Solution:

Every factorization is $10^n = (2^x 5^y)(2^{n-x} 5^{n-y}).$ If a factor is divisible by both $2$ and $5,$ it is a multiple of $10$ and ends in the digit $0.$ So the only possible zero-free factorization is $10^n = 2^n \cdot 5^n,$ and we need the least $n$ for which $2^n$ or $5^n$ contains a digit $0.$

The powers $2^1, \ldots, 2^8$ are $2, 4, 8, 16, 32, 64, 128, 256$ — no zeros. The powers $5^1, \ldots, 5^7$ are $5, 25, 125, 625, 3125, 15625, 78125$ — no zeros — but $5^8 = 390625$ contains a $0.$

Hence every factorization of $10^8$ contains a digit $0,$ while $10^7 = 2^7 \cdot 5^7 = 128 \cdot 78125$ does not, so the answer is $8.$

Solution:

Carrying out the reflections, $B = (v, u),$ $C = (-v, u),$ $D = (-v, -u),$ and $E = (v, -u).$ The points $B, C, D, E$ form a rectangle of width $2v$ and height $2u,$ with area $4uv,$ and $A = (u, v)$ sticks out to its right. Triangle $ABE$ has vertical base $BE$ of length $2u$ and horizontal height $u - v,$ so its area is $u(u - v).$

The pentagon's area is therefore $$4uv + u(u - v) = u^2 + 3uv = u(u + 3v) = 451 = 11 \cdot 41.$$ Since $0 \lt v \lt u,$ we have $u \lt u + 3v \lt 4u,$ which rules out the factorization $1 \cdot 451.$ So $u = 11$ and $u + 3v = 41,$ giving $v = 10,$ which indeed satisfies $v \lt u.$

Thus $u + v = 11 + 10 = 21.$

Solution:

By the binomial theorem, the coefficients of $x^2$ and $x^3$ are $\binom{2000}{2} a^2 b^{1998}$ and $\binom{2000}{3} a^3 b^{1997}.$ Setting them equal and cancelling $a^2 b^{1997}$ gives $$\binom{2000}{2} b = \binom{2000}{3} a, \qquad \text{so} \qquad b = \frac{1998}{3}\,a = 666a.$$

Since $\gcd(a, b) = 1,$ we must have $a = 1$ and $b = 666,$ so $a + b = 667.$

Solution:

Let the tiniest square (in the middle) have side $x$ and the small square just below and to its right have side $y.$ Chasing edge lengths through the figure, the remaining squares have sides $x + y,$ then $(x + y) + x = 2x + y,$ then $(x + y) + (2x + y) = 3x + 2y,$ then $(2x + y) + (3x + 2y) = 5x + 3y$ (the top-left square). The tall square on the right spans the previous three along its left edge minus overlaps, giving side $4x + 4y;$ the bottom-right square has side $(4x + 4y) + y = 4x + 5y;$ and the bottom-left square has side $x + (2x + y) + (5x + 3y) = 8x + 4y.$

Measuring the rectangle's height along its left and right sides, $$(8x + 4y) + (5x + 3y) = (4x + 5y) + (4x + 4y),$$ which simplifies to $5x = 2y.$ Taking the smallest positive integers, $x = 2$ and $y = 5,$ the nine squares have sides $2, 5, 7, 9, 16, 25, 28, 33, 36,$ and the rectangle is $(36 + 33) \times (36 + 25) = 69 \times 61.$ These dimensions are relatively prime (any common scaling would break that), and the areas check: $69 \cdot 61 = 4209$ equals the sum of the nine squares' areas.

The perimeter is $2(69 + 61) = 260.$

Solution:

Say the boxes hold $a$ and $b$ marbles with $a + b = 25,$ containing $p$ and $q$ black marbles. Then $\frac{pq}{ab} = \frac{27}{50},$ so $50pq = 27ab,$ and since $\gcd(27, 50) = 1,$ we need $50 \mid ab.$ Checking $a(25 - a)$ for $a = 1, \ldots, 12,$ only $\{a, b\} = \{20, 5\}$ and $\{10, 15\}$ give a multiple of $50.$

For sizes $20$ and $5:$ $pq = \frac{27 \cdot 100}{50} = 54,$ and since each box also holds a white marble, $p \le 19$ and $q \le 4,$ forcing $p = 18,$ $q = 3.$ The white counts are $2$ and $2,$ so the white-white probability is $\frac{2}{20} \cdot \frac{2}{5} = \frac{1}{25}.$ For sizes $10$ and $15:$ $pq = \frac{27 \cdot 150}{50} = 81,$ and $p \le 9,$ $q \le 14$ force $p = q = 9.$ The white counts are $1$ and $6,$ giving $\frac{1}{10} \cdot \frac{6}{15} = \frac{1}{25}$ again.

Either way the probability is $\frac{1}{25},$ so $m + n = 1 + 25 = 26.$

Solution:

The condition is $\frac{x + y}{2} = \sqrt{xy} + 2,$ that is, $x + y - 2\sqrt{xy} = 4,$ so $(\sqrt{y} - \sqrt{x})^2 = 4$ and (as $y \gt x$) $\sqrt{y} - \sqrt{x} = 2.$ Note $\sqrt{xy} = \frac{x + y - 4}{2}$ is rational, hence $\sqrt{y} + \sqrt{x} = \frac{y - x}{\sqrt{y} - \sqrt{x}} = \frac{y - x}{2}$ is rational too, so $\sqrt{x}$ and $\sqrt{y}$ are rational — and a rational square root of an integer is an integer.

Therefore $x = a^2$ and $y = (a + 2)^2$ for a positive integer $a.$ The constraint $y \lt 10^6$ means $a + 2 \le 999,$ so $a$ ranges over $1, 2, \ldots, 997,$ and each value gives a valid pair.

Hence there are $997$ ordered pairs.

Solution:

Let $t = z + \frac{1}{y}.$ Expanding the product of all three expressions, $$\left(x + \frac{1}{z}\right)\!\left(y + \frac{1}{x}\right)\!\left(z + \frac{1}{y}\right) = xyz + \frac{1}{xyz} + \left(x + \frac{1}{z}\right) + \left(y + \frac{1}{x}\right) + \left(z + \frac{1}{y}\right).$$ Since $xyz = 1,$ the left side is $5 \cdot 29 \cdot t = 145t$ and the right side is $2 + 5 + 29 + t = 36 + t.$

So $145t = 36 + t,$ giving $t = \frac{36}{144} = \frac{1}{4}.$ Thus $m + n = 1 + 4 = 5.$

Solution:

Held point down, the liquid forms a cone similar to the container with ratio $\frac{9}{12} = \frac{3}{4},$ so its volume is $\left(\frac{3}{4}\right)^3 = \frac{27}{64}$ of the container's volume.

Held point up, the empty space is a similar cone at the apex with $1 - \frac{27}{64} = \frac{37}{64}$ of the volume, so its height is $$12 \sqrt[3]{\frac{37}{64}} = \frac{12 \sqrt[3]{37}}{4} = 3\sqrt[3]{37}$$ inches. The liquid is therefore $12 - 3\sqrt[3]{37}$ inches deep.

Since $37$ is cube-free, $m + n + p = 12 + 3 + 37 = 52.$

Solution:

Let $a = \log_{10} x,$ $b = \log_{10} y,$ $c = \log_{10} z.$ Using $\log_{10} 2000 = 4 - \log_{10} 5,$ the first equation becomes $a + b - ab = \log_{10} 5,$ which factors as $(1 - a)(1 - b) = 1 - \log_{10} 5 = \log_{10} 2.$ Similarly the second equation gives $(1 - b)(1 - c) = 1 - (1 - \log_{10} 2) = \log_{10} 2,$ and the third gives $(1 - c)(1 - a) = 1.$

Dividing the first two (note $1 - b \ne 0$) yields $1 - a = 1 - c,$ and then the third equation gives $(1 - a)^2 = 1,$ so $1 - a = \pm 1.$ If $1 - a = 1,$ then $1 - b = \log_{10} 2,$ so $b = 1 - \log_{10} 2 = \log_{10} 5$ and $y = 5$ (indeed $(x, y, z) = (1, 5, 1)$ works). If $1 - a = -1,$ then $1 - b = -\log_{10} 2,$ so $b = \log_{10} 20$ and $y = 20$ (from $(x, y, z) = (100, 20, 100)$).

Therefore $y_1 + y_2 = 5 + 20 = 25.$

Solution:

Let $S = x_1 + x_2 + \cdots + x_{100}.$ The condition says $x_k = (S - x_k) - k,$ so $x_k = \frac{S - k}{2}$ for every $k.$ Summing over $k = 1, \ldots, 100,$ $$S = \frac{100S - (1 + 2 + \cdots + 100)}{2} = \frac{100S - 5050}{2},$$ so $98S = 5050$ and $S = \frac{2525}{49}.$

Then $$x_{50} = \frac{S - 50}{2} = \frac{2525 - 2450}{98} = \frac{75}{98},$$ which is in lowest terms, so $m + n = 75 + 98 = 173.$

Solution:

Write $a = 2^i 5^j$ and $b = 2^k 5^l$ with exponents between $0$ and $3.$ Coprimality means $\min(i, k) = 0$ and $\min(j, l) = 0,$ and these two constraints are independent. So as $(a, b)$ runs over all coprime pairs, the factor $2^{i - k}$ independently takes each value in $\{2^{-3}, \ldots, 2^{3}\}$ exactly once, and similarly for $5^{j - l}.$ Hence $$S = \left(1 + 2 + 4 + 8 + \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{8}\right) \left(1 + 5 + 25 + 125 + \tfrac{1}{5} + \tfrac{1}{25} + \tfrac{1}{125}\right) = \frac{127}{8} \cdot \frac{19531}{125}.$$

This equals $\frac{2480437}{1000} = 2480.437,$ so $\frac{S}{10} = 248.0437,$ and the greatest integer not exceeding it is $248.$

Solution:

Since $f(398 - x) = f(2158 - x)$ for all $x,$ substituting $t = 398 - x$ gives $f(t) = f(t + 1760);$ likewise $f(2158 - x) = f(3214 - x)$ gives period $1056.$ Combining, $f$ has period $\gcd(1760, 1056) = 352.$ Reducing $398$ mod $352,$ the symmetry $f(x) = f(398 - x)$ becomes $f(x) = f(46 - x).$

So $f$ is determined by residues mod $352,$ with residues $r$ and $46 - r$ forced to share a value. This pairing has exactly two fixed points, from $2r \equiv 46 \pmod{352}:$ $r = 23$ and $r = 199.$ Hence there are at most $\frac{352 - 2}{2} + 2 = 177$ classes, and since $0, 1, \ldots, 999$ covers every residue mod $352,$ the list contains at most $177$ different values.

This is achievable: $f(x) = \cos\frac{2\pi(x - 23)}{352}$ satisfies all three given symmetries (each of $398,$ $2158,$ $3214$ is $\equiv 46$ mod $352$), and two integers get equal values only when their residues are paired. So the answer is $177.$

Solution:

In six minutes the truck can drive $5$ miles on a highway or $1.4$ miles across the prairie, and an optimal route is a highway stretch followed by a straight prairie segment. Work in the first quadrant with the highways as axes. Driving to $(d, 0)$ takes $\frac{d}{50}$ hours, leaving a prairie range of $14\left(\frac{1}{10} - \frac{d}{50}\right) = 1.4\left(1 - \frac{d}{5}\right)$ miles. As $d$ runs from $0$ to $5,$ these disks shrink linearly to a point, so their union is the "cone": the convex hull of the disk of radius $1.4$ about the origin and the point $(5, 0),$ bounded by the tangent line from $(5, 0).$ The tangent length is $\sqrt{5^2 - 1.4^2} = 4.8,$ so the ratios are $7$–$24$–$25$ and the tangent line is $7x + 24y = 35.$ The $y$-axis gives the mirror-image region bounded by $24x + 7y = 35.$

The two tangent lines meet at $P = \left(\frac{35}{31}, \frac{35}{31}\right),$ which lies at distance $\frac{35\sqrt{2}}{31} \gt 1.4$ from the origin — outside the circle — so in the first quadrant the reachable set is exactly the (non-convex) quadrilateral with vertices $(0, 0),$ $(5, 0),$ $P,$ $(0, 5).$ Splitting it along the diagonal from the origin to $P$ gives two triangles, each with area $\frac{1}{2} \cdot 5 \cdot \frac{35}{31},$ for a quadrant area of $\frac{175}{31}.$

The full region is four copies, with area $\frac{700}{31}$ square miles. Since $\gcd(700, 31) = 1,$ the answer is $700 + 31 = 731.$

Solution:

Let $\angle A = \alpha,$ and scale so $AP = PQ = QB = BC = 1.$ In triangle $APQ,$ the equal sides $AP = PQ$ give $\angle AQP = \angle A = \alpha,$ so $\angle APQ = 180^\circ - 2\alpha$ and, by the law of sines, $AQ = \frac{\sin 2\alpha}{\sin \alpha} = 2\cos\alpha.$ In triangle $ABC,$ $\angle B = \angle C = 90^\circ - \frac{\alpha}{2},$ so $$AB = \frac{BC \sin C}{\sin A} = \frac{\cos(\alpha/2)}{\sin\alpha} = \frac{1}{2\sin(\alpha/2)}.$$

Since $AQ + QB = AB,$ $$2\cos\alpha + 1 = \frac{1}{2\sin(\alpha/2)} \quad\Longrightarrow\quad 4\sin\tfrac{\alpha}{2}\cos\alpha + 2\sin\tfrac{\alpha}{2} = 1.$$ By the product-to-sum identity, $4\sin\frac{\alpha}{2}\cos\alpha = 2\sin\frac{3\alpha}{2} - 2\sin\frac{\alpha}{2},$ so the equation collapses to $\sin\frac{3\alpha}{2} = \frac{1}{2}.$ Then $\alpha = 20^\circ$ or $\alpha = 100^\circ,$ but the latter makes $AQ = 2\cos\alpha$ negative, so $\alpha = 20^\circ.$

Now $\angle ACB = 80^\circ$ and $\angle APQ = 140^\circ,$ so $r = \frac{80}{140} = \frac{4}{7},$ and $\lfloor 1000r \rfloor = \left\lfloor \frac{4000}{7} \right\rfloor = 571.$

Solution:

Number the original positions $1$ (top) through $2000$ (bottom) and put them in a queue. Each step removes the front position (which receives the next label $1, 2, 3, \ldots$) and sends the new front to the back. So the card labelled $1999$ is the next-to-last card removed, and we must find which original position survives that long.

The first pass removes the odd positions $1, 3, \ldots, 1999$ (labels $1$ through $1000$) and, since it ends by sending $2000$ to the back, the next pass again starts by removing the front of the queue $2, 4, \ldots, 2000.$ Successive passes therefore remove $2, 6, \ldots, 1998$ (the positions $\equiv 2 \bmod 4$), then $4, 12, \ldots, 1996,$ then $8, 24, \ldots, 1992,$ then $16, 48, \ldots, 1968, 2000$ (the $63$ positions $\equiv 16 \bmod 32$). That last pass ran through an odd number ($125$) of cards, so the alternation shifts: the $62$ surviving multiples of $32$ now sit in the queue as $64, 96, \ldots, 1984, 32.$

Continuing the same removal pattern from that queue, the next rounds remove $64, 128, \ldots, 1984;$ then $96, 224, \ldots, 1888, 32;$ then $288, 544, \ldots, 1824;$ then $160, 672, 1184, 1696;$ then $416, 1440;$ and the final two cards removed are $928$ and $1952.$ So label $1999$ goes to the card at original position $928,$ which had $927$ cards above it.