Answer: 547

Solution:

Let $\angle BAC = \alpha.$ Since $AE = ED,$ triangle $AED$ is isosceles with $\angle ADE = \angle DAE = \alpha,$ so the exterior angle at $E$ gives $\angle DEB = 2\alpha.$ Since $ED = DB,$ triangle $EDB$ has $\angle DBE = \angle DEB = 2\alpha,$ hence $\angle EDB = 180^\circ - 4\alpha.$

The three angles at $D$ on segment $\overline{AC}$ sum to a straight angle: $\alpha + (180^\circ - 4\alpha) + \angle BDC = 180^\circ,$ so $\angle BDC = 3\alpha.$ Since $DB = BC,$ also $\angle BCD = \angle BDC = 3\alpha.$ But $AB = AC$ makes $\angle ABC = \angle ACB = 3\alpha,$ so the angle sum of $\triangle ABC$ gives $\alpha + 3\alpha + 3\alpha = 180^\circ,$ hence $\alpha = \frac{180}{7}$ degrees.

Then $\angle ABC = 3\alpha = \frac{540}{7}$ degrees, and $m + n = 540 + 7 = 547.$

Answer: 17

Solution:

Let $t = \log_2 x.$ Then $\log_4 x = \frac{t}{2}$ and $\log_8(2x) = \frac{1 + t}{3}.$ In a geometric progression the middle term squared equals the product of the outer terms: $$\left(\frac{t}{2}\right)^2 = \frac{1 + t}{3} \cdot t.$$

Since $t = 0$ gives no valid ratio, divide by $t:$ $\frac{t}{4} = \frac{1 + t}{3},$ so $3t = 4 + 4t$ and $t = -4.$ Thus $x = 2^{-4} = \frac{1}{16},$ and the progression is $-1,$ $-2,$ $-4$ with common ratio $2,$ which is positive as required.

Therefore $m + n = 1 + 16 = 17.$

Answer: 621

Solution:

Equating the two representations in base ten gives $121a + 11b + c = 512 + 64b + 8c + a,$ which simplifies to $$120a = 512 + 53b + 7c.$$ All of $a,$ $b,$ $c$ are base-eight digits, so $0 \le a, b, c \le 7$ (and $a \ge 1$ since it leads the base-eleven representation).

The right side is at least $512,$ so $a \ge 5.$ Since $N = 121a + 11b + c$ increases with $a,$ try $a = 5:$ then $53b + 7c = 88.$ Here $b = 0$ gives $7c = 88,$ impossible, and $b \ge 2$ overshoots, so $b = 1$ and $7c = 35,$ giving $c = 5.$

Thus $N = 121 \cdot 5 + 11 + 5 = 621,$ whose base-eight representation is $1155$ and base-eleven representation is $515,$ as required. The least such $N$ is $621.$

Answer: 93

Solution:

If removing the last four digits leaves $k \ge 1,$ then $N = 10000k + 2020,$ and the condition $k \mid N$ is equivalent to $k \mid 2020.$ Since $2020 = 2^2 \cdot 5 \cdot 101,$ there are $12$ choices of $k:$ $1,$ $2,$ $4,$ $5,$ $10,$ $20,$ $101,$ $202,$ $404,$ $505,$ $1010,$ $2020.$

Each member of $S$ has digit sum equal to the digit sum of $k$ plus $2 + 0 + 2 + 0 = 4.$ The digit sums of the twelve divisors are $1, 2, 4, 5, 1, 2, 2, 4, 8, 10, 2, 4,$ totaling $45.$

The answer is $45 + 12 \cdot 4 = 93.$

Answer: 52

Solution:

First count arrangements from which some card's removal leaves the rest ascending. Any such arrangement arises by choosing the card to remove ($6$ ways) and inserting it into one of the $6$ gaps of the other five cards written in increasing order, for $36$ constructions. But the fully sorted row $123456$ arises from all $6$ card choices, and each of the $5$ arrangements obtained by swapping two adjacent cards of the sorted row arises twice (move either card of the pair past the other). Every other construction gives a distinct arrangement.

So the ascending count is $1 + 5 + (36 - 6 - 10) = 26,$ and by symmetry there are $26$ descending arrangements. No arrangement is counted in both totals: that would require an ascending and a descending subsequence of five cards, needing at least $5 + 5 - 1 = 9$ cards.

The total is $26 + 26 = 52.$

Answer: 173

Solution:

A sphere of radius $r$ resting in a circular hole of radius $a$ has its center on the axis of the hole; since the center is at distance $r$ from every point of the hole's rim, it sits at distance $\sqrt{r^2 - a^2}$ from the plane of the board. So the two centers lie at depths $\sqrt{r^2 - 1}$ and $\sqrt{r^2 - 4}$ on the same side of the board, with horizontal separation $7.$

Tangency of the spheres means the centers are $2r$ apart: $$49 + \left(\sqrt{r^2 - 1} - \sqrt{r^2 - 4}\right)^2 = 4r^2.$$ Expanding gives $49 + 2r^2 - 5 - 2\sqrt{(r^2 - 1)(r^2 - 4)} = 4r^2,$ so $\sqrt{(r^2 - 1)(r^2 - 4)} = 22 - r^2.$ Squaring, $r^4 - 5r^2 + 4 = 484 - 44r^2 + r^4,$ hence $39r^2 = 480$ and $r^2 = \frac{160}{13}.$

Thus $m + n = 160 + 13 = 173.$

Answer: 81

Solution:

A committee with $k$ men has $k + 1$ women, so $$N = \sum_{k=0}^{11} \binom{11}{k}\binom{12}{k+1} = \sum_{k=0}^{11} \binom{11}{k}\binom{12}{11-k} = \binom{23}{11}$$ by Vandermonde's identity (both sides count ways to choose $11$ people from all $23$).

Now factor $\binom{23}{11} = \frac{23!}{11!\,12!}.$ The primes $13,$ $17,$ $19,$ $23$ each appear in the numerator but not the denominator. By Legendre's formula the exponent of $2$ is $19 - 8 - 10 = 1,$ of $3$ is $9 - 4 - 5 = 0,$ of $5$ is $4 - 2 - 2 = 0,$ of $7$ is $3 - 1 - 1 = 1,$ and of $11$ is $2 - 1 - 1 = 0.$ Hence $N = 2 \cdot 7 \cdot 13 \cdot 17 \cdot 19 \cdot 23.$

The sum of the primes dividing $N$ is $2 + 7 + 13 + 17 + 19 + 23 = 81.$

Answer: 103

Solution:

Work in the complex plane with $O$ at the origin and east along the positive real axis. Each day's displacement is the previous one multiplied by $z = \frac{1}{2}e^{i\pi/3},$ so $$P = 5\left(1 + z + z^2 + \cdots\right) = \frac{5}{1 - z}.$$

Since $z = \frac{1}{4} + \frac{\sqrt{3}}{4}i,$ we get $1 - z = \frac{3}{4} - \frac{\sqrt{3}}{4}i,$ whose squared magnitude is $\frac{9}{16} + \frac{3}{16} = \frac{3}{4}.$ Therefore $$OP^2 = \frac{25}{3/4} = \frac{100}{3},$$ and $m + n = 100 + 3 = 103.$

Answer: 77

Solution:

Write $20^9 = 2^{18} \cdot 5^9,$ so each $a_i = 2^{x_i} 5^{y_i}$ with $0 \le x_i \le 18$ and $0 \le y_i \le 9;$ there are $19 \cdot 10 = 190$ divisors, and the exponents of the two primes are chosen independently and uniformly. The chain $a_1 \mid a_2 \mid a_3$ holds exactly when $x_1 \le x_2 \le x_3$ and $y_1 \le y_2 \le y_3.$

Non-decreasing triples from a set of $k$ values correspond to multisets of size $3,$ counted by $\binom{k+2}{3}.$ So the probability is $$\frac{\binom{21}{3}}{19^3} \cdot \frac{\binom{12}{3}}{10^3} = \frac{1330}{6859} \cdot \frac{220}{1000} = \frac{70}{361} \cdot \frac{11}{50} = \frac{77}{1805}.$$

Since $1805 = 5 \cdot 19^2$ shares no factor with $77 = 7 \cdot 11,$ the fraction is in lowest terms and $m = 77.$

Answer: 407

Solution:

If a prime $p$ divides $n,$ then $p \mid n^n \mid m^m,$ so $p \mid m$ and hence $p \mid m + n.$ Since $\gcd(m + n, 210) = 1,$ no prime factor of $n$ is $2,$ $3,$ $5,$ or $7:$ every prime factor of $n$ is at least $11.$ Because $m$ is not a multiple of $n,$ some prime $p$ has $b = v_p(m) \lt a = v_p(n),$ where $v_p$ denotes the exponent of $p.$ Comparing exponents of $p$ in $n^n \mid m^m$ gives $bm \ge an,$ so $m \ge \frac{a}{b}n \ge 2n.$ In particular $a \ge 2,$ so $p^2 \mid n$ and $n \ge 11^2 = 121.$

Take $n = 121$ with $p = 11,$ $a = 2,$ $b = 1:$ then $m$ is a multiple of $11$ but not of $121,$ and $m \ge 2 \cdot 121 = 242.$ The candidates $m = 253, 264, 275$ give $m + n = 374 = 2 \cdot 11 \cdot 17,$ $385 = 5 \cdot 7 \cdot 11,$ $396 = 2^2 \cdot 3^2 \cdot 11,$ all sharing a factor with $210,$ while $m = 242 = 2 \cdot 11^2$ is a multiple of $121.$ But $m = 286 = 2 \cdot 11 \cdot 13$ works: $v_{11}(m^m) = 286 \ge 242 = v_{11}(n^n),$ so $n^n \mid m^m,$ and $m + n = 407 = 11 \cdot 37$ is coprime to $210.$

Any other admissible $n$ is at least $13^2 = 169,$ forcing $m + n \ge 3n \ge 507.$ Hence the least possible value is $407.$

Answer: 510

Solution:

The condition says the integers $f(2) = 4 + 2a + b$ and $f(4) = 16 + 4a + b$ are both roots of the monic quadratic $g.$ These two values are equal exactly when $a = -6.$

If $a = -6,$ then for any $b$ and any $c$ the choice $d = -f(2)^2 - c\,f(2)$ makes $f(2) = f(4)$ a root of $g,$ giving $21 \cdot 21 = 441$ triples. If $a \ne -6,$ the two distinct values must be the two roots of $g,$ so Vieta forces $c = -(f(2) + f(4)) = -(20 + 6a + 2b),$ and then $d = f(2)f(4)$ is an integer. The requirement $|c| \le 10$ becomes $-15 \le 3a + b \le -5.$

For each $a,$ count integers $b \in [-10, 10]$ with $-15 - 3a \le b \le -5 - 3a:$ the counts are $2, 5, 11, 11, 11, 11, 9, 6, 3$ for $a = -8, -7, -5, -4, -3, -2, -1, 0, 1$ respectively, and $0$ for all other $a \ne -6,$ totaling $69.$ The answer is $441 + 69 = 510.$

Answer: 270

Solution:

Work prime by prime. Since $149 - 2 = 147 = 3 \cdot 7^2,$ the lifting-the-exponent lemma gives $v_3(149^n - 2^n) = v_3(147) + v_3(n) = 1 + v_3(n)$ and $v_7(149^n - 2^n) = 2 + v_7(n)$ for every positive integer $n.$ Requiring at least $3$ and $7$ forces $3^2 \mid n$ and $7^5 \mid n.$

For $5$ we first need $149^n \equiv 2^n \pmod 5,$ i.e. $4^n \equiv 2^n,$ i.e. $2^n \equiv 1 \pmod 5,$ which requires $4 \mid n.$ Write $n = 4k.$ In $149^4 - 2^4 = (149 - 2)(149 + 2)(149^2 + 4),$ only the last factor is divisible by $5,$ and only once, since $149^2 + 4 = 22205 = 5 \cdot 4441.$ Lifting the exponent from the base $149^4, 2^4$ gives $v_5(149^n - 2^n) = 1 + v_5(k),$ so $5^4 \mid k,$ i.e. $4 \cdot 5^4 \mid n.$

The least valid $n$ is $2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5,$ which has $(2+1)(2+1)(4+1)(5+1) = 270$ positive divisors.

Answer: 36

Solution:

In triangle $ABD,$ the internal bisector of the angle at $B$ meets the circumcircle of $ABD$ again at the midpoint of arc $AD$ not containing $B,$ and that arc midpoint lies on the perpendicular bisector of $\overline{AD}$ — so $E$ is exactly that arc midpoint. The inscribed angles $\angle EAD$ and $\angle EBD$ subtend the same arc $ED,$ so $\angle EAD = \frac{B}{2}.$ Similarly $\angle FAD = \frac{C}{2},$ and $E, F$ lie on opposite sides of line $AD.$

Let $M$ be the midpoint of $\overline{AD}.$ In right triangles $AME$ and $AMF,$ $ME = AM\tan\frac{B}{2}$ and $MF = AM\tan\frac{C}{2},$ so $EF = AM\left(\tan\frac{B}{2} + \tan\frac{C}{2}\right),$ while the distance from $A$ to line $EF$ is $AM.$ Hence $[AEF] = \frac{1}{2}AM^2\left(\tan\frac{B}{2} + \tan\frac{C}{2}\right).$

Here $BD = 2$ and $DC = 3,$ so $AD^2 = AB \cdot AC - BD \cdot DC = 24 - 6 = 18$ and $AM^2 = \frac{9}{2}.$ The law of cosines gives $\cos B = \frac{1}{8}$ and $\cos C = \frac{3}{4},$ so $\tan\frac{B}{2} = \sqrt{\frac{1 - 1/8}{1 + 1/8}} = \frac{\sqrt{7}}{3}$ and $\tan\frac{C}{2} = \frac{1}{\sqrt{7}},$ with sum $\frac{10\sqrt{7}}{21}.$ The area is $\frac{1}{2} \cdot \frac{9}{2} \cdot \frac{10\sqrt{7}}{21} = \frac{15\sqrt{7}}{14},$ so $m + n + p = 15 + 7 + 14 = 36.$

Answer: 85

Solution:

Write $P(x) = x^2 + px + q$ with roots $r_1$ and $r_2.$ The solutions of $P(P(x)) = 0$ split into the two solutions of $P(x) = r_1$ and the two of $P(x) = r_2,$ and each pair sums to $-p.$

If $3$ and $4$ form one pair, then $a + b = -p = 3 + 4 = 7,$ so $(a + b)^2 = 49.$ This is achievable: $P(x) = (x - 3)(x - 4) + r_1$ with $r_1$ satisfying $r_1^2 - 6r_1 + 12 = 0,$ which has (complex) solutions, and the four roots are distinct.

Otherwise $3$ and $4$ lie in different pairs: $3 + a = 4 + b = -p = s,$ and $\{P(3), P(4)\} = \{r_1, r_2\}.$ The root sum gives $P(3) + P(4) = 25 + 7p + 2q = s,$ so with $p = -s$ we get $q = 4s - \frac{25}{2},$ and then $P(3) = s - \frac{7}{2}$ and $P(4) = \frac{7}{2}.$ The root product gives $\frac{7}{2}\left(s - \frac{7}{2}\right) = q = 4s - \frac{25}{2},$ whose solution is $s = \frac{1}{2}.$ Then $a + b = (s - 3) + (s - 4) = -6,$ so $(a + b)^2 = 36,$ with $a = -\frac{5}{2},$ $b = -\frac{7}{2}$ all distinct from $3$ and $4.$ The sum of all possible values is $49 + 36 = 85.$

Answer: 58

Solution:

Reflecting $H$ over line $BC$ lands on $\omega,$ so the circumcircle of $HBC$ is the reflection of $\omega$ over $BC.$ Take the circumcenter $O$ as the origin, so that $H = A + B + C$ as vectors. If $M$ is the midpoint of $\overline{BC},$ then $OM \perp BC,$ so the reflected center is $2M - O = B + C = H - A.$ Tangency at $H$ means $XY$ is perpendicular to the radius from $B + C$ to $H,$ which is the vector $A:$ the chord $XY$ is perpendicular to $OA.$

Place $A = (0, R)$ so that $XY$ is horizontal at height $h,$ with $H = (x_0, h).$ The half-chord length is $\sqrt{R^2 - h^2},$ and $HX = 2,$ $HY = 6$ give $\sqrt{R^2 - h^2} = 4$ with $|x_0| = 2.$ From $HA = 3:$ $4 + (R - h)^2 = 9,$ so $R - h = \sqrt{5}.$ Then $$16 = R^2 - h^2 = (R - h)(R + h) = \sqrt{5}\left(2R - \sqrt{5}\right),$$ giving $R = \frac{21}{2\sqrt{5}}.$

Now $B + C = H - A = (\pm 2, -\sqrt{5}),$ so $M = \left(\pm 1, -\frac{\sqrt{5}}{2}\right)$ and $OM = \frac{3}{2},$ whence $BC = 2\sqrt{R^2 - \frac{9}{4}} = 2\sqrt{\frac{99}{5}}.$ The distance from $A$ to line $BC$ (through $M,$ perpendicular to $OM$) is $\frac{|A \cdot M - OM^2|}{OM} = \frac{21/4 + 9/4}{3/2} = 5,$ using $A \cdot M = -\frac{\sqrt{5}R}{2} = -\frac{21}{4}.$ Hence $$[ABC] = \frac{1}{2} \cdot 2\sqrt{\frac{99}{5}} \cdot 5 = \sqrt{495} = 3\sqrt{55},$$ and $m + n = 3 + 55 = 58.$