Answer: 131

Solution:

The conditions say $N = \frac{78}{100}a = \frac{39}{50}a$ and $N = \frac{116}{100}b = \frac{29}{25}b$ for some integers $a$ and $b.$ Since $\gcd(39, 50) = 1,$ the first equation forces $50 \mid a,$ so $N$ is a multiple of $39;$ since $\gcd(29, 25) = 1,$ the second forces $25 \mid b,$ so $N$ is a multiple of $29.$

The least positive integer divisible by both is $N = 39 \cdot 29 = 1131,$ achieved with $a = 1450$ and $b = 975.$ The remainder upon division by $1000$ is $131.$

Answer: 25

Solution:

Assume the school has $100$ students. The Latin students are then $40$ freshmen, $30(0.8) = 24$ sophomores, $20(0.5) = 10$ juniors, and $10(0.2) = 2$ seniors, for a total of $76.$

The probability that a random Latin student is a sophomore is $\frac{24}{76} = \frac{6}{19},$ so $m + n = 6 + 19 = 25.$

Answer: 476

Solution:

Every number is congruent to its digit sum modulo $9,$ so $m = 17n$ must satisfy $17n \equiv 17 \pmod 9,$ that is $8n \equiv 8 \pmod 9,$ which gives $n \equiv 1 \pmod 9.$

Checking the candidates in increasing order: $n = 1, 10, 19$ give $17,$ $170,$ $323,$ with digit sums $8$ each, but $n = 28$ gives $17 \cdot 28 = 476$ with digit sum $4 + 7 + 6 = 17.$ So $m = 476.$

Answer: 18

Solution:

Dropping altitudes from the ends of the short base, each leg is the hypotenuse of a right triangle whose legs are the altitude $\log 16 = 4\log 2$ and half the difference of the bases, $\frac{1}{2}(\log 192 - \log 3) = \frac{1}{2}\log 64 = 3 \log 2.$ By the $3$-$4$-$5$ ratio, each leg has length $5 \log 2.$

The perimeter is $$\log 3 + \log 192 + 2 \cdot 5\log 2 = \log(3 \cdot 192) + \log 2^{10} = \log(2^6 3^2) + \log 2^{10} = \log 2^{16} 3^2,$$ so $p + q = 16 + 2 = 18.$

Answer: 90

Solution:

Each of the $n$ rows contains $n - 1$ horizontally adjacent pairs, so there are $n(n-1)$ horizontal pairs and likewise $n(n-1)$ vertical pairs. Out of $\binom{n^2}{2} = \frac{n^2(n^2-1)}{2}$ equally likely pairs, the probability of adjacency is $$\frac{2n(n-1) \cdot 2}{n^2(n^2 - 1)} = \frac{4}{n(n+1)}.$$

We need $n(n+1) \gt 4 \cdot 2015 = 8060.$ Since $89 \cdot 90 = 8010$ and $90 \cdot 91 = 8190,$ the least such $n$ is $90.$

Answer: 440

Solution:

Dividing by $2,$ the roots $r \le s \le t$ satisfy $r + s + t = a,$ $rs + rt + st = \frac{a^2 - 81}{2},$ and $rst = \frac{c}{2}.$ Therefore $$r^2 + s^2 + t^2 = (r + s + t)^2 - 2(rs + rt + st) = a^2 - (a^2 - 81) = 81.$$

The triples of positive integers whose squares sum to $81$ are $(1, 4, 8),$ $(4, 4, 7),$ and $(3, 6, 6),$ with $a = r + s + t$ equal to $13,$ $15,$ and $15.$ Since knowing $a$ still left Jon two choices, $a = 15,$ and the two polynomials come from $(4, 4, 7)$ and $(3, 6, 6).$

The corresponding values of $c = 2rst$ are $2 \cdot 4 \cdot 4 \cdot 7 = 224$ and $2 \cdot 3 \cdot 6 \cdot 6 = 216,$ with sum $224 + 216 = 440.$

Answer: 161

Solution:

By Heron's formula with $s = 27,$ the area of $ABC$ is $\sqrt{27 \cdot 2 \cdot 10 \cdot 15} = \sqrt{8100} = 90,$ so the altitude from $A$ to $\overline{BC}$ has length $h = \frac{2 \cdot 90}{25} = \frac{36}{5}.$

Since $\overline{PQ} \parallel \overline{BC},$ triangle $APQ$ is similar to $ABC$ with ratio $\frac{w}{25},$ so the distance from $A$ down to line $PQ$ is $\frac{w}{25}h,$ and the rectangle's height is $PS = h - \frac{w}{25}h.$ The area is $$w \cdot h\left(1 - \frac{w}{25}\right) = \frac{36}{5}\,w - \frac{36}{125}\,w^2.$$

Thus $\beta = \frac{36}{125},$ and $m + n = 36 + 125 = 161.$

Answer: 36

Solution:

If $a = 1$ or $b = 1,$ then $\frac{a^3b^3 + 1}{a^3 + b^3} = 1.$ So assume $a, b \ge 2.$ Clearing denominators, the hypothesis says $2ab + 2 \lt 3a + 3b,$ and multiplying by $2$ and rearranging gives $$(2a - 3)(2b - 3) = 4ab - 6a - 6b + 9 \lt 5.$$

For $a, b \ge 2$ both factors are positive odd integers, so up to symmetry the only options are $(a, b) = (2, 2)$ and $(2, 3)$ (both of which do satisfy the original inequality, while $(3, 3)$ gives the product $9$).

The values are $\frac{65}{16}$ for $(2, 2)$ and $\frac{8 \cdot 27 + 1}{8 + 27} = \frac{217}{35} = \frac{31}{5}$ for $(2, 3).$ The larger is $\frac{31}{5},$ so $p + q = 31 + 5 = 36.$

Answer: 384

Solution:

The displaced volume equals the volume of the part of the cube lying below the plane of the barrel's rim. By symmetry that region is a tetrahedron cut from the bottom corner of the cube: three mutually perpendicular edges of equal length $\ell$ along the cube's edges, capped by an equilateral triangle in the rim plane. The equilateral cross-section is inscribed in the rim circle of radius $4,$ so its side length is $4\sqrt{3},$ and therefore $\ell = \frac{4\sqrt{3}}{\sqrt{2}} = 2\sqrt{6}.$

Taking one of the right isosceles faces as the base, the volume is $$\frac{1}{3}\left(\frac{1}{2}\ell^2\right)\ell = \frac{\ell^3}{6} = \frac{(2\sqrt{6})^3}{6} = \frac{48\sqrt{6}}{6} = 8\sqrt{6}.$$

Thus $v = 8\sqrt{6}$ and $v^2 = 64 \cdot 6 = 384.$

Answer: 486

Solution:

Let $S_n$ be the number of quasi-increasing permutations of $1, \ldots, n.$ Insert $n$ into a quasi-increasing permutation of $1, \ldots, n - 1:$ the entry following $n$ must be at least $n - 2,$ so $n$ can go immediately before $n - 1,$ immediately before $n - 2,$ or at the very end — exactly $3$ positions, and each insertion keeps every other adjacent condition intact.

Conversely, deleting $n$ from a quasi-increasing permutation of $1, \ldots, n$ leaves a quasi-increasing permutation of $1, \ldots, n - 1,$ since the entries around the deleted $n$ satisfy $a_{k-1} \le n - 1 \le a_{k+1} + 2$ when $n \ge 3.$ So $S_n = 3S_{n-1}$ for $n \ge 3.$

Since $S_2 = 2,$ we get $S_7 = 2 \cdot 3^5 = 486.$

Answer: 23

Solution:

The central angle over $\overline{BC}$ is $\angle BOC = 2\angle A,$ and $OB = OC$ makes triangle $OBC$ isosceles, so $\angle OBC = 90^\circ - \angle A.$ In triangle $OBQ$ the angle at $O$ is $90^\circ,$ hence $$\angle BQP = 90^\circ - \angle OBQ = 90^\circ - (90^\circ - \angle A) = \angle A.$$

Triangles $BQP$ and $BAC$ share the angle at $B$ and have $\angle BQP = \angle BAC,$ so they are similar, giving $\frac{BP}{BC} = \frac{BQ}{BA}.$ Therefore $$BP = \frac{BQ \cdot BC}{BA} = \frac{4.5 \cdot 4}{5} = \frac{18}{5},$$ and $m + n = 18 + 5 = 23.$

Answer: 548

Solution:

The condition says every maximal run of identical letters has length at most $3.$ Let $s_n$ count the valid strings of length $n$ whose first letter is A; by symmetry the answer is $2s_{10}.$ Removing the first run (of length $1,$ $2,$ or $3$) leaves a valid shorter string beginning with B, so $$s_n = s_{n-1} + s_{n-2} + s_{n-3}.$$

Starting from $s_1 = 1,$ $s_2 = 2,$ $s_3 = 4,$ the sequence runs $7, 13, 24, 44, 81, 149, 274,$ so $s_{10} = 274.$

The number of valid strings is $2 \cdot 274 = 548.$

Answer: 628

Solution:

Multiplying each term by $2\sin\frac{1}{2}$ and using $2\sin k \sin\frac{1}{2} = \cos\left(k - \frac{1}{2}\right) - \cos\left(k + \frac{1}{2}\right),$ the sum telescopes: $$a_n = \frac{\cos\frac{1}{2} - \cos\left(n + \frac{1}{2}\right)}{2\sin\frac{1}{2}}.$$

So $a_n \lt 0$ exactly when $\cos\left(n + \frac{1}{2}\right) \gt \cos\frac{1}{2},$ which happens exactly when $n + \frac{1}{2}$ is within $\frac{1}{2}$ of a multiple of $2\pi:$ $$2\pi m - \tfrac{1}{2} \lt n + \tfrac{1}{2} \lt 2\pi m + \tfrac{1}{2}, \qquad \text{i.e.} \qquad 2\pi m - 1 \lt n \lt 2\pi m.$$ Each interval $(2\pi m - 1,\, 2\pi m)$ has length $1$ and contains exactly one integer, namely $\lfloor 2\pi m \rfloor.$

Hence the $100$th negative term has index $\lfloor 200\pi \rfloor.$ Since $3.14 \lt \pi \lt 3.145,$ we have $628 \lt 200\pi \lt 629,$ so the index is $628.$

Answer: 89

Solution:

The equations factor as $x^4y^4(x + y) = 810$ and $x^3y^3(x^3 + y^3) = 945.$ With $s = x + y$ and $p = xy,$ using $x^3 + y^3 = s(s^2 - 3p),$ they become $p^4 s = 810$ and $p^3 s\,(s^2 - 3p) = 945.$ Dividing, $$\frac{s^2 - 3p}{p} = \frac{945}{810} = \frac{7}{6}, \qquad \text{so} \qquad 6s^2 = 25p.$$

Substituting $p = \frac{6s^2}{25}$ into $p^4 s = 810$ gives $\left(\frac{6}{25}\right)^4 s^9 = 810,$ so $s^9 = 810 \cdot \frac{390625}{1296} = \frac{1953125}{8},$ which means $s^3 = \frac{125}{2}.$ Then $ps = \frac{6s^3}{25} = 15$ and $p^3 = \frac{216 s^6}{25^3} = \frac{216 \cdot 15625/4}{15625} = 54.$

Finally $$2x^3 + (xy)^3 + 2y^3 = 2(s^3 - 3ps) + p^3 = 2\left(\frac{125}{2} - 45\right) + 54 = 35 + 54 = 89.$$

Answer: 129

Solution:

Place line $BC$ on the $x$-axis, so the centers are $P = (0, 1)$ and $Q = (4, 4)$ (their distance is $1 + 4 = 5$), with $B = (0, 0),$ $C = (4, 0),$ and the tangency point $A = P + \frac{1}{5}(Q - P) = \left(\frac{4}{5}, \frac{8}{5}\right).$ The homothety centered at $A$ with ratio $-4$ carries $\mathcal{P}$ to $\mathcal{Q}$ and $D$ to $E,$ so $AE = 4\,AD.$ Since $[DBA] = \frac{1}{2} AD \cdot d(B, \ell)$ and $[ACE] = \frac{1}{2} AE \cdot d(C, \ell),$ the equal-area condition is $d(B, \ell) = 4\,d(C, \ell)$ with $B$ and $C$ on the same side of $\ell.$

Write $\ell$ as $u\left(x - \frac{4}{5}\right) + v\left(y - \frac{8}{5}\right) = 0.$ Its signed values at $B$ and $C$ are $-\frac{4u + 8v}{5}$ and $\frac{16u - 8v}{5},$ so the same-side ratio-$4$ condition reads $-(4u + 8v) = 4(16u - 8v),$ giving $24v = 68u,$ i.e. $v = \frac{17}{6}u.$ Taking $(u, v) = (6, 17),$ the line is $6x + 17y = 32.$

Then $d(B, \ell) = \frac{32}{\sqrt{325}},$ and the center $P = (0, 1)$ is at distance $\frac{15}{\sqrt{325}}$ from $\ell,$ so the chord gives $AD = 2\sqrt{1 - \frac{225}{325}} = \frac{20}{\sqrt{325}}.$ The common area is $$\frac{1}{2} \cdot \frac{20}{\sqrt{325}} \cdot \frac{32}{\sqrt{325}} = \frac{320}{325} = \frac{64}{65},$$ so $m + n = 64 + 65 = 129.$