Solution:

In base $b$ the two numbers are $17_b = b + 7$ and $97_b = 9b + 7.$ We need $b + 7 \mid 9b + 7,$ and since $b + 7$ certainly divides $9(b + 7) = 9b + 63,$ this is equivalent to $$b + 7 \mid (9b + 63) - (9b + 7) = 56.$$

For $b \gt 9$ we have $b + 7 \gt 16,$ so $b + 7$ must be $28$ or $56,$ giving $b = 21$ or $b = 49.$ The sum is $21 + 49 = 70.$

Solution:

Here $AB = 4 + 16 + 8 = 28$ and $AC = 13 + 52 + 26 = 91,$ so $D$ and $F$ lie $\frac{1}{7}$ of the way from $A$ along their sides while $E$ and $G$ lie $\frac{5}{7}$ of the way. Triangles sharing angle $A$ have areas proportional to the products of the adjacent sides, so $[ADF] = \frac{1}{49}[ABC]$ and $[AEG] = \frac{25}{49}[ABC].$ Therefore $$[DEGF] = [AEG] - [ADF] = \frac{24}{49}[ABC] = 288,$$ which gives $[ABC] = 588.$

Now set $\mathbf{b} = \overrightarrow{AB}$ and $\mathbf{c} = \overrightarrow{AC},$ so that $D = \frac{1}{7}\mathbf{b},$ $E = \frac{5}{7}\mathbf{b},$ $F = \frac{1}{7}\mathbf{c},$ $G = \frac{5}{7}\mathbf{c},$ and the reflections are $M = 2F - D = \frac{1}{7}(2\mathbf{c} - \mathbf{b})$ and $N = 2E - G = \frac{1}{7}(10\mathbf{b} - 5\mathbf{c}).$ The shoelace formula for $AFNBCEM$ sums cross products of consecutive vertices: the two terms at $A$ vanish, and $$F \times N = -\tfrac{10}{49}\,\mathbf{b} \times \mathbf{c}, \quad N \times B = \tfrac{5}{7}\,\mathbf{b} \times \mathbf{c}, \quad B \times C = \mathbf{b} \times \mathbf{c}, \quad C \times E = -\tfrac{5}{7}\,\mathbf{b} \times \mathbf{c}, \quad E \times M = \tfrac{10}{49}\,\mathbf{b} \times \mathbf{c}.$$

Everything cancels except the single term $\mathbf{b} \times \mathbf{c},$ so the heptagon's area is $\frac{1}{2}\left|\mathbf{b} \times \mathbf{c}\right| = [ABC] = 588.$

Solution:

Let $c \gt v \gt s \ge 1$ be the numbers of players choosing chocolate, vanilla, and strawberry, with $c + v + s = 9.$ Checking small values of $s$ shows the only possibilities are $(6, 2, 1),$ $(5, 3, 1),$ and $(4, 3, 2).$

Since the players are distinct, each triple of counts contributes a multinomial coefficient: $$\frac{9!}{6!\,2!\,1!} = 252, \qquad \frac{9!}{5!\,3!\,1!} = 504, \qquad \frac{9!}{4!\,3!\,2!} = 1260.$$ Thus $N = 252 + 504 + 1260 = 2016,$ and the remainder modulo $1000$ is $16.$

Solution:

The equation factors as $$12x^2 - xy - 6y^2 = (3x + 2y)(4x - 3y) = 0,$$ so every solution has $4x = 3y$ or $3x = -2y.$

Integer solutions of $4x = 3y$ are $(x, y) = (3t, 4t);$ the constraint $|4t| \le 100$ gives $-25 \le t \le 25,$ or $51$ pairs. Integer solutions of $3x = -2y$ are $(x, y) = (2t, -3t);$ the constraint $|3t| \le 100$ gives $-33 \le t \le 33,$ or $67$ pairs. The families overlap only at $(0, 0),$ so the count is $51 + 67 - 1 = 117.$

Solution:

The digits sum to $36.$ Divisibility by $11$ requires the alternating sum of digits to be a multiple of $11,$ so if the four digits in odd positions sum to $a,$ then $a - (36 - a) = 2a - 36$ must be a multiple of $11.$ Since $10 \le a \le 26,$ the only possibility is $a = 18:$ each block of four positions carries digit sum $18.$ The four-element subsets of $\{1, \ldots, 8\}$ with sum $18$ are $$\{1,2,7,8\},\ \{1,3,6,8\},\ \{1,4,5,8\},\ \{1,4,6,7\},\ \{2,3,5,8\},\ \{2,3,6,7\},\ \{2,4,5,7\},\ \{3,4,5,6\},$$ eight in all, and they come in complementary pairs.

Choose which of the $8$ subsets occupies the even positions (which include the units place); the complement fills the odd positions. If that subset contains $k$ of the even digits, then the units digit can be chosen in $k$ ways, the rest of the even positions in $3!$ ways, and the odd positions in $4!$ ways, for $144k$ numbers. Complementary subsets have $k$-values summing to $4,$ so over all $8$ choices $\sum k = 16.$ Hence $N = 144 \cdot 16 = 2304,$ and $N - 2025 = 279.$

Solution:

The circle is tangent to both parallel sides, so the height of the trapezoid is $2 \cdot 3 = 6.$ From the area, $\frac{r + s}{2} \cdot 6 = 72,$ so $r + s = 24.$ By the Pitot theorem the legs together also sum to $24,$ and since the trapezoid is isosceles each leg is $12.$

Dropping a perpendicular from an endpoint of the shorter base, the leg is the hypotenuse of a right triangle with legs $6$ and $\frac{|r - s|}{2}:$ $$144 = 36 + \left(\frac{r - s}{2}\right)^2,$$ so $(r - s)^2 = 432.$ Therefore $r^2 + s^2 = \frac{(r+s)^2 + (r-s)^2}{2} = \frac{576 + 432}{2} = 504.$

Solution:

There are $11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1 = 10395$ ways to pair the letters. Each word begins with the smaller letter of its pair, so the last word alphabetically is the pair whose smaller letter is largest.

Case 1: $G$ is the smaller letter of the last word. Then $G$ pairs with one of $H, I, J, K, L$ ($5$ ways), and no two of the remaining four late letters may pair together (such a pair would start with a letter after $G$). Those four letters must take distinct partners from $\{A, \ldots, F\},$ in $6 \cdot 5 \cdot 4 \cdot 3 = 360$ ways, and the two leftover early letters pair with each other. That gives $5 \cdot 360 = 1800$ pairings. Case 2: $G$ is the larger letter, paired with some $x$ before $G.$ Then none of $H, \ldots, L$ may pair together, so all five take partners among the other five early letters; the six smaller letters are then exactly $A$ through $F,$ and the largest is $F.$ So the last word is $FG,$ and $H, \ldots, L$ match with $A, \ldots, E$ in $5! = 120$ ways.

The probability is $\frac{1800 + 120}{10395} = \frac{1920}{10395} = \frac{128}{693},$ so $m + n = 128 + 693 = 821.$

Solution:

The first equation says $z$ lies on the circle of radius $5$ centered at $(25, 20).$ The second says $z$ is equidistant from $P_1 = (k + 4, 0)$ and $P_2 = (k, 3),$ i.e. it lies on the perpendicular bisector of $\overline{P_1 P_2}.$ The system has exactly one solution precisely when this line is tangent to the circle.

The midpoint is $\left(k + 2, \frac{3}{2}\right)$ and $\overline{P_1 P_2}$ has slope $-\frac{3}{4},$ so the bisector has slope $\frac{4}{3}:$ in standard form $8x - 6y - (8k + 7) = 0.$ Tangency requires $$\frac{|8 \cdot 25 - 6 \cdot 20 - 8k - 7|}{\sqrt{8^2 + 6^2}} = \frac{|73 - 8k|}{10} = 5,$$ so $8k = 73 \pm 50,$ giving $k = \frac{123}{8}$ or $k = \frac{23}{8}.$

The sum is $\frac{146}{8} = \frac{73}{4},$ so $m + n = 73 + 4 = 77.$

Solution:

A point $P$ lies on the image parabola exactly when its rotation by $-60^\circ,$ namely $$Q = \left(\frac{x}{2} + \frac{\sqrt{3}}{2}y,\ -\frac{\sqrt{3}}{2}x + \frac{y}{2}\right),$$ lies on the original parabola. So we need $P$ and $Q$ both on $y = x^2 - 4.$ The parabola is symmetric in $x \mapsto -x,$ so we look for $P = (x, y)$ whose rotated image is the mirror point $Q = (-x, y).$

Matching $y$-coordinates gives $-\frac{\sqrt{3}}{2}x + \frac{y}{2} = y,$ i.e. $y = -\sqrt{3}\,x,$ and then the $x$-coordinate works automatically: $\frac{x}{2} + \frac{\sqrt{3}}{2}(-\sqrt{3}x) = -x.$ Substituting $y = -\sqrt{3}\,x$ into $y = x^2 - 4$ gives $x^2 + \sqrt{3}\,x - 4 = 0,$ whose positive root is $x = \frac{-\sqrt{3} + \sqrt{19}}{2}.$ Then $$y = -\sqrt{3}\,x = \frac{3 - \sqrt{57}}{2} \lt 0,$$ so this point is in the fourth quadrant, on both curves.

The problem guarantees the fourth-quadrant intersection is unique, so its $y$-coordinate is $\frac{3 - \sqrt{57}}{2},$ giving $a + b + c = 3 + 57 + 2 = 62.$

Solution:

Fill the left block arbitrarily: $9!$ ways. Let $R_1, R_2, R_3$ be the sets of three digits in its rows. In the middle block, row $i$ must avoid $R_i$ (those digits already appear in row $i$), and the block's three rows must partition $\{1, \ldots, 9\}.$ Say its top row takes $j$ digits from $R_2$ and $3 - j$ from $R_3.$ Balancing the three rows then forces the middle row to take $3 - j$ digits from $R_1$ together with all $j$ remaining digits of $R_3,$ and the bottom row is determined. The number of content choices is $$\sum_{j=0}^{3} \binom{3}{j}\binom{3}{3-j}^2 = 1 + 27 + 27 + 1 = 56.$$

The right block's row contents are then forced (row $i$ takes whatever is missing from row $i$), and each of the six rows of the middle and right blocks can be ordered internally in $3!$ ways. The total is $$9! \cdot 56 \cdot 6^6 = (2^7 \cdot 3^4 \cdot 5 \cdot 7)(2^3 \cdot 7)(2^6 \cdot 3^6) = 2^{16} \cdot 3^{10} \cdot 5^1 \cdot 7^2.$$

Therefore $p \cdot a + q \cdot b + r \cdot c + s \cdot d = 2 \cdot 16 + 3 \cdot 10 + 5 \cdot 1 + 7 \cdot 2 = 81.$

Solution:

Since $f$ only takes values in $[-1, 1],$ any intersection has $-1 \le y \le 1$ and hence $x = 34y^2 \in [0, 34].$ On the rising pieces, $x \in [4k - 1, 4k + 1)$ with $f(x) = x - 4k,$ so $y = f(34y^2)$ becomes $34y^2 - y - 4k = 0;$ on the falling pieces, $x \in [4k + 1, 4k + 3)$ with $f(x) = 4k + 2 - x,$ giving $34y^2 + y - (4k + 2) = 0.$ In each case a root is valid exactly when it lies in $[-1, 1)$ (rising) or $(-1, 1]$ (falling), since then $x = 34y^2$ automatically falls in the correct interval.

For the rising pieces the roots are $\frac{1 \pm \sqrt{1 + 544k}}{68},$ and both are valid exactly when $\sqrt{1 + 544k} \le 67,$ i.e. for $k = 0, 1, \ldots, 8:$ nine quadratics, each contributing root sum $\frac{1}{34}$ by Vieta. For the falling pieces the roots are $\frac{-1 \pm \sqrt{544k + 273}}{68}.$ The root with the minus sign requires $\sqrt{544k + 273} \lt 67,$ which holds for $k = 0, \ldots, 7;$ those eight quadratics each contribute $-\frac{1}{34}.$ For $k = 8$ only the positive root $\frac{-1 + \sqrt{4625}}{68} = \frac{-1 + 5\sqrt{185}}{68}$ is valid.

The total is $$\frac{9}{34} - \frac{8}{34} + \frac{-1 + 5\sqrt{185}}{68} = \frac{1 + 5\sqrt{185}}{68},$$ and $185 = 5 \cdot 37$ is squarefree, so $a + b + c + d = 1 + 5 + 185 + 68 = 259.$

Solution:

Since $x - yz - (y - zx) = (x - y) + z(x - y) = (x - y)(1 + z),$ and similarly $y - zx - (z - xy) = (y - z)(1 + x),$ the conditions are $$(x - y)(1 + z) \lt 0 \qquad \text{and} \qquad (y - z)(1 + x) \lt 0.$$ Each condition offers two sign patterns, giving four combinations. The combination $x \gt y,$ $z \lt -1,$ $y \gt z,$ $x \lt -1$ is impossible on the plane: $x, z \lt -1$ forces $y \gt 77,$ contradicting $x \gt y.$ Two of the remaining combinations allow a coordinate to run off to infinity, producing the two unbounded regions.

The bounded region is $x \lt y,$ $y \lt z,$ $x \gt -1$ (the fourth constraint $z \gt -1$ is then automatic): the set $-1 \lt x \lt y \lt z$ on the plane. Its closure is the triangle whose vertices come from intersecting the boundary lines pairwise: $x = -1, x = y$ gives $(-1, -1, 77);$ $x = -1, y = z$ gives $(-1, 38, 38);$ and $x = y = z$ gives $(25, 25, 25).$

With $A = (-1, -1, 77),$ the edge vectors are $B - A = (0, 39, -39)$ and $C - A = (26, 26, -52),$ whose cross product is $-1014\,(1, 1, 1),$ of length $1014\sqrt{3}.$ The area is $\frac{1014\sqrt{3}}{2} = 507\sqrt{3},$ so $a + b = 507 + 3 = 510.$

Solution:

Adding chords one at a time, each new chord increases the region count by $1$ plus the number of existing chords it crosses inside the disk. Starting from one region, the expected total is $1 + 27 + E,$ where $E$ is the expected number of interior crossing pairs. The two diameters cross once. A random chord's endpoints land in one of the $6$ quadrant pairs, each with probability $\frac{1}{6}.$ The chord crosses the vertical diameter exactly when its endpoints have opposite $x$-signs, which happens for $4$ of the $6$ pairs, so it meets each diameter with probability $\frac{2}{3}$ and both diameters together $\frac{4}{3}$ times on average: the $25$ chords contribute $\frac{100}{3}$ expected crossings with the diameters.

For two random chords, condition on their quadrant pairs ($36$ equally likely ordered combinations). If one uses quadrants $1, 3$ and the other $2, 4,$ the endpoints always alternate, so they always cross: $2$ combinations. If the two pairs are adjacent and disjoint, such as $\{1, 2\}$ and $\{3, 4\},$ the chords never cross: $4$ combinations. In each of the other $30$ combinations, whether the endpoints alternate around the circle reduces to comparing independent uniform points inside shared quadrants — for example, a $\{1,3\}$ chord and a $\{1,2\}$ chord cross exactly when the two quadrant-$1$ points come in one specific order — and the probability is $\frac{1}{2}$ by symmetry. So two random chords cross with probability $$\frac{2 \cdot 1 + 4 \cdot 0 + 30 \cdot \frac{1}{2}}{36} = \frac{17}{36}.$$

The $\binom{25}{2} = 300$ chord pairs contribute $300 \cdot \frac{17}{36} = \frac{425}{3}$ expected crossings, so $E = 1 + \frac{100}{3} + \frac{425}{3} = 176$ and the expected number of regions is $1 + 27 + 176 = 204.$

Solution:

In triangle $ABC,$ the law of cosines with $\angle B = 60^\circ$ gives $AC^2 = 14^2 + 7^2 - 14 \cdot 7 = 147,$ so $AC = 7\sqrt{3};$ since $7^2 + 147 = 14^2,$ the angle at $C$ is right and $\angle BAC = 30^\circ.$ Likewise $AD = 13\sqrt{3},$ with a right angle at $D$ and $\angle DAE = 30^\circ.$ In triangle $ACD$ with $CD = 24,$ $$\cos \angle CAD = \frac{147 + 507 - 576}{2 \cdot 7\sqrt{3} \cdot 13\sqrt{3}} = \frac{1}{7}, \qquad \sin \angle CAD = \frac{4\sqrt{3}}{7}.$$

Split $f(X) = (BX + EX) + (AX + CX + DX) \ge BE + T,$ where $T$ is the minimum of $AX + CX + DX.$ Since $\angle BAE = 30^\circ + \angle CAD + 30^\circ,$ we get $\cos \angle BAE = \frac{1}{2} \cdot \frac{1}{7} - \frac{\sqrt{3}}{2} \cdot \frac{4\sqrt{3}}{7} = -\frac{11}{14},$ so $BE^2 = 14^2 + 26^2 + 2 \cdot 14 \cdot 26 \cdot \frac{11}{14} = 1444$ and $BE = 38.$ All angles of triangle $ACD$ are less than $120^\circ,$ so $T$ is attained at its Fermat point; erecting an equilateral triangle $ACP$ on side $AC$ away from $D,$ the standard rotation argument gives $T = PD,$ and since $\angle PAD = 60^\circ + \angle CAD$ also has cosine $-\frac{11}{14},$ $$T^2 = 147 + 507 + 2 \cdot 7\sqrt{3} \cdot 13\sqrt{3} \cdot \frac{11}{14} = 1083, \qquad T = 19\sqrt{3}.$$

Both bounds are tight simultaneously: let $F$ be the Fermat point of $ACD,$ so $\angle AFC = \angle AFD = 120^\circ.$ Since $\angle AFC + \angle ABC = 180^\circ,$ point $F$ lies on the circumcircle of $ABC,$ whence $\angle AFB = \angle ACB = 90^\circ;$ similarly $F$ lies on the circumcircle of $AED$ and $\angle AFE = \angle ADE = 90^\circ.$ Thus $\angle BFE = 180^\circ,$ so $F$ lies on segment $BE$ and $f(F) = BE + T = 38 + 19\sqrt{3}.$ The answer is $m + n + p = 38 + 19 + 3 = 60.$

Solution:

Since $(a + 3^6 t)^3 \equiv a^3 \pmod{3^7},$ the cube of $a$ modulo $3^7$ depends only on $a \bmod 3^6,$ and each residue occurs exactly once in $1 \le a \le 3^6.$ Moreover, the only cube roots of $1$ modulo $3^7$ are $1 + 3^6 t,$ which all agree modulo $3^6;$ hence cubing is a bijection from the $486$ units modulo $3^6$ onto the set of unit cubes modulo $3^7,$ which is exactly the set of units $\equiv \pm 1 \pmod 9.$ If all three of $a, b, c$ are prime to $3$ (or exactly one is), then modulo $9$ the sum of cubes is $\pm 1 \pm 1 \pm 1$ or $\pm 1,$ never $0:$ no solutions.

Exactly one multiple of $3,$ say $c = 3z:$ for each of the $486$ units $a$ and $243$ choices of $c,$ the requirement $b^3 \equiv -a^3 - 27z^3 \pmod{3^7}$ has a right side that is a unit $\equiv \pm 1 \pmod 9,$ hence has exactly one solution $b$ modulo $3^6.$ With $3$ choices for which variable is the multiple of $3,$ this case gives $3 \cdot 486 \cdot 243 = 354294$ triples.

All three multiples of $3:$ writing $a = 3x$ etc. with $x, y, z$ ranging modulo $3^5,$ the condition becomes $x^3 + y^3 + z^3 \equiv 0 \pmod{3^4},$ which depends only on the residues modulo $3^3,$ so the count is $9^3$ times the count modulo $27.$ Repeating the same analysis one level down: the two-unit case gives $3 \cdot 18 \cdot 9 = 486,$ and the all-divisible case reduces to $u + v + w \equiv 0 \pmod 3$ with $u, v, w$ modulo $9,$ giving $243;$ that is $486 + 243 = 729$ triples modulo $27,$ hence $729 \cdot 729 = 531441$ here. In total $N = 354294 + 531441 = 885735,$ whose remainder modulo $1000$ is $735.$