Answer: 196

Small Hint:

Count the complement: subsets that lie entirely inside $\{1,2,3,4,5\}$ or entirely inside $\{4,5,6,7,8\}$

Big Hint:

By inclusion-exclusion, the overlap consists of the subsets of $\{4, 5\},$ so subtract $2^5 + 2^5 - 2^2$ from $2^8$

Solution:

There are $2^8 = 256$ subsets in all. The ones to exclude are those contained in $\{1,2,3,4,5\}$ (there are $2^5 = 32$) or contained in $\{4,5,6,7,8\}$ (another $32$). A subset of both is exactly a subset of the intersection $\{4, 5\},$ and there are $2^2 = 4$ of those.

By inclusion-exclusion, $32 + 32 - 4 = 60$ subsets fail, so $256 - 60 = 196$ subsets have the required property.

Answer: 781

Small Hint:

$T_4$ must beat $T_1$ in its semifinal, then beat whichever of $T_2$ and $T_3$ wins the other semifinal

Big Hint:

Split into cases by the other semifinal: $T_2$ reaches the final with probability $\frac{2}{5},$ and $T_4$ beats $T_2$ or $T_3$ with probability $\frac{4}{6}$ or $\frac{4}{7}$

Solution:

To be champion, $T_4$ must first beat $T_1,$ which happens with probability $\frac{4}{4+1} = \frac{4}{5}.$ The other semifinal sends $T_2$ to the final with probability $\frac{2}{2+3} = \frac{2}{5}$ and $T_3$ with probability $\frac{3}{5};$ in the final, $T_4$ beats $T_2$ with probability $\frac{4}{4+2} = \frac{2}{3}$ and beats $T_3$ with probability $\frac{4}{4+3} = \frac{4}{7}.$

The probability that $T_4$ is champion is therefore $$\begin{aligned} &\frac{4}{5}\left(\frac{2}{5} \cdot \frac{2}{3} + \frac{3}{5} \cdot \frac{4}{7}\right) \\ &= \frac{4}{5} \cdot \frac{64}{105} \\ &= \frac{256}{525}. \end{aligned}$$ Since $256 = 2^8$ and $525 = 3 \cdot 5^2 \cdot 7,$ this is in lowest terms, and $p + q = 256 + 525 = 781.$

Answer: 409

Small Hint:

The points closer to $B$ than to $A$ lie on one side of the perpendicular bisector of $\overline{AB};$ likewise for $C$

Big Hint:

The bisectors are $x = 6$ and $y = \frac{2}{5}x + 1;$ find the area of the quadrilateral they cut off near $B$ and divide by $60$

Solution:

The points closer to $B$ than to $A$ lie to the right of the perpendicular bisector of $\overline{AB},$ the line $x = 6.$ The points closer to $B$ than to $C$ lie below the perpendicular bisector of $\overline{BC},$ which passes through the midpoint $(10, 5)$ with slope $\frac{2}{5}$ (the negative reciprocal of the slope $-\frac{5}{2}$ of $BC$): the line $y = \frac{2}{5}x + 1.$

Inside the triangle, the favorable region is the quadrilateral with vertices $(6, 0),$ $B(12, 0),$ the midpoint $(10, 5)$ of $\overline{BC},$ and $\left(6, \frac{17}{5}\right),$ where the two bisectors meet. Splitting it along the segment from $(6, 0)$ to $(10, 5),$ its area is $$\begin{aligned} &\frac{1}{2} \cdot \frac{17}{5} \cdot 4 + \frac{1}{2} \cdot 6 \cdot 5 \\ &= \frac{34}{5} + 15 \\ &= \frac{109}{5}. \end{aligned}$$

The triangle has area $\frac{1}{2} \cdot 12 \cdot 10 = 60,$ so the probability is $\frac{\frac{109}{5}}{60} = \frac{109}{300},$ and $p + q = 109 + 300 = 409.$

Answer: 222

Small Hint:

A base-three representation avoids the digit $0$ exactly when every digit is $1$ or $2;$ there are $2^k$ such $k$-digit numbers

Big Hint:

Since $2017 = 2202201_3,$ decide which $7$-digit strings of $1$s and $2$s stay below the bound by looking at the first two digits

Solution:

A positive integer has no $0$ in base three exactly when every digit is $1$ or $2.$ For $k = 1, 2, \ldots, 6$ there are $2^k$ such $k$-digit numbers, and all of them are at most $222222_3 = 728 \lt 2017.$

Since $2017 = 2202201_3,$ a seven-digit string of $1$s and $2$s is at most $2017$ exactly when it begins with $11,$ $12,$ or $21:$ any string beginning $22$ already beats $2202201_3$ at the third digit, since its digits are nonzero. That gives $3 \cdot 2^5 = 96$ seven-digit numbers.

The total is $2 + 4 + 8 + 16 + 32 + 64$ $+ 96 = 222.$

Answer: 791

Small Hint:

The six pairwise sums split into three complementary pairs, each pair adding up to the total $a + b + c + d$ of the set

Big Hint:

So $x + y = 3s - 1030,$ where the common pair total $s$ is a sum of two of the four given numbers; make $s$ as large as possible

Solution:

Let the set be $\{a, b, c, d\}$ with total $s = a + b + c + d.$ The six pairwise sums come in three complementary pairs: $$\begin{aligned} &(a+b) + (c+d) \\ &= (a+c) + (b+d) \\ &= (a+d) + (b+c) = s. \end{aligned}$$ No two of the pairings of $189,$ $320,$ $287,$ $234$ into two pairs give equal totals ($509 \ne 521,$ $476 \ne 554,$ $423 \ne 607$), so $x$ and $y$ are not paired with each other; each is paired with a given sum, and the remaining two given sums are paired together. Adding all six values, $x + y$ $= 3s - (189 + 320 + 287 + 234)$ $= 3s - 1030,$ where $s$ is the sum of two of the given numbers.

The largest choice is $s = 320 + 287 = 607,$ giving $x + y = 3 \cdot 607 - 1030 = 791.$ This is attained by the set $\{51.5,\ 137.5,\ 182.5,\ 235.5\},$ whose pairwise sums are $189,$ $234,$ $287,$ $320,$ $373,$ and $418,$ with $373 + 418 = 791.$

Answer: 195

Small Hint:

Set $n^2 + 85n + 2017 = m^2$ and multiply by $4$ so the left side becomes $(2n + 85)^2 + 843$

Big Hint:

Factor $4m^2 - (2n + 85)^2$ as a difference of squares; $843 = 3 \cdot 281$ has only two factorizations

Solution:

Suppose $n^2 + 85n + 2017 = m^2$ for a positive integer $m.$ Multiplying by $4$ and completing the square gives $(2n + 85)^2 + 843 = 4m^2,$ so $$\begin{aligned} &(2m - 2n - 85)(2m + 2n + 85) \\ &= 843 = 3 \cdot 281, \end{aligned}$$ where $281$ is prime. Both factors are positive with the second one larger, so either $2m - 2n - 85 = 1$ and $2m + 2n + 85 = 843,$ or $2m - 2n - 85 = 3$ and $2m + 2n + 85 = 281.$

The first system gives $m = 211$ and $n = 168,$ and indeed $168^2 + 85 \cdot 168 + 2017$ $= 44521 = 211^2.$ The second gives $m = 71$ and $n = 27,$ with $27^2 + 85 \cdot 27 + 2017$ $= 5041 = 71^2.$

The requested sum is $168 + 27 = 195.$

Answer: 501

Small Hint:

The equation says $kx = (x + 2)^2$ with the domain restrictions $kx \gt 0$ and $x \gt -2;$ treat $k \lt 0$ and $k \gt 0$ separately

Big Hint:

For every $k \lt 0$ the two graphs cross exactly once on $(-2, 0);$ for $k \gt 0$ the quadratic $x^2 + (4 - k)x + 4$ needs a repeated positive root

Solution:

The equation requires $x + 2 \gt 0$ and $kx \gt 0,$ and under those restrictions it is equivalent to $kx = (x + 2)^2,$ that is, $x^2 + (4 - k)x + 4 = 0.$

For $k \lt 0$ the restrictions force $-2 \lt x \lt 0.$ On this interval $kx$ decreases from $-2k \gt 0$ to $0$ while $(x + 2)^2$ increases from $0$ to $4,$ so the graphs cross exactly once. Hence every one of the $500$ negative values of $k$ works, while $k = 0$ makes $\log(kx)$ undefined.

For $k \gt 0$ the restrictions force $x \gt 0.$ The quadratic has root product $4,$ so any real roots have the same sign, and the discriminant $(4 - k)^2 - 16 = k(k - 8)$ is negative for $0 \lt k \lt 8.$ When $k \gt 8$ there are two distinct positive roots (root sum $k - 4 \gt 0$), giving two solutions; only $k = 8$ gives exactly one solution, the double root $x = 2.$ In total $500 + 1 = 501$ values of $k$ work.

Answer: 134

Small Hint:

Multiply by $6! = 720:$ the sum is an integer exactly when $720$ divides $n^6 + 6n^5 + 30n^4$ $+ 120n^3 + 360n^2$

Big Hint:

Checking mod $2$ and mod $3$ forces $n$ to be a multiple of $6;$ write $n = 6k,$ and only the condition that $5$ divides $k(k + 1)$ survives

Solution:

Multiplying by $6! = 720,$ the sum is an integer exactly when $$\begin{aligned} &n^6 + 6n^5 + 30n^4 \\ &{}+ 120n^3 + 360n^2 \\ &\text{is divisible by } 720. \end{aligned}$$ If $n$ were odd, every term except $n^6$ would be even, making the total odd. If $n$ is not divisible by $3,$ then modulo $3$ every term except $n^6$ vanishes while $n^6 \equiv 1 \pmod 3.$ So $n$ must be a multiple of $6.$

Write $n = 6k.$ Then $30n^4 = 720 \cdot 54k^4,$ $120n^3 = 720 \cdot 36k^3,$ and $360n^2 = 720 \cdot 18k^2$ are all divisible by $720,$ while $n^6 + 6n^5 = 6^6 k^5(k + 1).$ Since $6^6 = 2^6 3^6$ supplies the factors $2^4$ and $3^2$ of $720 = 2^4 \cdot 3^2 \cdot 5,$ the condition reduces to $k(k + 1)$ being divisible by $5,$ that is, $k \equiv 0$ or $4 \pmod 5.$

For $n = 6k \lt 2017$ we need $1 \le k \le 336.$ That range contains $67$ multiples of $5$ and $67$ values with $k \equiv 4 \pmod 5,$ so there are $67 + 67 = 134$ such $n.$

Answer: 13

Small Hint:

In such a hand exactly one number and exactly one color repeat; a card can be discarded only if a single card carries both repeats

Big Hint:

Count each type of hand: a rainbow $7$-card set plus any extra card, versus hands where the repeated number and repeated color sit on four distinct cards

Solution:

Since the eight cards cover all seven numbers and all seven colors, exactly one number and exactly one color appear twice. Sharon can discard a card exactly when a single card carries both the repeated number and the repeated color: that card is then the unique discardable one, while if no card carries both, removing any card loses a number or a color.

Hands of the first type consist of a rainbow set of seven cards — one of each number and each color, which is one of $7!$ permutation patterns — plus any of the remaining $42$ cards, and every such hand arises exactly once this way: $7! \cdot 42 = 211680$ hands. For the second type, choose the repeated number ($7$ ways) and the two colors of its cards ($\binom{7}{2} = 21$ ways); the repeated color must be one of the other $5$ colors, and the numbers of its two cards come from the remaining $6$ numbers ($\binom{6}{2} = 15$ ways); finally match the last four numbers to the last four colors ($4! = 24$ ways). That is $7 \cdot 21 \cdot 5 \cdot 15 \cdot 24 = 264600$ hands.

The probability is $\frac{211680}{211680 + 264600} = \frac{4}{9},$ so $p + q = 4 + 9 = 13.$

Answer: 546

Small Hint:

Use coordinates $A(0,0),$ $B(84,0),$ $C(84,42),$ $D(0,42);$ the lines $CM$ and $DN$ meet at $O(12, 24)$

Big Hint:

$[BCON] = 2184$ while $[BCO] = 1512,$ so $P$ lies on $\overline{CO};$ place it so that $[BPC] = 1092$

Solution:

Place $A = (0, 0),$ $B = (84, 0),$ $C = (84, 42),$ $D = (0, 42),$ so $M = (0, 21)$ and $N = (28, 0).$ Line $CM$ is $y = \frac{x}{4} + 21$ and line $DN$ is $y = 42 - \frac{3x}{2},$ which meet at $O = (12, 24).$

By the Shoelace Formula, quadrilateral $BCON$ has area $2184,$ so each half must have area $1092.$ Triangle $BCO$ alone has area $\frac{1}{2} \cdot 42 \cdot 72 = 1512 \gt 1092$ (base $\overline{BC},$ and $O$ is at horizontal distance $72$ from it), so the bisecting segment ends at a point $P$ on $\overline{CO}.$ For $[BPC] = 1092,$ the distance $d$ from $P$ to line $BC$ must satisfy $\frac{1}{2} \cdot 42 \cdot d = 1092,$ so $d = 52,$ giving $P$ the $x$-coordinate $84 - 52 = 32.$ Since $P$ lies on line $CO,$ which is $y = \frac{x}{4} + 21,$ we get $P = (32, 29).$

Triangle $CDP$ has base $CD = 84$ on the line $y = 42$ and height $42 - 29 = 13,$ so its area is $\frac{1}{2} \cdot 84 \cdot 13 = 546.$

Answer: 544

Small Hint:

The one-way roads work if and only if no town has all four of its roads inbound or all four outbound

Big Hint:

Count the bad assignments: $5 \cdot 2^6$ with an all-outbound town, the same with an all-inbound town, minus $5 \cdot 4 \cdot 2^3$ counted twice

Solution:

The assignment works if and only if no town has all four roads inbound or all four outbound. One direction is clear: an all-inbound town cannot be left, and an all-outbound town cannot be reached. Conversely, suppose every town has an inbound and an outbound road, yet town $B$ cannot be reached from town $A.$ Let $S$ be the set of towns reachable from $A$ (including $A$) and $T$ the set of towns from which $B$ is reachable (including $B$). These sets are disjoint, every outbound road of a town in $S$ stays inside $S,$ and every inbound road of a town in $T$ comes from inside $T.$ Since $A$ has an outbound road, $|S| \ge 2,$ and similarly $|T| \ge 2;$ as $|S| + |T| \le 5,$ one of the two sets has exactly two towns. If $S = \{A, X\},$ the outbound roads of $A$ and of $X$ must both stay inside $S,$ forcing the single road between them to point both ways — a contradiction (and $|T| = 2$ is symmetric).

Now count assignments with a bad town among the $2^{10} = 1024$ total. Choosing a town to be all-outbound ($5$ ways) and orienting the remaining $\binom{4}{2} = 6$ roads freely gives $5 \cdot 2^6 = 320$ assignments, and there can be at most one all-outbound town. Similarly $320$ assignments have an all-inbound town. Assignments with both are counted twice: choose the all-outbound town ($5$), the all-inbound town ($4$), and the other $3$ roads freely, $5 \cdot 4 \cdot 2^3 = 160.$ So $320 + 320 - 160 = 480$ assignments fail.

The number that work is $1024 - 480 = 544.$

Answer: 110

Small Hint:

In the complex plane the center of $C_{n+1}$ equals the center of $C_n$ plus $(r^n - r^{n+1})\,i^n$

Big Hint:

The centers converge to the geometric series sum $\frac{1-r}{1-ir},$ whose distance from the origin is $\frac{1-r}{\sqrt{1+r^2}}$

Solution:

Work in the complex plane with $C_0$ centered at $O_0 = 0$ and $A_0 = 1,$ and let $O_n$ be the center of $C_n.$ Inductively, $A_n = O_n + r^n i^n:$ this holds for $n = 0,$ and since $C_{n+1}$ is internally tangent to $C_n$ at $A_n,$ its center is $O_{n+1} = A_n - r^{n+1} i^n;$ then $A_n$ sits in direction $i^n$ from $O_{n+1},$ so rotating $90^\circ$ counterclockwise gives $A_{n+1} = O_{n+1} + r^{n+1} i^{n+1}.$

Therefore $O_{n+1} - O_n$ $= (r^n - r^{n+1})\, i^n$ $= (1 - r)(ir)^n.$ The circles are nested, and their radii shrink to $0,$ so the common point $B$ is the limit of the centers: $$B = (1 - r)\sum_{n=0}^{\infty} (ir)^n = \frac{1 - r}{1 - ir},$$ at distance $\frac{1 - r}{|1 - ir|} = \frac{1 - r}{\sqrt{1 + r^2}}$ from the origin.

For $r = \frac{11}{60}$ this equals $\frac{\frac{49}{60}}{\frac{\sqrt{3721}}{60}} = \frac{49}{61},$ so $m + n = 49 + 61 = 110.$

Answer: 245

Small Hint:

Count by apex: each vertex is the apex of $\lfloor \frac{n-1}{2} \rfloor$ isosceles triangles, but equilateral triangles get counted three times

Big Hint:

Write $f(n)$ with a floor function and compute $f(n+1) - f(n)$ separately in each residue class of $n$ modulo $6$