{128}},$ i.e. $2^{24} = a^{\left(a^{128} \cdot a^{a^{128}}\right)}.$ Setting $t = a^{a^{128}},$ the right side is $t^t$ and the left side is $\left(2^3\right)^{2^3},$ so by the strict monotonicity of $t^t$ we get $a^{a^{128}} = 8.$ Writing $c = \log_2 a \gt 0,$ this says $c \cdot 2^{128c} = 3,$ which is increasing in $c$ and satisfied by $c = \frac{3}{64}:$ indeed $\frac{3}{64} \cdot 2^6 = 3.$ So $a = 2^{3/64}.$

The second equation gives $x = a^{a^{256}}.$ Here $a^{256} = 2^{256 \cdot 3/64} = 2^{12} = 4096,$ so $$x = a^{4096} = 2^{4096 \cdot 3/64} = 2^{192}.$$

Clearly $2^{192} \equiv 0 \pmod{8}.$ By Euler's theorem $2^{100} \equiv 1 \pmod{125},$ so $2^{192} \equiv 2^{-8} \pmod{125},$ the inverse of $256 \equiv 6.$ Since $6 \cdot 21 = 126 \equiv 1 \pmod{125},$ we get $2^{192} \equiv 21 \pmod{125}.$ The unique residue mod $1000$ that is $0$ mod $8$ and $21$ mod $125$ is $896.$

Solution:

Place the right angle at the origin with vertices $(0, 0),$ $(5, 0),$ and $(0, 2\sqrt{3}),$ so the hypotenuse lies on the line $2\sqrt{3}\,x + 5y = 10\sqrt{3}.$ Let the equilateral triangle's side between the two legs have endpoints $(s\cos\theta, 0)$ and $(0, s\sin\theta),$ where $s$ is the side length. Its midpoint is $\frac{s}{2}(\cos\theta, \sin\theta),$ and moving a distance $\frac{\sqrt{3}}{2}s$ perpendicular to the side places the third vertex at $\frac{s}{2}\left(\cos\theta + \sqrt{3}\sin\theta,\ \sin\theta + \sqrt{3}\cos\theta\right).$

Substituting this vertex into the hypotenuse equation and simplifying gives $$s = \frac{20\sqrt{3}}{7\sqrt{3}\cos\theta + 11\sin\theta}.$$ The denominator is at most $\sqrt{(7\sqrt{3})^2 + 11^2} = \sqrt{268} = 2\sqrt{67},$ attained for an admissible $\theta,$ so the minimum side length satisfies $s^2 = \frac{(10\sqrt{3})^2}{67} = \frac{300}{67}.$

The minimum area is $\frac{\sqrt{3}}{4} \cdot \frac{300}{67} = \frac{75\sqrt{3}}{67},$ so $m + n + p = 75 + 67 + 3 = 145.$