Solution:

Since $20^{20} = 2^{40} \cdot 5^{20},$ a valid $m$ must have the form $2^a 5^b,$ and then $n = 2^{40 - 2a}\,5^{20 - 2b}$ is a positive integer exactly when $2a \le 40$ and $2b \le 20.$ Conversely every such choice works, and $n$ is uniquely determined by $m.$

So $a$ can be any of $0, 1, \ldots, 20$ and $b$ any of $0, 1, \ldots, 10,$ giving $21 \cdot 11 = 231$ ordered pairs.

Solution:

Let $Q = \left(\frac{5}{8}, \frac{3}{8}\right)$ and $P = (x, y).$ The slope condition $\frac{y - 3/8}{x - 5/8} \ge \frac{1}{2}$ becomes $y \ge \frac{x}{2} + \frac{1}{16}$ when $x \gt \frac{5}{8},$ and $y \le \frac{x}{2} + \frac{1}{16}$ when $x \lt \frac{5}{8}$ (multiplying by the negative quantity $x - \frac{5}{8}$ reverses the inequality).

For $x \gt \frac{5}{8},$ the region above the line $y = \frac{x}{2} + \frac{1}{16}$ inside the square is a trapezoid with parallel vertical sides of lengths $\frac{5}{8}$ (at $x = \frac{5}{8}$) and $\frac{7}{16}$ (at $x = 1$) and width $\frac{3}{8},$ with area $\frac{3}{8} \cdot \frac{5/8 + 7/16}{2} = \frac{51}{256}.$ For $x \lt \frac{5}{8},$ the region below the line is a trapezoid with parallel sides $\frac{1}{16}$ (at $x = 0$) and $\frac{3}{8}$ (at $x = \frac{5}{8}$) and width $\frac{5}{8},$ with area $\frac{5}{8} \cdot \frac{1/16 + 3/8}{2} = \frac{35}{256}.$

The probability is $\frac{51}{256} + \frac{35}{256} = \frac{86}{256} = \frac{43}{128},$ so $m + n = 43 + 128 = 171.$

Solution:

By the change-of-base formula, $$\log_{2^x} 3^{20} = \frac{20 \log 3}{x \log 2} \qquad \text{and} \qquad \log_{2^{x+3}} 3^{2020} = \frac{2020 \log 3}{(x + 3) \log 2}.$$ Cancelling the common factor $\frac{\log 3}{\log 2}$ leaves $\frac{20}{x} = \frac{2020}{x + 3}.$

Cross-multiplying gives $20x + 60 = 2020x,$ so $2000x = 60$ and $x = \frac{3}{100}.$ Thus $m + n = 3 + 100 = 103.$

Solution:

The vector $\overrightarrow{AB} = (0, 12)$ is vertical, while $\overrightarrow{A'B'} = (12, 0)$ is horizontal and of the same length, so the rotation turns directions by $90^\circ$ clockwise, and $m = 90.$

A $90^\circ$ clockwise rotation about $(a, b)$ sends $(p, q)$ to $(a + q - b,\ b - p + a).$ Applying this to $A = (0, 0)$ and setting the image equal to $A' = (24, 18)$ gives $a - b = 24$ and $a + b = 18,$ so $a = 21$ and $b = -3.$ Checking the other vertices: $B = (0, 12)$ maps to $(21 + 12 + 3,\ -3 + 21) = (36, 18) = B',$ and $C = (16, 0)$ maps to $(21 + 3,\ -3 - 16 + 21) = (24, 2) = C'.$

Therefore $m + x + y = 90 + 21 + (-3) = 108.$

Solution:

The base-sixteen representation of $g(n)$ needs a digit beyond $9$ exactly when $g(n) \ge 10.$ So we need the base-eight digit sum of $f(n)$ to be at least $10.$ Checking values in order, every number less than $31$ has base-eight digit sum at most $9,$ while $31 = 37_8$ has digit sum $10.$ Since the least $n$ achieving a given base-four digit sum increases with that sum, we want the least $n$ with $f(n) = 31.$

A base-four digit is at most $3,$ so a digit sum of $31$ requires at least $11$ digits, and the smallest $11$-digit choice is a leading $1$ followed by ten $3$s: $$N = 13333333333_4 = 4^{10} + (4^{10} - 1) = 2 \cdot 4^{10} - 1 = 2097151.$$

The remainder when $N = 2097151$ is divided by $1000$ is $151.$

Solution:

Computing terms exactly: $$t_3 = \frac{5 \cdot 21 + 1}{25 \cdot 20} = \frac{53}{250}, \qquad t_4 = \frac{5 \cdot \frac{53}{250} + 1}{25 \cdot 21} = \frac{103}{26250}, \qquad t_5 = \frac{5 \cdot \frac{103}{26250} + 1}{25 \cdot \frac{53}{250}} = \frac{5353}{5250} \cdot \frac{10}{53} = \frac{101}{525},$$ using $5353 = 53 \cdot 101.$ Then $t_6 = \frac{5 \cdot \frac{101}{525} + 1}{25 \cdot \frac{103}{26250}} = \frac{206}{105} \cdot \frac{1050}{103} = 20 = t_1$ and $t_7 = \frac{101}{101/21} = 21 = t_2.$

Since each term depends only on the two preceding terms, the sequence repeats with period $5.$ Because $2020$ is a multiple of $5,$ we get $t_{2020} = t_5 = \frac{101}{525}.$ As $101$ is prime and does not divide $525,$ the fraction is reduced, and $p + q = 101 + 525 = 626.$

Solution:

Put the origin at the point where the axes cross, and measure signed coordinates $u_1, u_2$ along the two axes. Along its axis, each cone has its apex at distance $8 - 3 = 5$ from the origin and its base plane at distance $3$ on the other side. Slicing a cone by any plane through its axis gives a triangle whose slant side, in coordinates $(u, w)$ with $w \ge 0$ the distance from the axis, is the line through $(5, 0)$ and $(-3, 3),$ namely $3u + 8w = 15.$

A sphere of radius $r$ centered at a point with axial coordinate $u$ and distance $\rho$ from the axis fits inside that cone only if its cross-section fits inside the triangle, so $r \le \frac{15 - 3u - 8\rho}{\sqrt{73}}.$ Since the two axes are perpendicular, the distance from the center to axis $1$ is at least $|u_2|,$ and vice versa. Adding the two constraints, $$2\sqrt{73}\,r \le 30 - 3(u_1 + u_2) - 8(|u_1| + |u_2|) \le 30,$$ because $3(u_1 + u_2) \ge -3(|u_1| + |u_2|) \ge -8(|u_1| + |u_2|).$ Hence $r \le \frac{15}{\sqrt{73}}.$

The sphere of radius $\frac{15}{\sqrt{73}}$ centered at the origin achieves this: its distance to each slant surface is $\frac{|3 \cdot 0 + 8 \cdot 0 - 15|}{\sqrt{3^2 + 8^2}} = \frac{15}{\sqrt{73}},$ and its distance $3$ to each base plane is larger. So the maximum of $r^2$ is $\frac{225}{73},$ and $m + n = 225 + 73 = 298.$

Solution:

Since $f_n(x) = f_{n-1}(|x - n|),$ the zeros of $f_n$ are exactly $x = n \pm z$ as $z$ runs over the nonnegative zeros of $f_{n-1}.$ The first few zero sets are $\{1\},$ $\{1, 3\},$ $\{0, 2, 4, 6\},$ $\{-2, 0, 2, \ldots, 10\}.$ Writing $T_k = \frac{k(k+1)}{2},$ we claim the zeros of $f_n$ are every other integer from $n - T_{n-1}$ through $T_n.$ Indeed, by induction the nonnegative zeros of $f_{n-1}$ are every other integer from $0$ or $1$ up to $T_{n-1},$ and applying $n \pm z$ yields every integer of the appropriate parity from $n - T_{n-1}$ through $n + T_{n-1} = T_n.$

This progression has $\frac{T_n + T_{n-1} - n}{2} + 1 = \frac{n(n-1)}{2} + 1$ terms, and its first and last terms sum to $(n - T_{n-1}) + T_n = 2n,$ so the sum of the zeros is $$S_n = n\left(\frac{n(n-1)}{2} + 1\right),$$ which is increasing in $n.$

Now $S_{100} = 100 \cdot 4951 = 495{,}100 \le 500{,}000,$ while $S_{101} = 101 \cdot 5051 = 510{,}151 \gt 500{,}000.$ The least such $n$ is $101.$

Solution:

Number the friends $1$ through $6$ in original seating order; we count orderings of $1, \ldots, 6$ in which no two consecutive integers are adjacent. Apply inclusion-exclusion over which of the five pairs $\{i, i+1\}$ are forced to sit together. If a chosen set of $k$ pairs forms $r$ maximal runs of consecutive integers, gluing each run into a block (orderable ascending or descending) gives $2^r (6 - k)!$ seatings containing all chosen adjacencies.

Tallying by $k:$ for $k = 0,$ $720.$ For $k = 1,$ five sets, each $2 \cdot 120,$ total $1200.$ For $k = 2,$ four sets form one run $(2 \cdot 24)$ and six form two runs $(4 \cdot 24),$ total $768.$ For $k = 3,$ three sets form one run $(2 \cdot 6),$ six form two runs $(4 \cdot 6),$ one forms three runs $(8 \cdot 6),$ total $228.$ For $k = 4,$ two sets form one run $(2 \cdot 2)$ and three form two runs $(4 \cdot 2),$ total $32.$ For $k = 5,$ one set, total $2.$

The count is $720 - 1200 + 768 - 228 + 32 - 2 = 90.$

Solution:

Let $K = \frac{n(n+1)}{2},$ so the sum of cubes is $K^2.$ Modulo $n + 5$ we have $n \equiv -5,$ hence $2K = n(n+1) \equiv (-5)(-4) = 20$ and $4K^2 \equiv 400.$ If $K^2 \equiv 17 \pmod{n+5},$ then $n + 5$ divides $400 - 4 \cdot 17 = 332 = 2^2 \cdot 83.$ Since a remainder of $17$ also forces $n + 5 \gt 17,$ the candidates are $n + 5 \in \{83, 166, 332\},$ i.e. $n \in \{78, 161, 327\}.$

Because we multiplied by $4,$ each candidate must be checked. For $n = 78:$ $K = 3081 \equiv 10 \pmod{83},$ and $10^2 = 100 \equiv 17 \pmod{83}.$ For $n = 161:$ $K = 13041 \equiv 93 \pmod{166},$ and $93^2 = 8649 = 52 \cdot 166 + 17.$ For $n = 327:$ $K = 53628 \equiv 176 \pmod{332},$ and $176^2 = 30976 = 93 \cdot 332 + 100,$ remainder $100 \ne 17,$ so this one fails.

The valid values are $n = 78$ and $n = 161,$ with sum $78 + 161 = 239.$

Solution:

Let $a$ be the common root of $P+Q$ and $P+R,$ let $b$ be that of $P+Q$ and $Q+R,$ and let $c$ be that of $P+R$ and $Q+R.$ Each sum is quadratic with leading coefficient $2,$ so $P+Q = 2(x-a)(x-b),$ $P+R = 2(x-a)(x-c),$ and $Q+R = 2(x-b)(x-c).$ Write $Q = x^2 + qx + 2$ and $R = x^2 + rx + s.$ By Vieta's formulas, the root sums are $a + b = \frac{3-q}{2},$ $a + c = \frac{3-r}{2},$ and $b + c = -\frac{q+r}{2},$ so $$2a = (a+b) + (a+c) - (b+c) = \frac{3-q}{2} + \frac{3-r}{2} + \frac{q+r}{2} = 3,$$ giving $a = \frac{3}{2}.$

The constant term of $P + Q$ is $-7 + 2 = -5,$ so $2ab = -5$ and $b = -\frac{5}{3}.$ Also $2ac = P(0) + R(0) = s - 7$ and $2bc = Q(0) + R(0) = s + 2.$ Subtracting, $2c(b - a) = 9,$ so $c = \frac{9}{2\left(-\frac{5}{3} - \frac{3}{2}\right)} = -\frac{27}{19}.$

Then $s = 2bc - 2 = 2 \cdot \left(-\frac{5}{3}\right)\left(-\frac{27}{19}\right) - 2 = \frac{90}{19} - \frac{38}{19} = \frac{52}{19},$ so $R(0) = \frac{52}{19}$ and $m + n = 52 + 19 = 71.$

Solution:

Use coordinates $(\text{column}, \text{row}).$ Square $200$ is in the top row, so $n \ge 200$ (hence $n \ge 201$ as $n$ is odd) and its center is $(200, 1).$ Square $2000$ is in the bottom row, so its column is $b = 2000 - (m-1)n$ with $1 \le b \le n,$ i.e. $(m-1)n \lt 2000 \le mn,$ and its center is $(b, m).$ Since $m$ and $n$ are odd, $b$ is even, so the midpoint $\left(\frac{200 + b}{2}, \frac{m+1}{2}\right)$ of the two centers has integer coordinates; its square number is $\frac{(m-1)n + 200 + b}{2} = \frac{2200}{2} = 1100.$ Its column $100 + \frac{b}{2}$ lies between $101$ and $n,$ so the line passes through the center of square $1100,$ and square $1099$ sits immediately to its left in the same row.

Square $1099$ lies only in that row, and the line crosses that row's horizontal strip in a segment centered (by symmetry) at the center of square $1100,$ extending $\frac{1}{2|s|}$ to each side, where $s = \frac{m-1}{b - 200}$ is the slope. So the line meets the interior of square $1099$ exactly when $\frac{1}{2|s|} \gt \frac{1}{2},$ that is $|s| \lt 1,$ i.e. $m - 1 \lt |b - 200| = |1800 - (m-1)n|$ (a vertical line, $b = 200,$ fails).

Since $n \ge 201$ and $(m-1)n \lt 2000,$ we need $m - 1 \lt 10,$ so $m \in \{3, 5, 7, 9\}.$ For $m = 3:$ odd $n \in [667, 999],$ $167$ values, excluding $|n - 900| \le 1$ (odd cases $899, 901$) leaves $165.$ For $m = 5:$ odd $n \in [401, 499],$ $50$ values, excluding $449, 451$ leaves $48.$ For $m = 7:$ odd $n \in [287, 333],$ $24$ values, excluding $299, 301$ leaves $22.$ For $m = 9:$ odd $n \in [223, 249],$ $14$ values, excluding $225$ leaves $13.$ The total is $165 + 48 + 22 + 13 = 248.$

Solution:

Let the tangent lengths from $A, B, C, D, E$ to the incircle be $a, b, c, d, e.$ Then $a + b = 5,$ $b + c = 6,$ $c + d = 6,$ $d + e = 6,$ and $e + a = 7.$ The middle equations give $d = b$ and $e = c,$ so $c + a = 7;$ with $a + b = 5$ and $b + c = 6$ this yields $a = 3,$ $b = d = 2,$ $c = e = 4.$ If $r$ is the inradius, the interior angle at a vertex with tangent length $t$ satisfies $\tan\frac{\theta}{2} = \frac{r}{t},$ and the half-angles sum to half of $540^\circ:$ $$\arctan\frac{r}{3} + 2\arctan\frac{r}{2} + 2\arctan\frac{r}{4} = 270^\circ.$$

Let $\beta = \arctan\frac{r}{2}$ and $\gamma = \arctan\frac{r}{4}.$ Then $\arctan\frac{r}{3} = 270^\circ - 2(\beta + \gamma),$ and since $\tan(270^\circ - \theta) = \cot\theta,$ we get $\frac{r}{3} = \cot 2(\beta + \gamma).$ With $T = \tan(\beta + \gamma) = \frac{r/2 + r/4}{1 - r^2/8} = \frac{6r}{8 - r^2},$ the identity $\frac{r}{3} = \frac{1 - T^2}{2T}$ becomes $2rT = 3(1 - T^2),$ and substituting $T$ and clearing denominators gives $$5r^4 - 84r^2 + 64 = 0,$$ so $r^2 = 16$ or $r^2 = \frac{4}{5}.$ For $r^2 = \frac{4}{5}$ every half-angle is well under $54^\circ,$ so the half-angle sum falls far short of $270^\circ;$ this root is extraneous. Hence $r = 4.$

The semiperimeter is $s = \frac{5 + 6 + 6 + 6 + 7}{2} = 15,$ so the area is $rs = 4 \cdot 15 = 60.$

Solution:

On $[k, k+1)$ with $k \ge 0$ an integer, write $x = k + t;$ then $f(x) = (k + t)t$ is strictly increasing from $0$ toward $k + 1,$ so $f$ maps $[k, k+1)$ bijectively onto $[0, k+1).$ Hence for any $c$ with $n \lt c \lt n + 1,$ the equation $f(w) = c$ has exactly one solution in $[k, k+1)$ for each integer $k \ge n,$ and no others.

The equation $f(z) = 17$ has one solution $z_n \in (n, n+1)$ for each $n \ge 17.$ In turn, $f(w) = z_n$ has one solution in $(k, k+1)$ for each $k \ge n.$ Finally, for $x \in [j, j+1)$ with $0 \le j \le 2019,$ $f$ maps $[j, j+1)$ bijectively onto $[0, j+1),$ so the number of solutions of $f(f(f(x))) = 17$ there equals the number of such $w$ with $w \lt j + 1,$ namely the number of pairs $(n, k)$ with $17 \le n \le k \le j,$ which is $\binom{j - 15}{2}.$ (The endpoint $x = 2020$ gives $f(x) = 0$ and is not a solution.)

By the hockey stick identity, $$N = \sum_{j=17}^{2019} \binom{j - 15}{2} = \sum_{a=2}^{2004} \binom{a}{2} = \binom{2005}{3} = \frac{2005 \cdot 2004 \cdot 2003}{6} = 1{,}341{,}349{,}010,$$ so the remainder when $N$ is divided by $1000$ is $10.$

Solution:

By the tangent-chord angle, $\angle TBC = A,$ so $\angle ABT = B + A = 180^\circ - C$ and $TX = TB \sin\angle ABT = 16 \sin C;$ similarly $TY = 16 \sin B.$ Also $\angle AXT = \angle AYT = 90^\circ,$ so $A, X, T, Y$ lie on a circle with diameter $AT,$ whence $XY = AT \sin A.$ Using the law of sines $\left(\sin A = \frac{11}{R},\ \sin B = \frac{AC}{2R},\ \sin C = \frac{AB}{2R}\right),$ the given condition becomes $$\frac{64\left(AB^2 + AC^2\right) + 121\,AT^2}{R^2} = 1143.$$

Place $B = (-11, 0)$ and $C = (11, 0).$ Since $TB = 16$ and $T$ lies on the perpendicular bisector of $BC,$ we get $T = (0, -\sqrt{135}).$ The circumcenter is $O = (0, k)$ with $OB \perp BT,$ which gives $121 - k\sqrt{135} = 0,$ so $k = \frac{121}{\sqrt{135}}$ and $R^2 = 121 + k^2 = \frac{30976}{135}.$ For $A = (x, y)$ on $\omega,$ expanding $x^2 + (y - k)^2 = R^2$ gives $x^2 + y^2 = \frac{242}{\sqrt{135}}\,y + 121.$ Therefore $$AB^2 + AC^2 = 2(x^2 + y^2) + 242 = \frac{484}{\sqrt{135}}\,y + 484, \qquad AT^2 = x^2 + y^2 + 2\sqrt{135}\,y + 135 = \frac{512}{\sqrt{135}}\,y + 256.$$

Substituting, $64(AB^2 + AC^2) + 121\,AT^2 = \frac{92928}{\sqrt{135}}\,y + 61952 = 1143 \cdot \frac{30976}{135}$ yields $y = \frac{291}{\sqrt{135}}.$ Then $AT^2 = \frac{512 \cdot 291}{135} + 256 = \frac{183552}{135},$ and $$XY^2 = AT^2 \sin^2 A = \frac{121\,AT^2}{R^2} = \frac{121 \cdot 183552}{30976} = \frac{183552}{256} = 717.$$