Solution:

An ordinary day has $60 \cdot 24 = 1440$ minutes, and $6{:}36$ AM comes $6 \cdot 60 + 36 = 396$ minutes after midnight. A metric day has $10 \cdot 100 = 1000$ metric minutes, so the equivalent metric time is $$\frac{396}{1440} \cdot 1000 = 275$$ metric minutes after midnight, which the new clock displays as $2{:}75.$

Therefore $100A + 10B + C = 275.$

Solution:

Working from the outside in, $\log_2(\cdot) = 0$ forces $\log_{2^a}(\log_{2^b}(2^{1000})) = 1,$ so $\log_{2^b}(2^{1000}) = 2^a,$ so $2^{1000} = (2^b)^{2^a} = 2^{b \cdot 2^a}.$ Hence $$b \cdot 2^a = 1000 = 2^3 \cdot 125.$$

Since $a \ge 1$ and $b$ is a positive integer, $2^a$ must be one of $2,$ $4,$ $8,$ giving $(a, b) = (1, 500),$ $(2, 250),$ $(3, 125).$ The sum of all possible values of $a + b$ is $501 + 252 + 128 = 881.$

Solution:

Burning the first $x$ centimeters takes $10(1 + 2 + \cdots + x) = 5x(x+1)$ seconds, so $T = 5 \cdot 119 \cdot 120 = 71400$ and $\frac{T}{2} = 35700.$

Setting $5x(x+1) = 35700$ gives $x(x+1) = 7140 = 84 \cdot 85,$ so at time $\frac{T}{2}$ the candle has burned down exactly $84$ centimeters. Its height is $h = 119 - 84 = 35,$ and $10h = 350.$

Solution:

The midpoint of $\overline{AB}$ is $M = \left(\frac{3}{2}, \sqrt{3}\right),$ and $AB = \sqrt{1 + 12} = \sqrt{13}.$ The third vertex lies at distance $\frac{\sqrt{3}}{2}\sqrt{13}$ from $M$ along a direction perpendicular to $\overrightarrow{AB} = (1, 2\sqrt{3});$ a unit perpendicular is $\frac{1}{\sqrt{13}}(2\sqrt{3}, -1).$ Taking the sign that lands in the first quadrant, $$C = M + \frac{\sqrt{3}}{2}\,(2\sqrt{3}, -1) = \left(\frac{9}{2}, \frac{\sqrt{3}}{2}\right)$$ (the other choice has negative $x$-coordinate).

The center of an equilateral triangle is its centroid, the average of the vertices: $$P = \left(\frac{1 + 2 + \frac{9}{2}}{3},\ \frac{0 + 2\sqrt{3} + \frac{\sqrt{3}}{2}}{3} \right) = \left(\frac{5}{2}, \frac{5\sqrt{3}}{6}\right).$$

Then $x \cdot y = \frac{5}{2} \cdot \frac{5\sqrt{3}}{6} = \frac{25\sqrt{3}}{12},$ so $p + q + r = 25 + 3 + 12 = 40.$

Solution:

Scale so the side length is $6,$ with $BD = DE = EC = 2.$ In triangle $AEC,$ the law of cosines gives $$AE^2 = 6^2 + 2^2 - 2 \cdot 6 \cdot 2 \cos 60^\circ = 28,$$ so $AE = 2\sqrt{7},$ and $AD = 2\sqrt{7}$ by symmetry.

Since $DE$ is one third of $BC$ and triangles $ADE$ and $ABC$ share the apex $A,$ we get $[ADE] = \frac{1}{3}[ABC] = \frac{1}{3} \cdot \frac{36\sqrt{3}}{4} = 3\sqrt{3}.$ On the other hand $[ADE] = \frac{1}{2} \cdot AD \cdot AE \cdot \sin(\angle DAE) = 14\sin(\angle DAE).$

Therefore $\sin(\angle DAE) = \frac{3\sqrt{3}}{14},$ and $a + b + c = 3 + 3 + 14 = 20.$

Solution:

The block $\{1000N, \ldots, 1000N + 999\}$ misses all squares exactly when some consecutive squares $x^2$ and $(x+1)^2$ jump over it, which requires $(x+1)^2 - x^2 = 2x + 1 \gt 1000,$ so $x \ge 500.$ In particular every block below $500^2 = 250000$ contains a square, so we search from there.

Write $x = 500 + a$ with $a \ge 0.$ Then $x^2 = 1000(250 + a) + a^2,$ so as long as $a^2 \lt 1000$ (that is, $a \le 31$), the square $x^2$ lies in block $250 + a;$ these cover blocks $250$ through $281.$ Block $251 + a$ is skipped exactly when $$(x+1)^2 = 1000(250 + a) + a^2 + 2a + 1001 \ge 1000(252 + a),$$ that is, $a^2 + 2a \ge 999.$ For $a \le 30$ this fails (so $(x+1)^2$ lands in block $251 + a$), and it first holds at $a = 31,$ since $961 + 62 = 1023.$

Indeed $531^2 = 281961$ and $532^2 = 283024$ straddle the block starting at $282000.$ The least such $N$ is $251 + 31 = 282.$

Solution:

Let $n$ clerks start and $k$ be reassigned at the end of each hour. In the final $10$ minutes each remaining clerk sorts $5$ files, so $$30n + 30(n - k) + 30(n - 2k) + 5(n - 3k) = 1775,$$ which simplifies to $95n - 105k = 1775,$ or $19n - 21k = 355.$

Modulo $19$ this reads $-2k \equiv 13,$ so $2k \equiv 6$ and $k \equiv 3 \pmod{19}.$ Taking $k = 3$ gives $n = \frac{355 + 63}{19} = 22;$ the next candidate, $k = 22,$ gives $n = 43,$ for which $n - 3k \lt 0.$ So $n = 22$ and $k = 3.$

In the first hour $22$ clerks sort $30 \cdot 22 = 660$ files, and in the next half hour $19$ clerks sort $15 \cdot 19 = 285,$ for a total of $660 + 285 = 945.$

Solution:

Let $r$ be the radius, and let each chord of length $22$ subtend central angle $\alpha$ and each chord of length $20$ subtend $\beta.$ The six central angles fill the circle: $4\alpha + 2\beta = 360^\circ,$ so $\frac{\beta}{2} = 90^\circ - \alpha$ and $\sin\frac{\beta}{2} = \cos\alpha.$

Half a $20$-chord gives $\sin\frac{\beta}{2} = \frac{10}{r},$ and the law of cosines on the isosceles triangle with legs $r$ and base $22$ gives $22^2 = 2r^2(1 - \cos\alpha),$ so $\cos\alpha = 1 - \frac{242}{r^2}.$ Equating, $$1 - \frac{242}{r^2} = \frac{10}{r} \quad\Longrightarrow\quad r^2 - 10r - 242 = 0,$$ so $r = 5 + \sqrt{267}$ (taking the positive root).

Therefore $p + q = 5 + 267 = 272.$

Solution:

First count colored tilings when $k$ colors are available. The first square's tile can be colored in $k$ ways, and each of the remaining $6$ squares either extends the current tile or starts a new tile in one of the $k$ colors, giving $k + 1$ choices per square. So there are $k(k+1)^6$ tilings.

With three colors that is $3 \cdot 4^6 = 12288$ tilings. By inclusion-exclusion over the unused colors, the number using all three colors is $$N = 3 \cdot 4^6 - 3 \cdot (2 \cdot 3^6) + 3 \cdot (1 \cdot 2^6) = 12288 - 4374 + 192 = 8106.$$

The remainder when $N$ is divided by $1000$ is $106.$

Solution:

The nearest point $B$ lies on segment $OA$ with $OB = \sqrt{13}$ and $AB = 4.$ Triangles $OKL$ and $BKL$ share the base $KL,$ and their heights are the distances from $O$ and $B$ to the line through $A.$ For any point $P$ on line $OA,$ that distance is $PA\sin\varphi,$ where $\varphi$ is the angle between the two lines, so $$\frac{[BKL]}{[OKL]} = \frac{AB}{AO} = \frac{4}{4 + \sqrt{13}}.$$

Since $OK = OL = \sqrt{13},$ we have $[OKL] = \frac{13}{2}\sin(\angle KOL) \le \frac{13}{2},$ with equality when $\angle KOL = 90^\circ.$ Such a chord lies at distance $\sqrt{13/2}$ from $O,$ which is less than $OA,$ so a line through $A$ can achieve it.

The maximum area is $$[BKL] = \frac{13}{2} \cdot \frac{4}{4 + \sqrt{13}} = \frac{26}{4 + \sqrt{13}} = \frac{26(4 - \sqrt{13})}{3} = \frac{104 - 26\sqrt{13}}{3},$$ so $a + b + c + d = 104 + 26 + 13 + 3 = 146.$

Solution:

Say $f(f(x)) = a$ for all $x,$ and let $S = \{x : f(x) = a\}.$ Picking any $t \in S,$ we get $a = f(f(t)) = f(a),$ so $a \in S.$ Every $x$ satisfies $f(x) \in S$ (because $f(f(x)) = a$), and if $x \notin S$ then $f(x) \ne a,$ so $f$ maps the complement of $S$ into $S \setminus \{a\}.$ Conversely, any $f$ built this way works.

If $|S| = k,$ we choose the constant $a$ in $7$ ways, the remaining $k - 1$ elements of $S$ in $\binom{6}{k-1}$ ways, and an image in $S \setminus \{a\}$ for each of the $7 - k$ other elements in $(k-1)^{7-k}$ ways. Hence $$N = 7\sum_{k=1}^{7} \binom{6}{k-1}(k-1)^{7-k} = 7\,(0 + 6 + 240 + 540 + 240 + 30 + 1) = 7 \cdot 1057 = 7399.$$

The remainder when $N$ is divided by $1000$ is $399.$

Solution:

A cubic with real coefficients has either three real roots or one real root and a conjugate pair. The only real numbers with modulus $20$ or $13$ are $\pm 20$ and $\pm 13,$ so in the all-real case the roots form a multiset of size $3$ from those $4$ values: $\binom{6}{3} = 20$ polynomials.

Otherwise the roots are $k \in \{\pm 20, \pm 13\}$ and a conjugate pair $r \pm si$ with $s \ne 0$ and $r^2 + s^2 = 400$ or $169.$ Expanding $$(z - k)\bigl(z^2 - 2rz + (r^2 + s^2)\bigr)$$ shows the coefficients are $-(2r + k),$ $r^2 + s^2 + 2rk,$ and $-(r^2 + s^2)k,$ which are all integers exactly when $2r$ is an integer. On the circle of radius $20$ we need $|r| \lt 20,$ allowing $2r \in \{-39, \ldots, 39\}:$ $79$ choices; on the circle of radius $13,$ $2r \in \{-25, \ldots, 25\}:$ $51$ choices. With $4$ choices of $k,$ that gives $4(79 + 51) = 520$ polynomials, each distinct since the roots determine the polynomial.

In total there are $20 + 520 = 540$ such polynomials.

Solution:

Let $AB = 2x$ and $AC = BC = y,$ so $BD = \frac{y}{4},$ $CD = \frac{3y}{4},$ and $\cos B = \frac{x}{y}$ (drop the altitude from $C$ to the midpoint of $\overline{AB}$). The law of cosines in triangle $ABD$ gives $$AD^2 = 4x^2 + \frac{y^2}{16} - 2 \cdot 2x \cdot \frac{y}{4} \cdot \frac{x}{y} = 3x^2 + \frac{y^2}{16}.$$

Both $CE$ and $BE$ are medians to $\overline{AD},$ in triangles $ACD$ and $ABD$ respectively. The median formula $4m^2 = 2b^2 + 2c^2 - a^2$ gives $$28 = 2y^2 + \frac{18y^2}{16} - AD^2 = \frac{49y^2}{16} - 3x^2, \qquad 36 = 8x^2 + \frac{2y^2}{16} - AD^2 = 5x^2 + \frac{y^2}{16}.$$ From the second equation $\frac{y^2}{16} = 36 - 5x^2;$ substituting into the first gives $49(36 - 5x^2) - 3x^2 = 28,$ so $248x^2 = 1736,$ $x^2 = 7,$ and then $y^2 = 16.$

The altitude from $C$ has length $\sqrt{y^2 - x^2} = 3,$ so the area is $\frac{1}{2} \cdot 2x \cdot 3 = 3\sqrt{7},$ and $m + n = 3 + 7 = 10.$

Solution:

For $k \le \frac{n}{2}$ the quotient $\lfloor n/k \rfloor$ is at least $2,$ so the remainder satisfies $f(n, k) \le n - 2k$ as well as $f(n, k) \le k - 1.$ Write $n = 3m + r$ with $r \in \{0, 1, 2\}.$ Dividing by $k = m + 1$ gives quotient $2$ and remainder $m + r - 2,$ so $F(n) \ge m + r - 2.$ Conversely, for $k \ge m + 1,$ $f(n, k) \le n - 2k \le m + r - 2,$ and for smaller $k$ the bound $f(n, k) \le k - 1$ finishes the job: when $r = 2$ it gives at most $m$ for $k \le m + 1;$ when $r = 1$ it gives at most $m - 1$ for $k \le m;$ and when $r = 0$ it gives at most $m - 2$ for $k \le m - 1,$ while $k = m$ divides $3m$ exactly, leaving remainder $0.$ Hence $$F(3m) = m - 2, \qquad F(3m + 1) = m - 1, \qquad F(3m + 2) = m.$$

Grouping $n = 20, \ldots, 100$ as triples $3m - 1,$ $3m,$ $3m + 1$ for $m = 7, \ldots, 33$ (note $F(3m - 1) = F(3(m-1) + 2) = m - 1$), each triple contributes $(m - 1) + (m - 2) + (m - 1) = 3m - 4,$ so $$\sum_{n=20}^{100} F(n) = \sum_{m=7}^{33} (3m - 4) = 3 \cdot \frac{(7 + 33) \cdot 27}{2} - 4 \cdot 27 = 1620 - 108 = 1512.$$

The requested remainder is $512.$

Solution:

Replacing each $\cos^2$ by $1 - \sin^2,$ the first equation becomes $\sin^2 A + \sin^2 B - 2 \sin A \sin B \cos C = \frac{1}{8}.$ By the law of sines, $\sin A = \frac{a}{2R}$ and so on, so the left side equals $$\frac{a^2 + b^2 - 2ab\cos C}{4R^2} = \frac{c^2}{4R^2} = \sin^2 C$$ by the law of cosines. Hence $\sin^2 C = 2 - \frac{15}{8} = \frac{1}{8}.$ The same argument turns the second equation into $\sin^2 A = 2 - \frac{14}{9} = \frac{4}{9},$ and shows the requested expression equals $2 - \sin^2 B.$

Since $A$ and $C$ are acute, $\cos A = \frac{\sqrt{5}}{3}$ and $\cos C = \frac{\sqrt{14}}{4},$ with $\sin A = \frac{2}{3}$ and $\sin C = \frac{\sqrt{2}}{4}.$ Then $$\sin B = \sin(A + C) = \frac{2}{3} \cdot \frac{\sqrt{14}}{4} + \frac{\sqrt{5}}{3} \cdot \frac{\sqrt{2}}{4} = \frac{2\sqrt{14} + \sqrt{10}}{12},$$ so $\sin^2 B = \frac{66 + 8\sqrt{35}}{144} = \frac{33 + 4\sqrt{35}}{72}.$

Therefore $2 - \sin^2 B = \frac{111 - 4\sqrt{35}}{72},$ and $p + q + r + s = 111 + 4 + 35 + 72 = 222.$