Solution:

Write $n = a^2 - b^2 = (a - b)(a + b).$ The factors $a - b$ and $a + b$ differ by the even number $2b,$ so they have the same parity. If both are odd, $n$ is odd; if both are even, $4 \mid n.$ Hence no integer $n \equiv 2 \pmod 4$ is a difference of two squares.

Conversely, every odd number $2k + 1$ equals $(k+1)^2 - k^2,$ and every multiple of $4,$ say $4k,$ equals $(k+1)^2 - (k-1)^2$ (with $k - 1 \ge 0$ since $k \ge 1$).

Between $1$ and $1000$ there are $500$ odd numbers and $250$ multiples of $4,$ for a total of $500 + 250 = 750.$

Solution:

A rectangle is determined by choosing two of the nine horizontal lines and two of the nine vertical lines, so $r = \binom{9}{2}^2 = 36^2 = 1296.$

A $k \times k$ square can be placed in $(9 - k)^2$ positions, so $$s = \sum_{k=1}^{8} (9 - k)^2 = 8^2 + 7^2 + \cdots + 1^2 = \frac{8 \cdot 9 \cdot 17}{6} = 204.$$

Then $\frac{s}{r} = \frac{204}{1296} = \frac{17}{108},$ which is in lowest terms, so $m + n = 17 + 108 = 125.$

Solution:

Let $a$ be the two-digit number and $b$ the three-digit number. The condition is $1000a + b = 9ab,$ which rearranges to $b(9a - 1) = 1000a.$ Since $\gcd(9a - 1, a) = 1,$ the number $9a - 1$ must divide $1000.$

For a two-digit $a,$ $9a - 1$ runs from $89$ to $890,$ and $9a - 1 \equiv 8 \pmod 9.$ The only divisor of $1000$ in that range congruent to $8$ modulo $9$ is $125,$ giving $a = 14$ and $b = \frac{1000 \cdot 14}{125} = 112,$ which is indeed a three-digit number. Check: $14112 = 9 \cdot 14 \cdot 112.$

The requested sum is $14 + 112 = 126.$

Solution:

Let the radius-$5$ circles have centers $P_1$ and $P_2,$ so $P_1P_2 = 5 + 5 = 10,$ and let $M$ be the midpoint. The radius-$8$ circle's center $Q$ satisfies $QP_1 = QP_2 = 13,$ so $Q$ lies on the perpendicular bisector of $\overline{P_1P_2}$ at distance $\sqrt{13^2 - 5^2} = 12$ from $M.$ Likewise the fourth circle, of radius $r,$ has its center $R$ on the same perpendicular bisector with $RP_1 = 5 + r,$ so $RM = \sqrt{(5+r)^2 - 25} = \sqrt{r^2 + 10r}.$

The small circle nestles in the space between the other three, so $R$ is between $M$ and $Q,$ and external tangency to the radius-$8$ circle gives $$12 - \sqrt{r^2 + 10r} = 8 + r.$$ Then $\sqrt{r^2 + 10r} = 4 - r,$ and squaring yields $r^2 + 10r = 16 - 8r + r^2,$ so $18r = 16$ and $r = \frac{8}{9}.$

Thus $m + n = 8 + 9 = 17.$

Solution:

Among fractions with numerator $1$ or $2,$ the closest neighbors of $\frac{2}{7} \approx 0.2857$ are $\frac{1}{4} = 0.25$ below (note $\frac{2}{8} = \frac{1}{4}$) and $\frac{1}{3} \approx 0.3333$ above (note $\frac{2}{6} = \frac{1}{3}$); no other candidate lies between them. So $\frac{2}{7}$ is the unique closest fraction to $r$ exactly when $r$ is closer to $\frac{2}{7}$ than to both $\frac{1}{4}$ and $\frac{1}{3},$ i.e. when $r$ lies strictly between the midpoints $$\frac{1}{2}\left(\frac{1}{4} + \frac{2}{7}\right) = \frac{15}{56} = 0.26785\ldots \qquad \text{and} \qquad \frac{1}{2}\left(\frac{2}{7} + \frac{1}{3}\right) = \frac{13}{42} = 0.30952\ldots.$$

The four-place decimals in that interval are $0.2679, 0.2680, \ldots, 0.3095,$ and there are $3095 - 2679 + 1 = 417$ of them.

Solution:

Since $B$ is outside the $n$-gon, the angles at $A_1$ — the interior angle $\angle A_nA_1A_2 = \frac{(n-2)180^\circ}{n},$ the equilateral angle $\angle A_2A_1B = 60^\circ,$ and $\angle BA_1A_n$ — fill a full revolution, so $$\angle BA_1A_n = 300^\circ - \frac{(n-2)180^\circ}{n} = 120^\circ + \frac{360^\circ}{n}.$$ Also $A_nA_1 = A_1B,$ since both equal the side of the $n$-gon.

For $A_n,$ $A_1,$ $B$ to be consecutive vertices of a regular $m$-gon, this angle must be the $m$-gon's interior angle: $120^\circ + \frac{360^\circ}{n} = 180^\circ - \frac{360^\circ}{m},$ which simplifies to $\frac{1}{m} = \frac{1}{6} - \frac{1}{n},$ so $m = \frac{6n}{n - 6} = 6 + \frac{36}{n - 6}.$

Thus $n - 6$ must divide $36,$ and the largest choice is $n - 6 = 36,$ i.e. $n = 42$ (with $m = 7$).

Solution:

Put the car at the origin at $t = 0,$ with east as the positive $x$-direction and north as the positive $y$-direction. At time $t$ the car is at $\left(\frac{2t}{3}, 0\right),$ and the storm center, moving southeast at speed $\frac{\sqrt{2}}{2}$ (components $\frac{1}{2}$ east and $\frac{1}{2}$ south), is at $\left(\frac{t}{2}, 110 - \frac{t}{2}\right).$

The car is on the storm boundary when the squared distance is $51^2:$ $$\left(\frac{2t}{3} - \frac{t}{2}\right)^2 + \left(110 - \frac{t}{2}\right)^2 = 51^2,$$ that is $\frac{t^2}{36} + \frac{t^2}{4} - 110t + 12100 - 2601 = 0,$ or $\frac{5}{18}t^2 - 110t + 9499 = 0.$

The roots are $t_1$ and $t_2,$ so by Vieta's formulas $t_1 + t_2 = \frac{110 \cdot 18}{5} = 396,$ and $\frac{1}{2}(t_1 + t_2) = 198.$

Solution:

Each row must contain two $1$'s and two $-1$'s, so identify each row with the pair of columns holding its $1$'s; each column must end up chosen by exactly two rows. There are $\binom{4}{2} = 6$ choices for row $1.$ Classify by how row $2$ overlaps row $1.$

If row $2$ uses the same pair ($1$ way), those two columns are full, so rows $3$ and $4$ must both use the complementary pair: $1$ completion. If row $2$ uses the complementary pair ($1$ way), every column has one $1$ so far, so rows $3$ and $4$ need only be a complementary pair themselves: $6$ choices for row $3,$ row $4$ forced, giving $6$ completions. If row $2$ shares exactly one column with row $1$ ($2 \cdot 2 = 4$ ways), one column is full, two have one $1,$ and one is empty; rows $3$ and $4$ must each take the empty column together with one of the two half-filled columns, so there are $2$ completions.

The total is $6\,(1 \cdot 1 + 1 \cdot 6 + 4 \cdot 2) = 6 \cdot 15 = 90.$

Solution:

From $2 \lt a^2 \lt 3$ we get $\sqrt{2} \lt a \lt \sqrt{3},$ so $0 \lt a^{-1} \lt 1$ and $\langle a^{-1}\rangle = a^{-1},$ while $\langle a^2\rangle = a^2 - 2.$ The condition becomes $a^{-1} = a^2 - 2,$ i.e. $a^3 - 2a - 1 = 0,$ which factors as $$(a + 1)(a^2 - a - 1) = 0.$$ Since $a \gt 0,$ we get $a = \frac{1 + \sqrt{5}}{2},$ the golden ratio, and indeed $a^2 = a + 1 \approx 2.618$ lies in $(2, 3).$

Using $a^2 = a + 1$ repeatedly: $a^4 = (a+1)^2 = 3a + 2,$ $a^8 = (3a+2)^2 = 9(a+1) + 12a + 4 = 21a + 13,$ and $a^{12} = a^8 a^4 = (21a + 13)(3a + 2) = 63(a+1) + 81a + 26 = 144a + 89.$ Also $a^{-1} = a - 1$ from $a^2 = a + 1.$

Therefore $a^{12} - 144a^{-1} = 144a + 89 - 144(a - 1) = 89 + 144 = 233.$

Solution:

Given any two distinct cards, there is exactly one card completing them to a complementary set: in each attribute, if the two cards agree, the third card must share that value, and if they differ, the third must take the one remaining value. The completing card is distinct from both (the two given cards differ somewhere, and in that attribute the third card differs from each).

So the $\binom{27}{2} = 351$ pairs of cards each extend to one complementary set, and each complementary set is produced by $\binom{3}{2} = 3$ of these pairs. The number of sets is $\frac{351}{3} = 117.$

Solution:

Multiply numerator and denominator by $2\sin\frac{1}{2}^\circ.$ Since $2\cos n^\circ \sin\frac{1}{2}^\circ = \sin\left(n + \frac{1}{2}\right)^\circ - \sin\left(n - \frac{1}{2}\right)^\circ$ and $2\sin n^\circ \sin\frac{1}{2}^\circ = \cos\left(n - \frac{1}{2}\right)^\circ - \cos\left(n + \frac{1}{2}\right)^\circ,$ both sums telescope: $$x = \frac{\sin 44.5^\circ - \sin 0.5^\circ}{\cos 0.5^\circ - \cos 44.5^\circ} = \frac{2\cos 22.5^\circ \sin 22^\circ}{2\sin 22.5^\circ \sin 22^\circ} = \cot 22.5^\circ,$$ using the sum-to-product identities in the last step.

By the half-angle formula, $\cot 22.5^\circ = \frac{1 + \cos 45^\circ}{\sin 45^\circ} = \sqrt{2} + 1.$ Hence $100x = 100\sqrt{2} + 100 = 241.42\ldots,$ and the greatest integer not exceeding it is $241.$

Solution:

Composing, $f(f(x)) = \frac{(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)},$ and this equals $x$ identically only if $c(a + d) = 0.$ Since $c \ne 0,$ we get $d = -a,$ so $f(x) = \frac{ax + b}{cx - a}.$

A fixed point satisfies $cx^2 - ax = ax + b,$ i.e. $cx^2 - 2ax - b = 0,$ whose roots are $19$ and $97.$ By Vieta's formulas, $19 + 97 = \frac{2a}{c},$ so $\frac{a}{c} = 58.$

Finally, $y$ is in the range exactly when $y = \frac{ax + b}{cx - a}$ has a solution, i.e. $x(cy - a) = ay + b.$ This solves for $x$ unless $cy - a = 0;$ and when $y = \frac{a}{c},$ the right side $\frac{a^2}{c} + b = \frac{a^2 + bc}{c}$ is nonzero (otherwise $f$ would be constant). So the unique number not in the range is $\frac{a}{c} = 58.$

Solution:

Let $f(t) = \bigl|\,||t| - 2| - 1\,\bigr|,$ so the equation is $f(x) + f(y) = 1.$ The function $f$ is even, and for $t \ge 0:$ on $[0, 2],$ $||t| - 2| - 1 = (2 - t) - 1 = 1 - t,$ so $f(t) = |t - 1|;$ on $[2, 4],$ $f(t) = |t - 3|;$ and for $t \gt 4,$ $f(t) = t - 3 \gt 1,$ which is too large. So on the relevant range, $f(t) = |t - a|$ where $a \in \{-3, -1, 1, 3\}$ is the nearest of those four values to $t.$

Therefore $S$ is the union of the $16$ taxicab circles $$|x - a| + |y - b| = 1, \qquad a, b \in \{-3, -1, 1, 3\},$$ which meet only at isolated points. Each is a square (diamond) with diagonal $2,$ hence side $\sqrt{2}$ and perimeter $4\sqrt{2}.$

The total length is $16 \cdot 4\sqrt{2} = 64\sqrt{2},$ so $a + b = 64 + 2 = 66.$

Solution:

By rotational symmetry we may fix $v$ and let $w = v e^{2\pi i k/1997}$ with $k$ uniform in $\{1, 2, \ldots, 1996\}.$ Then $$|v + w| = \left|1 + e^{2\pi i k/1997}\right| = 2\left|\cos\frac{\pi k}{1997}\right|.$$ Also $\left(2\cos 15^\circ\right)^2 = 2 + 2\cos 30^\circ = 2 + \sqrt{3},$ so the threshold is $\sqrt{2 + \sqrt{3}} = 2\cos\frac{\pi}{12}.$

The condition $\left|\cos\frac{\pi k}{1997}\right| \ge \cos\frac{\pi}{12}$ holds exactly when $\frac{\pi k}{1997}$ is within $\frac{\pi}{12}$ of $0$ or of $\pi,$ i.e. $k \le \frac{1997}{12} = 166.41\ldots$ or $k \ge \frac{11 \cdot 1997}{12} = 1830.58\ldots.$ That gives $166 + 166 = 332$ favorable values of $k.$

The probability is $\frac{332}{1996} = \frac{83}{499},$ and $499$ is prime, so $m + n = 83 + 499 = 582.$

Solution:

Place the rectangle with corners $(0, 0),$ $(11, 0),$ $(11, 10),$ $(0, 10).$ A maximal equilateral triangle can be enlarged unless it is pinned by the rectangle, and the extremal position has one vertex at a corner, say the origin, with the other two vertices $s(\cos\theta, \sin\theta)$ and $s\bigl(\cos(\theta + 60^\circ), \sin(\theta + 60^\circ)\bigr)$ touching the far sides $x = 11$ and $y = 10:$ $$s\cos\theta = 11, \qquad s\sin(\theta + 60^\circ) = 10.$$

Dividing, $11\sin(\theta + 60^\circ) = 10\cos\theta,$ and expanding the left side gives $\frac{11}{2}\sin\theta + \frac{11\sqrt{3}}{2}\cos\theta = 10\cos\theta,$ so $\tan\theta = \frac{20 - 11\sqrt{3}}{11}$ (about $4.9^\circ,$ a legal tilt). Then $$s^2 = \frac{121}{\cos^2\theta} = 121\left(1 + \tan^2\theta\right) = 121 + \left(20 - 11\sqrt{3}\right)^2 = 884 - 440\sqrt{3}.$$

The area is $\frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4}\left(884 - 440\sqrt{3}\right) = 221\sqrt{3} - 330 \approx 52.8,$ which indeed beats the untilted triangle of side $10.$ Thus $p + q + r = 221 + 3 + 330 = 554.$