Answer: 131

Small Hint:

Write $N = \frac{78}{100}a = \frac{116}{100}b$ and reduce both fractions to lowest terms

Big Hint:

In lowest terms $N = \frac{39}{50}a = \frac{29}{25}b,$ which forces $N$ to be a multiple of both $39$ and $29$

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 $a$ to be divisible by $50,$ so $N$ is a multiple of $39;$ since $\gcd(29, 25) = 1,$ the second forces $b$ to be divisible by $25,$ 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

Small Hint:

Suppose the school has $100$ students and count how many students in each class take Latin

Big Hint:

The Latin students number $40 + 24 + 10 + 2,$ and the answer comes from the sophomores’ share of that total

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

Small Hint:

A number is congruent to its digit sum modulo $9,$ so $m = 17n$ requires $17n \equiv 17 \pmod 9$

Big Hint:

That forces $n \equiv 1 \pmod 9;$ check the cases $n = 1,$ $n = 10,$ $n = 19,$ $n = 28,$ $\ldots$ in increasing order

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: $n = 1,$ $n = 10,$ and $n = 19$ give $17,$ $170,$ and $323,$ 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

Small Hint:

Each leg is the hypotenuse of a right triangle with legs $\log 16$ and half of $\log 192 - \log 3$

Big Hint:

Those legs are $4\log 2$ and $3\log 2,$ so each slanted side is $5\log 2$ by a $3$-$4$-$5$ triangle

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 $$\begin{aligned} &\log 3 + \log 192 + 2 \cdot 5\log 2 \\ &= \log(3 \cdot 192) + \log 2^{10} \\ &= \log(2^6 3^2) + \log 2^{10} \\ &= \log 2^{16} 3^2, \end{aligned}$$ so $p + q = 16 + 2 = 18.$

Answer: 90

Small Hint:

Count the adjacent pairs directly: $n(n-1)$ horizontal and $n(n-1)$ vertical

Big Hint:

Dividing by $\binom{n^2}{2}$ simplifies the probability to $\frac{4}{n(n+1)};$ make that less than $\frac{1}{2015}$

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

Small Hint:

If the roots are $r,$ $s,$ and $t,$ Vieta’s formulas give $r + s + t = a$ and $rs + rt + st = \frac{a^2 - 81}{2}$

Big Hint:

Then $r^2 + s^2 + t^2 = 81.$ List the positive integer triples and see which value of $a$ occurs twice.

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 $$\begin{aligned} &r^2 + s^2 + t^2 \\ &= (r + s + t)^2 - 2(rs + rt + st) \\ &= a^2 - (a^2 - 81) \\ &= 81. \end{aligned}$$

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

Small Hint:

Heron’s formula gives area $90$ for $ABC,$ so the altitude from $A$ to $\overline{BC}$ is $\frac{36}{5}$

Big Hint:

Triangle $APQ$ is similar to $ABC$ with ratio $\frac{w}{25},$ so the rectangle’s height is $\frac{36}{5}\left(1 - \frac{w}{25}\right)$

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 $$\begin{aligned} &w \cdot h\left(1 - \frac{w}{25}\right) \\ &= \frac{36}{5}\,w - \frac{36}{125}\,w^2. \end{aligned}$$

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

Answer: 36

Small Hint:

If $a = 1$ or $b = 1$ the second expression equals $1.$ Otherwise clear denominators and rearrange the inequality into a product of two factors.

Big Hint:

For $a, b \ge 2$ the condition becomes $(2a - 3)(2b - 3) \lt 5,$ leaving only $(2, 2)$ and $(2, 3)$ up to symmetry

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 $$\begin{aligned} &(2a - 3)(2b - 3) \\ &= 4ab - 6a - 6b + 9 \lt 5. \end{aligned}$$

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

Small Hint:

The displaced water fills the part of the cube below the plane of the barrel’s rim: a corner tetrahedron with three mutually perpendicular equal edges

Big Hint:

The rim cross-section is an equilateral triangle inscribed in the radius-$4$ circle, so its side is $4\sqrt{3}$ and the perpendicular edges have length $2\sqrt{6}$

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 $$\begin{aligned} &\frac{1}{3}\left(\frac{1}{2}\ell^2\right)\ell = \frac{\ell^3}{6} \\ &= \frac{(2\sqrt{6})^3}{6} \\ &= \frac{48\sqrt{6}}{6} = 8\sqrt{6}. \end{aligned}$$

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

Answer: 486

Small Hint:

Try building a quasi-increasing permutation of $1, \ldots, n$ by inserting $n$ into a quasi-increasing permutation of $1, \ldots, n-1$

Big Hint:

The $n$ can go immediately before $n-1,$ immediately before $n-2,$ or at the very end — always exactly $3$ places, so the count triples with each new $n$

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

Small Hint:

In right triangle $OBQ,$ $\angle OQB = 90^\circ - \angle OBC,$ and isosceles triangle $OBC$ gives $\angle OBC = 90^\circ - \angle A$

Big Hint:

So $\angle BQP = \angle BAC,$ making $\triangle BQP \sim \triangle BAC$ with $\frac{BP}{BC} = \frac{BQ}{BA}$

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 $$\begin{aligned} &\angle BQP = 90^\circ - \angle OBQ \\ &= 90^\circ - (90^\circ - \angle A) \\ &= \angle A. \end{aligned}$$

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 $$\begin{aligned} &BP = \frac{BQ \cdot BC}{BA} \\ &= \frac{4.5 \cdot 4}{5} = \frac{18}{5}, \end{aligned}$$ and $m + n = 18 + 5 = 23.$

Answer: 548

Small Hint:

The condition says every run of identical letters has length at most $3.$ Classify valid strings by the length of their first run.

Big Hint:

If $s_n$ counts valid length-$n$ strings starting with a given letter, then $s_n = s_{n-1} + s_{n-2} + s_{n-3}$

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

Small Hint:

Multiply the sum by $2\sin\frac{1}{2}$ and use a product-to-sum identity to make it telescope

Big Hint:

$a_n \lt 0$ exactly when $2\pi m - 1 \lt n \lt 2\pi m$ for some positive integer $m,$ and each such interval contains exactly one integer

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 $2\pi:$ $$\begin{aligned} &2\pi m - \tfrac{1}{2} \lt n + \tfrac{1}{2} \\ &\lt 2\pi m + \tfrac{1}{2}, \end{aligned}$$ i.e. $$2\pi m - 1 \lt n \lt 2\pi m.$$ 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

Small Hint:

Factor the equations as $x^4y^4(x + y) = 810$ and $x^3y^3(x^3 + y^3) = 945,$ then set $s = x + y$ and $p = xy$

Big Hint:

Dividing the equations gives $\frac{s^2 - 3p}{p} = \frac{7}{6},$ so $6s^2 = 25p;$ substitute back to find $s^3$ and $p^3$

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, $$\begin{aligned} &\frac{s^2 - 3p}{p} = \frac{945}{810} \\ &= \frac{7}{6}, \end{aligned}$$ so $$6s^2 = 25p.$$

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 $p^3 = \frac{216 s^6}{25^3} = \frac{\frac{216 \cdot 15625}{4}}{15625} = 54.$

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