Solution:

Since $2010^2 = 2^2 \cdot 3^2 \cdot 5^2 \cdot 67^2,$ it has $(2+1)^4 = 81$ positive divisors. A divisor is a perfect square exactly when each of its four exponents is $0$ or $2,$ giving $2^4 = 16$ perfect squares and $81 - 16 = 65$ non-squares.

The probability of picking one of each is $$p = \frac{16 \cdot 65}{\binom{81}{2}} = \frac{1040}{3240} = \frac{26}{81},$$ so $m + n = 26 + 81 = 107.$

Solution:

Work modulo $1000.$ Every factor from the third one on ends in at least three $9$s, so each is $\equiv -1 \pmod{1000}.$ There are $999$ factors in all, hence $997$ of them are $\equiv -1.$

The product is therefore $$\equiv 9 \cdot 99 \cdot (-1)^{997} \equiv -891 \equiv 109 \pmod{1000},$$ so the remainder is $109.$

Solution:

Substituting $y = \frac{3}{4}x$ into $x^y = y^x$ gives $$x^{\frac{3}{4}x} = \left(\tfrac{3}{4}x\right)^{x}.$$ Taking $x$th roots (the quantities here are positive), $x^{3/4} = \frac{3}{4}x,$ so dividing by $x$ yields $x^{-1/4} = \frac{3}{4},$ that is, $x = \left(\frac{4}{3}\right)^4 = \frac{256}{81}.$

Then $y = \frac{3}{4} \cdot \frac{256}{81} = \frac{64}{27},$ and $$x + y = \frac{256}{81} + \frac{192}{81} = \frac{448}{81}.$$ Since $\gcd(448, 81) = 1,$ the answer is $448 + 81 = 529.$

Solution:

Let $p(h)$ be the probability that one player flips $h$ heads. Splitting according to the two fair coins and the biased coin, $$p(0) = \tfrac{1}{4} \cdot \tfrac{3}{7} = \tfrac{3}{28}, \quad p(1) = \tfrac{2}{4} \cdot \tfrac{3}{7} + \tfrac{1}{4} \cdot \tfrac{4}{7} = \tfrac{10}{28}, \quad p(2) = \tfrac{1}{4} \cdot \tfrac{3}{7} + \tfrac{2}{4} \cdot \tfrac{4}{7} = \tfrac{11}{28}, \quad p(3) = \tfrac{1}{4} \cdot \tfrac{4}{7} = \tfrac{4}{28}.$$

Jackie's and Phil's flips are independent with the same distribution, so the probability that their head counts agree is $$\sum_h p(h)^2 = \frac{3^2 + 10^2 + 11^2 + 4^2}{28^2} = \frac{246}{784} = \frac{123}{392}.$$ Thus $m + n = 123 + 392 = 515.$

Solution:

Factoring, $$a^2 - b^2 + c^2 - d^2 = (a-b)(a+b) + (c-d)(c+d) \ge (a+b) + (c+d) = 2010,$$ since $a - b \ge 1$ and $c - d \ge 1.$ Equality holds, so $a - b = c - d = 1,$ that is, $b = a - 1$ and $d = c - 1.$ Then $2010 = a + (a-1) + c + (c-1)$ gives $a + c = 1006.$

The condition $b \gt c$ means $a - 1 \gt c = 1006 - a,$ so $a \ge 504,$ and $d \ge 1$ means $c \ge 2,$ so $a \le 1004.$ Every $a$ in this range works, via $(a, b, c, d) = (a,\, a-1,\, 1006-a,\, 1005-a).$

The count is $1004 - 504 + 1 = 501.$

Solution:

Completing the square, the condition reads $$(x-1)^2 + 1 \le P(x) \le 2(x-1)^2 + 1.$$ At $x = 1$ both bounds equal $1,$ so $P(1) = 1.$ The quadratic $P(x) - \left((x-1)^2 + 1\right)$ is nonnegative for all $x$ and vanishes at $x = 1,$ so $x = 1$ is a double root: $P(x) = a(x-1)^2 + 1$ for some constant $a.$

From $P(11) = 100a + 1 = 181$ we get $a = \frac{9}{5}.$ Then $$P(16) = \frac{9}{5} \cdot 225 + 1 = 405 + 1 = 406.$$

Solution:

Write $\mathcal{A} \cap \mathcal{B} = \{x\},$ $\mathcal{B} \cap \mathcal{C} = \{y\},$ and $\mathcal{C} \cap \mathcal{A} = \{z\}.$ Since $\mathcal{A} \cap \mathcal{B} \cap \mathcal{C} = \emptyset,$ the elements $x,$ $y,$ $z$ are distinct, and they can be chosen in $7 \cdot 6 \cdot 5 = 210$ ways.

Each of the remaining $4$ elements must not create any further pairwise intersections, so it can belong to exactly one of $\mathcal{A},$ $\mathcal{B},$ $\mathcal{C},$ or to none of them: $4$ choices each, for $4^4 = 256$ assignments.

Hence $N = 210 \cdot 256 = 53760,$ and the remainder upon division by $1000$ is $760.$

Solution:

Since $\lfloor x \rfloor$ and $\lfloor y \rfloor$ are integers whose squares sum to $25,$ the pair $(\lfloor x \rfloor, \lfloor y \rfloor)$ is one of the $12$ pairs $(\pm 5, 0),$ $(0, \pm 5),$ $(\pm 3, \pm 4),$ $(\pm 4, \pm 3).$ So $\mathcal{R}$ is the union of the $12$ unit squares whose lower-left corners are these points.

The map $(x, y) \mapsto (1 - x, 1 - y)$ permutes these squares, so $\mathcal{R}$ is symmetric under $180^\circ$ rotation about $Q = \left(\frac{1}{2}, \frac{1}{2}\right).$ The smallest enclosing disk is unique, so its center must be $Q.$ The farthest points of $\mathcal{R}$ from $Q$ are square corners such as $A = (4, 5)$ and $B = (5, 4),$ at distance $$\sqrt{\left(\tfrac{9}{2}\right)^2 + \left(\tfrac{7}{2}\right)^2} = \frac{\sqrt{130}}{2};$$ checking all twelve squares confirms no corner is farther.

Hence the minimum radius is $r = \frac{\sqrt{130}}{2},$ and $m + n = 130 + 2 = 132.$

Solution:

Adding $xyz$ to each equation gives $x^3 = 2 + xyz,$ $y^3 = 6 + xyz,$ and $z^3 = 20 + xyz.$ Let $P = xyz.$ Multiplying the three equations yields $$P^3 = (2 + P)(6 + P)(20 + P) = P^3 + 28P^2 + 172P + 240,$$ so $28P^2 + 172P + 240 = 0,$ i.e. $7P^2 + 43P + 60 = 0,$ whose roots are $P = -\frac{15}{7}$ and $P = -4.$ Each root is achievable: the cube roots of $2 + P,$ $6 + P,$ $20 + P$ then really do have product $P.$

Adding the original equations, $x^3 + y^3 + z^3 = 28 + 3P,$ which is maximized by the larger root $P = -\frac{15}{7}:$ $$a^3 + b^3 + c^3 = 28 - \frac{45}{7} = \frac{151}{7}.$$ Thus $m + n = 151 + 7 = 158.$

Solution:

Write each coefficient as $a_i = 10b_i + c_i$ with digits $b_i, c_i \in \{0, 1, \ldots, 9\};$ every integer $0 \le a_i \le 99$ splits this way uniquely. Setting $m = b_3 b_2 b_1 b_0$ and $n = c_3 c_2 c_1 c_0$ (read as base-$10$ numbers), the condition becomes $$2010 = 10m + n.$$

Conversely, any nonnegative integers $m, n$ with $10m + n = 2010$ satisfy $m \le 201$ and $n \le 2010,$ so each has at most four digits; those digits recover the $b_i$ and $c_i,$ hence the $a_i.$ So representations correspond exactly to choices of $m \in \{0, 1, \ldots, 201\}$ with $n = 2010 - 10m,$ and $N = 202.$

Solution:

The condition $|8 - x| + y \le 10$ means $y \le x + 2$ for $x \le 8$ and $y \le 18 - x$ for $x \ge 8.$ Intersecting with the half-plane $3y - x \ge 15$ leaves the triangle with vertices $A = \left(\frac{9}{2}, \frac{13}{2}\right)$ and $B = \left(\frac{39}{4}, \frac{33}{4}\right)$ on the line $3y - x = 15,$ and apex $C = (8, 10).$

Side $AB$ lies on the axis of revolution, and the foot $D$ of the perpendicular from $C$ to the line, namely $(8.7, 7.9),$ lies between $A$ and $B.$ So the solid is two cones sharing a base of radius $CD$ with heights summing to $AB,$ and its volume is $\frac{1}{3}\pi \cdot CD^2 \cdot AB.$ Here $$CD = \frac{|3 \cdot 10 - 8 - 15|}{\sqrt{10}} = \frac{7}{\sqrt{10}}, \qquad AB = \sqrt{\left(\tfrac{21}{4}\right)^2 + \left(\tfrac{7}{4}\right)^2} = \frac{7\sqrt{10}}{4}.$$

The volume is $\frac{1}{3}\pi \cdot \frac{49}{10} \cdot \frac{7\sqrt{10}}{4} = \frac{343\pi}{12\sqrt{10}},$ so $m + n + p = 343 + 12 + 10 = 365.$

Solution:

First, $m = 243$ works. Suppose $S = \{3, 4, \ldots, 243\}$ were partitioned into $T$ and $U$ with neither containing a product, and say $3 \in T.$ Then $9 = 3 \cdot 3$ must lie in $U,$ so $81 = 9 \cdot 9$ must lie in $T,$ and then $243 = 3 \cdot 81$ must lie in $U.$ Now consider $27:$ if $27 \in T,$ then $3 \cdot 27 = 81$ puts a product in $T;$ if $27 \in U,$ then $9 \cdot 27 = 243$ puts one in $U.$ Either way we reach a contradiction.

For $m = 242,$ the partition $T = \{3, \ldots, 8\} \cup \{81, \ldots, 242\}$ and $U = \{9, \ldots, 80\}$ avoids products: two elements of $\{3, \ldots, 8\}$ multiply to something in $[9, 64] \subseteq U,$ any product involving an element of $\{81, \ldots, 242\}$ is at least $3 \cdot 81 = 243 \gt 242,$ and two elements of $U$ multiply to at least $81 \gt 80.$

Hence the smallest such $m$ is $243.$

Solution:

Here $AB = 84 + 168 = 252,$ so the semicircle has center $O$ (the midpoint of $\overline{AB}$) and radius $126.$ Since $AN = AO = ON = 126,$ triangle $AON$ is equilateral, so $\angle AON = 60^\circ$ and sector $AON$ is exactly one third of the semicircle. Likewise, if $Q$ is the foot of the perpendicular from $U$ to $\overline{DC},$ then $AU : UB = 1 : 2$ makes rectangle $AUQD$ one third of rectangle $ABCD.$

The part of $\mathcal{R}$ on the $A$-side of $\ell$ equals (sector $AON$) $-$ $[NUO]$ $+$ (rectangle $AUQD$) $+$ $[UQT].$ Since this must be one third of $\mathcal{R},$ we need $[NUO] = [UQT].$ Let $P$ be the foot of the perpendicular from $N$ to $\overline{AB}.$ In the $30$-$60$-$90$ triangle $NOP,$ $OP = 63$ and $NP = 63\sqrt{3},$ so $UP = 84 - 63 = 21$ and $UO = 126 - 84 = 42.$ Triangles $NUP$ and $TUQ$ are similar right triangles (vertical angles at $U$), so $$\frac{[UQT]}{[NUP]} = \left(\frac{UQ}{NP}\right)^2, \qquad \frac{[NUO]}{[NUP]} = \frac{UO}{UP} = 2,$$ the latter because both triangles have height $NP$ over bases $UO$ and $UP.$

Setting the two triangle areas equal gives $\left(\frac{UQ}{NP}\right)^2 = 2,$ so $$DA = UQ = NP\sqrt{2} = 63\sqrt{3} \cdot \sqrt{2} = 63\sqrt{6},$$ and $m + n = 63 + 6 = 69.$

Solution:

Each term $\lfloor \log_{10}(kn) \rfloor$ is nondecreasing in $n,$ so $f$ is nondecreasing and we just locate where it passes $300.$ For $n = 100:$ the products $kn$ run from $100$ to $10^4,$ giving $\lfloor \log_{10} \rfloor = 2$ for $k \le 9,$ $3$ for $10 \le k \le 99,$ and $4$ for $k = 100,$ so $f(100) = 9 \cdot 2 + 90 \cdot 3 + 4 = 292.$

For $n = 109:$ since $9 \cdot 109 = 981 \lt 1000$ and $91 \cdot 109 = 9919 \lt 10^4,$ the terms are $2$ for $k \le 9,$ $3$ for $10 \le k \le 91,$ and $4$ for $92 \le k \le 100:$ $$f(109) = 9 \cdot 2 + 82 \cdot 3 + 9 \cdot 4 = 300.$$ For $n = 110:$ now $91 \cdot 110 = 10010 \ge 10^4,$ so ten terms equal $4$ and $f(110) = 18 + 81 \cdot 3 + 10 \cdot 4 = 301 \gt 300.$

By monotonicity, the largest valid $n$ is $109.$

Solution:

Let $k = \frac{AM}{CM}.$ Triangles $ABM$ and $CBM$ share the altitude from $B,$ so $\frac{[ABM]}{[CBM]} = k.$ Since the inradius of a triangle is its area divided by its semiperimeter, equal inradii force $\frac{12 + AM + BM}{13 + CM + BM} = k$ as well. From $AM + CM = 15$ we get $AM = \frac{15k}{k+1}$ and $CM = \frac{15}{k+1};$ since $AM = k \cdot CM,$ the perimeter equation simplifies to $BM(1 - k) = 13k - 12,$ so $$BM = \frac{13k - 12}{1 - k},$$ and $BM \gt 0$ forces $\frac{12}{13} \lt k \lt 1.$

Stewart's theorem on cevian $\overline{BM}$ gives $AB^2 \cdot CM + BC^2 \cdot AM = AC\left(BM^2 + AM \cdot CM\right),$ so $$BM^2 = \frac{144 + 169k}{k + 1} - \frac{225k}{(k+1)^2}.$$ Setting this equal to $\frac{(13k-12)^2}{(1-k)^2}$ and clearing denominators yields $(169k^2 + 88k + 144)(1-k)^2 = (13k-12)^2(k+1)^2,$ which simplifies to $4k\left(69k^2 - 112k + 44\right) = 0.$

The roots are $k = 0,$ $k = \frac{2}{3},$ and $k = \frac{22}{23},$ and only $k = \frac{22}{23}$ exceeds $\frac{12}{13}$ (then $AM = \frac{22}{3},$ $CM = \frac{23}{3},$ $BM = 10$). Hence $p + q = 22 + 23 = 45.$