Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$

Solution: Define a function $f(n)$ on the non-negative integers, as \[f(n) = \frac{a_n^2 + a_{n+1}^2}{a_na_{n+1}} = \frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_{n}}\] We want $\left\lfloor f(2006) \right\rfloor$
Consider the relation $a_{n+1}a_{n-1}=a_{n}^{2}+2007$ . Dividing through by $a_{n}a_{n-1}$ , we get \[.\phantom{------------} \frac{a_{n+1}}{a_{n}} = \frac{a_{n}}{a_{n-1}} + \frac{2007}{a_{n}a_{n-1}} \phantom{------------} (1)\] and dividing through by $a_{n}a_{n+1}$ , we get \[.\phantom{------------} \frac{a_{n-1}}{a_{n}} = \frac{a_{n}}{a_{n+1}} + \frac{2007}{a_{n}a_{n+1}} \phantom{------------} (2)\] Adding LHS of $(1)$ with RHS of $(2)$ (and vice-versa), we get \[\frac{a_{n+1}}{a_{n}} + \frac{a_{n}}{a_{n+1}} + \frac{2007}{a_{n}a_{n+1}} = \frac{a_{n}}{a_{n-1}} + \frac{a_{n-1}}{a_{n}} + \frac{2007}{a_{n}a_{n-1}}\] i.e. \[f(n)+ \frac{2007}{a_{n}a_{n+1}} = f(n-1) + \frac{2007}{a_{n}a_{n-1}}\] Summing over $n=1$ to $n=2006$ , we notice that most of the terms on each side cancel against the corresponding term on the other side. We are left with \[f(2006) + \frac{2007}{a_{2006}a_{2007}} = f(0) + \frac{2007}{a_{1}a_{0}}\] We have $f(0) = 2$ , and $2007/a_0a_1 = 2007/9 = 223$ . So \[f(2006) = 2 + 223 - \frac{2007}{a_{2006}a_{2007}} = 224 + \left( 1 - \frac{2007}{a_{2006}a_{2007}}\right)\] Since all the $a_i$ are positive, $(1)$ tells us that the ratio $a_{n+1}/a_n$ of successive terms is increasing. Since this ratio starts with $a_1/a_0 = 1$ , this means that the sequence $(a_n)$ is increasing. Since $a_3=672$ already, we must have $a_{2006}a_{2007} > 672^2 > 2007$ . It follows that $\left\lfloor f(2006) \right\rfloor = \boxed{224}$

Answer:  224
Problem: sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$

Solution: We are given that
$a_{n+1}a_{n-1}= a_{n}^{2}+2007$ $a_{n-1}^{2}+2007 = a_{n}a_{n-2}$
Add these two equations to get
This is an invariant . Defining $b_{i}= \frac{a_{i}}{a_{i-1}}$ for each $i \ge 2$ , the above equation means
$b_{n+1}+\frac{1}{b_{n}}= b_{n}+\frac{1}{b_{n-1}}$
We can thus calculate that $b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225$ . Using the equation $a_{2007}a_{2005}=a_{2006}^{2}+2007$ and dividing both sides by $a_{2006}a_{2005}$ , notice that $b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}> \frac{a_{2006}}{a_{2005}}= b_{2006}$ . This means that
$b_{2007}+\frac{1}{b_{2007}}< b_{2007}+\frac{1}{b_{2006}}= 225$ . It is only a tiny bit less because all the $b_i$ are greater than $1$ , so we conclude that the floor of $\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\frac{1}{b_{2007}}$ is $\boxed{224}$

Answer:  224
Problem: sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$

Solution: We will try to manipulate $\frac{a_0^2+a_1^2}{a_0a_1}$ to get $\frac{a_1^2+a_2^2}{a_1a_2}$
$\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_0+\frac{a_1^2}{a_0}}{a_1} = \frac{a_0+\frac{a_1^2+2007}{a_0}-\frac{2007}{a_0}}{a_1}$ Using the recurrence relation, $= \frac{a_0+a_2-\frac{2007}{a_0}}{a_1}  = \frac{a_0a_2+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2}$ Applying the relation to $a_0a_2$ $= \frac{a_1^2+2007+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2} = \frac{a_1^2+a_2^2}{a_1a_2}+\frac{2007}{a_1a_2}-\frac{2007}{a_0a_1}$
We can keep on using this method to get that $\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{2005}a_{2006}}+\frac{2007}{a_{2005}a_{2006}}-\ldots-\frac{2007}{a_{0}a_{1}}$
This telescopes to $\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{0}a_{1}}$
or $\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \frac{a_0^2+a_1^2}{a_0a_1}+\frac{2007}{a_{0}a_{1}}-\frac{2007}{a_{2006}a_{2007}}$
Finding the first few values, we notice that they increase rapidly, so $\frac{2007}{a_{2006}a_{2007}} < 1$ . Calculating the other values, $\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\frac{2007}{a_{2006}a_{2007}}$
The greatest number that does not exceed this is $\boxed{224}$

Answer:  224
Problem: sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$ $a_{n+1}a_{n-1}=a_{n}^{2}+2007$
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$

Solution: Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that $a_0 = a_1 = 0$ , and solving for $a_2$ and $a_3$ using the given relation we get $a_2 = 672 = 3(224)$ and $a_3 = 3(224^{2} + 223)$ , respectively. It will be clear why I decided to factor these expressions as I did momentarily.
Next, let's see what the expression $\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}$ looks like for small values of $n$ . For $n = 1$ , we get $\frac{1 + 224^2}{224}$ , the floor of which is clearly $224$ because the $1$ in the numerator is insignificant. Repeating the procedure for $n + 1$ is somewhat messier, but we end up getting $\frac{224^4 + 224^2\cdot223\cdot2 + 224^2 + 223^2}{224^3 + 224\cdot223}$ . It's not too hard to see that $224^4$ is much larger than the sum of the remaining terms in the numerator, and that $224^3$ is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than $224^3$ , while the second-largest term in the denominator is smaller than $224^2$ . Thus, the floor of this expression will come out to be $224$ as well.
Now we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time $n$ increases by $1$ , the degrees of both the numerator and denominator increase by $2$ , because we are squaring the $n+1th$ term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation $a_{n+1}^2 = (\frac{a_{n}^2 + 2007}{a_{n-1}})^2$ ). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the $\approx224:1$ ratio between the two.
For the non-greatest terms in the expression to offset this ratio for values of $n$ in the ballpark of $2006$ , they would have to have massive coefficients, because or else they are dwarfed by the additional $224$ attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to $\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }$ for all $k\geq2$ , whose $\lim_{k\to\infty}=$ $\boxed{224}$

Answer:  224
Problem: Let $ABC$ be an equilateral triangle , and let $D$ and $F$ be points on sides $BC$ and $AB$ , respectively, with $FA = 5$ and $CD = 2$ .  Point $E$ lies on side $CA$ such that angle $DEF = 60^{\circ}$ .  The area of triangle $DEF$ is $14\sqrt{3}$ .  The two possible values of the length of side $AB$ are $p \pm q \sqrt{r}$ , where $p$ and $q$ are rational, and $r$ is an integer not divisible by the square of a prime .  Find $r$

Solution: AIME I 2007-15.png
Denote the length of a side of the triangle $x$ , and of $\overline{AE}$ as $y$ . The area of the entire equilateral triangle is $\frac{x^2\sqrt{3}}{4}$ . Add up the areas of the triangles using the $\frac{1}{2}ab\sin C$ formula (notice that for the three outside triangles, $\sin 60 = \frac{\sqrt{3}}{2}$ ): $\frac{x^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(5 \cdot y + (x - 2)(x - 5) + 2(x - y)) + 14\sqrt{3}$ . This simplifies to $\frac{x^2\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(5y + x^2 - 7x + 10 + 2x - 2y + 56)$ . Some terms will cancel out, leaving $y = \frac{5}{3}x - 22$
$\angle FEC$ is an exterior angle to $\triangle AEF$ , from which we find that $60 + \angle CED = 60 + \angle AFE$ , so $\angle CED = \angle AFE$ . Similarly, we find that $\angle EDC = \angle AEF$ . Thus, $\triangle AEF \sim \triangle CDE$ . Setting up a ratio of sides, we get that $\frac{5}{x-y} = \frac{y}{2}$ . Using the previous relationship between $x$ and $y$ , we can solve for $x$
Use the quadratic formula , though we only need the root of the discriminant . This is $\sqrt{(7 \cdot 66)^2 - 4 \cdot 10 \cdot (66^2 + 90)} = \sqrt{49 \cdot 66^2 - 40 \cdot 66^2 - 4 \cdot 9 \cdot 100}$ $= \sqrt{9 \cdot 4 \cdot 33^2 - 9 \cdot 4 \cdot 100} = 6\sqrt{33^2 - 100}$ . The answer is $\boxed{989}$

Answer:  989
Problem: Let $ABC$ be an equilateral triangle , and let $D$ and $F$ be points on sides $BC$ and $AB$ , respectively, with $FA = 5$ and $CD = 2$ .  Point $E$ lies on side $CA$ such that angle $DEF = 60^{\circ}$ .  The area of triangle $DEF$ is $14\sqrt{3}$ .  The two possible values of the length of side $AB$ are $p \pm q \sqrt{r}$ , where $p$ and $q$ are rational, and $r$ is an integer not divisible by the square of a prime .  Find $r$

Solution: First of all, assume $EC=x,BD=m, ED=a, EF=b$ , then we can find $BF=m-3, AE=2+m-x$ It is not hard to find $ab*sin60^{\circ}*\frac{1}{2}=14\sqrt{3}, ab=56$ , we apply LOC on $\triangle{DEF}, \triangle{BFD}$ , getting that $(m-3)^2+m^2-m(m-3)=a^2+b^2-ab$ , leads to $a^2+b^2=m^2-3m+65$ Apply LOC on $\triangle{CED}, \triangle{AEF}$ separately, getting $4+x^2-2x=a^2; 25+(2+m-x)^2-5(2+m-x)=b^2.$ Add those terms together and use the equality $a^2+b^2=m^2-3m+65$ , we can find: $2x^2-(2m+1)x+2m-42=0$
According to basic angle chasing, $\angle{A}=\angle{C}; \angle{AFE}=\angle{CED}$ , so $\triangle{AFE}\sim \triangle{CED}$ , the ratio makes $\frac{5}{x}=\frac{2+m-x}{2}$ , getting that $x^2-(2+m)x+10=0$ Now we have two equations with $m$ , and $x$ values for both equations must be the same, so we can solve for $x$ in two equations. $x=\frac{2m+1 \pm \sqrt{4m^2+4m+1-16m+336}}{4}; x=\frac{4+2m \pm \sqrt{4m^2+16m-144}}{4}$ , then we can just use positive sign to solve, simplifies to $3+\sqrt{4m^2+16m-144}=\sqrt{4m^2-12m+337}$ , getting $m=\frac{211-3\sqrt{989}}{10}$ , since the triangle is equilateral, $AB=BC=2+m=\frac{231-3\sqrt{989}}{10}$ , and the desired answer is $\boxed{989}$

Answer:  989
Problem: A mathematical organization is producing a set of commemorative license plates. Each plate contains a sequence of five characters chosen from the four letters in AIME and the four digits in 2007. No character may appear in a sequence more times than it appears among the four letters in AIME or the four digits in 2007. A set of plates in which each possible sequence appears exactly once contains N license plates. Find $\frac{N}{10}$

Solution: There are 7 different characters that can be picked, with 0 being the only number that can be repeated twice.
Thus, $N = 2520 + 1200 = 3720$ , and $\frac{N}{10} = \boxed{372}$

Answer:  372
Problem: Find the number of ordered triples $(a,b,c)$ where $a$ $b$ , and $c$ are positive integers $a$ is a factor of $b$ $a$ is a factor of $c$ , and $a+b+c=100$

Solution: Denote $x = \frac{b}{a}$ and $y = \frac{c}{a}$ . The last condition reduces to $a(1 + x + y) = 100$ . Therefore, $1 + x + y$ is equal to one of the 9 factors of $100 = 2^25^2$
Subtracting the one, we see that $x + y = \{0,1,3,4,9,19,24,49,99\}$ . There are exactly $n - 1$ ways to find pairs of $(x,y)$ if $x + y = n$ . Thus, there are $0 + 0 + 2 + 3 + 8 + 18 + 23 + 48 + 98 = \boxed{200}$ solutions of $(a,b,c)$

Answer:  200
Problem: Square $ABCD$ has side length $13$ , and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$ . Find $EF^{2}$ [asy]unitsize(0.2 cm);  pair A, B, C, D, E, F;  A = (0,13); B = (13,13); C = (13,0); D = (0,0); E = A + (12*12/13,5*12/13); F = D + (5*5/13,-5*12/13);  draw(A--B--C--D--cycle); draw(A--E--B); draw(C--F--D);  dot("$A$", A, W); dot("$B$", B, dir(0)); dot("$C$", C, dir(0)); dot("$D$", D, W); dot("$E$", E, N); dot("$F$", F, S);[/asy]

Solution: Drawing $EF$ , it clearly passes through the center of $ABCD$ . Letting this point be $P$ , we note that $AEBP$ and $CFDP$ are congruent cyclic quadrilaterals, and that $AP=BP=CP=DP=\frac{13}{\sqrt{2}}.$ Now, from Ptolemy's, $13\cdot EP=\frac{13}{\sqrt{2}}(12+5)\implies EP=\frac{17\sqrt{2}}{2}$ . Since $EF=EP+FP=2\cdot EP$ , the answer is $(17\sqrt{2})^2=\boxed{578}.$

Answer:  578
Problem: Square $ABCD$ has side length $13$ , and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$ . Find $EF^{2}$ [asy]unitsize(0.2 cm);  pair A, B, C, D, E, F;  A = (0,13); B = (13,13); C = (13,0); D = (0,0); E = A + (12*12/13,5*12/13); F = D + (5*5/13,-5*12/13);  draw(A--B--C--D--cycle); draw(A--E--B); draw(C--F--D);  dot("$A$", A, W); dot("$B$", B, dir(0)); dot("$C$", C, dir(0)); dot("$D$", D, W); dot("$E$", E, N); dot("$F$", F, S);[/asy]

Solution: We first see that the whole figure is symmetrical and reflections across the center that we will denote as $O$ bring each half of the figure to the other half. Thus we consider a single part of the figure, namely $EO.$
First note that $\angle BAO = 45^{\circ}$ since $O$ is the center of square $ABCD.$ Also note that $\angle EAB = \arccos{\left(\frac{12}{13}\right)}$ or $\arcsin{\left(\frac{5}{13}\right)}.$ Finally, we know that $AO =\frac{13\sqrt{2}}{2}.$ Now we apply laws of cosines on $\bigtriangleup AEO.$
We have $EO^2 = 12^2 + (\frac{13\sqrt{2}}{2})^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}.$ We know that $\angle EAO = 45^{\circ} + \arccos{\left(\frac{12}{13}\right)}.$ Thus we have $\cos{\angle EAO} = \cos\left(45^{\circ} + \arccos{\left(\frac{12}{13}\right)}\right)$ which applying the cosine sum identity yields $\cos{45^{\circ}}\cos{\arccos\frac{12}{13}} - \sin{45^{\circ}}\sin\arcsin{\frac{5}{12}} =\frac{12\sqrt{2}}{26} -\frac{5\sqrt{2}}{26} =\frac{7\sqrt{2}}{26}.$
Note that we are looking for $4EO^2$ so we multiply $EO^2 = 12^2 + \left(\frac{13\sqrt{2}}{2}\right)^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}$ by $4$ obtaining $4EO^2 = 576 + 338 - 8 \cdot\left(\frac{13\sqrt{2}}{2}\right) \cdot 12 \cdot\frac{7\sqrt{2}}{26} = 576 + 338 - 4  \cdot 12 \cdot  7 = \boxed{578}.$

Answer:  578
Problem: The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, $100$ workers can produce $300$ widgets and $200$ whoosits. In two hours, $60$ workers can produce $240$ widgets and $300$ whoosits. In three hours, $50$ workers can produce $150$ widgets and $m$ whoosits. Find $m$

Solution: Suppose that it takes $x$ hours for one worker to create one widget, and $y$ hours for one worker to create one whoosit.
Therefore, we can write that (note that two hours is similar to having twice the number of workers, and so on):
Solve the system of equations with the first two equations to find that $(x,y) = \left(\frac{1}{7}, \frac{2}{7}\right)$ . Substitute this into the third equation to find that $1050 = 150 + 2m$ , so $m = \boxed{450}$

Answer:  450
Problem: An integer is called parity-monotonic if its decimal representation $a_{1}a_{2}a_{3}\cdots a_{k}$ satisfies $a_{i}<a_{i+1}$ if $a_{i}$ is odd , and $a_{i}>a_{i+1}$ if $a_{i}$ is even . How many four-digit parity-monotonic integers are there?

Solution: This problem can be solved via recursion since we are "building a string" of numbers with some condition. We want to create a new string by adding a new digit at the front so we avoid complications( $0$ can't be at the front and no digit is less than $9$ ). There are $4$ options to add no matter what(try some examples if you want) so the recursion is $S_n=4S_{n-1}$ where $S_n$ stands for the number of such numbers with $n$ digits. Since $S_1=10$ the answer is $\boxed{640}$

Answer:  640
Problem: Given a real number $x,$ let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x.$ For a certain integer $k,$ there are exactly $70$ positive integers $n_{1}, n_{2}, \ldots, n_{70}$ such that $k=\lfloor\sqrt[3]{n_{1}}\rfloor = \lfloor\sqrt[3]{n_{2}}\rfloor = \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor$ and $k$ divides $n_{i}$ for all $i$ such that $1 \leq i \leq 70.$
Find the maximum value of $\frac{n_{i}}{k}$ for $1\leq i \leq 70.$

Solution: Obviously $k$ is positive. Then, we can let $n_1$ equal $k^3$ and similarly let $n_i$ equal $k^3 + (i - 1)k$
The wording of this problem (which uses "exactly") tells us that $k^3+69k<(k+1)^3 = k^3 + 3k^2 + 3k+1 \leq k^3+70k$ . Taking away $k^3$ from our inequality results in $69k<3k^2+3k+1\leq 70k$ . Since $69k$ $3k^2+3k+1$ , and $70k$ are all integers, this inequality is equivalent to $69k\leq 3k^2+3k<70k$ . Since $k$ is positive, we can divide the inequality by $k$ to get $69 \leq 3k+3 < 70$ . Clearly the only $k$ that satisfies is $k=22$
Then, $\frac{n_{70}}{k}=k^2+69=484+69=\boxed{553}$ . (Remember we set $n_i$ equal to $k^3 + (i - 1)k$ !)

Answer:  553
Problem: rectangular piece of paper measures 4 units by 5 units. Several lines are drawn parallel to the edges of the paper. A rectangle determined by the intersections of some of these lines is called basic if
Given that the total length of all lines drawn is exactly 2007 units, let $N$ be the maximum possible number of basic rectangles determined. Find the remainder when $N$ is divided by 1000.

Solution: Denote the number of horizontal lines drawn as $x$ , and the number of vertical lines drawn as $y$ . The number of basic rectangles is $(x - 1)(y - 1)$ $5x + 4y = 2007 \Longrightarrow y = \frac{2007 - 5x}{4}$ . Substituting, we find that $(x - 1)\left(-\frac 54x + \frac{2003}4\right)$
FOIL this to get a quadratic, $-\frac 54x^2 + 502x - \frac{2003}4$ . Use $\frac{-b}{2a}$ to find the maximum possible value of the quadratic: $x = \frac{-502}{-2 \cdot \frac 54} = \frac{1004}5 \approx 201$ . However, this gives a non-integral answer for $y$ . The closest two values that work are $(199,253)$ and $(203,248)$
We see that $252 \cdot 198 = 49896 > 202 \cdot 247 = 49894$ . The solution is $\boxed{896}$

Answer:  896
Problem: Rectangle $ABCD$ is given with $AB=63$ and $BC=448.$ Points $E$ and $F$ lie on $AD$ and $BC$ respectively, such that $AE=CF=84.$ The inscribed circle of triangle $BEF$ is tangent to $EF$ at point $P,$ and the inscribed circle of triangle $DEF$ is tangent to $EF$ at point $Q.$ Find $PQ.$

Solution: Several Pythagorean triples exist amongst the numbers given. $BE = DF = \sqrt{63^2 + 84^2} = 21\sqrt{3^2 + 4^2} = 105$ . Also, the length of $EF = \sqrt{63^2 + (448 - 2\cdot84)^2} = 7\sqrt{9^2 + 40^2} = 287$
Use the Two Tangent Theorem on $\triangle BEF$ . Since both circles are inscribed in congruent triangles, they are congruent; therefore, $EP = FQ = \frac{287 - PQ}{2}$ . By the Two Tangent theorem, note that $EP = EX = \frac{287 - PQ}{2}$ , making $BX = 105 - EX = 105 - \left[\frac{287 - PQ}{2}\right]$ . Also, $BX = BY$ $FY = 364 - BY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right]$
Finally, $FP = FY = 364 - \left[105 - \left[\frac{287 - PQ}{2}\right]\right] = \frac{805 - PQ}{2}$ . Also, $FP = FQ + PQ = \frac{287 - PQ}{2} + PQ$ . Equating, we see that $\frac{805 - PQ}{2} = \frac{287 + PQ}{2}$ , so $PQ = \boxed{259}$

Answer:  259
Problem: Let $S$ be a set with six elements . Let $\mathcal{P}$ be the set of all subsets of $S.$ Subsets $A$ and $B$ of $S$ , not necessarily distinct, are chosen independently and at random from $\mathcal{P}$ . The probability that $B$ is contained in one of $A$ or $S-A$ is $\frac{m}{n^{r}},$ where $m$ $n$ , and $r$ are positive integers $n$ is prime , and $m$ and $n$ are relatively prime . Find $m+n+r.$ (The set $S-A$ is the set of all elements of $S$ which are not in $A.$

Solution: Let $|S|$ denote the number of elements in a general set $S$ . We use complementary counting.
There is a total of $2^6$ elements in $P$ , so the total number of ways to choose $A$ and $B$ is $(2^6)^2 = 2^{12}$
Note that the number of $x$ -element subset of $S$ is $\binom{6}{x}$ . In general, for $0 \le |A| \le 6$ , in order for $B$ to be in neither $A$ nor $S-A$ $B$ must have at least one element from both $A$ and $S-A$ . In other words, $B$ must contain any subset of $A$ and $S-A$ except for the empty set $\{\}$ . This can be done in $\binom{6}{|A|} (2^{|A|} - 1)(2^{6-|A|} - 1)$ ways. As $|A|$ ranges from $0$ to $6$ , we can calculate the total number of unsuccessful outcomes to be \[\sum_{|A| = 0}^{6} \binom{6}{|A|} (2^{|A|} - 1)(2^{6-|A|} - 1) = 2702.\] So our desired answer is \[1 - \dfrac{2702}{2^{12}} = \dfrac{697}{2^{11}} \Rightarrow \boxed{710}.\]

Answer:  710
Problem: Let $S$ be a set with six elements . Let $\mathcal{P}$ be the set of all subsets of $S.$ Subsets $A$ and $B$ of $S$ , not necessarily distinct, are chosen independently and at random from $\mathcal{P}$ . The probability that $B$ is contained in one of $A$ or $S-A$ is $\frac{m}{n^{r}},$ where $m$ $n$ , and $r$ are positive integers $n$ is prime , and $m$ and $n$ are relatively prime . Find $m+n+r.$ (The set $S-A$ is the set of all elements of $S$ which are not in $A.$

Solution: To begin with, we note that there are $2^6$ subsets of $S$ (which we can assume is $\{1,2,3,4,5,6\}$ ), including the null set. This gives a total of $(2^6)^2 = 2^{12}$ total possibilities for A and B.
Case 1: B is contained in A only.
If B has $0$ elements, which occurs in $\binom{6}{0}$ ways, A can be anything, giving us $\binom{6}{0} \cdot 2^6$ . If B has $1$ element, A must contain that element, and then the remaining 5 are free to be in A or not in A. This gives us $\binom{6}{1} \cdot 2^5$ . Summing, we end up with the binomial expansion of $(2 + 1)^6 = 3^6$
Case 2: B is contained in S-A only.
By symmetry, this case is the same as Case 1, once again giving us $3^6$ possibilities.
Case 3: B is contained in both.
We claim here that B can only be the null set. For contradiction, assume that there exists some element $x$ in B which satisfies this restriction. Then, A must contain $x$ as well, but we also know that $S-A$ contains $x$ , contradiction. Hence, B is the null set, whereas A can be anything. This gives us $2^6$ possibilities.
Since we have overcounted Case 3 in both of the other two cases, our final count is $2 \cdot 3^6 - 2^6$ . This gives us the probability $\frac{2 \cdot 3^6 - 2^6}{2^{12}}$ . Upon simplifying, we end up with $\frac{697}{2^{11}}$ , giving the desired answer of $\boxed{710}$ .
- Spacesam

Answer:  710