Solution:

Let the number be $10a + b$ with tens digit $a$ and units digit $b.$ Since $a \mid 10a + b,$ we need $a \mid b,$ so $b = ka$ for some positive integer $k.$ Since $b \mid 10a + b,$ we need $b \mid 10a,$ that is $ka \mid 10a,$ so $k \mid 10.$ Because $b = ka \le 9,$ only $k = 1,$ $2,$ and $5$ are possible.

For $k = 1$ the numbers are $11, 22, \ldots, 99,$ with sum $11 \cdot 45 = 495.$ For $k = 2$ they are $12, 24, 36, 48,$ with sum $120.$ For $k = 5$ the only one is $15.$

The total is $495 + 120 + 15 = 630.$

Solution:

Let $\mathcal{S}$ have $n$ elements with mean $x,$ so the elements sum to $nx.$ The two conditions say $$\frac{nx + 1}{n + 1} = x - 13 \qquad \text{and} \qquad \frac{nx + 2001}{n + 1} = x + 27.$$

Clearing denominators, $nx + 1 = (n+1)x - 13(n+1)$ gives $x - 13(n+1) = 1,$ and $nx + 2001 = (n+1)x + 27(n+1)$ gives $x + 27(n+1) = 2001.$ Subtracting the first equation from the second yields $40(n+1) = 2000,$ so $n + 1 = 50.$

Then $x = 1 + 13 \cdot 50 = 651.$

Solution:

Expand $\left(\frac{1}{2} - x\right)^{2001}$ by the binomial theorem. Its leading term $(-x)^{2001} = -x^{2001}$ cancels the $x^{2001}$ in the equation, so what remains is a polynomial of degree $2000:$ $$2001 \cdot \frac{1}{2}\,x^{2000} - \binom{2001}{2}\frac{1}{4}\,x^{1999} + \cdots = 0.$$

By Vieta's formulas, the sum of the $2000$ roots is $$\frac{\binom{2001}{2}/4}{2001/2} = \frac{2001 \cdot 2000/8}{2001/2} = \frac{2000}{4} = 500.$$

Solution:

Since $\angle A = 60^\circ$ and $\angle B = 45^\circ,$ we have $\angle C = 75^\circ.$ In triangle $ATC,$ angle $TAC = 30^\circ$ (half of angle $A$), so $\angle ATC = 180^\circ - 30^\circ - 75^\circ = 75^\circ.$ Thus triangle $ACT$ is isosceles with $AC = AT = 24.$

Drop the altitude $CH$ to $\overline{AB}.$ Triangle $ACH$ is $30$-$60$-$90,$ so $AH = 12$ and $CH = 12\sqrt{3}.$ Triangle $BCH$ is $45$-$45$-$90,$ so $BH = CH = 12\sqrt{3}.$

The area is $\frac{1}{2} \cdot CH \cdot AB = \frac{1}{2} \cdot 12\sqrt{3}\,(12 + 12\sqrt{3}) = 216 + 72\sqrt{3}.$ Then $a + b + c = 216 + 72 + 3 = 291.$

Solution:

Since one altitude lies along the $y$-axis, the other two vertices are symmetric: $(x, y)$ and $(-x, y)$ with $x \gt 0.$ The side from $(0,1)$ to $(x,y)$ makes a $120^\circ$ angle with the positive $x$-axis, so it lies on the line $y = -\sqrt{3}\,x + 1.$

Substituting into $x^2 + 4y^2 = 4$ gives $x^2 + 4(1 - \sqrt{3}x)^2 = 4,$ which simplifies to $13x^2 - 8\sqrt{3}\,x = 0,$ so $x = \frac{8\sqrt{3}}{13}.$

The side length is $2x = \frac{16\sqrt{3}}{13},$ whose square is $\frac{768}{169}.$ Since $\gcd(768, 169) = 1,$ the answer is $768 + 169 = 937.$

Solution:

The rolls must form a non-decreasing sequence. Every multiset of four values from $\{1, \ldots, 6\}$ can be arranged in non-decreasing order in exactly one way, so the number of successful outcomes equals the number of such multisets. By stars and bars (4 stars and 5 dividers), that count is $\binom{9}{4} = 126.$

The probability is $\frac{126}{6^4} = \frac{126}{1296} = \frac{7}{72},$ so $m + n = 7 + 72 = 79.$

Solution:

Since $\overline{DE} \parallel \overline{BC},$ triangles $ADE$ and $ABC$ are similar, and the ratio equals the ratio of their heights from $A.$ The line $DE$ passes through the incenter, which sits at height $r$ (the inradius) above $BC,$ so the ratio is $\frac{h - r}{h} = 1 - \frac{r}{h},$ where $h$ is the height from $A$ to $\overline{BC}.$

If $K$ is the area and $s = \frac{21 + 22 + 20}{2} = \frac{63}{2}$ the semiperimeter, then $r = \frac{K}{s}$ and $h = \frac{2K}{20},$ so $$\frac{r}{h} = \frac{20}{2s} = \frac{20}{63}.$$

Therefore $DE = 20\left(1 - \frac{20}{63}\right) = 20 \cdot \frac{43}{63} = \frac{860}{63},$ which is in lowest terms, and $m + n = 860 + 63 = 923.$

Solution:

Suppose $N$ has base-$7$ digits $d_k \ldots d_1 d_0.$ The condition is $\sum d_i \, 10^i = 2 \sum d_i \, 7^i,$ that is $\sum d_i \,(10^i - 2 \cdot 7^i) = 0.$ The coefficients $10^i - 2 \cdot 7^i$ for $i = 0, 1, 2, 3$ are $-1,$ $-4,$ $2,$ $314.$ If there were a digit $d_3 \ge 1,$ the positive contribution would be at least $314,$ but the negative terms total at most $4 \cdot 6 + 6 = 30.$ So $N$ has at most three base-$7$ digits.

For three digits the condition reads $2 d_2 = 4 d_1 + d_0.$ To maximize $N = 49 d_2 + 7 d_1 + d_0,$ take $d_2 = 6,$ so $4 d_1 + d_0 = 12;$ the largest value of $7 d_1 + d_0$ comes from $d_1 = 3,$ $d_0 = 0.$

Thus $N = 49 \cdot 6 + 7 \cdot 3 = 315,$ whose base-$7$ representation is $630 = 2 \cdot 315.$

Solution:

Each corner triangle's area is a product of side fractions: $[ADF] = p(1-r)[ABC],$ $[BED] = q(1-p)[ABC],$ and $[CFE] = r(1-q)[ABC],$ using the formula $\frac{1}{2}xy\sin\theta$ on the shared angles. Subtracting, $$\frac{[DEF]}{[ABC]} = 1 - p(1-r) - q(1-p) - r(1-q) = 1 - (p+q+r) + (pq+qr+rp).$$

From the given values, $pq + qr + rp = \frac{(2/3)^2 - 2/5}{2} = \frac{4/9 - 2/5}{2} = \frac{1}{45}.$

Therefore the ratio is $1 - \frac{2}{3} + \frac{1}{45} = \frac{16}{45},$ and $m + n = 16 + 45 = 61.$

Solution:

The midpoint is a lattice point exactly when the two chosen points agree in parity in each coordinate. Count ordered pairs (allowing equality) coordinate by coordinate. For $x \in \{0, 1, 2\}$ there are $2$ even and $1$ odd values, giving $2^2 + 1^2 = 5$ same-parity ordered pairs. For $y \in \{0, \ldots, 3\}:$ $2^2 + 2^2 = 8.$ For $z \in \{0, \ldots, 4\}:$ $3^2 + 2^2 = 13.$

That gives $5 \cdot 8 \cdot 13 = 520$ ordered pairs, including the $60$ pairs where the two points are equal, so $520 - 60 = 460$ ordered pairs of distinct points, or $230$ unordered pairs. The total number of unordered pairs is $\binom{60}{2} = 1770.$

The probability is $\frac{230}{1770} = \frac{23}{177},$ and since $177 = 3 \cdot 59,$ this is in lowest terms. Thus $m + n = 23 + 177 = 200.$

Solution:

Let $P_i$ sit in column $c_i.$ Then $x_i = (i-1)N + c_i$ and $y_i = 5(c_i - 1) + i.$ The five conditions become $$c_1 = 5c_2 - 3, \quad N + c_2 = 5c_1 - 4, \quad 2N + c_3 = 5c_4 - 1,$$ $$3N + c_4 = 5c_5, \quad 4N + c_5 = 5c_3 - 2.$$

Substituting $c_1 = 5c_2 - 3$ into the second equation gives $N = 24c_2 - 19.$ Eliminating $c_3$ and $c_4$ from the last three equations yields $124 c_5 = 89N + 7.$ Substituting $N = 24c_2 - 19$ and reducing, $31 \mid 534 c_2 - 421,$ i.e. $7 c_2 \equiv 18 \pmod{31},$ whose smallest positive solution is $c_2 = 7.$

Then $N = 24 \cdot 7 - 19 = 149,$ and back-substituting gives valid columns $(c_1, \ldots, c_5) = (32, 7, 107, 45, 86),$ all at most $149.$ So the smallest possible $N$ is $149.$

Solution:

Connecting the incenter to the four vertices splits the tetrahedron into four pyramids of height $r$ over the faces, so $V = \frac{1}{3} r S,$ i.e. $r = \frac{3V}{S},$ where $S$ is the total surface area.

Here $V = \frac{1}{6} \cdot 6 \cdot 4 \cdot 2 = 8.$ The three faces on the coordinate planes are right triangles with areas $\frac{1}{2} \cdot 6 \cdot 4 = 12,$ $\frac{1}{2} \cdot 6 \cdot 2 = 6,$ and $\frac{1}{2} \cdot 4 \cdot 2 = 4.$ For face $ABC,$ the cross product of $\overrightarrow{AB} = (-6, 4, 0)$ and $\overrightarrow{AC} = (-6, 0, 2)$ is $(8, 12, 24),$ with length $4\sqrt{2^2 + 3^2 + 6^2} = 28,$ so that face has area $14.$

Then $S = 12 + 6 + 4 + 14 = 36$ and $r = \frac{24}{36} = \frac{2}{3},$ giving $m + n = 2 + 3 = 5.$

Solution:

A chord subtending a $\theta$-degree arc in a circle of radius $R$ has length $2R\sin\frac{\theta}{2}.$ Write $t = \frac{d}{2},$ so the three chords are $2R\sin t = 22,$ $2R \sin 2t,$ and $2R \sin 3t.$ Using $\sin 2t = 2 \sin t \cos t$ and $\sin 3t = \sin t\,(4\cos^2 t - 1),$ the chords of the $2d$- and $3d$-degree arcs are $22 \cdot 2\cos t = 44 \cos t$ and $22\,(4\cos^2 t - 1).$

The condition "the $2d$-chord is $20$ longer than the $3d$-chord" becomes $44\cos t = 22\,(4\cos^2 t - 1) + 20,$ which simplifies to $44 \cos^2 t - 22 \cos t - 1 = 0.$ From this, $4\cos^2 t = 2\cos t + \frac{1}{11},$ so the $3d$-chord equals $22\left(2\cos t + \frac{1}{11} - 1\right) = 44 \cos t - 20.$

Solving the quadratic, $\cos t = \frac{22 + \sqrt{484 + 176}}{88} = \frac{11 + \sqrt{165}}{44}$ (the $+$ root since $d \lt 120$ means $t \lt 60^\circ,$ so $\cos t \gt \frac{1}{2}$). Then the $3d$-chord is $44\cos t - 20 = 11 + \sqrt{165} - 20 = -9 + \sqrt{165},$ giving $m + n = 9 + 165 = 174.$

Solution:

Write $1$ for a house that gets mail and $0$ for one that does not. Valid patterns are binary strings of length $19$ with no two consecutive $1$s and no three consecutive $0$s. Let $A_n,$ $B_n,$ $C_n$ count valid length-$n$ strings ending in $1,$ in exactly one $0,$ and in exactly two $0$s. A $1$ may follow either kind of $0$-ending, a single $0$ may follow a $1,$ and a second $0$ may follow a single $0:$ $$A_n = B_{n-1} + C_{n-1}, \qquad B_n = A_{n-1}, \qquad C_n = B_{n-1}.$$

Starting from $A_1 = B_1 = 1,$ $C_1 = 0$ and iterating, the totals $A_n + B_n + C_n$ run $2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, 49, 65, 86, 114, 151, 200, 265, 351.$

For $n = 19$ the count is $351.$

Solution:

Pass to the dual cube: the octahedron's faces correspond to a cube's vertices, and two faces share an edge exactly when the corresponding cube vertices are adjacent. Following the numbers $1, 2, \ldots, 8$ and back to $1$ traces a closed $8$-step circuit through all the cube's vertices, and the requirement is that every step is a diagonal (an edge of one of the two inscribed tetrahedra, or one of the $4$ long space diagonals). There are $16$ such diagonals.

Each vertex lies on exactly one long diagonal, so the circuit cannot take two long diagonals in a row, and switching between the two tetrahedra is possible only via a long diagonal. Hence the circuit uses either $4$ long diagonals alternating with tetrahedron edges, or $2$ long diagonals separated by $3$-edge paths in each tetrahedron. In the first case, choosing a pair of opposite edges in each tetrahedron ($3 \cdot 2$ ways) gives $6$ octagons, each traceable as $8 \cdot 2$ permutations: $96.$ In the second case, a $3$-edge path in one tetrahedron can be chosen in $4! = 24$ ways, and the return path through the other tetrahedron is then forced up to $2$ choices, giving $8 \cdot 24 \cdot 2 = 384$ permutations.

So $96 + 384 = 480$ of the $8! = 40320$ labelings work, and the probability is $\frac{480}{40320} = \frac{1}{84}.$ Thus $m + n = 1 + 84 = 85.$