Answer: 722

Solution:

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

Therefore $B - A = -38 + 4(1 + 2 + \cdots + 19) = -38 + 4 \cdot 190 = 722.$

Answer: 139

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 $m + n = 55 + 84 = 139.$

Answer: 307

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 $16 \cdot 307 + 1 = 4913 = 17^3.$

Answer: 507

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 $x^2 = 169 \cdot 3 = 507.$

Answer: 341

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 $m + n = 26 + 315 = 341.$

Answer: 58

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 $$\left(45^\circ + \frac{15\alpha}{4}\right) - \left(180^\circ - \frac{3\alpha}{2}\right) = \frac{21\alpha}{4} - 135^\circ = 12^\circ,$$ 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 $\angle BAG = \angle BAE + \angle EAG = 30^\circ + 28^\circ = 58^\circ.$

Answer: 539

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 $$2s = BC = BJ + JC = x\left(\frac{1}{\sqrt{5}} + \frac{\sqrt{5}}{2}\right) = \frac{7x}{2\sqrt{5}},$$ 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, $\frac{x}{y} = \frac{2s}{6s/7} = \frac{7}{3}.$

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

Answer: 695

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 $695.$

Answer: 494

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 $494.$

Answer: 72

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 $|f(0)| = 72.$

Answer: 108

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 $$\frac{b}{a} = 2\left(\frac{b}{8}\right)^2 - 1, \qquad \text{so} \qquad a = \frac{32b}{b^2 - 32}.$$

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 $48 + 48 + 12 = 108.$

Answer: 431

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 $p + q = 288 + 143 = 431.$

Answer: 91

Solution:

Let $P = \sin 1^\circ \sin 3^\circ \cdots \sin 89^\circ$ and $Q = \sin 2^\circ \sin 4^\circ \cdots \sin 88^\circ,$ so the desired product is $\frac{1}{P^2}.$ Then $PQ = \prod_{k=1}^{89} \sin k^\circ,$ and multiplying this by itself in reverse order, using $\sin(90 - k)^\circ = \cos k^\circ,$ gives $$P^2Q^2 = \prod_{k=1}^{89} \sin k^\circ \cos k^\circ.$$

Multiply by $2^{89}$ and use $2\sin k^\circ \cos k^\circ = \sin 2k^\circ:$ $$2^{89} P^2 Q^2 = \prod_{k=1}^{89} \sin 2k^\circ = \left(\prod_{k=1}^{44} \sin 2k^\circ\right)\left(\prod_{k=46}^{89} \sin 2k^\circ\right) = Q \cdot Q,$$ since $\sin 90^\circ = 1$ and $\sin(180 - x)^\circ = \sin x^\circ$ turns the second half into $Q$ as well.

Because $Q \ne 0,$ it follows that $P^2 = 2^{-89},$ so $\prod_{k=1}^{45} \csc^2(2k-1)^\circ = 2^{89}.$ Since $89$ is prime, the only representation $m^n$ with $m, n \gt 1$ is $m = 2,$ $n = 89,$ and $m + n = 91.$

Answer: 483

Solution:

On the strip $m \le x \lt m + 1$ we have $\lfloor\sqrt{x}\rfloor = k = \lfloor\sqrt{m}\rfloor,$ so the region above it is a trapezoid under $y = kx$ with area $A(m+1) - A(m) = \frac{(2m+1)k}{2}:$ an integer when $k$ is even, a half-integer when $k$ is odd. Hence as $n$ grows by $1,$ the integrality of $A(n)$ is unchanged while $k$ is even and flips at every step while $k$ is odd.

Consider the block of $2k + 1$ values $k^2 \lt n \le (k+1)^2.$ Starting from $A(1) = 0,$ the statuses of $A(k^2)$ cycle with period $4:$ integer for $k \equiv 0, 1$ and non-integer for $k \equiv 2, 3 \pmod 4$ (an odd block flips the status an odd number of times, an even block preserves it). Counting integer values of $A(n)$ inside each block: for $k = 4j - 3$ the block alternates, beginning and ending with non-integers, giving $4j - 3;$ for $k = 4j - 2$ every value is a non-integer, giving $0;$ for $k = 4j - 1$ it alternates, beginning and ending with integers, giving $4j;$ for $k = 4j$ all $8j + 1$ values are integers.

For $j = 1, \ldots, 7,$ covering $2 \le n \le 29^2,$ the four blocks contribute $(4j - 3) + 0 + 4j + (8j + 1) = 16j - 2$ integers, totaling $\sum_{j=1}^{7}(16j - 2) = 434.$ Then the block $k = 29$ contributes $29$ integers for $841 \lt n \le 900,$ the block $k = 30$ contributes none, and for $k = 31$ the alternation over $961 \lt n \le 1000$ begins with an integer at $n = 962$ and gives $20$ more. The total is $434 + 29 + 20 = 483.$

Answer: 53

Solution:

Stand the block on the face containing $A$ and $B,$ and let $O'$ be the center of that face, $M$ the midpoint of $\overline{AB},$ and $O$ the center of the cylinder. The cutting plane meets the bottom face in chord $\overline{AB}$ and, by symmetry through $O,$ meets the top face in the reflected chord, so the cut face projects vertically onto the region $R'$ between chord $\overline{AB}$ and its mirror image through $O'$ (shaded below). Each $120^\circ$ circular segment cut off has area $\frac{1}{3}\pi \cdot 6^2 - \frac{1}{2} \cdot 6 \cdot 6 \sin 120^\circ = 12\pi - 9\sqrt{3},$ so $R'$ has area $36\pi - 2\left(12\pi - 9\sqrt{3}\right) = 12\pi + 18\sqrt{3}.$

Since $\overset{\frown}{AB} = 120^\circ,$ triangle $AO'B$ gives $O'M = 6\cos 60^\circ = 3,$ and $OO' = 4,$ so $OM = 5.$ The cut face is planar and tilted from the horizontal only in the direction of $\overline{O'M},$ at the angle $\theta$ with $\cos\theta = \frac{O'M}{OM} = \frac{3}{5}.$ Undoing the projection therefore multiplies areas by $\frac{5}{3},$ so the unpainted face has area $\frac{5}{3}\left(12\pi + 18\sqrt{3}\right) = 20\pi + 30\sqrt{3}.$ Thus $a + b + c = 20 + 30 + 3 = 53.$