Solution:

Let the six counts be $a, a+d, \ldots, a+5d$ with common difference $d \ge 0.$ The greatest count is double the least, so $a + 5d = 2a,$ which gives $a = 5d.$ The total is $$6a + 15d = 30d + 15d = 45d = 990,$$ so $d = 22.$

The greatest number of apples is $a + 5d = 10d = 220.$

Solution:

A four-digit base-eight number lies between $512$ and $4095,$ so a base-eight palindrome less than $1000$ with four digits must have leading (and trailing) digit $1:$ it has the form $\overline{1bb1}_{\text{eight}} = 512 + 64b + 8b + 1 = 513 + 72b.$ Keeping this below $1000$ requires $b \le 6,$ giving the candidates $513, 585, 657, 729, 801, 873, 945.$

Checking from the top, the only one of these that is also a palindrome in base ten is $585 = 1111_{\text{eight}}.$ Every base-eight palindrome with at most three digits is at most $777_{\text{eight}} = 511 \lt 585,$ so the answer is $585.$

Solution:

Let $\omega$ denote the common angle and $L = AB = AC.$ Since $\angle PAB = \omega,$ we have $\angle PAC = 90^\circ - \omega,$ and with $\angle PCA = \omega$ the angles of triangle $APC$ give $\angle APC = 90^\circ.$ Hence in right triangle $APC,$ $$L = AC = \frac{AP}{\sin\omega} = \frac{10}{\sin\omega}.$$

In triangle $ABP,$ the angle at $A$ is $\omega$ and the angle at $B$ is $45^\circ - \omega,$ so $\angle APB = 135^\circ.$ The law of sines gives $\frac{AP}{\sin(45^\circ - \omega)} = \frac{AB}{\sin 135^\circ},$ that is, $10 \sin 135^\circ = L \sin(45^\circ - \omega).$ Substituting $L = \frac{10}{\sin\omega}$ and expanding yields $$\sin\omega = \sqrt{2}\,\sin(45^\circ - \omega) = \cos\omega - \sin\omega,$$ so $\tan\omega = \frac{1}{2}$ and $\sin^2\omega = \frac{1}{5}.$

Therefore $L^2 = \frac{100}{\sin^2\omega} = 500,$ and the area is $\frac{1}{2}L^2 = 250.$

Solution:

Subtracting the second equation from the first gives $xy - yz + 4z - 4x = 0,$ which factors as $(y - 4)(x - z) = 0.$ So $y = 4$ or $x = z.$

If $y = 4:$ the first equation becomes $4x + 4z = 60,$ so $x + z = 15,$ and the second becomes $4z + 4x = 60$ again while the third gives $zx = 44.$ Then $x$ and $z$ are roots of $t^2 - 15t + 44 = (t - 4)(t - 11),$ so $x \in \{4, 11\}.$

If $x = z:$ the first equation reads $x(y + 4) = 60,$ so $y = \frac{60}{x} - 4,$ and the third reads $x^2 + 4y = 60.$ Substituting, $$x^2 + \frac{240}{x} - 16 = 60 \quad\Longrightarrow\quad x^3 - 76x + 240 = 0 = (x - 4)(x - 6)(x + 10),$$ so $x \in \{4, 6, -10\},$ each with real $y$ and $z.$ Hence $S = \{-10, 4, 6, 11\}$ and the sum of squares is $100 + 16 + 36 + 121 = 273.$

Solution:

Write $r = \frac{a}{b}$ in lowest terms and let $g = \gcd(b, 55).$ Then $55r$ in lowest terms is $\frac{55a/g}{b/g}$ (no further cancellation is possible since $\gcd(a, b) = 1$ and $55$ is squarefree). The condition is $$a + b = \frac{55a}{g} + \frac{b}{g}.$$ If $g = 1$ this forces $a = 55a,$ impossible; if $g = 55$ it forces $b = \frac{b}{55},$ impossible.

If $g = 5:$ $a + b = 11a + \frac{b}{5}$ gives $\frac{4b}{5} = 10a,$ so $2b = 25a.$ Since $\gcd(a, b) = 1,$ we need $a = 2$ and $b = 25,$ so $r = \frac{2}{25}$ (indeed $2 + 25 = 27 = 22 + 5$ from $55r = \frac{22}{5}$). If $g = 11:$ $a + b = 5a + \frac{b}{11}$ gives $\frac{10b}{11} = 4a,$ so $5b = 22a,$ forcing $a = 5,$ $b = 22$ and $r = \frac{5}{22}$ (with $5 + 22 = 27 = 25 + 2$ from $55r = \frac{25}{2}$).

Hence $S = \left\{\frac{2}{25}, \frac{5}{22}\right\}$ and the sum is $\frac{2}{25} + \frac{5}{22} = \frac{44 + 125}{550} = \frac{169}{550},$ already in lowest terms. The answer is $169 + 550 = 719.$

Solution:

Place the region as $[0,1]^2 \cup \bigl([0,1] \times [1,2]\bigr) \cup \bigl([1,2] \times [0,1]\bigr),$ so it is the $2 \times 2$ square with the top-right unit square removed. Both coordinates of the midpoint are averages of numbers in $[0, 2],$ so the midpoint always lies in the $2 \times 2$ square; it fails to lie in the region exactly when it lands in the missing square, i.e. when $x_A + x_B \gt 2$ and $y_A + y_B \gt 2.$

If neither point is in the right square, then $x_A + x_B \le 2;$ if neither is in the top square, then $y_A + y_B \le 2.$ So failure requires one point in the top square and the other in the right square, which happens with probability $2 \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{2}{9}.$ In that case, one $x$-coordinate is uniform on $[0,1]$ and the other on $[1,2],$ so $x_A + x_B \gt 2$ with probability $\frac{1}{2},$ and independently $y_A + y_B \gt 2$ with probability $\frac{1}{2}.$

The failure probability is $\frac{2}{9} \cdot \frac{1}{4} = \frac{1}{18},$ so the desired probability is $\frac{17}{18}$ and $m + n = 17 + 18 = 35.$

Solution:

The twelve vertices lie on a circle, and a rectangle inscribed in a circle must have its diagonals pass through the center. So the rectangles with vertices among the twelve are exactly the pairs of distinct diameters, where the diameters join the $6$ antipodal pairs of vertices. A monochromatic rectangle appears exactly when two antipodal pairs are each colored solidly in the same color.

Each antipodal pair is independently both red ($1$ way), both blue ($1$ way), or mixed ($2$ ways). A coloring is valid exactly when at most one pair is both red and at most one pair is both blue. Counting by the numbers of solid red and solid blue pairs: $$2^6 + 6 \cdot 2^5 + 6 \cdot 2^5 + 6 \cdot 5 \cdot 2^4 = 64 + 192 + 192 + 480 = 928.$$

Solution:

Let $P(x) = x^3 + x + 1,$ so the product is $\prod_{k=0}^{6} P(\omega^k),$ where $\omega^0, \ldots, \omega^6$ are all seventh roots of unity. Since $x^7 - 1 = \prod_k (x - \omega^k),$ writing $P(x) = (x - \beta_1)(x - \beta_2)(x - \beta_3)$ and swapping the order of the double product gives $$\prod_{k=0}^{6} P(\omega^k) = \prod_{j=1}^{3} \prod_{k=0}^{6} (\omega^k - \beta_j) = \prod_{j=1}^{3} \bigl(-(\beta_j^7 - 1)\bigr) = \prod_{j=1}^{3} (1 - \beta_j^7).$$

For a root $\beta$ of $P,$ repeatedly using $\beta^3 = -\beta - 1$ gives $\beta^4 = -\beta^2 - \beta,$ $\beta^5 = -\beta^2 + \beta + 1,$ $\beta^6 = \beta^2 + 2\beta + 1,$ and $\beta^7 = 2\beta^2 - 1.$ Hence $1 - \beta^7 = 2(1 - \beta)(1 + \beta),$ and $$\prod_{j} (1 - \beta_j^7) = 2^3 \prod_j (1 - \beta_j) \prod_j (1 + \beta_j) = 8 \cdot P(1) \cdot \bigl(-P(-1)\bigr) = 8 \cdot 3 \cdot 1 = 24.$$

So the requested product equals $24.$

Solution:

Since the tangent to $\omega_1$ at $A$ is parallel to the chord $XP,$ the point $A$ is the midpoint of arc $XP,$ so the perpendicular from $A$ to line $XY$ lands at the midpoint of $\overline{XP};$ similarly the perpendicular from $B$ lands at the midpoint of $\overline{PY}.$ As $X$ and $Y$ are on opposite sides of $P,$ the parallel sides of the trapezoid are $XY = 10 + 14 = 24$ and $AB = \frac{10}{2} + \frac{14}{2} = 12.$

Line $PQ$ is the radical axis, so its intersection $M$ with the tangent line satisfies $MA^2 = MP \cdot MQ = MB^2:$ $M$ is the midpoint of $\overline{AB},$ and with $MA = 6$ and $MQ = MP + 5,$ $$36 = MP(MP + 5), \qquad MP = 4.$$

Set up coordinates along $AB:$ the feet of $A$ and $B$ are the midpoints of $XP$ and $PY,$ so $P$ lies $5$ units from the first foot, while $M$ lies $6$ units from $A.$ Hence the horizontal offset between $M$ and $P$ is $6 - 5 = 1,$ and the height $h$ of the trapezoid satisfies $h^2 = MP^2 - 1 = 15.$ The area is $$\frac{24 + 12}{2}\sqrt{15} = 18\sqrt{15},$$ so $m + n = 18 + 15 = 33.$

Solution:

The condition says adjacent cells have different residues mod $3.$ Each residue class among $1, \ldots, 12$ has exactly $4$ members, so $N = K \cdot (4!)^3,$ where $K$ is the number of ways to fill the grid with residues $0, 1, 2,$ each used $4$ times, with adjacent cells different.

A column is an ordered pair $(a, b)$ of distinct residues. If the current column is $(a, b)$ and $e$ is the third residue, the next column must be one of $(b, a),$ $(b, e),$ $(e, a):$ each of the three unordered pairs $\{a,b\},$ $\{b,e\},$ $\{a,e\}$ occurs in exactly one allowed orientation. So a residue pattern is determined by the sequence of six unordered pairs together with the orientation of the first column. Since each residue must appear $4$ times, each of the three pairs must be used exactly twice, giving $\frac{6!}{2!\,2!\,2!} = 90$ sequences and $K = 2 \cdot 90 = 180.$

Therefore $N = 180 \cdot 24^3 = 2{,}488{,}320 = 2^{11} \cdot 3^5 \cdot 5,$ which has $12 \cdot 6 \cdot 2 = 144$ positive divisors.

Solution:

The $32$ subsets split into $16$ complementary pairs $\{X, X^{\mathsf{c}}\},$ and no collection can contain both members of a pair (they are disjoint). A collection of $16$ pairwise-intersecting subsets must therefore contain exactly one member of every pair; in particular it contains $\{1,2,3,4,5\}$ and not $\emptyset.$

If some singleton $\{x\}$ is chosen, every member must meet $\{x\},$ i.e. contain $x.$ Exactly one set in each complementary pair contains $x,$ so the collection must be exactly the $16$ subsets containing $x:$ this gives $5$ collections. Otherwise no singleton is chosen, so all five $4$-element sets are in the collection. Any two $3$-element subsets of a $5$-element set intersect, a $4$-element set is disjoint only from its complement, and a chosen $2$-element set and a chosen $3$-element set are disjoint only if they are complements, which cannot both be chosen. So the only remaining condition is that the chosen $2$-element sets pairwise intersect.

Viewing $2$-element sets as edges of $K_5,$ a pairwise-intersecting collection of edges either has all edges through one common vertex or is a triangle. The number of such edge families is: the empty family ($1$), triangles ($\binom{5}{3} = 10$), and nonempty families within a star, $5(2^4 - 1) - 10 = 65$ (subtracting the $10$ single edges counted at both endpoints). That is $1 + 10 + 65 = 76$ collections, for a total of $5 + 76 = 81.$

Solution:

Place $B = (0, 0),$ $C = (14, 0),$ $A = (5, 12),$ so $M = (7, 0)$ and $AM = \sqrt{4 + 144} = 2\sqrt{37}.$ By power of the point $M$ in the circumcircle, $MA \cdot MP = MB \cdot MC = 49,$ so $MP = \frac{49}{2\sqrt{37}}$ and extending $A \to M$ by that length gives $P = \left(\frac{567}{74}, -\frac{147}{37}\right).$ The direction of $\overrightarrow{BP}$ is proportional to $(27, -14),$ and the direction of $\overrightarrow{CP}$ is proportional to $-(67, 42).$

Write $Q = (5 + 2t,\ 12 - 12t)$ for $t \in (0, 1),$ so that $AQ = t \cdot AM.$ Using $\tan\theta = \frac{|u \times v|}{u \cdot v}$ for the angle between rays, $$\tan\angle PBQ = \frac{394 - 296t}{222t - 33}, \qquad \tan\angle PCQ = \frac{1182 - 888t}{370t + 99},$$ and the second numerator is exactly $3(394 - 296t).$ Setting the two tangents equal cancels this common factor and leaves $370t + 99 = 3(222t - 33),$ so $296t = 198$ and $t = \frac{99}{148}.$

Then $AQ = \frac{99}{148} \cdot 2\sqrt{37} = \frac{99}{148}\sqrt{148} = \frac{99}{\sqrt{148}},$ and since $\gcd(99, 148) = 1,$ the answer is $99 + 148 = 247.$

Solution:

Let $s = \sec A$ and $t = \tan A.$ The hypothesis $\tan A = 2\cos A$ says $\tan A \sec A = 2,$ i.e. $st = 2,$ and always $s^2 - t^2 = 1.$ Then $(s^2 + t^2)^2 = (s^2 - t^2)^2 + 4s^2t^2 = 17,$ so $u = s^2$ and $v = t^2$ satisfy $u + v = \sqrt{17},$ $uv = 4.$ The sums $w_m = u^m + v^m$ obey $w_{m+1} = \sqrt{17}\,w_m - 4w_{m-1}$ with $w_0 = 2,$ $w_1 = \sqrt{17};$ by induction $w_m$ is a positive integer for even $m$ and an integer times $\sqrt{17}$ for odd $m.$

For even $n = 2m,$ $s^n + t^n = w_m,$ an integer exactly when $m$ is even, i.e. $4 \mid n.$ For odd $n,$ $(s^n + t^n)^2 = w_n + 2 (st)^n = w_n + 2^{n+1}$ is irrational, so $s^n + t^n$ is not an integer. Thus write $n = 4j$ and $x_j = w_{2j}.$ Since $u^2 + v^2 = 9$ and $u^2 v^2 = 16,$ the integers $x_j$ satisfy $$x_{j+1} = 9x_j - 16x_{j-1}, \qquad x_0 = 2, \quad x_1 = 9,$$ giving $9, 49, 297, 1889, \ldots$ whose units digits repeat with period three: $9, 9, 7.$ The units digit is $7$ when $3 \mid j$ and $9$ otherwise.

The valid $n \le 1000$ are $n = 4j$ with $1 \le j \le 250$ and $3 \nmid j:$ there are $250 - 83 = 167$ of them.

Solution:

Give the cube coordinates so that $A$ is the origin, the edges lie along the axes, and the edge length is $s:$ then $B = (s, 0, 0),$ $C = (0, s, s),$ $D = (s, s, s)$ satisfy the description ($\overline{AB} \parallel \overline{CD}$ are edges and $\overline{AC},$ $\overline{BD}$ are face diagonals). Height above $\mathcal{P}$ is a linear function $h(x, y, z) = u_1 x + u_2 y + u_3 z$ for some unit vector $u.$ The plane of rectangle $ABDC$ has normal direction $(0, 1, -1),$ and perpendicularity to $\mathcal{P}$ means the vertical direction $u$ lies in that plane, so $u_2 = u_3.$ The heights of $B$ and $C$ give $s u_1 = 2$ and $s(u_2 + u_3) = 8,$ so $su_2 = su_3 = 4,$ and $|u| = 1$ forces $s^2 = 2^2 + 4^2 + 4^2 = 36.$ Thus $s = 6$ and $u = \frac{1}{3}(1, 2, 2)$ (and indeed $h(D) = 10$).

The water is the region of $[0, 6]^3$ where $h \le 7,$ i.e. $x + 2y + 2z \le 21.$ For fixed $x = a,$ the slice is $\{(y, z) \in [0,6]^2 : y + z \le \tfrac{21 - a}{2}\},$ and since $\frac{21 - a}{2}$ lies between $6$ and $12,$ its area is $36 - \frac{1}{2}\left(12 - \frac{21 - a}{2}\right)^2 = 36 - \frac{(3 + a)^2}{8}.$

Integrating, $$V = \int_0^6 \left(36 - \frac{(3 + a)^2}{8}\right) da = 216 - \frac{9^3 - 3^3}{24} = 216 - \frac{702}{24} = \frac{747}{4},$$ so $m + n = 747 + 4 = 751.$

Solution:

Write $a_n = 23 b_n,$ where $b_n$ is the unique integer in $[1, 2^n]$ with $23 b_n \equiv 1 \pmod{2^n},$ i.e. the inverse of $23$ mod $2^n.$ Reducing mod $2^n$ shows $b_{n+1} \in \{b_n,\ b_n + 2^n\},$ so $a_n = a_{n+1}$ exactly when $b_{n+1} = b_n,$ which happens exactly when the binary digit of weight $2^n$ in $b_{n+1}$ is $0.$

Since $23 \cdot 89 = 2047 = 2^{11} - 1,$ setting $T_k = 1 + 2^{11} + 2^{22} + \cdots + 2^{11(k-1)}$ gives $23 \cdot 89\,T_k = 2^{11k} - 1,$ so $$23\left(2^{11k} - 89\,T_k\right) \equiv 1 \pmod{2^{11k}},$$ and every $b_n$ with $n \le 11k$ is the reduction of $2^{11k} - 89\,T_k$ mod $2^n.$ In binary, $89 = 1011001_{\text{two}}$ occupies positions $0, 3, 4, 6$ of each $11$-bit block of $89\,T_k.$ Since $89\,T_k$ is odd, $2^{11k} - 89\,T_k = \bigl(2^{11k} - 1 - 89\,T_k\bigr) + 1$ keeps digit $0$ equal to $1$ and complements every digit in positions $1$ through $11k - 1.$

So for $n \ge 1,$ the digit of weight $2^n$ is $0$ exactly when $n \equiv 0, 3, 4, 6 \pmod{11}.$ Among $1 \le n \le 1000 = 90 \cdot 11 + 10,$ the residue $0$ occurs $90$ times and the residues $3,$ $4,$ $6$ occur $91$ times each, for a total of $90 + 3 \cdot 91 = 363.$