Solution:

Let the terms be $p,$ $p + d,$ $\ldots,$ $p + 4d.$ If $d$ were odd, consecutive terms would have opposite parity, so some term other than the first would be even and greater than $2$ — impossible. If $d$ were not a multiple of $3,$ then $p,$ $p + d,$ $p + 2d$ would cover all residues mod $3,$ so some term would be divisible by $3;$ that term would have to be $3$ itself, forcing $p = 3,$ but then $p + 3d = 3(1 + d)$ is composite. Hence $6 \mid d.$

With $d \ge 6$ the fifth term is at least $p + 24.$ Trying $p = 5$ and $d = 6$ gives $5, 11, 17, 23, 29,$ all prime, and no smaller fifth term is possible since $p \ge 5$ (the starts $p = 2$ and $p = 3$ fail as above). The answer is $29.$

Solution:

A parallelogram is symmetric under the $180^\circ$ rotation about its center, so any line through the center cuts it into two pieces that are swapped by that rotation, hence congruent. The center is the midpoint of a diagonal: $$\left(\frac{10 + 28}{2}, \frac{45 + 153}{2}\right) = (19, 99).$$

The line through the origin and $(19, 99)$ has slope $\frac{99}{19},$ and $\gcd(99, 19) = 1,$ so $m + n = 99 + 19 = 118.$

Solution:

Suppose $n^2 - 19n + 99 = k^2.$ Multiplying by $4$ and completing the square gives $(2n - 19)^2 + 35 = (2k)^2,$ so $$\bigl(2k - (2n - 19)\bigr)\bigl(2k + (2n - 19)\bigr) = 35.$$ The two factors sum to $4k \gt 0,$ so both are positive: the factor pairs are $(1, 35),$ $(5, 7),$ $(7, 5),$ and $(35, 1).$

Subtracting the first factor from the second gives $2(2n - 19) = 34,$ $2,$ $-2,$ or $-34,$ so $n = 18,$ $10,$ $9,$ or $1.$ Each indeed makes the expression a perfect square ($81,$ $9,$ $9,$ $81$), and the sum is $18 + 10 + 9 + 1 = 38.$

Solution:

The whole configuration is unchanged by rotating $90^\circ$ about $O,$ which cycles the octagon side $AB$ to $CD,$ $EF,$ $GH,$ and it is also unchanged by the reflection that swaps the two squares, which carries those sides to $BC,$ $DE,$ $FG,$ $HA.$ So all eight sides of the octagon have the same length, $\frac{43}{99}.$

Segments from $O$ to the eight vertices cut the octagon into $8$ triangles. Each has base $\frac{43}{99}$ lying on a side of one of the unit squares, so its height from $O$ is the distance from the center to that side, namely $\frac{1}{2}.$ The area is $$8 \cdot \frac{1}{2} \cdot \frac{43}{99} \cdot \frac{1}{2} = \frac{86}{99}.$$

Since $\gcd(86, 99) = 1,$ the answer is $86 + 99 = 185.$

Solution:

If the last digit of $x$ is at most $7,$ adding $2$ changes no other digit, so $T(x) = 2.$ Otherwise there is carrying. If $x$ ends in the digit $8$ preceded by exactly $m \ge 0$ nines, then $x + 2$ replaces $\ldots a\underbrace{9 \cdots 9}_{m}8$ by $\ldots (a{+}1)\underbrace{0 \cdots 0}_{m}0,$ so $S(x + 2) - S(x) = 1 - 9m - 8$ and $T(x) = 9m + 7.$ If $x$ ends in exactly $m \ge 1$ nines, then $x + 2$ replaces $\ldots a\underbrace{9 \cdots 9}_{m}$ by $\ldots (a{+}1)\underbrace{0 \cdots 0}_{m - 1}1,$ so $S(x + 2) - S(x) = 2 - 9m$ and $T(x) = 9m - 2.$

Both carrying families give exactly the values $7, 16, 25, \ldots,$ that is, $9j + 7$ for $j \ge 0,$ and every such value occurs. So the possible values of $T$ are $2$ together with all $9j + 7.$ Requiring $9j + 7 \le 1999$ gives $j \le 221,$ which is $222$ values, and $T = 2$ adds one more, for a total of $223.$

Solution:

Follow the four edges. Sides $AB$ and $DC$ lie on the lines $y = \frac{x}{3}$ and $y = 3x,$ which map to the lines $v = \frac{u}{\sqrt{3}}$ and $v = \sqrt{3}\,u$ — rays from the origin at angles $30^\circ$ and $60^\circ.$ Sides $AD$ and $BC$ lie on $x + y = 1200$ and $x + y = 2400,$ which map to arcs of the circles $u^2 + v^2 = 1200$ and $u^2 + v^2 = 2400.$

So the image is the part of the annulus between radii $\sqrt{1200}$ and $\sqrt{2400}$ lying between the $30^\circ$ and $60^\circ$ rays, one twelfth of the full annulus: $$k = \frac{30}{360}\,\pi\,(2400 - 1200) = 100\pi \approx 314.16.$$

The greatest integer not exceeding $k$ is $314.$

Solution:

The switch labeled $d$ is advanced exactly once for each step $i$ whose label is a multiple of $d.$ The multiples of $2^x 3^y 5^z$ among the labels are the $2^{x'} 3^{y'} 5^{z'}$ with $x \le x' \le 9,$ etc., so that switch advances $(10 - x)(10 - y)(10 - z)$ times. It returns to position $A$ exactly when this count is a multiple of $4.$

Write $a = 10 - x,$ $b = 10 - y,$ $c = 10 - z,$ each ranging over $1$ through $10.$ We count the triples where $abc$ is not divisible by $4:$ either all three are odd, or exactly one is even but not divisible by $4.$ Among $1, \ldots, 10$ there are $5$ odd values and $3$ values ($2, 6, 10$) that are twice an odd number. That gives $5^3 = 125$ triples of the first kind and $3 \cdot 3 \cdot 5^2 = 225$ of the second, or $350$ in all.

Therefore $1000 - 350 = 650$ switches end in position $A.$

Solution:

$\mathcal{T}$ is the triangle with vertices $(1,0,0),$ $(0,1,0),$ $(0,0,1).$ Because $\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1 = x + y + z,$ whenever two of the inequalities $x \ge \frac{1}{2},$ $y \ge \frac{1}{3},$ $z \ge \frac{1}{6}$ hold, the third can hold only on a boundary segment of zero area. So $\mathcal{S}$ is, up to measure zero, the union of the three regions where a specific pair of inequalities holds.

The region with $x \ge \frac{1}{2}$ and $y \ge \frac{1}{3}$ becomes, after substituting $x = \frac{1}{2} + x'$ and $y = \frac{1}{3} + y',$ a copy of $\mathcal{T}$ with coordinate sum $1 - \frac{1}{2} - \frac{1}{3} = \frac{1}{6},$ i.e. a triangle similar to $\mathcal{T}$ with ratio $\frac{1}{6}$ and area $\left(\frac{1}{6}\right)^2$ of $\mathcal{T}.$ Likewise the pairs $\{x, z\}$ and $\{y, z\}$ give similar triangles with ratios $\frac{1}{3}$ and $\frac{1}{2}.$

The ratio of areas is $$\frac{1}{36} + \frac{1}{9} + \frac{1}{4} = \frac{1 + 4 + 9}{36} = \frac{7}{18},$$ so $m + n = 7 + 18 = 25.$

Solution:

The condition is $|f(z) - z| = |f(z)|$ for all $z,$ that is, $|(a - 1 + bi)z| = |(a + bi)z|.$ Dividing by $|z|$ (for $z \ne 0$) gives $|a - 1 + bi| = |a + bi|,$ so $$(a - 1)^2 + b^2 = a^2 + b^2,$$ which forces $a = \frac{1}{2}.$

Since $|a + bi| = 8,$ we have $a^2 + b^2 = 64,$ so $b^2 = 64 - \frac{1}{4} = \frac{255}{4}.$ As $\gcd(255, 4) = 1,$ the answer is $255 + 4 = 259.$

Solution:

There are $\binom{10}{2} = 45$ segments, so $\binom{45}{4} = 148995$ equally likely choices. Two distinct triangles share at most one edge, so together they use at least $5$ segments; hence a set of $4$ segments contains at most one triangle, and the favorable sets are counted exactly once by choosing a triangle and then a fourth segment: $$\binom{10}{3} \cdot (45 - 3) = 120 \cdot 42 = 5040.$$

The probability is $\frac{5040}{148995} = \frac{16}{473},$ already in lowest terms ($473 = 11 \cdot 43$), so $m + n = 16 + 473 = 489.$

Solution:

Multiply the sum by $2 \sin 2.5^\circ$ and apply $2 \sin 5k^\circ \sin 2.5^\circ = \cos(5k - 2.5)^\circ - \cos(5k + 2.5)^\circ,$ so the sum telescopes: $$2 \sin 2.5^\circ \sum_{k=1}^{35} \sin 5k^\circ = \cos 2.5^\circ - \cos 177.5^\circ = 2\cos 2.5^\circ.$$

Hence the sum equals $\frac{\cos 2.5^\circ}{\sin 2.5^\circ} = \cot 2.5^\circ = \tan 87.5^\circ = \tan \frac{175}{2}^\circ.$ Since $\gcd(175, 2) = 1$ and $\frac{175}{2} \lt 90,$ we get $m + n = 175 + 2 = 177.$

Solution:

Tangent segments from a point are equal, so the tangent lengths from $A,$ $B,$ $C$ are $23,$ $27,$ and some $z.$ Then the sides are $50,$ $23 + z,$ $27 + z,$ the semiperimeter is $s = 50 + z,$ and Heron's formula gives area $\sqrt{(50 + z) \cdot z \cdot 23 \cdot 27}.$

The area also equals $rs = 21(50 + z).$ Squaring $21(50 + z) = \sqrt{621 z (50 + z)}$ and dividing by $50 + z$ gives $441(50 + z) = 621 z,$ so $180 z = 22050$ and $z = \frac{245}{2}.$

The perimeter is $2s = 2\left(50 + \frac{245}{2}\right) = 345.$

Solution:

There are $\binom{40}{2} = 780$ games, hence $2^{780}$ equally likely outcomes. If all $40$ win totals are distinct, they must be exactly $0, 1, \ldots, 39.$ In that case the team with $39$ wins beat everyone, the team with $38$ wins beat everyone except that team, and so on: the assignment of totals to teams determines every game. Conversely each of the $40!$ assignments arises from exactly one outcome, so the probability is $\frac{40!}{2^{780}}.$

By Legendre's formula the power of $2$ dividing $40!$ is $20 + 10 + 5 + 2 + 1 = 38.$ In lowest terms the denominator is therefore $n = 2^{780 - 38} = 2^{742},$ so $\log_2 n = 742.$

Solution:

Let $\omega = \angle PAB = \angle PBC = \angle PCA.$ In triangle $ABP,$ the angles at $A$ and $B$ are $\omega$ and $B - \omega,$ so $\angle APB = 180^\circ - B$ and the law of sines gives $BP = \frac{c \sin\omega}{\sin B}.$ In triangle $BCP,$ the angles at $B$ and $C$ are $\omega$ and $C - \omega,$ so $\angle BPC = 180^\circ - C$ and $BP = \frac{a \sin(C - \omega)}{\sin C}.$

Equating and substituting $a = 2R\sin A,$ $c = 2R\sin C$ yields $\sin^2 C \sin\omega = \sin A \sin B \sin(C - \omega).$ Expanding $\sin(C - \omega)$ and dividing by $\sin A \sin B \sin C \sin\omega,$ $$\frac{\sin C}{\sin A \sin B} = \cot\omega - \cot C,$$ and since $\sin C = \sin(A + B) = \sin A \cos B + \cos A \sin B,$ the left side is $\cot A + \cot B.$ Hence $\cot\omega = \cot A + \cot B + \cot C.$

Using $\cot A = \frac{b^2 + c^2 - a^2}{4K}$ and its analogues, where $K$ is the area, $$\cot\omega = \frac{a^2 + b^2 + c^2}{4K} = \frac{169 + 196 + 225}{4 \cdot 84} = \frac{590}{336} = \frac{295}{168},$$ since the $13$-$14$-$15$ triangle has area $84.$ So $\tan\omega = \frac{168}{295},$ which is in lowest terms, and $m + n = 168 + 295 = 463.$

Solution:

The midpoints are $M_1 = (17, 0),$ $M_2 = (25, 12),$ and $M_3 = (8, 12).$ Folding the three corner triangles up along the sides of the midpoint triangle brings the corners together at one apex $Q$ (each pair of glued half-sides has equal length). The apex keeps its folded distances: $QM_1 = 17$ (half of the side of length $34$ that $M_1$ bisects), $QM_2 = 15$ (half of $30$), and $QM_3 = 4\sqrt{13}$ (half of $\sqrt{16^2 + 24^2} = 8\sqrt{13}$).

Keep the midpoint triangle in the plane $z = 0$ and let $Q = (x, y, z).$ Subtracting $|Q - M_3|^2 = 208$ from $|Q - M_2|^2 = 225$ gives $(x - 25)^2 - (x - 8)^2 = 17,$ so $x = 16;$ subtracting $|Q - M_2|^2 = 225$ from $|Q - M_1|^2 = 289$ gives $2x + 3y = 68,$ so $y = 12.$ Then $z^2 = 289 - (16 - 17)^2 - 12^2 = 144,$ so the apex is at height $z = 12.$

The base is the midpoint triangle, with area one quarter of the original triangle's $\frac{1}{2} \cdot 34 \cdot 24 = 408,$ i.e. $102.$ The volume is $$\frac{1}{3} \cdot 102 \cdot 12 = 408.$$