Answer: 191

Solution:

The $14$ positions split into $7$ diametrically opposite pairs. Only the set of positions occupied by the men matters, and all $\binom{14}{5} = 2002$ five-element sets are equally likely. Every man stands opposite a woman exactly when no pair contains two men, so choose which $5$ of the $7$ pairs contain a man ($\binom{7}{5} = 21$ ways) and which position of each chosen pair the man occupies ($2^5 = 32$ ways), for $21 \cdot 32 = 672$ favorable sets.

The probability is $\frac{672}{2002} = \frac{48}{143},$ so $m + n = 48 + 143 = 191.$

Answer: 881

Solution:

Let $x = \log_b n.$ The first equation says $\sqrt{x} = \log_b n^{1/2} = \frac{x}{2},$ so $x = \frac{x^2}{4},$ giving $x = 0$ or $x = 4.$ If $x = 0$ then $n = 1,$ and the second equation would read $0 = \log_b b = 1,$ impossible; so $x = 4.$

The second equation says $bx = \log_b b + \log_b n = 1 + x,$ so $4b = 5$ and $b = \frac{5}{4}.$ Then $$n = b^4 = \left(\frac{5}{4}\right)^4 = \frac{625}{256},$$ which is in lowest terms, so $j + k = 625 + 256 = 881.$

Answer: 607

Solution:

Since no two of the $40$ lines are parallel, every two lines cross, giving $\binom{40}{2} = 780$ pairs of lines, and each pair meets at exactly one point. A point where exactly $k$ lines meet accounts for exactly $\binom{k}{2}$ of these pairs.

The given points account for $3\binom{3}{2} + 4\binom{4}{2} + 5\binom{5}{2} + 6\binom{6}{2} = 9 + 24 + 50 + 90 = 173$ pairs of lines. Each remaining pair meets at a point where exactly $2$ lines intersect, one point per pair, so there are $780 - 173 = 607$ such points.

Answer: 12

Solution:

Since $13! = 2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13,$ a valid $m = 2^x 3^y 5^z 7^w 11^u 13^v$ must leave every exponent of $\frac{13!}{m}$ even: $x \in \{0, 2, 4, 6, 8, 10\},$ $y \in \{1, 3, 5\},$ $z \in \{0, 2\},$ and $w = u = v = 1.$

The choices are independent, so the sum of all such $m$ factors as $$(1 + 4 + \cdots + 1024)(3 + 27 + 243)(1 + 25) \cdot 7 \cdot 11 \cdot 13 = 1365 \cdot 273 \cdot 26 \cdot 1001.$$ Since $1365 = 3 \cdot 5 \cdot 7 \cdot 13,$ $273 = 3 \cdot 7 \cdot 13,$ $26 = 2 \cdot 13,$ and $1001 = 7 \cdot 11 \cdot 13,$ the sum equals $2^1 3^2 5^1 7^3 11^1 13^4,$ and $a + b + c + d + e + f = 1 + 2 + 1 + 3 + 1 + 4 = 12.$

Answer: 106

Solution:

Let the circle have center $O$ and radius $R,$ with $A = (R, 0),$ $B = (0, R),$ $C = (-R, 0),$ $D = (0, -R),$ and $P = (R\cos\theta, R\sin\theta).$ Then $PA^2 = 2R^2(1 - \cos\theta)$ and $PC^2 = 2R^2(1 + \cos\theta),$ so $PA \cdot PC = 2R^2|\sin\theta| = 56.$ In the same way $PB \cdot PD = 2R^2|\cos\theta| = 90.$

Squaring and adding, $4R^4 = 56^2 + 90^2 = 11236,$ so $2R^2 = \sqrt{11236} = 106.$ The square has diagonal $2R,$ hence area $\frac{(2R)^2}{2} = 2R^2 = 106.$

Answer: 51

Solution:

Whatever Alice guesses, the deck evolves the same way; only the immediate success probability depends on her guess, so it is optimal to guess a color with the most cards remaining. Let $E(r, b)$ be the expected number of correct guesses from a state with $r$ red and $b$ black cards left. Then $E(r, 0) = r,$ $E(0, b) = b,$ and $$E(r, b) = \frac{\max(r, b)}{r + b} + \frac{r\,E(r-1, b) + b\,E(r, b-1)}{r + b}.$$

By symmetry $E(r, b) = E(b, r).$ Computing upward: $E(1,1) = \frac{3}{2},$ $E(2,1) = \frac{2}{3} + \frac{2 \cdot \frac{3}{2} + 2}{3} = \frac{7}{3},$ $E(2,2) = \frac{1}{2} + \frac{7}{3} = \frac{17}{6},$ $E(3,1) = \frac{3}{4} + \frac{3 \cdot \frac{7}{3} + 3}{4} = \frac{13}{4},$ $E(3,2) = \frac{3}{5} + \frac{3 \cdot \frac{17}{6} + 2 \cdot \frac{13}{4}}{5} = \frac{18}{5},$ and finally $E(3,3) = \frac{1}{2} + \frac{18}{5} = \frac{41}{10}.$

So the expected number of correct guesses is $\frac{41}{10},$ and $m + n = 41 + 10 = 51.$

Answer: 49

Solution:

Write $r_k$ for the remainder of $n$ modulo $k,$ and note $r_2 \equiv r_4 \pmod 2,$ $r_2 \equiv r_6 \pmod 2,$ and $r_3 \equiv r_6 \pmod 3.$ If $r_2 = 0:$ then $r_4$ is even and different from $r_2,$ so $r_4 = 2;$ then $r_6$ is even and avoids $\{0, 2\},$ so $r_6 = 4,$ which gives $r_3 = 1;$ finally $r_5$ avoids $\{0, 1, 2, 4\},$ so $r_5 = 3.$ These say $n \equiv -2$ modulo each of $2, 3, 4, 5, 6,$ i.e. $n \equiv 58 \pmod{60}.$

If $r_2 = 1:$ similarly $r_4 = 3,$ then $r_6 = 5,$ giving $r_3 = 2,$ and $r_5$ avoids $\{1, 2, 3, 5\},$ so $r_5 \in \{0, 4\}.$ The choice $r_5 = 4$ gives $n \equiv -1 \pmod{60},$ i.e. $n \equiv 59;$ the choice $r_5 = 0$ gives $n \equiv 11 \pmod{12}$ with $5 \mid n,$ i.e. $n \equiv 35 \pmod{60}.$

Below $1000$ there are $17$ integers congruent to $35,$ $16$ congruent to $58,$ and $16$ congruent to $59$ modulo $60,$ for a total of $17 + 16 + 16 = 49.$

Answer: 125

Solution:

The distances from an interior point to the parallel lines $DA$ and $BC$ add up to the distance between them, the height of the rhombus. So the height is $9 + 16 = 25,$ and the incircle, tangent to both lines, has radius $\frac{25}{2}.$ Center the incircle at the origin with $DA:\ y = \frac{25}{2}$ and $BC:\ y = -\frac{25}{2}.$ Then $P$ has $y$-coordinate $\frac{25}{2} - 9 = \frac{7}{2},$ and $x^2 + \left(\frac{7}{2}\right)^2 = \left(\frac{25}{2}\right)^2$ gives $x = \pm 12.$

Let $\alpha = \angle BAD.$ Line $AB$ is tangent to the incircle and makes angle $\alpha$ with the horizontal, so (orienting the figure suitably) it is $x \sin\alpha + y \cos\alpha = -\frac{25}{2},$ and interior points satisfy $x\sin\alpha + y\cos\alpha + \frac{25}{2} \gt 0.$ The condition $\operatorname{dist}(P, AB) = 5$ reads $x \sin\alpha + \frac{7}{2}\cos\alpha + \frac{25}{2} = 5.$ For $x = 12$ the left side exceeds $\frac{25}{2},$ so $x = -12,$ and the equation becomes $24\sin\alpha - 7\cos\alpha = 15.$

Substituting $7\cos\alpha = 24\sin\alpha - 15$ into $\sin^2\alpha + \cos^2\alpha = 1$ yields $625\sin^2\alpha - 720\sin\alpha + 176 = 0,$ so $\sin\alpha = \frac{4}{5}$ or $\frac{44}{125}.$ The root $\frac{44}{125}$ makes $\cos\alpha = \frac{24\sin\alpha - 15}{7}$ negative, contradicting $\angle BAD \lt 90^\circ.$ So $\sin\alpha = \frac{4}{5},$ the side length is $\frac{25}{\sin\alpha} = \frac{125}{4},$ and the perimeter is $4 \cdot \frac{125}{4} = 125.$

Answer: 738

Solution:

Since $p(m) - p(2)$ does not involve $c,$ each valid pair $(a, b)$ contributes $41$ polynomials. Factoring, $$p(x) - p(2) = (x - 2)\bigl(x^2 + (a+2)x + (2a + b + 4)\bigr),$$ so we need the quadratic factor $q(x)$ to have exactly one integer root different from $2.$ If $q$ has any integer root, its other root is also an integer (their sum $-(a+2)$ is an integer); if $q$ has no integer root, then no $m$ exists at all. So either $q$ has roots $2$ and $k$ with $k \ne 2,$ or a double root $k \ne 2.$

Roots $2$ and $k:$ Vieta's formulas give $a = -4 - k$ and $b = 4k + 4.$ The constraint $-20 \le b \le 20$ forces $-6 \le k \le 4$ (and then $a$ is automatically in range), so excluding $k = 2$ leaves $10$ pairs. Double root $k:$ here $a = -2k - 2$ and $b = k^2 + 4k,$ and $b \le 20$ forces $-6 \le k \le 2,$ all valid for $a,$ so excluding $k = 2$ leaves $8$ pairs.

That is $18$ pairs $(a, b),$ hence $18 \cdot 41 = 738$ polynomials.

Answer: 944

Solution:

Ignoring the floors, $\sum_{n=1}^{2023} \frac{n^2 - na}{5} = \frac{1}{5}\left(\sum n^2 - a\sum n\right)$ vanishes exactly when $a = \frac{\sum n^2}{\sum n} = \frac{2 \cdot 2023 + 1}{3} = 1349,$ an integer. For any other integer $a$ the raw sum has absolute value at least $\frac{1}{5}\sum n = \frac{2023 \cdot 1012}{5} \approx 409455,$ while taking floors changes the total by less than $2023,$ so only $a = 1349$ can put $U$ strictly between $-1000$ and $1000.$

With $a = 1349,$ each term is $\frac{n^2 - 1349n - r_n}{5}$ with $r_n = (n^2 - 1349n) \bmod 5,$ so $U = -\frac{1}{5}\sum r_n.$ Since $1349 \equiv 4 \pmod 5,$ we have $n^2 - 1349n \equiv n(n+1) \pmod 5,$ whose residues for $n \equiv 0, 1, 2, 3, 4$ are $0, 2, 1, 2, 0,$ summing to $5$ per block of five. With $2023 = 5 \cdot 404 + 3,$ the leftover terms $n \equiv 1, 2, 3$ contribute $2 + 1 + 2 = 5,$ so $\sum r_n = 405 \cdot 5 = 2025.$

So $U = -\frac{2025}{5} = -405,$ which indeed lies strictly between $-1000$ and $1000,$ and $a + U = 1349 - 405 = 944.$

Answer: 235

Solution:

First, the number of subsets of a block of $m$ consecutive integers containing no two consecutive elements is the Fibonacci number $F_{m+2}$ (with $F_1 = F_2 = 1$): conditioning on whether the last element is used gives the Fibonacci recursion, and the counts start $1, 2, 3, 5, \ldots$

Suppose the unique consecutive pair is $\{k, k+1\}$ for some $1 \le k \le 9.$ The remaining elements must exclude $k - 1$ and $k + 2$ (either would create a second consecutive pair) and must contain no consecutive pair within $\{1, \ldots, k-2\}$ or within $\{k+3, \ldots, 10\},$ blocks of sizes $k - 2$ and $8 - k.$ So the count for this $k$ is $F_k \cdot F_{10-k}.$

Summing over $k = 1, \ldots, 9:$ $$\sum_{k=1}^{9} F_k F_{10-k} = 34 + 21 + 26 + 24 + 25 + 24 + 26 + 21 + 34 = 235.$$

Answer: 75

Solution:

Place $B = (0, 0),$ $C = (55, 0),$ $A = \left(\frac{55}{2}, \frac{55\sqrt{3}}{2}\right),$ so that $D = (7, 0),$ $E = (40, 15\sqrt{3}),$ and $F = \left(\frac{15}{2}, \frac{15\sqrt{3}}{2}\right).$ Let $\theta$ be the common angle and let $\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3$ be the unit vectors in the directions $C \to A,$ $A \to B,$ $B \to C$ — the directions from $E$ toward $A,$ from $F$ toward $B,$ and from $D$ toward $C.$ Splitting $P - E$ into its component along $\mathbf{u}_1$ and its component perpendicular to side $CA,$ whose length is the distance $d_1$ from $P$ to line $CA,$ the angle condition gives $(P - E)\cdot\mathbf{u}_1 = d_1\cot\theta;$ similarly $(P - F)\cdot\mathbf{u}_2 = d_2\cot\theta$ and $(P - D)\cdot\mathbf{u}_3 = d_3\cot\theta.$

Now add all three relations. Since $\mathbf{u}_1 + \mathbf{u}_2 + \mathbf{u}_3 = \mathbf{0}$ (the directed sides of a triangle close up), $P$ drops out, and by Viviani's theorem $d_1 + d_2 + d_3$ equals the height $\frac{55\sqrt{3}}{2}.$ With $\mathbf{u}_1 = \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right),$ $\mathbf{u}_2 = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right),$ and $\mathbf{u}_3 = (1, 0),$ we get $E \cdot \mathbf{u}_1 = \frac{5}{2},$ $F \cdot \mathbf{u}_2 = -15,$ and $D \cdot \mathbf{u}_3 = 7,$ so $$\frac{55\sqrt{3}}{2}\,\cot\theta = -\left(\frac{5}{2} - 15 + 7\right) = \frac{11}{2}.$$

Hence $\cot\theta = \frac{11}{55\sqrt{3}} = \frac{1}{5\sqrt{3}},$ so $\tan^2(\angle AEP) = \left(5\sqrt{3}\right)^2 = 75.$

Answer: 125

Solution:

A rhombus with diagonals $\sqrt{21}$ and $\sqrt{31}$ has side $\sqrt{\frac{21}{4} + \frac{31}{4}} = \sqrt{13},$ so the three edge vectors $\mathbf{u},$ $\mathbf{v},$ $\mathbf{w}$ all have squared length $13.$ In the face spanned by $\mathbf{u}$ and $\mathbf{v}$ the diagonals are $\mathbf{u} \pm \mathbf{v},$ with $|\mathbf{u} \pm \mathbf{v}|^2 = 26 \pm 2\,\mathbf{u}\cdot\mathbf{v};$ matching $\{21, 31\}$ gives $\mathbf{u}\cdot\mathbf{v} = \pm\frac{5}{2},$ and likewise for the other two pairs.

The squared volume is the Gram determinant $$V^2 = \det\begin{pmatrix} 13 & x & y \\ x & 13 & z \\ y & z & 13 \end{pmatrix} = 2197 - 13(x^2 + y^2 + z^2) + 2xyz = 2197 - \frac{975}{4} + 2xyz$$ with $x, y, z \in \left\{\pm\frac{5}{2}\right\}.$ Negating an edge vector flips the signs of two of $x, y, z,$ so only the sign of $xyz$ matters: $2xyz = \pm\frac{125}{4},$ giving $V^2 = \frac{7938}{4}$ or $\frac{7688}{4}.$

The ratio of the volumes is $\sqrt{\frac{7938}{7688}} = \sqrt{\frac{3969}{3844}} = \frac{63}{62},$ already in lowest terms, so $m + n = 63 + 62 = 125.$

Answer: 608

Solution:

Record the hands as an ordered pair $(a, b) \in \mathbb{Z}_{12} \times \mathbb{Z}_{12};$ each movement replaces $(a, b)$ by $(a+1, b)$ or $(a, b+1),$ so a valid sequence is a closed tour through all $144$ positions — equivalently, a choice, at each position, of which hand moves next. Sort the positions into $12$ rows according to $b,$ and let $S_b \subseteq \mathbb{Z}_{12}$ be the set of $a$-values at which the tour leaves row $b$ (a $b$-move). A $b$-move from row $b - 1$ enters row $b$ at the same $a$-value, and the tour then runs through consecutive $a$-values until its next exit. For these runs to cover row $b$ exactly once they must partition $\mathbb{Z}_{12},$ which forces each entry point to sit one step past an exit: $S_{b-1} = S_b + 1.$ In particular every $S_b$ has the same size $c,$ and $S_b = S_0 - b.$

Leaving row $b$ at its $i$-th exit (in cyclic order) leads to a run ending at the $(i+1)$-st exit of row $b + 1:$ each $b$-move advances the row index by $1$ modulo $12$ and the exit index by $1$ modulo $c.$ The tour therefore closes after $\operatorname{lcm}(12, c)$ $b$-moves, while a full tour must use all $12c$ exits, so the tour is a single cycle through all $144$ positions precisely when $\gcd(c, 12) = 1.$ Conversely, every choice of $S_0$ with $\gcd(|S_0|, 12) = 1$ yields exactly one valid movement sequence from the starting position.

Hence $$N = \binom{12}{1} + \binom{12}{5} + \binom{12}{7} + \binom{12}{11} = 12 + 792 + 792 + 12 = 1608,$$ and the remainder when $N$ is divided by $1000$ is $608.$

Answer: 349

Solution:

Write $z = a + bi$ with $a^2 + b^2 = p,$ so $p = 2$ or $p \equiv 1 \pmod 4,$ and the pair $\{|a|, |b|\}$ is then unique. Replacing $z$ by $\pm z,$ $\pm\bar{z},$ $\pm iz,$ $\pm i\bar{z}$ only changes the real and imaginary parts of $z^3$ by signs and swaps, so we may take $a \gt b \gt 0,$ and the two candidate side lengths are $|\operatorname{Re} z^3|$ and $|\operatorname{Im} z^3|.$ Expanding $z^3 = (a^3 - 3ab^2) + (3a^2b - b^3)i$ and factoring, $$\operatorname{Re} z^3 + \operatorname{Im} z^3 = (a - b)(p + 4ab), \qquad \operatorname{Re} z^3 - \operatorname{Im} z^3 = (a + b)(p - 4ab).$$ The triangle exists exactly when $\bigl||\operatorname{Re}| - |\operatorname{Im}|\bigr| \lt p \lt |\operatorname{Re}| + |\operatorname{Im}|,$ and those two quantities are, in some order, the absolute values above. Since $a \gt b$ forces $(a - b)(p + 4ab) \gt p,$ the whole condition reduces to $(a + b)\,|p - 4ab| \lt p.$

Because $a + b \gt \sqrt{a^2 + b^2} = \sqrt{p},$ this requires $|p - 4ab| \lt \sqrt{p} \lt 32:$ the products $4ab$ and $a^2 + b^2$ must nearly coincide. Checking the primes $p \equiv 1 \pmod 4$ below $1000$ from the top down, each with its unique representation (for instance $997 = 31^2 + 6^2$ gives $(a+b)|p - 4ab| = 37 \cdot 253,$ far too big), the condition fails for every prime greater than $349$ and holds for $349 = 18^2 + 5^2,$ where $(a + b)\,|p - 4ab| = 23 \cdot |349 - 360| = 253 \lt 349.$

Indeed for $z = 18 + 5i$ we get $z^3 = 4482 + 4735i,$ and the lengths $349,$ $4482,$ $4735$ form a valid triangle. The answer is $349.$