Solution:

Each pair of consecutive digits must be one of the two-digit squares $16,$ $25,$ $36,$ $49,$ $64,$ $81.$ So each digit determines its successor uniquely if one exists: $1 \to 6,$ $2 \to 5,$ $3 \to 6,$ $4 \to 9,$ $6 \to 4,$ $8 \to 1,$ while $5$ and $9$ end the number.

Following these chains from each possible starting digit, the longest strings are $25,$ $3649,$ and $81649.$ The five-digit chain $8 \to 1 \to 6 \to 4 \to 9$ beats everything else, so $N = 81649,$ whose leftmost three digits are $816.$

Solution:

Let $s$ and $f$ be the numbers of students studying Spanish and French. Since every student studies at least one language, the number studying both is $s + f - 2001.$ The bounds $1600.8 \lt s \lt 1700.85$ force $1601 \le s \le 1700,$ and $600.3 \lt f \lt 800.4$ force $601 \le f \le 800.$

The overlap is smallest when $s + f$ is smallest, giving $m = 1601 + 601 - 2001 = 201,$ and largest when $s + f$ is largest, giving $M = 1700 + 800 - 2001 = 499.$ Both extremes are achievable, so $M - m = 499 - 201 = 298.$

Solution:

For $n \ge 6,$ substitute the recurrence for $x_{n-1}:$ $$x_n = (x_{n-2} - x_{n-3} + x_{n-4} - x_{n-5}) - x_{n-2} + x_{n-3} - x_{n-4} = -x_{n-5}.$$ Hence $x_{n+10} = -x_{n+5} = x_n,$ so the sequence has period $10.$

Since $531,$ $753,$ and $975$ leave remainders $1,$ $3,$ and $5$ upon division by $10,$ we get $x_{531} = x_1 = 211,$ $x_{753} = x_3 = 420,$ and $x_{975} = x_5 = x_4 - x_3 + x_2 - x_1 = 523 - 420 + 375 - 211 = 267.$

The sum is $211 + 420 + 267 = 898.$

Solution:

Points on the two lines can be written $P = (8t, 15t)$ and $Q = (10u, 3u).$ Since $R = (8, 6)$ is the midpoint of $\overline{PQ},$ $$8t + 10u = 16 \qquad \text{and} \qquad 15t + 3u = 12.$$

The second equation gives $u = 4 - 5t;$ substituting into the first, $8t + 40 - 50t = 16,$ so $t = \frac{4}{7}$ and $u = \frac{8}{7}.$ Thus $P = \left(\frac{32}{7}, \frac{60}{7}\right)$ and $Q = \left(\frac{80}{7}, \frac{24}{7}\right).$

Then $PQ = \sqrt{\left(\frac{48}{7}\right)^2 + \left(\frac{36}{7}\right)^2} = \frac{12}{7}\sqrt{4^2 + 3^2} = \frac{60}{7},$ so $m + n = 60 + 7 = 67.$

Solution:

Suppose a ten-element set $\{a_1 \lt a_2 \lt \cdots \lt a_{10}\}$ has no triangle. Then every three elements fail the strict triangle inequality; in particular $a_{i+2} \ge a_{i+1} + a_i$ for each $i.$ Starting from $a_1 \ge 4$ and $a_2 \ge 5,$ this forces $a_3 \ge 9,$ $a_4 \ge 14,$ $a_5 \ge 23,$ $a_6 \ge 37,$ $a_7 \ge 60,$ $a_8 \ge 97,$ $a_9 \ge 157,$ and $a_{10} \ge 254.$

So if $n \le 253,$ no ten-element subset of $\{4, 5, \ldots, n\}$ can avoid triangles, since its largest element would have to be at least $254.$ Conversely, taking equality throughout, the subset $\{4, 5, 9, 14, 23, 37, 60, 97, 157, 254\}$ of $\{4, 5, \ldots, 254\}$ has no triangle.

Therefore the largest possible value is $n = 253.$

Solution:

Center the circle at the origin and let $ABCD$ have side $s,$ so the circle is $x^2 + y^2 = \frac{s^2}{2}$ and side $\overline{CD}$ lies on the line $y = \frac{s}{2}.$ The small square sits on $\overline{CD},$ outside $ABCD:$ if its side is $t,$ then by symmetry $G = \left(\frac{t}{2}, \frac{s}{2} + t\right),$ which must lie on the circle.

Substituting, $\frac{t^2}{4} + \left(\frac{s}{2} + t\right)^2 = \frac{s^2}{2},$ which expands to $5t^2 + 4st - s^2 = 0,$ or $(5t - s)(t + s) = 0.$ Since $t \gt 0,$ we get $t = \frac{s}{5}.$

The ratio of areas is $\frac{t^2}{s^2} = \frac{1}{25},$ so $m = 1,$ $n = 25,$ and $10n + m = 251.$

Solution:

The right angle is at $P,$ so place $P = (0, 0),$ $Q = (0, 90),$ $R = (120, 0).$ The inradius of a right triangle is half the sum of the legs minus the hypotenuse: $r_1 = \frac{90 + 120 - 150}{2} = 30,$ so $C_1$ has center $(30, 30).$ The tangent line to $C_1$ perpendicular to $\overline{PR}$ (on the side toward $R$) is $x = 60,$ and the tangent perpendicular to $\overline{PQ}$ (toward $Q$) is $y = 60.$

Triangle $RST$ is similar to triangle $RPQ$ with ratio $\frac{RS}{RP} = \frac{60}{120} = \frac{1}{2},$ so its inradius is $15$ and its incircle $C_2$ is centered at $(60 + 15, 15) = (75, 15).$ Triangle $QUV$ is similar to triangle $QPR$ with ratio $\frac{QU}{QP} = \frac{30}{90} = \frac{1}{3},$ so its inradius is $10$ and $C_3$ is centered at $(10, 60 + 10) = (10, 70).$

The squared distance is $65^2 + 55^2 = 4225 + 3025 = 7250 = 10 \cdot 725,$ so $n = 725.$

Solution:

Applying $f(3x) = 3f(x)$ six times gives $f(2001) = 3^6 f\left(\frac{2001}{729}\right),$ and $\frac{2001}{729}$ lies in $[1, 3],$ so $$f(2001) = 729\left(1 - \left|\tfrac{2001}{729} - 2\right|\right) = 729 - |2001 - 1458| = 729 - 543 = 186.$$

For $x \in [3^k, 3^{k+1}],$ we have $f(x) = 3^k f\left(\frac{x}{3^k}\right) = 3^k\left(1 - \left|\frac{x}{3^k} - 2\right|\right),$ a tent whose maximum value is $3^k.$ To achieve $186$ we need $3^k \ge 186,$ so $k \ge 5,$ and the smallest solutions lie in $[243, 729],$ where $f(x) = 243 - |x - 486|.$

Setting $243 - |x - 486| = 186$ gives $|x - 486| = 57,$ so $x = 429$ or $x = 543.$ The smallest $x$ is $429.$

Solution:

Compute the probability that the grid does contain an all-red $2$-by-$2$ block by inclusion-exclusion over the four possible positions. One block forces $4$ cells; two blocks sharing an edge force $6$ cells ($4$ such pairs), while the two diagonal pairs force $7;$ any three blocks force $8$ cells, and all four force all $9.$

Each configuration of forced red cells has probability $\left(\frac{1}{2}\right)^{\text{cells}},$ so the probability of at least one red block is $$4 \cdot \frac{1}{16} - \left(4 \cdot \frac{1}{64} + 2 \cdot \frac{1}{128}\right) + 4 \cdot \frac{1}{256} - \frac{1}{512} = \frac{128 - 40 + 8 - 1}{512} = \frac{95}{512}.$$

The desired probability is $1 - \frac{95}{512} = \frac{417}{512},$ and $417 = 3 \cdot 139$ is coprime to $512,$ so $m + n = 417 + 512 = 929.$

Solution:

Factor $10^j - 10^i = 10^i(10^{j-i} - 1).$ Since $1001 = 7 \cdot 11 \cdot 13$ is coprime to $10^i,$ we need $1001 \mid 10^{j-i} - 1.$ The multiplicative order of $10$ is $6$ modulo $7,$ $2$ modulo $11,$ and $6$ modulo $13,$ so $10^k \equiv 1 \pmod{1001}$ exactly when $k$ is a multiple of $6.$ Distinct pairs $(i, j)$ give distinct values, so we just count the pairs.

For $j - i = 6d$ with $1 \le d \le 16,$ the index $i$ can be $0, 1, \ldots, 99 - 6d,$ giving $100 - 6d$ choices. The total is $$\sum_{d=1}^{16} (100 - 6d) = 94 + 88 + \cdots + 4 = \frac{(94 + 4) \cdot 16}{2} = 784.$$

Solution:

Swapping wins and losses is a probability-preserving symmetry, so the probability $P$ of more wins than losses equals the probability of more losses than wins, giving $P = \frac{1 - p_0}{2},$ where $p_0$ is the probability of equally many wins and losses.

An outcome with $k$ wins, $k$ losses, and $6 - 2k$ ties can be arranged in $\frac{6!}{k! \, k! \, (6 - 2k)!}$ ways: $1,$ $30,$ $90,$ $20$ for $k = 0, 1, 2, 3,$ totaling $141.$ Each of the $3^6 = 729$ outcome sequences is equally likely, so $p_0 = \frac{141}{729} = \frac{47}{243}.$

Therefore $P = \frac{1}{2}\left(1 - \frac{47}{243}\right) = \frac{98}{243},$ and $m + n = 98 + 243 = 341.$

Solution:

Attaching a tetrahedron over the midpoint triangle of a face replaces that face by $6$ equilateral triangles of half the side length: the $3$ corner triangles plus $3$ exposed faces of the new tetrahedron. So all faces of $\mathcal{P}_i$ are congruent, with side $\left(\frac{1}{2}\right)^i$ times the original, and $\mathcal{P}_i$ has $4 \cdot 6^i$ faces.

Passing from $\mathcal{P}_i$ to $\mathcal{P}_{i+1}$ glues one regular tetrahedron onto each face; each is similar to $\mathcal{P}_0$ with ratio $\left(\frac{1}{2}\right)^{i+1},$ hence has volume $\left(\frac{1}{8}\right)^{i+1}.$ The volume added is $$4 \cdot 6^i \left(\frac{1}{8}\right)^{i+1} = \frac{1}{2}\left(\frac{3}{4}\right)^i.$$

Therefore the volume of $\mathcal{P}_3$ is $1 + \frac{1}{2} + \frac{3}{8} + \frac{9}{32} = \frac{69}{32},$ and $m + n = 69 + 32 = 101.$

Solution:

Extend $\overline{AB}$ beyond $B$ and $\overline{DC}$ beyond $C$ to meet at $P.$ Since $\angle PAD = \angle PDA,$ triangle $APD$ is isosceles with $PA = PD.$ Also $\angle PBD = 180^\circ - \angle ABD$ and $\angle PCB = 180^\circ - \angle BCD,$ so $\angle PBD = \angle PCB.$

Triangles $PCB$ and $PBD$ share angle $P$ and have $\angle PCB = \angle PBD,$ so they are similar, giving $$\frac{PC}{PB} = \frac{PB}{PD} = \frac{CB}{BD} = \frac{3}{5}.$$ Since $PB = PA - 8 = PD - 8,$ the middle ratio reads $\frac{PD - 8}{PD} = \frac{3}{5},$ so $PD = 20$ and $PB = 12.$ Then $PC = \frac{3}{5} \cdot 12 = \frac{36}{5}.$

Finally $CD = PD - PC = 20 - \frac{36}{5} = \frac{64}{5},$ which is in lowest terms, so $m + n = 64 + 5 = 69.$

Solution:

Write $\operatorname{cis}\theta = \cos\theta + i\sin\theta.$ The equation says $z^8(z^{20} - 1) = 1.$ Taking absolute values and using $|z| = 1$ gives $|z^{20} - 1| = 1,$ so $z^{20}$ is at distance $1$ from both $0$ and $1:$ it is $\operatorname{cis}(\pm 60^\circ).$

If $z^{20} = \operatorname{cis} 60^\circ,$ then $z^{20} - 1 = \operatorname{cis} 120^\circ,$ so $z^8 = \operatorname{cis}(-120^\circ)$ and $$z^4 = \frac{z^{20}}{(z^8)^2} = \operatorname{cis}(60^\circ + 240^\circ) = \operatorname{cis} 300^\circ,$$ which means $4\theta \equiv 300^\circ,$ i.e. $\theta \equiv 75^\circ \pmod{90^\circ}.$ Conversely every such $\theta$ works, since then $z^{20} = (z^4)^5 = \operatorname{cis} 60^\circ$ and $z^8 = (z^4)^2 = \operatorname{cis}(-120^\circ).$ The case $z^{20} = \operatorname{cis}(-60^\circ)$ similarly gives exactly $\theta \equiv 15^\circ \pmod{90^\circ}.$

So the $2n = 8$ angles in increasing order are $15, 75, 105, 165, 195, 255, 285, 345,$ and $\theta_2 + \theta_4 + \theta_6 + \theta_8 = 75 + 165 + 255 + 345 = 840.$

Solution:

Place $A = (0,0,0)$ and $E = (8,8,8),$ so that $I = (6,8,8),$ $J = (8,6,8),$ $K = (8,8,6),$ and $\overline{AE}$ has direction $(1,1,1).$ The line through $I$ in that direction leaves the cube at $L = (0,2,2);$ similarly $J$ and $K$ lead to $M = (2,0,2)$ and $N = (2,2,0).$ The tunnel wall through $I$ and $J$ is the plane $2z = x + y + 2,$ which also contains $L$ and $M$ and crosses the $z$-axis at $O = (0,0,1);$ the other two walls behave symmetrically, crossing the $y$- and $x$-axes at $(0,1,0)$ and $(1,0,0).$

Now add up the surface. Each of the three cube faces at $E$ loses a right triangle with legs $2$ (such as $IEJ$), leaving area $64 - 2 = 62.$ Each of the three faces at $A$ loses a quadrilateral of area $2:$ on the face $z = 0$ its vertices are $(0,0,0),$ $(1,0,0),$ $(2,2,0),$ $(0,1,0).$ Each tunnel wall is a pentagon like $ILOMJ:$ the rectangle $ILMJ$ with $IJ = 2\sqrt{2}$ and $IL = 6\sqrt{3}$ has area $12\sqrt{6},$ and the isosceles triangle $LOM$ with base $LM = 2\sqrt{2}$ and height $\sqrt{3}$ adds $\sqrt{6},$ for $13\sqrt{6}$ per wall.

The total surface area is $6 \cdot 62 + 3 \cdot 13\sqrt{6} = 372 + 39\sqrt{6},$ so $m + n + p = 372 + 39 + 6 = 417.$