Small Hint:

Compute $B - A$ by pairing each product in $B$ with the product in $A$ that shares its even factor

Big Hint:

Each pair contributes $(2k+1)(2k)$ $- (2k-1)(2k) = 4k,$ and the leftover terms are $1$ and $-39$

Solution:

Subtract term by term: $$\begin{aligned} &B - A = (1 - 39) \\ &\quad {}+ (2 \times 3 - 1 \times 2) \\ &\quad {}+ (4 \times 5 - 3 \times 4) + \cdots \\ &\quad {}+ (38 \times 39 - 37 \times 38). \end{aligned}$$ Each parenthesized difference has the form $(2k+1)(2k)$ $- (2k-1)(2k) = 4k$ for $k = 1, 2, \ldots, 19.$

Therefore $$\begin{aligned} B - A &= -38 \\ &\quad {}+ 4(1 + 2 + \cdots + 19) \\ &= -38 + 4 \cdot 190 = 722. \end{aligned}$$

Small Hint:

All $\binom{9}{3}$ sleeper triples are equally likely; count the favorable ones by which country supplies the pair

Big Hint:

Exactly two from a country of size $s$ means $\binom{s}{2}$ choices for the pair times $9 - s$ choices for the third sleeper

Solution:

There are $\binom{9}{3} = 84$ equally likely sets of three sleepers. Exactly two sleepers come from the same country when one country supplies exactly two of them and the third sleeper comes from a different country: $\binom{4}{2}(2 + 3) = 30$ ways with the pair from the United States, $\binom{3}{2}(2 + 4) = 18$ with the pair from Canada, and $\binom{2}{2}(3 + 4) = 7$ with the pair from Mexico.

The probability is $\frac{30 + 18 + 7}{84} = \frac{55}{84},$ already in lowest terms, so $m + n = 55 + 84 = 139.$

Small Hint:

Write $16p + 1 = n^3$ and factor $n^3 - 1$

Big Hint:

$n^2 + n + 1$ is always odd, so the factor $16$ must come entirely from $n - 1;$ primality then forces $n - 1 = 16$

Solution:

Write $16p + 1 = n^3,$ so $16p = n^3 - 1$ $= (n - 1)(n^2 + n + 1).$ Since $16p + 1$ is odd, $n$ is odd, and $n^2 + n + 1$ is odd as well. Therefore all four factors of $2$ must divide $n - 1:$ write $n - 1 = 16k,$ which gives $p = k(n^2 + n + 1).$ For $p$ to be prime we need $k = 1,$ so $n = 17.$

Then $p = 17^2 + 17 + 1 = 307,$ which is indeed prime, and $16 \cdot 307 + 1 = 4913 = 17^3.$

Small Hint:

Put $B$ at the origin with $A = (-16, 0)$ and $C = (4, 0),$ and write $D$ and $E$ using equilateral-triangle altitudes

Big Hint:

Compute the midpoints $M$ and $N$ and the three distances $BM,$ $MN,$ $NB;$ triangle $BMN$ turns out equilateral

Solution:

Place $B = (0, 0),$ $A = (-16, 0),$ and $C = (4, 0).$ Each equilateral triangle has its apex above the midpoint of its base at height $\frac{\sqrt{3}}{2}$ times the side, so $D = (-8, 8\sqrt{3})$ and $E = (2, 2\sqrt{3}).$ The midpoints are $M = (-7, \sqrt{3})$ and $N = (-2, 4\sqrt{3}).$

Now $BM^2 = 49 + 3 = 52,$ $BN^2 = 4 + 48 = 52,$ and $MN^2 = 25 + 27 = 52,$ so $\triangle BMN$ is equilateral with side $\sqrt{52}.$ Its area is $x = \frac{\sqrt{3}}{4} \cdot 52 = 13\sqrt{3},$ so $x^2 = 169 \cdot 3 = 507.$

Small Hint:

Splitting the $10$ socks into a sequence of daily pairs is symmetric: any reordering of the days is equally likely

Big Hint:

So compute instead the probability of a match on Monday and then no match on Tuesday or Wednesday, tracking how many complete pairs remain each day

Solution:

Imagine dealing all ten socks out two per day for five days; every assignment of unordered pairs to days is equally likely, and permuting the days does not change this distribution. Swapping Monday and Wednesday therefore shows that the desired probability (mismatch, mismatch, match) equals the probability of a match on Monday followed by mismatches on Tuesday and Wednesday.

That pattern is easy to compute in order. Monday matches with probability $\frac{1}{9}$ (the second sock must be the first sock’s mate). The remaining $8$ socks then form $4$ complete pairs, so Tuesday mismatches with probability $1 - \frac{4}{\binom{8}{2}} = \frac{6}{7}.$ Tuesday’s mismatch breaks two pairs, leaving $2$ complete pairs among the $6$ remaining socks, so Wednesday mismatches with probability $1 - \frac{2}{\binom{6}{2}} = \frac{13}{15}.$

The probability is $\frac{1}{9} \cdot \frac{6}{7} \cdot \frac{13}{15} = \frac{26}{315},$ so $m + n = 26 + 315 = 341.$

Small Hint:

Let $\alpha$ be the common central angle at $C$ in the second circle; then $\angle ACE = 5\alpha$ is simultaneously an inscribed angle of the first circle

Big Hint:

Express every arc of both circles in terms of $\alpha;$ the condition $\angle ABD - \angle AHG = 12^\circ$ pins down $\alpha$

Solution:

Let $\alpha = \angle ECF$ $= \angle FCG$ $= \angle GCH$ $= \angle HCI$ $= \angle ICA,$ the common central angle of the second circle, so $\angle ACE = 5\alpha.$ Since $C$ also lies on the first circle, $\angle ACE$ is an inscribed angle there, so the arc $AE$ not containing $C$ measures $10\alpha,$ and each of the four equal arcs $AB,$ $BC,$ $CD,$ $DE$ measures $\frac{360^\circ - 10\alpha}{4} = 90^\circ - \frac{5\alpha}{2}.$

Angle $ABD$ subtends the arc $AD$ not containing $B,$ which is $360^\circ - 3\left(90^\circ - \frac{5\alpha}{2}\right),$ so $\angle ABD = 45^\circ + \frac{15\alpha}{4}.$ Angle $AHG$ subtends the second circle’s arc $AG$ not containing $H,$ which is $360^\circ - 3\alpha,$ so $\angle AHG = 180^\circ - \frac{3\alpha}{2}.$ The given condition reads $$\begin{aligned} &\left(45^\circ + \frac{15\alpha}{4}\right) - \left(180^\circ - \frac{3\alpha}{2}\right) \\ &= \frac{21\alpha}{4} - 135^\circ = 12^\circ, \end{aligned}$$ so $\alpha = 28^\circ.$

Finally, $\angle BAE$ subtends the first circle’s arc $BCDE = 3\left(90^\circ - \frac{5\alpha}{2}\right) = 60^\circ,$ giving $\angle BAE = 30^\circ,$ and $\angle EAG$ subtends the second circle’s arc $EFG = 2\alpha,$ giving $\angle EAG = 28^\circ.$ Hence $\angle BAG = \angle BAE + \angle EAG$ $= 30^\circ + 28^\circ = 58^\circ.$

Small Hint:

Every right triangle cut off by the tilted squares ($HBJ,$ $JFC,$ $MAN,$ $NKH$) is similar to $\triangle CDE,$ with legs in ratio $1 : 2$

Big Hint:

Write $BC = BJ + JC$ to get $FGHJ$’s side from $AE;$ then $AH = AN + NH$ gives $KLMN$’s side in exactly the same way

Solution:

Let $AE = s,$ so the big square has side $2s$ and $CE = s\sqrt{5}.$ The right triangles $CDE,$ $JFC,$ $HBJ,$ $NKH,$ and $MAN$ are all similar, with legs in ratio $1 : 2.$ Let $x$ be the side of $FGHJ.$ In $\triangle HBJ$ the hypotenuse is $HJ = x,$ so $BJ = \frac{x}{\sqrt{5}}$ and $HB = \frac{2x}{\sqrt{5}};$ in $\triangle JFC$ the longer leg is $JF = x,$ so the hypotenuse is $JC = \frac{x\sqrt{5}}{2}.$ Then $$\begin{aligned} 2s = BC &= BJ + JC \\ &= x\left(\frac{1}{\sqrt{5}} + \frac{\sqrt{5}}{2}\right) \\ &= \frac{7x}{2\sqrt{5}}, \end{aligned}$$ so $x = \frac{4\sqrt{5}\,s}{7}.$

Next, $AH = 2s - HB = 2s - \frac{8s}{7}$ $= \frac{6s}{7}.$ The identical decomposition along $\overline{AB}$ for the square $KLMN$ of side $y$ gives $\frac{6s}{7} = AH = AN + NH$ $= y\left(\frac{1}{\sqrt{5}} + \frac{\sqrt{5}}{2}\right).$ Dividing the two equations, $\frac{x}{y} = \frac{2s}{\frac{6s}{7}} = \frac{7}{3}.$

The areas are therefore in ratio $\left(\frac{7}{3}\right)^2 = \frac{49}{9},$ so the area of $FGHJ$ is $99 \cdot \frac{49}{9} = 539.$

Small Hint:

Each carry in an addition lowers the digit sum by $9,$ so $s(n + 864) = s(n) + 18 - 9c$ where $c$ counts the carries

Big Hint:

Exactly two carries must occur, and the hundreds place always carries, so try each choice of the second carrying place and minimize the hundreds digit

Solution:

Each carry in an addition replaces $10$ in one place by $1$ in the next, lowering the digit sum by $9.$ Hence $s(n + 864) = s(n) + s(864) - 9c$ $= 20 + 18 - 9c,$ where $c$ is the number of carries, and $s(n + 864) = 20$ forces $c = 2.$ For a three-digit candidate $n$ with digits $t,$ $u,$ $v$ summing to $20:$ since $u + v \le 18,$ we have $t \ge 2,$ so the hundreds place always carries ($t + 8 \ge 10$), and exactly one of the units and tens places carries.

If the units carry and the tens do not, the tens computation $u + 6 + 1$ must stay below $10,$ so $u \le 2;$ then $t = 20 - u - v$ $\ge 20 - 2 - 9 = 9,$ forcing $n = 929.$ If the tens carry and the units do not, then $v + 4 \le 9$ gives $v \le 5,$ so $t = 20 - u - v$ $\ge 20 - 9 - 5 = 6,$ and $t = 6,$ $u = 9,$ $v = 5$ works: $n = 695.$

Indeed $s(695) = 20$ and $695 + 864 = 1559$ with $s(1559) = 20,$ so the smallest such $n$ is $695.$

Small Hint:

If two consecutive terms are ever equal, the sequence hits $0$ two steps later; if they differ by $1,$ it hits $0$ within four steps

Big Hint:

Count triples whose consecutive entries are equal or differ by $1,$ correcting for overlaps, then check what happens when a difference of $2$ meets a $1$

Solution:

If $a_{k-1} = a_k$ then $a_{k+2} = a_{k+1}|a_k - a_{k-1}| = 0,$ and if $|a_k - a_{k-1}| = 1$ then $a_{k+2} = a_{k+1},$ so $a_{k+4} = 0.$ Hence every triple of one of the forms $(j,j,k),$ $(j,k,k),$ $(j,j\pm1,k),$ $(j,k,k\pm1)$ produces a $0.$ These forms contain $100 + 100 + 4 \cdot 90 = 560$ triples, but triples fitting two forms are counted twice: the $10$ of the form $(j,j,j),$ the $9$ in each of the six families $(j,j,j\pm1),$ $(j,j\pm1,j),$ $(j,j\pm1,j\pm1)$ (matching signs), and the $8$ in each of $(j,j+1,j+2)$ and $(j,j-1,j-2).$ That leaves $560 - 10 - 54 - 16 = 480$ triples.

A few other triples also work: if $(a_1, a_2, a_3) = (j, j\pm2, 1),$ then $a_4 = 2$ and $|a_4 - a_3| = 1,$ so $a_8 = 0.$ These $16$ triples include $(3,1,1)$ and $(4,2,1),$ which were already counted, so they add $14$ new ones, for $480 + 14 = 494.$

No other triple reaches $0:$ if both consecutive differences are at least $2$ and $a_3 \ge 2,$ then $a_4 = a_3|a_2 - a_1| \ge 2a_3 \gt a_3$ and $|a_4 - a_3| \ge a_3 \ge 2,$ so inductively the terms grow forever and no factor ever vanishes. If instead $a_3 = 1$ with $|a_2 - a_1| \ge 3,$ then $a_4 \ge 3$ and $|a_4 - a_3| \ge 2,$ and the same growth takes over. The count is $494.$

Small Hint:

Each of $f(x) - 12$ and $f(x) + 12$ is a cubic, so each has exactly three of $1,$ $2,$ $3,$ $5,$ $6,$ $7$ as roots

Big Hint:

The two cubics differ by a constant, so the two root triples have equal sums and equal pairwise-product sums; only one split works

Solution:

Each of $f(x) - 12$ and $f(x) + 12$ is a cubic, so each vanishes at exactly three of $1,$ $2,$ $3,$ $5,$ $6,$ $7.$ Writing them as $c(x - r_1)(x - r_2)(x - r_3)$ and $c(x - s_1)(x - s_2)(x - s_3),$ the two cubics differ by the constant $24,$ so their $x^2$ and $x$ coefficients agree: the root triples have equal sums and equal sums of pairwise products. The only partition of $\{1,2,3,5,6,7\}$ into two triples of equal sum is $\{2,3,7\}$ and $\{1,5,6\}$ (each summing to $12$), and indeed both have pairwise-product sum $41.$

Replacing $f$ by $-f$ if necessary (which does not change $|f(0)|$), we have $f(x) = c(x-2)(x-3)(x-7)$ $+ 12$ $= c(x-1)(x-5)(x-6) - 12.$ Setting $x = 0$ gives $-42c + 12 = -30c - 12,$ so $c = 2$ and $f(0) = -42 \cdot 2 + 12 = -72.$ Thus $|f(0)| = 72.$

Small Hint:

Let $M$ be the midpoint of $BC:$ right triangles at $M$ give $\cos\angle ABM = \frac{BM}{AB}$ and $\cos\angle IBM = \frac{BM}{8}$

Big Hint:

Since $\angle IBM$ is half of $\angle ABM,$ the double-angle formula relates $AB$ to $BC;$ integrality and $BM \lt 8$ leave only a few cases to test

Solution:

Let $M$ be the midpoint of $\overline{BC};$ by symmetry $A,$ $I,$ and $M$ are collinear with $AM \perp BC.$ With $a = AB$ and $b = BM,$ right triangles $ABM$ and $IBM$ give $\cos\angle ABM = \frac{b}{a}$ and $\cos\angle IBM = \frac{b}{8}.$ Since $BI$ bisects $\angle ABM,$ the double-angle formula yields $$\frac{b}{a} = 2\left(\frac{b}{8}\right)^2 - 1,$$ so $$a = \frac{32b}{b^2 - 32}.$$

Writing $c = BC = 2b,$ this becomes $a = \frac{64c}{c^2 - 128}.$ We need $c^2 \gt 128,$ so $c \ge 12,$ while $\cos\angle IBM = \frac{b}{8} \lt 1$ forces $c \lt 16.$ Testing $c = 12, 13, 14, 15,$ only $c = 12$ makes $a$ an integer, namely $a = \frac{768}{16} = 48.$

The triangle with sides $48,$ $48,$ $12$ satisfies all the conditions, and its perimeter is $48 + 48 + 12 = 108.$

Small Hint:

Exactly $\binom{2015-j}{999}$ of the subsets have least element $j$

Big Hint:

Interpret $\sum j\binom{2015-j}{999}$ as counting $1001$-element subsets of a $2016$-element set by their second-smallest element

Solution:

A $1000$-element subset has least element $j$ exactly when it contains $j$ together with $999$ larger elements, so $\binom{2015-j}{999}$ of the subsets have least element $j.$ The mean is therefore $\frac{\sum_{j} j\binom{2015-j}{999}}{\binom{2015}{1000}}.$

The numerator counts something concrete: to build a $1001$-element subset of $\{0, 1, \ldots, 2015\}$ whose second-smallest element is $j,$ choose its smallest element from $\{0, \ldots, j-1\}$ ($j$ ways) and its top $999$ elements from $\{j+1, \ldots, 2015\}.$ Summing over $j$ produces every $1001$-element subset exactly once, so $\sum_j j\binom{2015-j}{999} = \binom{2016}{1001}.$

Hence the mean is $\frac{\binom{2016}{1001}}{\binom{2015}{1000}} = \frac{2016}{1001} = \frac{288}{143},$ which is in lowest terms, and $p + q = 288 + 143 = 431.$