Answer: 37

Solution:

Say Gary purchased an amount $x$ and drank an amount $y,$ wasting $x - y.$ In the second scenario he would have purchased $\frac{x}{2}$ and drunk $2y,$ wasting $\frac{x}{2} - 2y.$ The condition is $$\frac{x}{2} - 2y = \frac{2}{9}(x - y).$$

Multiplying by $18$ gives $9x - 36y = 4x - 4y,$ so $5x = 32y$ and $\frac{y}{x} = \frac{5}{32}.$ Since $\gcd(5, 32) = 1,$ the answer is $5 + 32 = 37.$

Answer: 810

Solution:

Let the side length be $s,$ and place $B = (0, 0),$ $C = (s, 0),$ $A = (0, s),$ $D = (s, s).$ Write $E = (a, s)$ and $F = (b, 0).$ Then $BE^2 = a^2 + s^2,$ $FD^2 = (s - b)^2 + s^2,$ and $EF^2 = (a - b)^2 + s^2.$

From $BE = FD$ we get $a = s - b,$ so $a - b = 2a - s.$ Then $EF = BE$ gives $(2a - s)^2 = a^2,$ whose solutions are $a = \frac{s}{3}$ and $a = s$ (the latter collapses $E$ and $F$ onto the corners $D$ and $B,$ making the three segments coincide). So $a = \frac{s}{3}.$

Now $900 = BE^2 = \frac{s^2}{9} + s^2 = \frac{10s^2}{9},$ so the area is $s^2 = \frac{9}{10} \cdot 900 = 810.$

Answer: 143

Solution:

The interior angles of an $18$-gon sum to $180 \cdot 16 = 2880$ degrees. If the smallest angle is $a$ and the common difference is $d,$ then $18a + 153d = 2880,$ i.e. $2a + 17d = 320.$ Since $a$ and $d$ are integers, $17d$ must be even, so $d$ is even, and $d \ge 2$ because the sequence is increasing.

Convexity requires the largest angle $a + 17d = \frac{320 + 17d}{2}$ to be less than $180,$ so $17d \lt 40$ and $d \le 2.$ Thus $d = 2$ and $a = \frac{320 - 34}{2} = 143.$

Answer: 51

Solution:

By the angle bisector theorem, $\frac{BD}{DC} = \frac{AB}{AC} = \frac{20}{11}.$ Use mass points: place mass $11$ at $B$ and mass $20$ at $C,$ so that $D,$ which divides $\overline{BC}$ with $BD : DC = 20 : 11,$ is their balance point and carries mass $31.$ Placing mass $31$ at $A$ makes the balance point of $A$ and $D$ exactly the midpoint $M$ of $\overline{AD}.$

The center of mass of the whole system therefore lies on line $BM,$ and it also lies on the segment from $B$ to the balance point of $A$ and $C.$ That balance point is precisely where line $BM$ crosses $\overline{AC},$ namely $P,$ and it satisfies $31 \cdot PA = 20 \cdot CP.$

Hence $\frac{CP}{PA} = \frac{31}{20},$ which is in lowest terms, and $m + n = 31 + 20 = 51.$

Answer: 542

Solution:

Group the series into blocks of $2011$ consecutive terms. Each term of the second block is $r^{2011}$ times the corresponding term of the first block, so the block sums form a geometric sequence with ratio $r^{2011}.$ The first block sums to $200$ and the second block sums to $380 - 200 = 180,$ so $r^{2011} = \frac{180}{200} = \frac{9}{10}.$

The third block then sums to $180 \cdot \frac{9}{10} = 162,$ so the sum of the first $6033$ terms is $380 + 162 = 542.$

Answer: 80

Solution:

The condition $a + d \gt b + c$ is equivalent to $d - c \gt b - a.$ There are $\binom{10}{4} = 210$ quadruples in all, and the involution $(a, b, c, d) \mapsto (11 - d,\, 11 - c,\, 11 - b,\, 11 - a)$ exchanges the outer gaps $b - a$ and $d - c.$ So the quadruples with $d - c \gt b - a$ and those with $d - c \lt b - a$ are equinumerous, and the answer is $\frac{210 - T}{2},$ where $T$ counts quadruples with $d - c = b - a.$

If $b - a = d - c = k$ and $c - b = j,$ the quadruple is determined by $(a, j, k)$ with $a, j, k \ge 1$ and $a + 2k + j \le 10.$ For $k = 1, 2, 3, 4$ the pairs $(a, j)$ with $a + j \le 8, 6, 4, 2$ number $28,$ $15,$ $6,$ $1,$ so $T = 50.$

Therefore the number of interesting quadruples is $\frac{210 - 50}{2} = 80.$

Answer: 3

Solution:

Break the row into maximal single-color runs. If there are $k$ runs, there are exactly $k - 1$ different-color neighbor pairs. Since runs alternate colors and the five green marbles form at most $5$ runs, there are at most $6$ red runs, hence at most $11$ runs and at most $10$ different-color pairs. With $n$ red marbles there are $n + 4$ neighbor pairs in all, and the requirement says half of them are different-color pairs, so $n + 4 \le 20,$ i.e. $n \le 16.$ Thus $m = 16.$

With $16$ reds and $21$ marbles, the count of different-color pairs must be exactly $10,$ so there are exactly $11$ runs: the colors must alternate as red–green–red–$\cdots$–red with $6$ red runs and $5$ single green marbles between them. The arrangements correspond to compositions of $16$ into $6$ positive parts, of which there are $\binom{15}{5} = 3003.$

Hence $N = 3003,$ and the remainder upon division by $1000$ is $3.$

Answer: 784

Solution:

The zeroes are $z_j = 8\left(\cos\frac{\pi j}{6} + i\sin\frac{\pi j}{6}\right)$ for $j = 1, \ldots, 12,$ and $\operatorname{Re}(iz_j) = -\operatorname{Im}(z_j).$ Since the choices are independent, the maximum real part of the sum is $\sum_j 8\max\left(\cos\frac{\pi j}{6},\, -\sin\frac{\pi j}{6}\right).$

Comparing the two values, $-\sin\frac{\pi j}{6}$ is larger exactly for $j = 5, \ldots, 10.$ The cosines kept, for $j = 1, 2, 3, 4, 11, 12,$ sum to $\frac{\sqrt{3}}{2} + \frac{1}{2} + 0 - \frac{1}{2} + \frac{\sqrt{3}}{2} + 1 = 1 + \sqrt{3},$ and the values $-\sin\frac{\pi j}{6}$ kept, for $j = 5, \ldots, 10,$ sum to $-\frac{1}{2} + 0 + \frac{1}{2} + \frac{\sqrt{3}}{2} + 1 + \frac{\sqrt{3}}{2} = 1 + \sqrt{3}.$

The maximum is $8\left(2 + 2\sqrt{3}\right) = 16 + 16\sqrt{3} = 16 + \sqrt{768},$ so $m + n = 16 + 768 = 784.$

Answer: 559

Solution:

Let $r = x_1x_3x_5 + x_2x_4x_6$ and let $s$ be the cyclic sum in question. Expanding $(x_1 + x_4)(x_2 + x_5)(x_3 + x_6)$ produces eight triple products, which are exactly the six terms of $s$ together with the two terms of $r.$ So $r + s = (x_1 + x_4)(x_2 + x_5)(x_3 + x_6),$ and by AM-GM this is at most $\left(\frac{1}{3}\right)^3 = \frac{1}{27}.$

Therefore $$s \le \frac{1}{27} - r \le \frac{1}{27} - \frac{1}{540} = \frac{20 - 1}{540} = \frac{19}{540}.$$ Equality needs $x_1 + x_4 = x_2 + x_5 = x_3 + x_6 = \frac{1}{3}$ with $r = \frac{1}{540}:$ take $x_1 = x_3 = \frac{3}{10},$ $x_5 = \frac{1}{60},$ $x_2 = \frac{19}{60},$ $x_4 = x_6 = \frac{1}{30}.$ Then $r = \frac{9}{6000} + \frac{19}{54000} = \frac{100}{54000} = \frac{1}{540},$ as required.

So the maximum is $\frac{19}{540},$ and $p + q = 19 + 540 = 559.$

Answer: 57

Solution:

Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{CD}.$ The segment from the center to a chord's midpoint is perpendicular to the chord, so $OM = \sqrt{25^2 - 15^2} = 20$ and $ON = \sqrt{25^2 - 7^2} = 24,$ with $MN = 12.$

Since $P$ lies on both chords, $\angle OMP = \angle ONP = 90^\circ,$ so $M$ and $N$ lie on the circle with diameter $OP.$ In triangle $OMN,$ the law of cosines gives $$\cos \angle MON = \frac{20^2 + 24^2 - 12^2}{2 \cdot 20 \cdot 24} = \frac{832}{960} = \frac{13}{15},$$ so $\sin \angle MON = \frac{2\sqrt{14}}{15}.$ In the circle through $O,$ $M,$ $P,$ $N,$ the extended law of sines says the chord $MN$ equals the diameter $OP$ times $\sin \angle MON,$ so $$OP = \frac{12}{\frac{2\sqrt{14}}{15}} = \frac{90}{\sqrt{14}}, \qquad OP^2 = \frac{8100}{14} = \frac{4050}{7}.$$

Then $m + n = 4050 + 7 = 4057,$ which leaves remainder $57$ upon division by $1000.$

Answer: 73

Solution:

Expanding $D_n$ along the first row gives $10 D_{n-1}$ minus $3$ times a cofactor whose first column is $(3, 0, \ldots, 0);$ expanding that cofactor down its first column leaves $3 D_{n-2}.$ Hence $$D_n = 10 D_{n-1} - 9 D_{n-2},$$ with $D_1 = 10$ and $D_2 = 100 - 9 = 91.$

The characteristic equation $k^2 = 10k - 9$ has roots $9$ and $1,$ and fitting the initial values gives $D_n = \frac{9^{n+1} - 1}{8}.$ Therefore $8D_n + 1 = 9^{n+1},$ and $$\sum_{n=1}^{\infty} \frac{1}{8D_n + 1} = \sum_{n=1}^{\infty} \frac{1}{9^{n+1}} = \frac{1/81}{1 - \frac{1}{9}} = \frac{1}{72}.$$

Thus $\frac{p}{q} = \frac{1}{72}$ and $p + q = 73.$

Answer: 97

Solution:

Only the pattern of countries in the nine chairs matters, and all $\frac{9!}{3!\,3!\,3!} = 1680$ patterns are equally likely. The condition fails for some delegate exactly when both of his neighbors are compatriots, which happens exactly when some country's three delegates occupy three consecutive chairs. Let $A_i$ be the set of patterns in which country $i$'s delegates are consecutive.

There are $9$ triples of consecutive chairs, so $|A_i| = 9\binom{6}{3} = 180,$ choosing which $3$ of the remaining $6$ chairs go to one of the other countries. For two countries, after placing the first block ($9$ ways) the remaining six chairs form an arc containing $4$ triples of consecutive chairs, so $|A_i \cap A_j| = 9 \cdot 4 = 36.$ For all three, the circle must split into three consecutive triples ($3$ ways) assigned to the countries in $3!$ orders: $|A_1 \cap A_2 \cap A_3| = 18.$ By inclusion-exclusion, $$|A_1 \cup A_2 \cup A_3| = 3 \cdot 180 - 3 \cdot 36 + 18 = 450.$$

The probability is $1 - \frac{450}{1680} = 1 - \frac{15}{56} = \frac{41}{56},$ so $m + n = 41 + 56 = 97.$

Answer: 96

Solution:

Place $A = (0, 0),$ $B = (12, 0),$ $C = (12, 12),$ $D = (0, 12),$ and $P = (p, p)$ with $p \gt 6.$ Since $O_1$ is equidistant from $A$ and $B,$ it lies on $x = 6;$ setting $O_1 = (6, k)$ and equating $O_1A^2 = O_1P^2$ gives $k = p - 6,$ so $O_1 = (6,\, p - 6).$ Similarly $O_2,$ equidistant from $C$ and $D,$ is $O_2 = (6,\, p + 6).$

The vectors from $P$ to the centers are $(6 - p,\, -6)$ and $(6 - p,\, 6),$ so $$\cos 120^\circ = \frac{(6 - p)^2 - 36}{(6 - p)^2 + 36} = -\frac{1}{2},$$ which gives $3(p - 6)^2 = 36,$ so $(p - 6)^2 = 12$ and $p = 6 + 2\sqrt{3}.$

Then $AP = p\sqrt{2} = 6\sqrt{2} + 2\sqrt{6} = \sqrt{72} + \sqrt{24},$ so $a + b = 72 + 24 = 96.$

Answer: 440

Solution:

For each $m \in \{2, 3, 5\},$ the condition $a_{n+m} \equiv a_n \pmod{m}$ means the residue of $a_n$ modulo $m$ depends only on $n \bmod m,$ defining a map $\sigma_m$ from residues to residues. Each residue class of positions has $\frac{30}{m}$ members, and so does each residue class of values; if $\sigma_m$ sent two position classes to the same value class, that class's $\frac{30}{m}$ values would have to fill $\frac{60}{m}$ positions, which is impossible. So each $\sigma_m$ is a permutation of the residues modulo $m.$

Conversely, by the Chinese remainder theorem each position $n \in \{1, \ldots, 30\}$ corresponds to a unique triple $(n \bmod 2,\, n \bmod 3,\, n \bmod 5),$ and likewise for values. Any choice of permutations $(\sigma_2, \sigma_3, \sigma_5)$ therefore determines a unique valid permutation of $1, \ldots, 30,$ sending the position triple to the prescribed value triple.

Hence $N = 2! \cdot 3! \cdot 5! = 1440,$ and the remainder upon division by $1000$ is $440.$

Answer: 850

Solution:

For $x \in [n, n + 1)$ the right-hand side is $\sqrt{P(n)},$ which must be an integer, so $P(n) = n^2 - 3n - 9$ must be a perfect square. For $n = 5, 6, \ldots, 14$ the values are $1, 9, 19, 31, 45, 61, 79, 99, 121, 145:$ only $n = 5, 6, 13$ give squares, with $\sqrt{P(n)} = 1, 3, 11$ respectively.

$P$ is increasing on $[5, 15],$ so for $x \in [n, n + 1)$ we automatically have $\sqrt{P(x)} \ge \sqrt{P(n)},$ and $\left\lfloor \sqrt{P(x)} \right\rfloor = \sqrt{P(n)} = m$ holds exactly when $P(x) \lt (m + 1)^2,$ i.e. $x \lt \frac{3 + \sqrt{45 + 4(m+1)^2}}{2}.$ For $m = 1, 3, 11$ the cutoffs are $\frac{3 + \sqrt{61}}{2},$ $\frac{3 + \sqrt{109}}{2},$ $\frac{3 + \sqrt{621}}{2},$ each lying inside the corresponding unit interval, so the successful subintervals have lengths $\frac{\sqrt{61} - 7}{2},$ $\frac{\sqrt{109} - 9}{2},$ $\frac{\sqrt{621} - 23}{2}.$

The interval $[5, 15]$ has length $10,$ so the probability is $$\frac{1}{10} \cdot \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{2} = \frac{\sqrt{61} + \sqrt{109} + \sqrt{621} - 39}{20},$$ giving $a + b + c + d + e = 61 + 109 + 621 + 39 + 20 = 850.$