Small Hint:

Let $\angle BAC = \alpha$ and chase angles along the chain of isosceles triangles $AED,$ $EDB,$ and $DBC.$

Big Hint:

The exterior angle gives $\angle DEB = 2\alpha;$ then the three angles at $D$ along $\overline{AC}$ force $\angle BDC = 3\alpha = \angle BCD.$

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 $m + n = 540 + 7 = 547.$

Small Hint:

Set $t = \log_2 x;$ the three terms become $\frac{1+t}{3},$ $\frac{t}{2},$ and $t.$

Big Hint:

The middle term squared equals the product of its neighbors; solve for $t,$ discarding $t = 0.$

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

Small Hint:

Set the base-ten values equal: $121a + 11b + c$ $= 512 + 64b + 8c + a.$

Big Hint:

Simplify to $120a = 512 + 53b + 7c$ with digits $a, b, c \le 7.$ The right side forces $a \ge 5,$ and smaller $a$ means smaller $N.$

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 $621.$

Small Hint:

Write $N = 10000k + 2020;$ then $k$ divides $N$ exactly when $k$ divides $2020.$

Big Hint:

Each of the $12$ divisors $k$ of $2020 = 2^2 \cdot 5 \cdot 101$ contributes the digit sum of $k$ plus $2 + 0 + 2 + 0.$

Solution:

If removing the last four digits leaves $k \ge 1,$ then $N = 10000k + 2020,$ and the condition that $k$ divides $N$ is equivalent to $k$ dividing $2020.$ 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 $45 + 12 \cdot 4 = 93.$

Small Hint:

Count arrangements fixable to ascending order: insert one card into the other five written in increasing order. Beware of building the same arrangement twice.

Big Hint:

The sorted row arises from all $6$ insertions, and each adjacent swap arises twice. No arrangement works for both ascending and descending.

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 $26 + 26 = 52.$

Small Hint:

A sphere of radius $r$ sitting in a hole of radius $a$ has its center at distance $\sqrt{r^2 - a^2}$ from the plane of the board.

Big Hint:

The centers are $2r$ apart: $49$ $+ \left(\sqrt{r^2 - 1} - \sqrt{r^2 - 4}\right)^2$ $= 4r^2.$ Isolate the radical and square.

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: $$\begin{aligned} &49 + \left(\sqrt{r^2 - 1} - \sqrt{r^2 - 4}\right)^2 \\ &= 4r^2. \end{aligned}$$ 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 $m + n = 160 + 13 = 173.$

Small Hint:

With $k$ men and $k + 1$ women the count is $\sum_k \binom{11}{k}\binom{12}{k+1}.$

Big Hint:

Since $\binom{12}{k+1} = \binom{12}{11-k},$ Vandermonde’s identity collapses the sum to $\binom{23}{11}.$ Factor that binomial coefficient into primes.

Solution:

A committee with $k$ men has $k + 1$ women, so $$\begin{aligned} N &= \sum_{k=0}^{11} \binom{11}{k}\binom{12}{k+1} \\ &= \sum_{k=0}^{11} \binom{11}{k}\binom{12}{11-k} \\ &= \binom{23}{11} \end{aligned}$$ 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 $2 + 7 + 13 + 17 + 19 + 23 = 81.$

Small Hint:

Use complex numbers: the total displacement is $5\left(1 + z + z^2 + \cdots\right)$ with $z = \frac{1}{2}e^{\frac{i\pi}{3}}.$

Big Hint:

The series sums to $\frac{5}{1 - z};$ compute $|1 - z|^2$ from $z = \frac{1}{4} + \frac{\sqrt{3}}{4}i.$

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 $z = \frac{1}{2}e^{\frac{i\pi}{3}},$ so $$\begin{aligned} P &= 5\left(1 + z + z^2 + \cdots\right) \\ &= \frac{5}{1 - z}. \end{aligned}$$

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 $$OP^2 = \frac{25}{\frac{3}{4}} = \frac{100}{3},$$ and $m + n = 100 + 3 = 103.$

Small Hint:

Since $20^9 = 2^{18} \cdot 5^9,$ requiring $a_1$ to divide $a_2$ and $a_2$ to divide $a_3$ says each prime’s exponents form a non-decreasing triple.

Big Hint:

Non-decreasing triples are multisets: $\binom{21}{3}$ choices for the exponents of $2$ and $\binom{12}{3}$ for $5.$

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 conditions that $a_1$ divides $a_2$ and $a_2$ divides $a_3$ hold 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 $$\begin{aligned} \frac{\binom{21}{3}}{19^3} \cdot \frac{\binom{12}{3}}{10^3} &= \frac{1330}{6859} \cdot \frac{220}{1000} \\ &= \frac{70}{361} \cdot \frac{11}{50} \\ &= \frac{77}{1805}. \end{aligned}$$

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

Small Hint:

Every prime dividing $n$ must divide $m,$ hence divides $m + n$ — so all prime factors of $n$ are at least $11.$

Big Hint:

Try $n = 11^2$ and $m = 11t$ with $t$ not divisible by $11;$ then $n^n$ dividing $m^m$ needs $m \ge 242.$ Scan upward for $\gcd(m + n, 210) = 1,$ then use the resulting upper bound to rule out larger deficient exponents and larger $n.$

Solution:

If a prime $p$ divides $n,$ then $p$ divides $n^n,$ which in turn divides $m^m,$ so $p$ divides $m$ and hence $p$ divides $m + n.$ 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 $p.$ Since $n^n$ divides $m^m,$ comparing exponents of $p$ gives $bm \ge an,$ so $m \ge \frac{a}{b}n.$ In particular $a \ge 2,$ so $p^2$ divides $n$ 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 $n^n$ divides $m^m,$ and $m + n = 407 = 11 \cdot 37$ is coprime to $210.$

It remains to prove that nothing smaller works. Suppose $m + n \lt 407.$ Then $n \lt 407.$ For the deficient prime above, $a \ge 3$ would imply $n \ge 11^3 \gt 407,$ so $a = 2$ and $b = 1.$ Thus $m \ge 2n,$ and $m + n \lt 407$ implies $n \le 135.$ The only integer in this range divisible by the square of a prime at least $11$, with no prime factors below $11,$ is $n = 121.$ The preceding check exhausts every possible $m$ for this $n$ below $286,$ so the least possible value is $407.$

Small Hint:

$f(2)$ and $f(4)$ must both be roots of $g.$ Treat $f(2) = f(4)$ and $f(2) \ne f(4)$ separately.

Big Hint:

$f(2) = f(4)$ forces $a = -6$ with $b, c$ free; otherwise Vieta pins $c = -(20 + 6a + 2b),$ which must lie in $[-10, 10].$

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 $441 + 69 = 510.$

Small Hint:

Handle $3^3,$ $5^5,$ $7^7$ separately. Since $149 - 2 = 147 = 3 \cdot 7^2,$ lifting the exponent applies directly at $3$ and $7.$

Big Hint:

At $5,$ first $4^n \equiv 2^n \pmod 5$ requires $4$ to divide $n.$ Then $v_5(149^4 - 2^4) = 1,$ so lifting the exponent forces $5^4$ to divide $\frac{n}{4}.$

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 $n$ to be divisible by $3^2$ and by $7^5.$

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 $4$ to divide $n.$ 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 $5^4$ divides $k,$ i.e. $4 \cdot 5^4$ divides $n.$

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.