Solution:

An integer is a multiple of $4$ exactly when its last two digits form a multiple of $4,$ so we need $4 \mid 10b + c$ and $4 \mid 10b + a.$ In particular $a$ and $c$ are even, and subtracting the two conditions shows $4 \mid a - c.$ The even nonzero digits split by remainder mod $4$ into $\{2, 6\}$ and $\{4, 8\},$ so $a$ and $c$ must both come from the same one of these sets: $4$ ordered pairs $(a, c)$ from each.

If $c \equiv 2 \pmod 4,$ then $10b + c \equiv 2b + 2 \pmod 4$ requires $b$ odd ($5$ choices), and the condition on $10b + a$ holds automatically since $a \equiv c \pmod 4.$ If $c \equiv 0 \pmod 4,$ then $b$ must be even ($5$ choices).

The count is $4 \cdot 5 + 4 \cdot 5 = 40.$

Solution:

If the sequence has $n$ terms, the amounts added are the first $n$ odd numbers, whose sum is $n^2.$ Thus $n^2 = 836 - 715 = 121,$ so $n = 11.$

The average of the $11$ terms is $\frac{715}{11} = 65,$ which equals the middle (sixth) term of the arithmetic sequence. The first and last terms also average to $65,$ so they add to $130.$

The requested sum is $65 + 130 = 195.$

Solution:

Choose the one person served correctly ($9$ ways); by symmetry say they ordered beef. The remaining meals — $2$ beef, $3$ chicken, and $3$ fish — must go to the other $8$ people ($2$ beef, $3$ chicken, and $3$ fish orderers) with nobody matched. Track where the $2$ leftover beef meals go: to chicken or fish orderers.

If both go to the same group, say to two of the three chicken orderers ($3 + 3 = 6$ ways counting both groups), then the third chicken orderer must receive fish, the three fish orderers must take the three chicken meals, and the two beef orderers take the remaining fish: everything is forced. If one goes to a chicken orderer and one to a fish orderer ($3 \cdot 3 = 9$ ways), the other two chicken orderers must take fish and the other two fish orderers must take chicken, leaving one chicken and one fish meal to split between the two beef orderers ($2$ ways).

The total is $9\,(6 + 9 \cdot 2) = 216.$

Solution:

Walking a mile takes Sparky $10$ minutes, Butch $15,$ and Sundance $24.$ The horse advances along the same route as the men and is ridden over each mile by exactly one of them, so if Butch walks $x$ of the $n$ miles and rides the other $n - x,$ then Sundance rides those $x$ miles and walks the remaining $n - x.$

When they meet at a milepost they have been traveling for the same amount of time, so $$15x + 10(n - x) = 10x + 24(n - x),$$ which simplifies to $19x = 14n.$ Since the handoffs happen at mileposts, $x$ and $n$ are integers, and the smallest positive solution is $x = 14,$ $n = 19.$

Then $t = 15 \cdot 14 + 10 \cdot 5 = 260$ minutes, so $n + t = 19 + 260 = 279.$

Solution:

We must count pairs of elements of $B$ differing by $1,$ say $m$ and $m + 1.$ Adding $1$ to a binary number turns its trailing block $011\cdots1$ (a zero followed by $k$ ones) into $100\cdots0,$ changing the number of ones by $1 - k.$ Both numbers have exactly eight ones precisely when $k = 1:$ $m$ ends in $01,$ $m + 1$ ends in $10,$ and the two numbers agree everywhere else.

The shared first eleven digits then consist of the remaining seven ones and four zeros, and since leading zeros are allowed, every arrangement gives a valid pair: $\binom{11}{4} = 330.$ Each pair produces the answer $1$ exactly once, so the count is $330.$

Solution:

Substituting, $z = w^{11} = (z^{13})^{11} = z^{143},$ and $z \ne 0,$ so $z^{142} = 1.$ Conversely, any $142$nd root of unity $z$ works with $w = z^{13},$ since then $w^{11} = z^{143} = z.$

Hence $z = \cos\frac{2k\pi}{142} + i\sin\frac{2k\pi}{142}$ for some integer $k,$ and the imaginary part of $z$ is $\sin\frac{k\pi}{71}.$ Since $71$ is prime, for every $k$ with $1 \le k \le 70$ the fraction $\frac{k}{71}$ is already in lowest terms, matching the required form $\sin\left(\frac{m\pi}{n}\right)$ with $m \lt n.$ Thus $n = 71.$

Solution:

Group the sixteen circles into rings: the center, the inner ring of five, the middle ring of five, and the outer ring of five, holding $p,$ $q,$ $r,$ and $s$ coins in total, respectively. The center has $5$ neighbors (the inner ring); each inner student has $3$ (the center and two middle students); each middle student has $4$ (two inner and two outer); each outer student has $4$ (two middle and two outer). A student with $k$ neighbors sends $\frac{1}{k}$ of their coins to each neighbor.

Summing the trades over each ring (for example, the outer ring receives a quarter of each middle student's coins twice over, which totals $\frac{r}{2}$) gives $$p = \frac{q}{3}, \qquad q = p + \frac{r}{2}, \qquad r = \frac{2q}{3} + \frac{s}{2}, \qquad s = \frac{r}{2} + \frac{s}{2}.$$

The first equation gives $q = 3p,$ the second then gives $r = 2(q - p) = 4p,$ and the last gives $s = r = 4p.$ The total is $p + 3p + 4p + 4p = 12p = 3360,$ so the center student had $p = 280$ coins.

Solution:

Extend the cutting plane. In the bottom face, line $DM$ meets line $CB$ extended beyond $B$ at a point $K;$ since $MB \parallel DC$ and $MB = \frac{1}{2}DC,$ segment $MB$ is a midline of triangle $KDC,$ so $B$ is the midpoint of $\overline{CK}$ and $CK = 2.$ The plane also cuts edge $BF$ at a point $P,$ and the piece of the cube cut off past the plane is the pyramid $KDCN$ with the small pyramid $KMBP$ sliced away.

Pyramid $KDCN$ has base $DCN,$ a right triangle with legs $DC = 1$ and $CN = \frac{1}{2},$ and its apex $K$ is at distance $CK = 2$ from the plane of that base, so its volume is $\frac{1}{3} \cdot \frac{1}{4} \cdot 2 = \frac{1}{6}.$ Pyramid $KMBP$ is similar to $KDCN$ with ratio $\frac{KB}{KC} = \frac{1}{2},$ so its volume is $\frac{1}{8} \cdot \frac{1}{6} = \frac{1}{48}.$

The smaller piece therefore has volume $\frac{1}{6} - \frac{1}{48} = \frac{7}{48},$ and the larger piece has volume $1 - \frac{7}{48} = \frac{41}{48},$ giving $p + q = 41 + 48 = 89.$

Solution:

Write $x = 2^a,$ $y = 2^b,$ $z = 2^c.$ Then $\log_x(2y) = \frac{b + 1}{a},$ $\log_{2x}(4z) = \frac{c + 2}{a + 1},$ and $\log_{2x^4}(8yz) = \frac{b + c + 3}{4a + 1},$ so the condition is $$\frac{2(b + 1)}{a} = \frac{2(c + 2)}{a + 1} = \frac{b + c + 3}{4a + 1} \ne 0.$$

From the first two, $\frac{b + 1}{a} = \frac{c + 2}{a + 1},$ and equal ratios also equal their mediant $\frac{b + c + 3}{2a + 1}.$ Comparing with the third expression gives $\frac{2(b + c + 3)}{2a + 1} = \frac{b + c + 3}{4a + 1}.$ The common value is nonzero, so $b + c + 3 \ne 0,$ and thus $2(4a + 1) = 2a + 1,$ giving $a = -\frac{1}{6}.$ Then $\frac{b + 1}{-1/6} = \frac{c + 2}{5/6}$ yields $c + 2 = -5(b + 1),$ that is, $5b + c = -7.$

Therefore $xy^5z = 2^{a + 5b + c} = 2^{-1/6 - 7} = \frac{1}{2^{43/6}},$ so $p + q = 43 + 6 = 49.$

Solution:

A square $n^2$ ends in $256$ exactly when $1000 \mid n^2 - 256 = (n - 16)(n + 16).$ Modulo $8:$ $n^2 \equiv 0 \pmod 8$ forces $4 \mid n.$ Modulo $125:$ the factors $n \pm 16$ differ by $32,$ so $5$ divides at most one of them, and hence $125$ must divide a single factor: $n \equiv \pm 16 \pmod{125}.$ Because $16$ is a multiple of $4,$ the two conditions combine to $n \equiv \pm 16 \pmod{500}.$

So $\mathcal{S}$ consists of the numbers $(500m \pm 16)^2,$ whose square roots in increasing order are $16,$ $484,$ $516,$ $984,$ $1016, \ldots.$ The tenth smallest element of $\mathcal{S}$ is $(500 \cdot 5 - 16)^2 = 2484^2.$

The corresponding element of $\mathcal{T}$ is $\frac{2484^2 - 256}{1000} = \frac{2468 \cdot 2500}{1000} = 6170,$ whose remainder upon division by $1000$ is $170.$

Solution:

Each jump changes $x + y$ by $+9$ or $-15$ and changes $x - y$ by $\pm 5.$ Starting from $(0, 0),$ every reachable point therefore has $x + y = 3j$ and $x - y = 5k$ for integers $j$ and $k;$ moreover $x = \frac{3j + 5k}{2}$ must be an integer, so $j$ and $k$ have the same parity. Since $|x| + |y| = \max(|x + y|,\, |x - y|),$ the condition $|x| + |y| \le 100$ becomes $|j| \le 33$ and $|k| \le 20.$

Conversely, every such point is reachable: a single jump moves between neighboring lines $x - y = 5k$ (changing $j$ by $3$ or $-5,$ which flips its parity), and two-jump combinations translate by $(9, 9)$ or $(-15, -15),$ which combine — two of the former plus one of the latter — into the shift $(3, 3),$ moving $j$ by $2$ along a fixed line. Together these reach every pair $(j, k)$ of equal parity.

Counting: even $j$ ($33$ values) pairs with even $k$ ($21$ values), and odd $j$ ($34$ values) with odd $k$ ($20$ values), so $M = 33 \cdot 21 + 34 \cdot 20 = 1373.$ The remainder is $373.$

Solution:

The trisectors make $\angle ACD = \angle DCE = \angle ECB = 30^\circ.$ In triangle $DCB,$ ray $CE$ bisects the $60^\circ$ angle $DCB,$ so the angle bisector theorem gives $\frac{CD}{CB} = \frac{DE}{EB} = \frac{8}{15}.$ Scale the triangle so that $CD = 8$ and $CB = 15.$

By the Law of Cosines in triangle $DCB,$ $$BD^2 = 8^2 + 15^2 - 2 \cdot 8 \cdot 15 \cos 60^\circ = 169,$$ so $BD = 13.$ Applying the Law of Cosines again in the same triangle, $8^2 = 13^2 + 15^2 - 2 \cdot 13 \cdot 15 \cos B,$ which gives $\cos B = \frac{11}{13}.$

Then $\sin B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13},$ so $\tan B = \frac{4\sqrt{3}}{11}$ and $m + n + p = 4 + 11 + 3 = 18.$

Solution:

Let $O$ be the common center and label the triangle $ABC$ with $OA = 3,$ $OB = 4,$ $OC = 5.$ Rotate the plane by $60^\circ$ about $A$ so that $B$ maps to $C,$ and let $P$ be the image of $O.$ Then triangle $AOP$ is equilateral, so $OP = OA = 3,$ and $PC,$ the image of $OB,$ has length $4.$

Triangle $OPC$ has sides $3,$ $4,$ $5,$ so $\angle OPC = 90^\circ.$ In the configuration giving the largest triangle, $O$ lies inside $ABC$ and $\angle APC = \angle APO + \angle OPC = 60^\circ + 90^\circ = 150^\circ,$ so by the Law of Cosines $$s^2 = AC^2 = 3^2 + 4^2 - 2 \cdot 3 \cdot 4\cos 150^\circ = 25 + 12\sqrt{3}.$$ (If $O$ lies outside the triangle, the triangle fits in a half-disk of radius $5,$ so its altitude is at most $5$ and $s^2 \le \frac{100}{3},$ which is smaller.)

The area is $\frac{\sqrt{3}}{4}\,s^2 = \frac{\sqrt{3}}{4}\left(25 + 12\sqrt{3}\right) = 9 + \frac{25}{4}\sqrt{3},$ so $a + b + c + d = 9 + 25 + 4 + 3 = 41.$

Solution:

Since $P(z)$ has no $z^2$ term, $a + b + c = 0.$ Say the right angle is at $b;$ then the hypotenuse joins $a$ and $c,$ so $h = |a - c|,$ and $b = -(a + c).$ The midpoint $d = \frac{a + c}{2}$ of the hypotenuse is the circumcenter of the right triangle, so $|b - d| = \frac{h}{2}.$ Since $b - d = -\frac{3}{2}(a + c),$ this gives $|a - c| = 3\,|a + c|.$

By the parallelogram law, $|a|^2 + |c|^2 = \frac{|a - c|^2 + |a + c|^2}{2},$ and $|b|^2 = |a + c|^2,$ so $$250 = \frac{9\,|a + c|^2 + |a + c|^2}{2} + |a + c|^2 = 6\,|a + c|^2.$$

Therefore $h^2 = |a - c|^2 = 9\,|a + c|^2 = \frac{9 \cdot 250}{6} = 375.$

Solution:

Condition (1) requires the seats $a, 2a, \ldots, na$ to be pairwise distinct modulo $n,$ which happens if and only if $\gcd(a, n) = 1.$ For condition (2), the two mathematicians from seats $i$ and $j$ have gap counts before the break determined by $\pm(i - j) \bmod n$ and after the break by $\pm a(i - j) \bmod n,$ so the requirement is $a(i - j) \not\equiv \pm(i - j) \pmod{n}$ for all $i \ne j.$ Equivalently, $(a - 1)(i - j) \not\equiv 0$ and $(a + 1)(i - j) \not\equiv 0 \pmod{n}$ for every nonzero residue $i - j,$ which holds exactly when $a - 1$ and $a + 1$ are also relatively prime to $n.$

So $n$ is possible if and only if some $a$ satisfies $\gcd\big((a - 1)\,a\,(a + 1),\, n\big) = 1.$ Any three consecutive integers include a multiple of $2$ and a multiple of $3,$ so no $a$ works when $\gcd(n, 6) \gt 1.$ Conversely, if $\gcd(n, 6) = 1,$ then $a = 3$ works, since $2 \cdot 3 \cdot 4 = 24$ has only the prime factors $2$ and $3.$

The valid $n$ with $1 \lt n \lt 1000$ are those congruent to $\pm 1 \pmod 6,$ namely $6k \pm 1$ for $1 \le k \le 166,$ and there are $2 \cdot 166 = 332$ of them.