Solution:

Group the terms four at a time. For $k = 1, 2, \ldots, 25,$ the block ending at $(4k)^2$ is $$(4k)^2 + (4k-1)^2 - (4k-2)^2 - (4k-3)^2 = 2(8k - 2) + 2(8k - 4) = 32k - 12,$$ using the difference of squares $a^2 - b^2 = (a + b)(a - b)$ with $a - b = 2$ twice.

Summing over $k = 1$ to $25,$ $$N = 32 \cdot \frac{25 \cdot 26}{2} - 12 \cdot 25 = 10400 - 300 = 10100,$$ so the remainder when $N$ is divided by $1000$ is $100.$

Solution:

Let Rudolph bike at $r$ miles per minute. He rests after each of miles $1$ through $49,$ so his total time is $\frac{50}{r} + 49 \cdot 5 = \frac{50}{r} + 245$ minutes. Jennifer bikes at $\frac{3r}{4}$ miles per minute and rests after each of miles $2, 4, \ldots, 48,$ so her total time is $\frac{50}{3r/4} + 24 \cdot 5 = \frac{200}{3r} + 120$ minutes.

Setting the times equal gives $$\frac{50}{r} + 245 = \frac{200}{3r} + 120, \qquad \text{so} \qquad 125 = \frac{200 - 150}{3r} = \frac{50}{3r},$$ and $r = \frac{2}{15}.$ The common time is $\frac{50}{2/15} + 245 = 375 + 245 = 620$ minutes.

Solution:

Every slice is $1$ cm wide and parallel to a face, so after each cut the remaining cheese is still a rectangular block, with one dimension shortened by $1.$ If the ten slices shorten the three dimensions by $p,$ $q,$ and $r$ with $p + q + r = 10,$ the remaining block measures $(10 - p) \times (13 - q) \times (14 - r),$ and these dimensions sum to $37 - 10 = 27.$

By the AM-GM inequality, a product of positive numbers with fixed sum $27$ is greatest when all three are equal to $9,$ which is achieved by taking $1$ slice from the $10$ cm dimension, $4$ from the $13$ cm dimension, and $5$ from the $14$ cm dimension. The maximum volume is $9^3 = 729$ cubic cm.

Solution:

In base $3,$ $2008 = 2202101_3,$ that is, $$2008 = 2 \cdot 3^6 + 2 \cdot 3^5 + 2 \cdot 3^3 + 3^2 + 3^0.$$ To convert the digits $2$ into coefficients $\pm 1,$ use $2 \cdot 3^k = 3^{k+1} - 3^k.$ The two adjacent digits $2$ collapse neatly: $2 \cdot 3^6 + 2 \cdot 3^5 = (3^7 - 3^6) + (3^6 - 3^5) = 3^7 - 3^5,$ and $2 \cdot 3^3 = 3^4 - 3^3.$

Therefore $$2008 = 3^7 - 3^5 + 3^4 - 3^3 + 3^2 + 3^0,$$ which has distinct exponents and coefficients $\pm 1,$ as required. The sum of the exponents is $7 + 5 + 4 + 3 + 2 + 0 = 21.$

Solution:

Extend legs $\overline{AB}$ and $\overline{DC}$ until they meet at a point $E.$ Since $\angle A + \angle D = 37^\circ + 53^\circ = 90^\circ,$ triangle $EAD$ has a right angle at $E.$ Because $\overline{BC} \parallel \overline{AD},$ triangle $EBC$ is the image of triangle $EAD$ under a homothety centered at $E,$ so the midpoint $M$ of $\overline{BC}$ maps to the midpoint $N$ of $\overline{AD};$ in particular $E,$ $M,$ and $N$ are collinear.

The median to the hypotenuse of a right triangle is half the hypotenuse, so $EN = \frac{2008}{2} = 1004$ and $EM = \frac{1000}{2} = 500.$ Therefore $$MN = EN - EM = 1004 - 500 = 504.$$

Solution:

Dividing the recurrence by $a_{n-1}$ gives $$\frac{a_n}{a_{n-1}} = 1 + \frac{a_{n-1}}{a_{n-2}},$$ so the consecutive-term ratio increases by exactly $1$ each step. For $\{a_n\}$ the first ratio is $\frac{a_1}{a_0} = 1,$ so $\frac{a_n}{a_{n-1}} = n$ and $a_n = n!.$ The same computation applies to $\{b_n\},$ whose first ratio is $\frac{b_1}{b_0} = 3,$ so $\frac{b_n}{b_{n-1}} = n + 2$ and $b_n = \frac{(n+2)!}{2}.$

Therefore $$\frac{b_{32}}{a_{32}} = \frac{34!/2}{32!} = \frac{34 \cdot 33}{2} = 561.$$

Solution:

The cubic has no $x^2$ term, so $r + s + t = 0$ by Vieta's formulas. Hence $r + s = -t,$ $s + t = -r,$ and $t + r = -s,$ and the desired sum is $$(-t)^3 + (-r)^3 + (-s)^3 = -(r^3 + s^3 + t^3).$$

Whenever $r + s + t = 0,$ the identity $r^3 + s^3 + t^3 - 3rst = (r + s + t)(r^2 + s^2 + t^2 - rs - st - tr)$ gives $r^3 + s^3 + t^3 = 3rst.$ By Vieta's formulas, $rst = -\frac{2008}{8} = -251,$ so $r^3 + s^3 + t^3 = -753,$ and the answer is $-(-753) = 753.$

Solution:

By the product-to-sum identity, $2\cos(k^2 a)\sin(ka) = \sin(k^2 a + ka) - \sin(k^2 a - ka) = \sin(k(k+1)a) - \sin((k-1)k a).$ Summing over $k = 1$ to $n,$ the terms telescope, leaving $$\sin(n(n+1)a) = \sin\frac{n(n+1)\pi}{2008}.$$

A sine is an integer only when it is $-1,$ $0,$ or $1,$ that is, when its argument is a multiple of $\frac{\pi}{2}.$ So we need $\frac{n(n+1)}{2008}$ to be a multiple of $\frac{1}{2},$ i.e. $1004 \mid n(n+1),$ where $1004 = 4 \cdot 251$ and $251$ is prime.

Since $n$ and $n + 1$ are coprime, $251$ must divide one of them, so $n \ge 250.$ For $n = 250$ the product $250 \cdot 251$ is not divisible by $4.$ For $n = 251$ the product $251 \cdot 252$ is divisible by $4 \cdot 251 = 1004.$ The smallest such $n$ is $251.$

Solution:

Identify the plane with the complex plane, so a move sends $z$ to $\omega z + 10$ with $\omega = e^{i\pi/4}.$ Starting from $z_0 = 5$ and iterating, $$z_{150} = 5\omega^{150} + 10\left(\omega^{149} + \omega^{148} + \cdots + \omega + 1\right).$$

Since $\omega^8 = 1$ and $150 = 8 \cdot 18 + 6,$ we get $\omega^{150} = \omega^6 = -i.$ In the geometric sum, every block of $8$ consecutive powers adds to $0,$ so the $150$ terms reduce to $1 + \omega + \cdots + \omega^5 = -\omega^6 - \omega^7 = i - \left(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\right).$ Therefore $$z_{150} = -5i + 10\left(-\frac{\sqrt{2}}{2} + \left(1 + \frac{\sqrt{2}}{2}\right)i\right) = -5\sqrt{2} + \left(5 + 5\sqrt{2}\right)i.$$

Thus $|p| + |q| = 5\sqrt{2} + 5 + 5\sqrt{2} = 5 + 10\sqrt{2} \approx 19.14,$ and the greatest integer less than or equal to this is $19.$

Solution:

The squared distance between two points of the array is $a^2 + b^2,$ where $a$ and $b$ are the coordinate differences, each in $\{0, 1, 2, 3\}$ and not both zero. The possible values are $1, 2, 4, 5, 8, 9, 10, 13, 18$ — only $9$ values — so a growing path has at most $10$ points, and a path with $10$ points must use all nine distances in increasing order. Label its points $P_1, \ldots, P_{10}$ so that $P_1 P_2 = 1$ and $P_9 P_{10} = \sqrt{18}.$

Since $\sqrt{18}$ is realized only by opposite corners, there are $4$ ordered choices of $(P_{10}, P_9).$ Next, $P_8 P_9 = \sqrt{13}$ leaves $2$ choices for $P_8,$ the two neighbors of $P_{10},$ symmetric across the main diagonal. From there the distances $\sqrt{10}, 3, \sqrt{8}, \sqrt{5}, 2, \sqrt{2}$ force $P_7, P_6, \ldots, P_2$ uniquely (for $P_7$ the alternative corner choice fails because the point needed next for $P_6$ would coincide with $P_9$ or $P_{10}$). Finally $P_1$ must be at distance $1$ from $P_2,$ and $3$ of its neighbors are unused. One of the resulting paths is shown below.

Hence $m = 10$ and $r = 4 \cdot 2 \cdot 3 = 24,$ so $mr = 240.$

Solution:

Place $B = (0, 0)$ and $C = (56, 0);$ the altitude from $A$ has length $\sqrt{100^2 - 28^2} = 96,$ so $A = (28, 96).$ Then $\sin B = \frac{24}{25},$ $\cos B = \frac{7}{25},$ and $$\tan\frac{B}{2} = \frac{\sin B}{1 + \cos B} = \frac{3}{4} = \tan\frac{C}{2}.$$ A circle of radius $r$ tangent to $\overline{BC}$ and to a slanted side has its center on the bisector from that base vertex, at height $r$ and horizontal distance $\frac{r}{\tan(C/2)} = \frac{4r}{3}$ from the vertex. Thus $P = \left(56 - \frac{64}{3},\, 16\right)$ and $Q = \left(\frac{4q}{3},\, q\right),$ where $q$ is the radius of circle $Q.$

External tangency means $PQ = q + 16:$ $$\left(\frac{104 - 4q}{3}\right)^2 + (16 - q)^2 = (16 + q)^2.$$ Since $(16 + q)^2 - (16 - q)^2 = 64q,$ this becomes $(104 - 4q)^2 = 576q,$ i.e. $(26 - q)^2 = 36q,$ which simplifies to $q^2 - 88q + 676 = 0,$ so $q = 44 \pm 6\sqrt{35}.$

The root $44 + 6\sqrt{35} \approx 79.5$ would make circle $Q$ extend outside the triangle, so $q = 44 - 6\sqrt{35}.$ Here $m = 44,$ $n = 6,$ and $k = 35 = 5 \cdot 7,$ giving $m + nk = 44 + 210 = 254.$

Solution:

Suppose the first pole gets $b$ blue and $g$ green flags, the second the remaining $10 - b$ blue and $9 - g$ green. On a pole with $b$ blue flags, the green flags must occupy distinct gaps among the $b + 1$ gaps around the blues, in $\binom{b+1}{g}$ ways. Temporarily ignoring the requirement that each pole be nonempty, the total is $$\sum_{b=0}^{10} \sum_{g=0}^{9} \binom{b+1}{g}\binom{11-b}{9-g} = \sum_{b=0}^{10} \binom{12}{9} = 11 \cdot 220 = 2420,$$ where the inner sum collapses by Vandermonde's identity, since $(b + 1) + (11 - b) = 12.$

The arrangements that leave a pole empty put all $19$ flags on one pole, in $\binom{11}{9} = 55$ ways for each choice of pole. Hence $$N = 2420 - 2 \cdot 55 = 2310,$$ and the remainder when $N$ is divided by $1000$ is $310.$

Solution:

The hexagon's sides lie at distance $\frac{1}{2}$ from the origin, with one side on the line $\mathrm{Re}\,z = \frac{1}{2},$ so $R$ is the union of the six half-planes obtained by rotating $\mathrm{Re}\,z \gt \frac{1}{2}$ by multiples of $60^\circ.$ If $w = u + vi = \frac{1}{z},$ then $\mathrm{Re}\,z = \mathrm{Re}\,\frac{1}{w} = \frac{u}{u^2 + v^2} \gt \frac{1}{2}$ is equivalent to $u^2 + v^2 \lt 2u,$ i.e. $(u - 1)^2 + v^2 \lt 1.$ So each half-plane maps onto an open unit disk, and $S$ is the union of six unit disks centered at the sixth roots of unity.

Cut the plane into six $60^\circ$ wedges by the rays at angles $30^\circ + 60^\circ k;$ by symmetry, within each wedge $S$ coincides with the disk whose center lies in that wedge. The rays at $\pm 30^\circ$ meet the circle $|w - 1| = 1$ at $\left(\frac{3}{2}, \pm\frac{\sqrt{3}}{2}\right),$ so the piece of $S$ in that wedge consists of two triangles with vertices at $0,$ the center $1,$ and one of these points — each isosceles with two sides $1$ and apex angle $120^\circ,$ area $\frac{\sqrt{3}}{4}$ — together with the $120^\circ$ sector of the disk between them, area $\frac{\pi}{3}.$

Each wedge therefore contributes $\frac{\pi}{3} + \frac{\sqrt{3}}{2},$ and the total area is $$6\left(\frac{\pi}{3} + \frac{\sqrt{3}}{2}\right) = 2\pi + 3\sqrt{3} = 2\pi + \sqrt{27}.$$ Thus $a = 2,$ $b = 27,$ and $a + b = 29.$

Solution:

Draw the rectangle with vertices $A = (0, 0),$ $B = (a, 0),$ $C = (a, b),$ $D = (0, b),$ and let $E = (x, 0)$ on $\overline{AB}$ and $F = (a, b - y)$ on $\overline{BC}.$ Then $DE^2 = b^2 + x^2,$ $DF^2 = a^2 + y^2,$ and $EF^2 = (a - x)^2 + (b - y)^2,$ so the system says exactly that triangle $DEF$ is equilateral, with the constraints keeping $E$ and $F$ on those two sides.

Let $\theta = \angle ADE,$ so $x = b\tan\theta$ and $DE = \frac{b}{\cos\theta}.$ Since $\angle EDF = 60^\circ$ and the corner angle at $D$ is $90^\circ,$ we get $\angle CDF = 30^\circ - \theta,$ so $y = a\tan(30^\circ - \theta)$ and $DF = \frac{a}{\cos(30^\circ - \theta)}.$ Setting $DE = DF$ gives $$\frac{a}{b} = \frac{\cos(30^\circ - \theta)}{\cos\theta} = \cos 30^\circ + \sin 30^\circ \tan\theta,$$ which is increasing in $\theta.$ The requirement $y \ge 0$ forces $\theta \le 30^\circ.$

The maximum is therefore at $\theta = 30^\circ,$ where $\frac{a}{b} = \frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{3}} = \frac{2}{\sqrt{3}},$ attained with $y = 0$ and $x = \frac{b}{\sqrt{3}} \lt a.$ Hence $\rho^2 = \frac{4}{3},$ and $m + n = 4 + 3 = 7.$

Solution:

Condition (i) says $n^2 = (m + 1)^3 - m^3 = 3m^2 + 3m + 1$ for some integer $m.$ Multiplying by $4$ and rearranging, $4n^2 - 1 = 12m^2 + 12m + 3,$ i.e. $$(2n - 1)(2n + 1) = 3(2m + 1)^2.$$ The factors on the left are consecutive odd numbers, hence coprime, so one of them is a perfect square and the other is $3$ times a square. If $2n - 1 = 3k^2,$ then $2n + 1 \equiv 2 \pmod 3$ would be a perfect square, which is impossible. Hence $2n - 1 = k^2$ with $k$ odd.

Writing $k = 2a + 1$ gives $n = 2a^2 + 2a + 1.$ Condition (ii) says $2n + 79 = 4a^2 + 4a + 81 = d^2,$ so $$(d - 2a - 1)(d + 2a + 1) = d^2 - (2a + 1)^2 = 80.$$ The two factors have the same parity, so both are even: the pairs $(2, 40),$ $(4, 20),$ $(8, 10)$ give $2a + 1 = 19,$ $8,$ $1,$ of which the odd values yield $a = 9$ (so $n = 181$) and $a = 0$ (so $n = 1$).

For $n = 181:$ indeed $181^2 = 32761 = 105^3 - 104^3$ (here $2n + 1 = 363 = 3 \cdot 11^2,$ as required), and $2n + 79 = 441 = 21^2.$ So the largest such $n$ is $181.$