Answer: 114

Solution:

Say each stripe used $s$ ounces of paint. Blue was used only on the blue stripe, so $s$ ounces of blue were used. Since the three leftovers are equal and the colors started $34$ and $58$ ounces apart, red use exceeded blue use by $164 - 130 = 34$ ounces and white use exceeded blue use by $188 - 130 = 58$ ounces. That extra red and white is exactly the pink stripe, so $s = 34 + 58 = 92.$

Bill therefore had $130 - 92 = 38$ ounces of each color left, for a total of $3 \cdot 38 = 114$ ounces.

Answer: 469

Solution:

By the power rule for exponents, $$a^{(\log_3 7)^2} = \left(a^{\log_3 7}\right)^{\log_3 7} = 27^{\log_3 7} = \left(3^{\log_3 7}\right)^3 = 7^3 = 343.$$

In the same way, $b^{(\log_7 11)^2} = 49^{\log_7 11} = \left(7^{\log_7 11}\right)^2 = 11^2 = 121,$ and $c^{(\log_{11} 25)^2} = \left(\sqrt{11}\right)^{\log_{11} 25} = \left(11^{\log_{11} 25}\right)^{1/2} = 25^{1/2} = 5.$

The sum is $343 + 121 + 5 = 469.$

Answer: 141

Solution:

Let $AD = h,$ and place $A = (0, 0),$ $B = (100, 0),$ $C = (100, h),$ $D = (0, h),$ so $E = \left(0, \frac{h}{2}\right).$ Line $AC$ has slope $\frac{h}{100}$ and line $BE$ has slope $\frac{h/2}{-100} = -\frac{h}{200}.$ Perpendicularity gives $$\frac{h}{100} \cdot \left(-\frac{h}{200}\right) = -1,$$ so $h^2 = 20000$ and $h = 100\sqrt{2} \approx 141.42.$

The greatest integer less than $AD$ is $141.$

Answer: 89

Solution:

Let $c$ be the number of children. The grape counts $n,$ $n - 2,$ $\ldots,$ $n + 2 - 2c$ form an arithmetic sequence, so the total is $c$ times the average of the first and last terms: $$c \cdot \frac{n + (n + 2 - 2c)}{2} = c(n + 1 - c) = 2009 = 7^2 \cdot 41.$$ Thus $c \mid 2009$ and $n = \frac{2009}{c} + c - 1.$

The last-place child ate $n + 2 - 2c = \frac{2009}{c} + 1 - c \ge 0$ grapes, which forces $c(c - 1) \le 2009,$ ruling out $c = 49,$ $287,$ and $2009.$ The remaining divisors give $n = 2009$ for $c = 1,$ $n = 287 + 6 = 293$ for $c = 7,$ and $n = 49 + 40 = 89$ for $c = 41.$

The smallest possible value is $n = 89.$

Answer: 32

Solution:

Place the center of circle $A$ at the origin with the triangle's vertices at $(0, 10)$ and $\left(\pm 5\sqrt{3}, -5\right).$ A circle internally tangent to $A$ at a vertex has its center on the radius to that vertex, so circle $B$ has center $(0, 7)$ and circles $C$ and $D$ have centers $\left(\mp 4\sqrt{3}, -4\right)$ (at distance $10 - 2 = 8$ from the origin).

By symmetry the center of circle $E,$ of radius $r,$ lies on the $y$-axis at $(0, y).$ External tangency to $B$ gives $7 - y = r + 3,$ so $y = 4 - r.$ External tangency to $C$ gives $$\left(4\sqrt{3}\right)^2 + (4 - r + 4)^2 = (r + 2)^2,$$ that is, $48 + (8 - r)^2 = (r + 2)^2,$ which simplifies to $112 - 16r = 4r + 4,$ so $r = \frac{27}{5}.$

Then $m + n = 27 + 5 = 32.$

Answer: 750

Solution:

Count the complement: subsets $a_1 \lt a_2 \lt a_3 \lt a_4 \lt a_5$ with no two consecutive. Setting $b_i = a_i - (i - 1)$ turns each such subset into five distinct numbers $b_1 \lt b_2 \lt \cdots \lt b_5$ in $\{1, \ldots, 10\},$ and this map is reversible, so there are $\binom{10}{5} = 252$ subsets with no two consecutive numbers.

Therefore $m = \binom{14}{5} - \binom{10}{5} = 2002 - 252 = 1750,$ and the remainder upon division by $1000$ is $750.$

Answer: 401

Solution:

The $i$th term is $\frac{(2i-1)!!}{(2i)!!}$ with odd numerator, and $(2i)!! = 2^i \cdot i!.$ Because $\binom{2i}{i} = \frac{(2i)!}{i!\,i!} = \frac{2^i (2i-1)!!}{i!}$ is an integer, every odd prime power dividing $i!$ also divides $(2i-1)!!.$ Hence in lowest terms the $i$th term has denominator exactly $2^{a_i}$ where $a_i = i + e_i$ and $e_i$ is the exponent of $2$ in $i!.$ The $a_i$ strictly increase, so over the common denominator $2^{a_{2009}}$ every term except the last contributes an even numerator while the last contributes an odd one. The sum in lowest terms therefore has denominator exactly $2^{a_{2009}},$ so $b = 1.$

By Legendre's formula, $$e_{2009} = 1004 + 502 + 251 + 125 + 62 + 31 + 15 + 7 + 3 + 1 = 2001,$$ so $a = 2009 + 2001 = 4010.$ Then $\frac{ab}{10} = \frac{4010 \cdot 1}{10} = 401.$

Answer: 41

Solution:

The probability that a player's first six appears on roll $k$ is $p_k = \left(\frac{5}{6}\right)^{k-1} \cdot \frac{1}{6}.$ The probability of a tie is $$\sum_{k=1}^{\infty} p_k^2 = \frac{1}{36} \cdot \frac{1}{1 - \frac{25}{36}} = \frac{1}{11}.$$

The probability that Linda needs exactly one more roll than Dave is $\sum_{k=1}^{\infty} p_k p_{k+1} = \frac{5}{6} \sum_{k=1}^{\infty} p_k^2 = \frac{5}{66},$ and by symmetry the same holds with the players swapped.

The total probability is $\frac{1}{11} + 2 \cdot \frac{5}{66} = \frac{6 + 10}{66} = \frac{8}{33},$ so $m + n = 8 + 33 = 41.$

Answer: 0

Solution:

If $(x, y, z)$ is a positive solution of $4x + 3y + 2z = 2009,$ then $(x - 1, y - 1, z - 1)$ is a nonnegative solution of $4x + 3y + 2z = 2000,$ and conversely, since $4 + 3 + 2 = 9.$ So $m$ equals the number of nonnegative solutions of $4x + 3y + 2z = 2000,$ and $m - n$ counts the nonnegative solutions of that equation in which at least one variable is $0.$

If $x = 0:$ $3y + 2z = 2000$ forces $y$ even, $0 \le y \le 666,$ giving $334$ solutions. If $y = 0:$ $2x + z = 1000$ with $0 \le x \le 500$ gives $501.$ If $z = 0:$ $4x + 3y = 2000$ forces $y \equiv 0 \pmod 4,$ $0 \le y \le 664,$ giving $167.$ The solutions $(0, 0, 1000)$ and $(500, 0, 0)$ are each counted twice, so $$m - n = 334 + 501 + 167 - 2 = 1000.$$

The remainder upon division by $1000$ is $0.$

Answer: 96

Solution:

Since $5^2 + 12^2 = 13^2,$ angle $B$ is right. Place $A = (0, 0),$ $B = (5, 0),$ $C = (5, 12).$ The condition at $A$ says $\angle BAD = \angle CAD,$ so $D$ lies on the bisector of angle $BAC.$ Using the half-angle formula with $\tan \angle BAC = \frac{12}{5},$ $$\tan \frac{\angle BAC}{2} = \frac{\sin \angle BAC}{1 + \cos \angle BAC} = \frac{12/13}{1 + 5/13} = \frac{2}{3},$$ so $D$ lies on the line $y = \frac{2}{3}x.$

The condition at $C$ says $CB$ bisects angle $ACD,$ so ray $CD$ is the reflection of ray $CA$ over line $CB,$ which is the vertical line $x = 5.$ The reflection of $A$ is $(10, 0),$ so $D$ lies on the line through $C = (5, 12)$ and $(10, 0),$ namely $5y = 120 - 12x.$

Solving $y = \frac{2}{3}x$ and $5y = 120 - 12x$ gives $x = \frac{180}{23},$ $y = \frac{120}{23}.$ Then $$AD = \frac{60}{23}\sqrt{3^2 + 2^2} = \frac{60\sqrt{13}}{23},$$ so $p + q + r = 60 + 23 + 13 = 96.$

Answer: 125

Solution:

The inequality $|\log m - \log k| \lt \log n$ is equivalent to $\frac{m}{n} \lt k \lt mn.$ Write $m = nq + r$ with $0 \le r \lt n;$ since $m \ge n \ge 2$ (for $n = 1$ no $k$ works), $q \ge 1.$ The integers $k$ in the interval are $q + 1, q + 2, \ldots, mn - 1,$ so there are $mn - q - 1 = 50$ of them, that is, $mn - q = 51,$ or $$q(n^2 - 1) + nr = 51.$$

For $n \ge 8$ the left side is at least $63,$ so $2 \le n \le 7.$ Checking each case, only $n = 2,$ $r = 0,$ $q = 17$ (so $m = 34$) and $n = 3,$ $r = 1,$ $q = 6$ (so $m = 19$) work. These give $mn = 68$ and $mn = 57;$ indeed $17 \lt k \lt 68$ and $\frac{19}{3} \lt k \lt 57$ each contain exactly $50$ integers.

The sum of all possible values of $mn$ is $68 + 57 = 125.$

Answer: 803

Solution:

Let $S = \sum_{i=1}^{k} (a_i + b_i).$ The $2k$ chosen elements are distinct positive integers, so $S \ge 1 + 2 + \cdots + 2k = k(2k + 1).$ The $k$ sums are distinct integers at most $2009,$ so $S \le 2009 + 2008 + \cdots + (2010 - k) = \frac{k(4019 - k)}{2}.$ Combining, $$k(2k + 1) \le \frac{k(4019 - k)}{2} \implies 4k + 2 \le 4019 - k \implies k \le \frac{4017}{5} = 803.4,$$ so $k \le 803.$

To achieve $k = 803,$ take the pairs $(i,\, 1206 + i)$ for $1 \le i \le 401,$ whose sums are the even numbers $1208, 1210, \ldots, 2008,$ together with the pairs $(a,\, a + 403)$ for $402 \le a \le 803,$ whose sums are the odd numbers $1207, 1209, \ldots, 2009.$ The elements used are $1$–$803,$ $805$–$1607$ with no repeats, and all $803$ sums are distinct and at most $2009.$

The maximum is $k = 803.$

Answer: 672

Solution:

Put the circle in the complex plane with center $0,$ $A = -2,$ $B = 2,$ and $C_i = 2\omega^i$ for $i = 1, \ldots, 6,$ where $\omega = e^{i\pi/7}.$ Then $AC_i = 2\,|\omega^i + 1|$ and $BC_i = 2\,|\omega^i - 1|,$ so $$AC_i \cdot BC_i = 4\,\bigl|\omega^{2i} - 1\bigr|.$$

As $i$ runs over $1, \ldots, 6,$ the numbers $\omega^{2i} = e^{2\pi i \cdot i/7}$ run over all six nontrivial $7$th roots of unity $\zeta^j.$ Since $\prod_{j=1}^{6} (x - \zeta^j) = 1 + x + \cdots + x^6,$ plugging in $x = 1$ gives $\prod_{j=1}^{6} \bigl|1 - \zeta^j\bigr| = 7.$ Therefore $$n = \prod_{i=1}^{6} 4\,\bigl|\omega^{2i} - 1\bigr| = 4^6 \cdot 7 = 28672.$$

The remainder when $n$ is divided by $1000$ is $672.$

Answer: 983

Solution:

Write $a_n = 2^n \sin\theta_n$ with $\theta_0 = 0,$ so $\sqrt{4^n - a_n^2} = 2^n\,|\cos\theta_n|.$ Let $\theta = \arcsin\frac{3}{5},$ so $\cos\theta = \frac{4}{5}.$ The recursion becomes $$a_{n+1} = 2^{n+1}\left(\cos\theta \sin\theta_n + \sin\theta\,|\cos\theta_n|\right) = 2^{n+1}\sin(\theta_n \pm \theta),$$ with the plus sign when $\cos\theta_n \ge 0$ and the minus sign when $\cos\theta_n \lt 0.$

Since $\frac{1}{2} \lt \frac{3}{5} \lt \frac{\sqrt{2}}{2},$ we have $30^\circ \lt \theta \lt 45^\circ.$ The angles $\theta,$ $2\theta$ have positive cosine, so the sequence of angles runs $0, \theta, 2\theta, 3\theta.$ But $90^\circ \lt 3\theta \lt 135^\circ$ has negative cosine, so $\theta_4 = 2\theta,$ and from then on the angle alternates between $3\theta$ and $2\theta.$ In particular $\theta_n = 2\theta$ for every even $n \ge 2.$

With $\sin 2\theta = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25},$ $$a_{10} = 2^{10} \sin 2\theta = 1024 \cdot \frac{24}{25} = \frac{24576}{25} = 983.04,$$ so the answer is $983.$

Answer: 14

Solution:

Let chords $BC$ and $AC$ meet $\overline{MN}$ at $P$ and $Q,$ and set $x = \frac{CM}{CN}.$ Since $\angle MBN = 90^\circ$ (angle in a semicircle) and $MB = \frac{3}{5},$ we get $BN = \frac{4}{5};$ also $AM = AN = \frac{\sqrt{2}}{2}.$ Because $P$ lies on both $MN$ and $BC,$ the ratio $\frac{MP}{PN}$ equals the ratio of the distances from $M$ and $N$ to line $BC,$ i.e. $\frac{[BMC]}{[BNC]}.$ In cyclic quadrilateral $MBNC$ the angles $BMC$ and $BNC$ are supplementary, so their sines are equal and $$\frac{MP}{PN} = \frac{BM \cdot MC}{BN \cdot NC} = \frac{3x}{4}, \qquad \frac{MQ}{QN} = \frac{AM \cdot MC}{AN \cdot NC} = x.$$

Since $MN = 1,$ these give $MP = \frac{3x}{3x + 4}$ and $MQ = \frac{x}{x + 1},$ so $$d = MQ - MP = \frac{x}{x + 1} - \frac{3x}{3x + 4} = \frac{x}{3x^2 + 7x + 4} = \frac{1}{3x + \frac{4}{x} + 7}.$$

As $C$ ranges over the far semicircle, $x$ takes every positive value. By AM-GM, $3x + \frac{4}{x} \ge 2\sqrt{12} = 4\sqrt{3},$ with equality at $x = \frac{2}{\sqrt{3}}.$ Hence the largest value of $d$ is $$\frac{1}{7 + 4\sqrt{3}} = 7 - 4\sqrt{3},$$ since $(7 + 4\sqrt{3})(7 - 4\sqrt{3}) = 1.$ Then $r + s + t = 7 + 4 + 3 = 14.$