Solution:

Each summand is $10^k - 1,$ so $$N = \sum_{k=1}^{321} \left(10^k - 1\right) = \underbrace{11\ldots1}_{321}0 - 321.$$

The subtraction changes only the last four digits: $1110 - 321 = 789,$ so those four digits become $0789.$ Thus $N$ consists of $318$ ones followed by $0789,$ and the digit sum is $318 + 0 + 7 + 8 + 9 = 342.$

Solution:

There are $20 \cdot 19 = 380$ equally likely ordered pairs $(J, B)$ with $B \neq J.$ The condition $B \ge J + 2$ allows $19 - J$ choices of $B$ for each $J \le 18,$ so the number of favorable pairs is $$\sum_{J=1}^{18} (19 - J) = 18 + 17 + \cdots + 1 = 171.$$

The probability is $\frac{171}{380} = \frac{9}{20},$ so $m + n = 9 + 20 = 29.$

Solution:

Since $15^2 + 20^2 = 25^2,$ the triangle is right-angled at $R,$ and its area is $\frac{1}{2} \cdot 15 \cdot 20 = 150.$ Also $\sin P = \frac{20}{25} = \frac{4}{5}$ and $\sin Q = \frac{15}{25} = \frac{3}{5}.$

The hexagon is the triangle minus three corner triangles, each with two sides of length $5:$ at $P,$ area $\frac{1}{2} \cdot 5 \cdot 5 \cdot \frac{4}{5} = 10;$ at $Q,$ area $\frac{1}{2} \cdot 5 \cdot 5 \cdot \frac{3}{5} = \frac{15}{2};$ at $R,$ area $\frac{1}{2} \cdot 5 \cdot 5 = \frac{25}{2}.$

Therefore the hexagon has area $150 - 10 - \frac{15}{2} - \frac{25}{2} = 120.$

Solution:

At every moment there are $11$ players in the game, any of whom may be removed, while the bench shrinks by one with each substitution. So the first substitution can be made in $11 \cdot 11$ ways, the second in $11 \cdot 10$ ways, and the third in $11 \cdot 9$ ways.

Summing over $0, 1, 2,$ or $3$ substitutions, $$n = 1 + 11 \cdot 11 + 11^2 \cdot 11 \cdot 10 + 11^3 \cdot 11 \cdot 10 \cdot 9 = 1 + 121 + 13310 + 1317690 = 1331122.$$ The remainder upon division by $1000$ is $122.$

Solution:

Coordinates never increase, so the first axis point reached is $(0,0)$ exactly when the particle reaches $(1, 1)$ and then takes the diagonal step. Every path from $(4,4)$ to $(1,1)$ automatically stays off the axes, since its coordinates remain at least $1.$

A path from $(4,4)$ to $(1,1)$ with $d$ diagonal steps also has $3 - d$ left steps and $3 - d$ down steps, for $6 - d$ steps in all, and there are $\frac{(6-d)!}{d!\,(3-d)!\,(3-d)!}$ orderings: $20, 30, 12, 1$ for $d = 0, 1, 2, 3.$ Since a path with $6 - d$ steps has probability $\left(\frac{1}{3}\right)^{6-d},$ the probability of reaching $(1,1)$ and then stepping to $(0,0)$ is $$\frac{1}{3}\left(\frac{20}{3^6} + \frac{30}{3^5} + \frac{12}{3^4} + \frac{1}{3^3}\right) = \frac{1}{3} \cdot \frac{20 + 90 + 108 + 27}{3^6} = \frac{245}{3^7}.$$

Since $245 = 5 \cdot 7^2$ is not divisible by $3,$ we get $m + n = 245 + 7 = 252.$

Solution:

Let $F$ be the foot of the perpendicular from $L$ to $\overline{KN},$ so $O$ lies on segment $LF.$ In right triangle $KLN$ (right angle at $L$), the altitude $LF$ to the hypotenuse gives the geometric mean relation $KF \cdot KN = KL^2 = 28^2 = 784.$

Triangles $KFO$ and $KMN$ share angle $K,$ and $\angle KFO = 90^\circ = \angle KMN,$ so they are similar. Hence $\frac{KF}{KM} = \frac{KO}{KN},$ that is, $KO \cdot KM = KF \cdot KN = 784.$ With $KO = 8$ this gives $KM = 98,$ so $$MO = KM - KO = 98 - 8 = 90.$$

Solution:

The equations say $x \cdot \gcd(x,y)^2 = 10^{60}$ and $y \cdot \operatorname{lcm}(x,y)^2 = 10^{570},$ so $x$ and $y$ are products of the primes $2$ and $5$ only. Fix one of these primes and let $a$ and $b$ be its exponents in $x$ and $y.$ Since the gcd takes the smaller exponent and the lcm the larger, $$a + 2\min(a, b) = 60, \qquad b + 2\max(a, b) = 570.$$

If $a \gt b,$ then $a + 2b = 60$ and $b + 2a = 570;$ adding gives $a + b = 210,$ and subtracting gives $a - b = 510,$ forcing $b \lt 0,$ impossible. So $a \le b,$ and the equations become $3a = 60$ and $3b = 570,$ giving $a = 20$ and $b = 190$ for both primes.

Thus $x = 2^{20} 5^{20}$ and $y = 2^{190} 5^{190},$ so $m = 40,$ $n = 380,$ and $3m + 2n = 120 + 760 = 880.$

Solution:

Let $u = \sin^2 x$ and $v = \cos^2 x,$ so $u + v = 1,$ and set $p = uv.$ Expanding $(u+v)^5$ gives $u^5 + v^5 = 1 - 5p(u+v)^3 + 5p^2(u+v) = 1 - 5p + 5p^2,$ so the hypothesis reads $$1 - 5p + 5p^2 = \frac{11}{36} \quad\Longrightarrow\quad 36p^2 - 36p + 5 = 0,$$ with roots $p = \frac{1}{6}$ and $p = \frac{5}{6}.$ Since $p = \sin^2 x \cos^2 x = \frac{1}{4}\sin^2 2x \le \frac{1}{4},$ we must have $p = \frac{1}{6}.$

Similarly $u^6 + v^6 = (u^2 + v^2)^3 - 3p^2(u^2+v^2) = (1 - 2p)^3 - 3p^2(1 - 2p) = 1 - 6p + 9p^2 - 2p^3.$ Substituting $p = \frac{1}{6},$ $$u^6 + v^6 = 1 - 1 + \frac{1}{4} - \frac{1}{108} = \frac{26}{108} = \frac{13}{54},$$ so $m + n = 13 + 54 = 67.$

Solution:

Since $7 = 2 + 5 = 3 + 4,$ one of $\tau(n), \tau(n+1)$ equals $3$ or $5,$ and $\tau = 3$ means a prime square $p^2$ while $\tau = 5$ means a prime fourth power $p^4.$ So one of $n, n+1$ lies in $\{4, 9, 25, 49, 121, 169, 289, 361, \ldots\} \cup \{16, 81, 625, \ldots\},$ and its neighbor must have $\tau = 4$ (for a square) or be prime (for a fourth power).

Checking neighbors in increasing order: $n = 8$ works $(\tau(8) = 4,$ $\tau(9) = 3);$ $n = 9$ works $(\tau(10) = 4);$ $n = 16$ works $(\tau(16) = 5,$ $17$ prime$);$ $n = 25$ works $(\tau(26) = 4).$ Then $49, 81, 169,$ and $289$ all fail: $\tau(48) = 10,$ $\tau(50) = 6,$ $\tau(80) = 10,$ $82$ is not prime, $\tau(168) = 16,$ $\tau(170) = 8,$ $\tau(288) = 18,$ $\tau(290) = 8.$ Next, $n = 121$ works $(\tau(122) = 4)$ and $n = 361$ works $(\tau(362) = 4).$

The six least solutions are $8, 9, 16, 25, 121, 361,$ with sum $540.$

Solution:

The polynomial's $2019$ roots are the numbers $z_j,$ each repeated three times. By Vieta's formulas, the coefficient of $x^{2018}$ is minus the sum of all roots: $3\sum_j z_j = -20,$ so $\sum_j z_j = -\frac{20}{3}.$

The coefficient of $x^{2017}$ is the sum over unordered pairs of roots. A pair may use two copies from one triple ($\binom{3}{2} = 3$ pairs for each $j$) or copies from two different triples ($3 \cdot 3 = 9$ pairs for each $j \lt k$). Writing $S = \sum_{j \lt k} z_j z_k,$ $$19 = 3\sum_j z_j^2 + 9S = 3\left[\left(-\tfrac{20}{3}\right)^2 - 2S\right] + 9S = \frac{400}{3} + 3S,$$ so $3S = 19 - \frac{400}{3} = -\frac{343}{3}$ and $S = -\frac{343}{9}.$

Hence $|S| = \frac{343}{9},$ which is in lowest terms, and $m + n = 343 + 9 = 352.$

Solution:

Let $BC = a$ and $AB = AC = b.$ Place $B = \left(-\frac{a}{2}, 0\right),$ $C = \left(\frac{a}{2}, 0\right),$ $A = (0, h)$ with $h = \sqrt{b^2 - \frac{a^2}{4}},$ so the semiperimeter is $s = b + \frac{a}{2}$ and the area is $K = \frac{ah}{2}.$ The inradius and exradii are $r = \frac{K}{s} = \frac{ah}{a + 2b},$ $r_A = \frac{K}{s - a} = \frac{ah}{2b - a},$ and $r_B = \frac{K}{s - b} = h.$ The incenter is $I = (0, r)$ and the $A$-excircle has center $(0, -r_A).$ The $B$-excircle touches line $BC$ at distance $s$ from $B,$ that is, at $x = b,$ so its center is $(b, h).$

Internal tangency with the $A$-excircle: the center distance is $r + r_A,$ so the radius $\rho$ of $\omega$ satisfies $\rho - r_A = r + r_A,$ i.e. $\rho = r + 2r_A.$ External tangency with the $B$-excircle requires $b^2 + (h - r)^2 = (\rho + h)^2 = (h + r + 2r_A)^2,$ which rearranges to $b^2 = 4(r + r_A)(h + r_A).$ Since $$r + r_A = \frac{4abh}{4b^2 - a^2} \qquad \text{and} \qquad h + r_A = \frac{2bh}{2b - a},$$ and $h^2 = \frac{4b^2 - a^2}{4},$ the condition becomes $b^2 = \frac{8ab^2}{2b - a},$ that is, $2b - a = 8a,$ so $2b = 9a.$

For integer sides, $a = 2t$ and $b = 9t$ for a positive integer $t,$ giving perimeter $20t.$ The minimum is $20,$ achieved by the triangle with sides $9, 9, 2.$

Solution:

Since $f(w) - w = w(w - 20),$ the two legs at $f(z)$ are $$f(z) - z = z(z - 20), \qquad f(f(z)) - f(z) = f(z)\bigl(f(z) - 20\bigr) = z(z - 19)(z - 20)(z + 1),$$ using $f(z) = z(z - 19)$ and $f(z) - 20 = (z - 20)(z + 1).$ They are perpendicular exactly when their quotient $(z - 19)(z + 1)$ is purely imaginary and nonzero.

Write $z = x + 11i.$ The real part of $(z - 19)(z + 1) = z^2 - 18z - 19$ is $x^2 - 121 - 18x - 19,$ so we need $x^2 - 18x - 140 = 0,$ giving $x = 9 \pm \sqrt{221}.$ The form $m + \sqrt{n} + 11i$ with $m, n$ positive integers requires $x = 9 + \sqrt{221}.$

Hence $m + n = 9 + 221 = 230.$

Solution:

Points $D, E$ lie beyond $B$ on ray $AB,$ and $F$ lies on the opposite side of line $AB$ from $C.$ Since $ACFD$ and $BCFE$ are cyclic, the inscribed angles give $\angle FDA = \angle FCA$ and $\angle FEB = \angle FCB.$ Writing $\alpha = \angle FCA$ and $\beta = \angle FCB,$ triangle $DEF$ has angles $\angle FDE = 180^\circ - \alpha$ and $\angle FED = \beta,$ so $\angle DFE = \alpha - \beta = \angle ACB.$ From triangle $ABC,$ $\cos \angle ACB = \frac{25 + 36 - 16}{2 \cdot 5 \cdot 6} = \frac{3}{4},$ so the law of cosines in triangle $DFE$ gives $$DE^2 = 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cdot \tfrac{3}{4} = 32, \qquad DE = 4\sqrt{2}.$$

In triangle $DFE,$ $\cos \angle FDE = \frac{4 + 32 - 49}{2 \cdot 2 \cdot 4\sqrt{2}} = -\frac{13\sqrt{2}}{32},$ so $\alpha$ is acute with $\cos\alpha = \frac{13\sqrt{2}}{32}$ and $\sin\alpha = \sqrt{1 - \frac{338}{1024}} = \frac{7\sqrt{14}}{32}.$ Let $G$ be the intersection of line $CF$ with line $AB.$ In triangle $ACG,$ $\angle GAC = \angle BAC$ has $\cos \angle BAC = \frac{16 + 36 - 25}{2 \cdot 4 \cdot 6} = \frac{9}{16},$ $\sin \angle BAC = \frac{5\sqrt{7}}{16},$ and $\angle ACG = \alpha,$ so $\sin(\angle BAC + \alpha) = \frac{5\sqrt{7}}{16} \cdot \frac{13\sqrt{2}}{32} + \frac{9}{16} \cdot \frac{7\sqrt{14}}{32} = \frac{\sqrt{14}}{4}$ and $$AG = \frac{AC \sin \alpha}{\sin(\angle BAC + \alpha)} = \frac{6 \cdot \frac{7\sqrt{14}}{32}} {\frac{\sqrt{14}}{4}} = \frac{21}{4}.$$

Line $CF$ is the radical axis of the two circles, so $GA \cdot GD = GB \cdot GE.$ With $x = BD$ and $BE = x + DE = x + 4\sqrt{2}:$ $$\frac{21}{4}\left(x - \frac{5}{4}\right) = \frac{5}{4}\left(x - \frac{5}{4} + 4\sqrt{2}\right),$$ since $GD = 4 + x - \frac{21}{4}$ and $GB = \frac{21}{4} - 4.$ This gives $16\left(x - \frac{5}{4}\right) = 20\sqrt{2},$ so $x = \frac{5 + 5\sqrt{2}}{4}$ and $BE = \frac{5 + 21\sqrt{2}}{4}.$ Therefore $a + b + c + d = 5 + 21 + 2 + 4 = 32.$

Solution:

Suppose an odd prime $p$ divides $2019^8 + 1.$ Then $2019^8 \equiv -1 \pmod{p},$ so $2019^{16} \equiv 1$ while $2019^8 \not\equiv 1:$ the multiplicative order of $2019$ modulo $p$ is exactly $16.$ Since the order divides $p - 1,$ we need $p \equiv 1 \pmod{16},$ and the smallest such primes are $17$ and $97.$

Modulo $17:$ $2019 \equiv 13,$ and $13^2 = 169 \equiv -1,$ so $2019^8 \equiv (-1)^4 = 1$ and $2019^8 + 1 \equiv 2 \neq 0.$ Modulo $97:$ $2019 \equiv -18,$ and squaring repeatedly, $$2019^2 \equiv 324 \equiv 33, \qquad 2019^4 \equiv 33^2 = 1089 \equiv 22, \qquad 2019^8 \equiv 22^2 = 484 \equiv -1 \pmod{97}.$$

So $97$ divides $2019^8 + 1,$ and it is the least odd prime factor: $97.$

Solution:

Since $A$ lies on both $\omega$ and $\omega_1$ and internally tangent circles meet only at their point of tangency, $\omega_1$ is tangent to $\omega$ at $A;$ likewise $\omega_2$ is tangent at $B.$ Let $Z$ be the intersection of the tangent lines to $\omega$ at $A$ and $B.$ Each tangent line is also tangent to the corresponding inner circle, so the powers of $Z$ with respect to $\omega_1$ and $\omega_2$ are $ZA^2$ and $ZB^2,$ which are equal. Hence $Z$ lies on the radical axis $PQ,$ and along the line through $Z, X, P, Q, Y:$ $$ZP \cdot ZQ = ZA^2 = ZX \cdot ZY,$$ the last equality because $ZA$ is tangent to $\omega.$

Because $ZA = ZB,$ the point $Z$ lies on the perpendicular bisector of $\overline{AB};$ if $M$ is the midpoint of $\overline{AB},$ then $ZA^2 - ZP^2 = MA^2 - MP^2 = 4^2 - 1^2 = 15.$ Also the power of $P$ in $\omega$ gives $XP \cdot PY = AP \cdot PB = 15.$ Set $s = ZP$ and $u = ZX,$ so $ZY = u + 11.$ The relations become $$u(u + 11) = s^2 + 15, \qquad (s - u)(u + 11 - s) = 15.$$ Expanding the second and substituting the first yields $u = s - \frac{11}{2} + \frac{15}{s},$ and substituting back gives $\left(s + \frac{15}{s}\right)^2 - \frac{121}{4} = s^2 + 15,$ so $\frac{225}{s^2} = \frac{61}{4}.$

Finally $ZQ = \frac{ZA^2}{ZP} = s + \frac{15}{s},$ so $PQ = ZQ - ZP = \frac{15}{s}$ and $PQ^2 = \frac{225}{s^2} = \frac{61}{4}.$ Therefore $m + n = 61 + 4 = 65.$