Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $S$ be the set of points in the Cartesian plane that satisfy
If a model of $S$ were built from wire of negligible thickness, then the total length of wire required would be $a\sqrt{b}$ , where $a$ and $b$ are positive integers and $b$ is not divisible by the square of any prime number. Find $a+b$

Solution: Since $0 \le \Big|\big||x| - 2\big| - 1\Big| \le 1$ and $0 \le \Big|\big||y| - 2\big| - 1\Big| \le 1$ $- 1 \le \big||x| - 2\big| - 1 \le 1$ $0 \le \big||x| - 2\big| \le 2$ $- 2 \le |x| - 2 \le 2$ $- 4 \le x \le 4$ Also $- 4 \le y \le 4$
Define $f(a) = \Big|\big||a| - 2\big| - 1\Big|$
[this is also true for horizontal reflection, with $3 \le y \le 4$ , etc]
So it is only necessary to find the length of the function at $3 \le x \le 4$ and $3 \le y \le 4$ $\Big|\big||x| - 2\big| - 1\Big| + \Big|\big||y| - 2\big| - 1\Big| = 1$ $x - 3 + y - 3 = 1$ $y = - x + 7$ (Length = $\sqrt {2}$
This graph is reflected over the line y=3, the quantity of which is reflected over y=2,
So a total of $6$ doublings = $2^6$ $64$ , the total length = $64 \cdot \sqrt {2} = a\sqrt {b}$ , and $a + b = 64 + 2 = \boxed{066}$

Answer:  066
Problem: Let $S$ be the set of points in the Cartesian plane that satisfy
If a model of $S$ were built from wire of negligible thickness, then the total length of wire required would be $a\sqrt{b}$ , where $a$ and $b$ are positive integers and $b$ is not divisible by the square of any prime number. Find $a+b$

Solution: We make use of several consecutive substitutions. 
Let $||x| - 2|= x_1$ and similarly with $y$ .
Therefore, our graph is $|x_1 - 1| + |y_1 - 1| = 1$ . This is a rhombus with perimeter $4\sqrt{2}$ . Now, we make use of the following fact for a function of two variables $x$ and $y$ : Suppose we have $f(x, y) = c$ . Then $f(|x|, |y|)$ is equal to the graph of $f(x, y)$ reflected across the y axis and x axis, and the reflection across the y axis across the x axis, therefore the perimeter of  of $f(|x|, |y|)$ is 4 times the perimeter of $f(x, y)$ . Now, we continue making substitutions at each absolute value sign ( $|x| - 2 = x_2$ and finally $x = x_3$ , similarly for y as well. ),  noting that the constants don't matter as they just translate the graph and each absolute value sign increases the perimeter 4 times as much. Therefore, the length is $4^2 \times 4\sqrt{2} = 64\sqrt{2} = a\sqrt{b}$ , and $a+b = \boxed{066}$

Answer:  066
Problem: Let $S$ be the set of points in the Cartesian plane that satisfy
If a model of $S$ were built from wire of negligible thickness, then the total length of wire required would be $a\sqrt{b}$ , where $a$ and $b$ are positive integers and $b$ is not divisible by the square of any prime number. Find $a+b$

Solution: In a similar manner to the other posted solutions, we want to find the length of just a piece of this monstrosity and then figure out what fraction of the entire function that piece represents. It's possible to go right for the smallest possible piece that'll help us by letting $a = \big||x|-2\big|-1$ and $b = \big||y|-2\big|-1$ . We then get that $b = 1 - a$ (obviously the other way would work too). One specific "case" here is the one where all the numbers in the absolute value bars are positive, and thus where $b = 1 - x - 2 = -x - 1$ . Since we're dealing with compounded absolute values, we can infer that the piece we just found has to be bounded by the two axes; the length of this piece also won't be affected by the constants that are hidden in $b$ . Using the Pythagorean Theorem / distance formula, that tiny length comes out to be $\sqrt{2}$ . 
Because we're working with linear terms only, the presence of each absolute value bar will double the length of the function's graph at each step as we work backward to our original equation. As there are six absolute value bars to be added back in (three on each side), we find that there are $2^6$ of these pieces of length $\sqrt{2}$ , making our answer $2 + 64 = \boxed{066}$

Answer:  066
Problem: Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ .  Let $\frac{m}{n}$ be the probability that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $m$ and $n$ are relatively prime positive integers .  Find $m+n$

Solution: Define $\theta = 2\pi/1997$ . By De Moivre's Theorem the roots are given by
Now, let $v$ be the root corresponding to $m\theta=2m\pi/1997$ , and let $w$ be the root corresponding to $n\theta=2n\pi/ 1997$ . Then \begin{align*} |v+w|^2 &= \left(\cos(m\theta) + \cos(n\theta)\right)^2 + \left(\sin(m\theta) + \sin(n\theta)\right)^2 \\ &= 2 + 2\cos\left(m\theta\right)\cos\left(n\theta\right) + 2\sin\left(m\theta\right)\sin\left(n\theta\right) \end{align*} The cosine difference identity simplifies that to \[|v+w|^2 = 2+2\cos((m-n)\theta)\]
We need $|v+w|^2 \ge 2+\sqrt{3}$ , which simplifies to \[\cos((m-n)\theta) \ge \frac{\sqrt{3}}{2}\] Thus, \[|m - n| \le \frac{\pi}{6} \cdot \frac{1997}{2 \pi} = \left\lfloor \frac{1997}{12} \right\rfloor =166\]
Therefore, $m$ and $n$ cannot be more than $166$ away from each other.  This means that for a given value of $m$ , there are $332$ values for $n$ that satisfy the inequality; $166$ of them $> m$ , and $166$ of them $< m$ .  Since $m$ and $n$ must be distinct, $n$ can have $1996$ possible values.  Therefore, the probability is $\frac{332}{1996}=\frac{83}{499}$ .  The answer is then $499+83=\boxed{582}$

Answer:  582
Problem: Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ .  Let $\frac{m}{n}$ be the probability that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $m$ and $n$ are relatively prime positive integers .  Find $m+n$

Solution: The solutions of the equation $z^{1997} = 1$ are the $1997$ th roots of unity and are equal to $\text{cis}(\theta_k)$ , where $\theta_k = \tfrac {2\pi k}{1997}$ for $k = 0,1,\ldots,1996.$ Thus, they are located at uniform intervals on the unit circle in the complex plane.
The quantity $|v+w|$ is unchanged upon rotation around the origin, so, WLOG, we can assume $v=1$ after rotating the axis till $v$ lies on the real axis. Let $w=\text{cis}(\theta_k)$ . Since $w\cdot \overline{w}=|w|^2=1$ and $w+\overline{w}=2\text{Re}(w) = 2\cos\theta_k$ , we have \[|v + w|^2  =  (1+w)(1+\overline{w}) = 2+2\cos\theta_k\] We want $|v + w|^2\ge 2 + \sqrt {3}.$ From what we just obtained, this is equivalent to \[\cos\theta_k\ge \frac {\sqrt {3}}2 \qquad \Leftrightarrow \qquad -\frac {\pi}6\le \theta_k \le  \frac {\pi}6\] which is satisfied by $k = 166,165,\ldots, - 165, - 166$ (we don't include 0 because that corresponds to $v$ ).  So out of the $1996$ possible $k$ $332$ work.  Thus, $m/n = 332/1996 = 83/499.$ So our answer is $83 + 499 = \boxed{582}.$

Answer:  582
Problem: Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ .  Let $\frac{m}{n}$ be the probability that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $m$ and $n$ are relatively prime positive integers .  Find $m+n$

Solution: We can solve a geometrical interpretation of this problem.
Without loss of generality, let $u = 1$ . We are now looking for a point exactly one unit away from $u$ such that the point is at least $\sqrt{2 + \sqrt{3}}$ units away from the origin. Note that the "boundary" condition is when the point will be exactly $\sqrt{2+\sqrt{3}}$ units away from the origin; these points will be the intersections of the circle centered at $(1,0)$ with radius $1$ and the circle centered at $(0,0)$ with radius $\sqrt{2+\sqrt{3}}$ . The equations of these circles are $(x-1)^2 = 1$ and $x^2 + y^2 = 2 + \sqrt{3}$ . Solving for $x$ yields $x = \frac{\sqrt{3}}{2}$ . Clearly, this means that the real part of $v$ is greater than $\frac{\sqrt{3}}{2}$ . Solving, we note that $332$ possible $v$ s exist, meaning that $\frac{m}{n} = \frac{332}{1996} = \frac{83}{499}$ . Therefore, the answer is $83 + 499 = \boxed{582}$

Answer:  582
Problem: Let $v$ and $w$ be distinct, randomly chosen roots of the equation $z^{1997}-1=0$ .  Let $\frac{m}{n}$ be the probability that $\sqrt{2+\sqrt{3}}\le\left|v+w\right|$ , where $m$ and $n$ are relatively prime positive integers .  Find $m+n$

Solution: Since $z^{1997}=1$ , the roots will have magnitude $1$ . Thus, the roots can be written as $\cos(\theta)+i\sin(\theta)$ and $\cos(\omega)+i\sin(\omega)$ for some angles $\theta$ and $\omega$ . We rewrite the requirement as $\sqrt{2+\sqrt3}\le|\cos(\theta)+\cos(\omega)+i\sin(\theta)+i\sin(\omega)|$ , which can now be easily manipulated to $2+\sqrt{3}\le(\cos(\theta)+\cos(\omega))^2+(\sin(\theta)+\sin(\omega))^2$
WLOG, let $\theta = 0$ . Thus, our inequality becomes $2+\sqrt{3}\le(1+\cos(\omega))^2+(\sin(\omega))^2$ $2+\sqrt{3}\le2+2\cos(\omega)$ , and finally $\cos(\omega)\ge\frac{\sqrt{3}}{2}$ . Obviously, $\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}$ , and thus it follows that, on either side of a given point, $\frac{1997}{12}\approx166$ points will work. The probability is $\frac{166\times2}{1996} = \frac{83}{499}$ , and thus our requested sum is $\boxed{582}$ ~SigmaPiE

Answer:  582
Problem: The sides of rectangle $ABCD$ have lengths $10$ and $11$ . An equilateral triangle is drawn so that no point of the triangle lies outside $ABCD$ . The maximum possible area of such a triangle can be written in the form $p\sqrt{q}-r$ , where $p$ $q$ , and $r$ are positive integers, and $q$ is not divisible by the square of any prime number. Find $p+q+r$

Solution: Consider points on the complex plane $A (0,0),\ B (11,0),\ C (11,10),\ D (0,10)$ . Since the rectangle is quite close to a square, we figure that the area of the equilateral triangle is maximized when a vertex of the triangle coincides with that of the rectangle. Set one vertex of the triangle at $A$ , and the other two points $E$ and $F$ on $BC$ and $CD$ , respectively. Let $E (11,a)$ and $F (b, 10)$ . Since it's equilateral, then $E\cdot\text{cis}60^{\circ} = F$ , so $(11 + ai)\left(\frac {1}{2} + \frac {\sqrt {3}}{2}i\right) = b + 10i$ , and expanding we get $\left(\frac {11}{2} - \frac {a\sqrt {3}}{2}\right) + \left(\frac {11\sqrt {3}}{2} + \frac {a}{2}\right)i = b + 10i$
We can then set the real and imaginary parts equal, and solve for $(a,b) = (20 - 11\sqrt {3},22 - 10\sqrt {3})$ . Hence a side $s$ of the equilateral triangle can be found by $s^2 = AE^2 = a^2 + AB^2 = 884 - 440\sqrt{3}$ . Using the area formula $\frac{s^2\sqrt{3}}{4}$ , the area of the equilateral triangle is $\frac{(884-440\sqrt{3})\sqrt{3}}{4} = 221\sqrt{3} - 330$ . Thus $p + q + r = 221 + 3 + 330 = \boxed{554}$

Answer:  554
Problem: The sides of rectangle $ABCD$ have lengths $10$ and $11$ . An equilateral triangle is drawn so that no point of the triangle lies outside $ABCD$ . The maximum possible area of such a triangle can be written in the form $p\sqrt{q}-r$ , where $p$ $q$ , and $r$ are positive integers, and $q$ is not divisible by the square of any prime number. Find $p+q+r$

Solution: Since $\angle{BAD}=90$ and $\angle{EAF}=60$ , it follows that $\angle{DAF}+\angle{BAE}=90-60=30$ . Rotate triangle $ADF$ $60$ degrees clockwise. Note that the image of $AF$ is $AE$ . Let the image of $D$ be $D'$ . Since angles are preserved under rotation, $\angle{DAF}=\angle{D'AE}$ . It follows that $\angle{D'AE}+\angle{BAE}=\angle{D'AB}=30$ . Since $\angle{ADF}=\angle{ABE}=90$ , it follows that quadrilateral $ABED'$ is cyclic with circumdiameter $AE=s$ and thus circumradius $\frac{s}{2}$ . Let $O$ be its circumcenter. By Inscribed Angles, $\angle{BOD'}=2\angle{BAD'}=60$ . By the definition of circle, $OB=OD'$ . It follows that triangle $OBD'$ is equilateral. Therefore, $BD'=r=\frac{s}{2}$ . Applying the Law of Cosines to triangle $ABD'$ $\frac{s}{2}=\sqrt{10^2+11^2-(2)(10)(11)(\cos{30})}$ . Squaring and multiplying by $\sqrt{3}$ yields $\frac{s^2\sqrt{3}}{4}=221\sqrt{3}-330\implies{p+q+r=221+3+330=\boxed{554}$

Answer:  554
Problem: In a magic square , the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of the entries of a magic square. Find $x$

Solution: Let's make a table.
\[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table}}\\\hline x&19&96\\\hline 1&a&b\\\hline c&d&e\\\hline \end{array}\]
\begin{eqnarray*} x+19+96=x+1+c\Rightarrow c=19+96-1=114,\\ 114+96+a=x+1+114\Rightarrow a=x-95 \end{eqnarray*}
\[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table in progress}}\\\hline x&19&96\\\hline 1&x-95&b\\\hline 114&d&e\\\hline \end{array}\]
\begin{eqnarray*}19+x-95+d=x+d-76=115+x\Rightarrow  d=191,\\ 114+191+e=x+115\Rightarrow e=x-190 \end{eqnarray*} \[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table in progress}}\\\hline x&19&96\\\hline 1&x-95&b\\\hline 114&191&x-190\\\hline \end{array}\]
\[3x-285=x+115\Rightarrow 2x=400\Rightarrow x=\boxed{200}\]

Answer:  200
Problem: In a magic square , the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of the entries of a magic square. Find $x$

Solution: Use the table from above. Obviously $c = 114$ . Hence $a+e = 115$ . Similarly, $1+a = 96 + e \Rightarrow a = 95+e$
Substitute that into the first to get $2e = 20 \Rightarrow e=10$ , so $a=105$ , and so the value of $x$ is just $115+x = 210 + 105 \Rightarrow x = \boxed{200}$

Answer:  200
Problem: In a magic square , the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of the entries of a magic square. Find $x$

Solution: \[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table}}\\\hline x&19&96\\\hline 1&a&b\\\hline c&d&e\\\hline \end{array}\] The formula \[e=\frac{1+19}{2}\] can be used. Therefore, $e=10$ . Similarly, \[96=\frac{1+d}{2}\] So $d=191$
Now we have this table: \[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table}}\\\hline x&19&96\\\hline 1&a&b\\\hline c&191&10\\\hline \end{array}\] By property of magic squares, observe that \[x+a+10=19+a+191\] The $a$ 's cancel! We now have \[x+10=19+191\] Thus $x=\boxed{200}.$

Answer:  200
Problem: For each real number $x$ , let $\lfloor x \rfloor$ denote the greatest integer that does not exceed x. For how many positive integers $n$ is it true that $n<1000$ and that $\lfloor \log_{2} n \rfloor$ is a positive even integer?

Solution: For integers $k$ , we want $\lfloor \log_2 n\rfloor = 2k$ , or $2k \le \log_2 n < 2k+1 \Longrightarrow 2^{2k} \le n < 2^{2k+1}$ . Thus, $n$ must satisfy these inequalities (since $n < 1000$ ):
There are $4$ for the first inequality, $16$ for the second, $64$ for the third, and $256$ for the fourth, so the answer is $4+16+64+256=\boxed{340}$

Answer:  340
Problem: Find the smallest positive integer $n$ for which the expansion of $(xy-3x+7y-21)^n$ , after like terms have been collected, has at least 1996 terms.

Solution: Using Simon's Favorite Factoring Trick , we rewrite as $[(x+7)(y-3)]^n = (x+7)^n(y-3)^n$ . Both binomial expansions will contain $n+1$ non-like terms; their product will contain $(n+1)^2$ terms, as each term will have an unique power of $x$ or $y$ and so none of the terms will need to be collected. Hence $(n+1)^2 \ge 1996$ , the smallest square after $1996$ is $2025 = 45^2$ , so our answer is $45 - 1 = \boxed{044}$

Answer:  044
Problem: Find the smallest positive integer $n$ for which the expansion of $(xy-3x+7y-21)^n$ , after like terms have been collected, has at least 1996 terms.

Solution: Notice that the coefficients in the problem statement have no effect on how many unique terms there will be in the expansion. Therefore this problem is synonymous with finding the amount of terms in the expansion of $(xy+x+y+1)^n$ (we do this to simplify the problem).
If we expand the exponent the expression becomes $\underbrace{(xy+x+y+1)\cdot(xy+x+y+1)\cdots(xy+x+y+1)}_{n\;times}$
This is equivalent to starting off with a $x^0y^0$ terms and choosing between $4$ options $n$ different times:
$\bullet$ Adding nothing to either exponent (choosing $1$ ).
$\bullet$ Adding $1$ to the $x$ exponent (choosing $x$ ).
$\bullet$ Adding $1$ to the $y$ exponent (choosing $y$ ).
$\bullet$ Adding $1$ to the $x$ exponent and adding $1$ to the $y$ exponent (choosing $xy$ ).
Doing this $n$ times, you can see that you end up with a term in the form $k\cdot x^i\cdot y^j$ where $k$ is some coefficient (which we don't care about) and $0\le i,j\le n$ and $i,j \in \mathbb{N}$
Repeating this for all possible combinations of choices yields $n+1$ options for each of $i$ and $j$ which means there are a total of $(n+1)^2$ possible terms in the form $x^i\cdot y^j$ . Therefore $(xy+x+y+1)^n$ has $(n+1)^2$ terms.
$(n+1)^2 \ge 1996$ which yields $n=44$ $\boxed{044}$

Answer:  044
Problem: A wooden cube , whose edges are one centimeter long, rests on a horizontal surface. Illuminated by a point source of light that is $x$ centimeters directly above an upper vertex , the cube casts a shadow on the horizontal surface. The area of the shadow, which does not include the area beneath the cube is 48 square centimeters. Find the greatest integer that does not exceed $1000x$

Solution: [asy] import three; size(250);defaultpen(0.7+fontsize(9)); real unit = 0.5;  real r = 2.8; triple O=(0,0,0), P=(0,0,unit+unit/(r-1)); dot(P); draw(O--P); draw(O--(unit,0,0)--(unit,0,unit)--(0,0,unit)); draw(O--(0,unit,0)--(0,unit,unit)--(0,0,unit)); draw((unit,0,0)--(unit,unit,0)--(unit,unit,unit)--(unit,0,unit)); draw((0,unit,0)--(unit,unit,0)--(unit,unit,unit)--(0,unit,unit));  draw(P--(r*unit,0,0)--(r*unit,r*unit,0)--(0,r*unit,0)--P); draw(P--(r*unit,r*unit,0)); draw((r*unit,0,0)--(0,0,0)--(0,r*unit,0)); draw(P--(0,0,unit)--(unit,0,unit)--(unit,0,0)--(r*unit,0,0)--P,dashed+blue+linewidth(0.8)); label("$x$",(0,0,unit+unit/(r-1)/2),WSW); label("$1$",(unit/2,0,unit),N); label("$1$",(unit,0,unit/2),W); label("$1$",(unit/2,0,0),N); label("$6$",(unit*(r+1)/2,0,0),N); label("$7$",(unit*r,unit*r/2,0),SW); [/asy] (Figure not to scale) The area of the square shadow base is $48 + 1 = 49$ , and so the sides of the shadow are $7$ . Using the similar triangles in blue, $\frac {x}{1} = \frac {1}{6}$ , and $\left\lfloor 1000x \right\rfloor = \boxed{166}$

Answer:  166
Problem: In a five-team tournament, each team plays one game with every other team. Each team has a $50\%$ chance of winning any game it plays. (There are no ties.) Let $\dfrac{m}{n}$ be the probability that the tournament will produce neither an undefeated team nor a winless team, where $m$ and $n$ are relatively prime integers. Find $m+n$

Solution: We can use complementary counting : finding the probability that at least one team wins all games or at least one team loses all games.
No more than 1 team can win or lose all games, so at most one team can win all games and at most one team can lose all games.
Now we use PIE
The probability that one team wins all games is $5\cdot \left(\frac{1}{2}\right)^4=\frac{5}{16}$
Similarity, the probability that one team loses all games is $\frac{5}{16}$
The probability that one team wins all games and another team loses all games is $\left(5\cdot \left(\frac{1}{2}\right)^4\right)\left(4\cdot \left(\frac{1}{2}\right)^3\right)=\frac{5}{32}$
$\frac{5}{16}+\frac{5}{16}-\frac{5}{32}=\frac{15}{32}$
Since this is the opposite of the probability we want, we subtract that from 1 to get $\frac{17}{32}$
$17+32=\boxed{049}$

Answer:  049
Problem: In a five-team tournament, each team plays one game with every other team. Each team has a $50\%$ chance of winning any game it plays. (There are no ties.) Let $\dfrac{m}{n}$ be the probability that the tournament will produce neither an undefeated team nor a winless team, where $m$ and $n$ are relatively prime integers. Find $m+n$

Solution: There are $\dbinom{5}{2} = 10$ games in total, and every game can either end in a win or a loss. Therefore, there are $2^{10} = 1024$ possible outcomes.
Now, computing this probability directly seems a little hard, so let's compute the complement -- the probability that there is an undefeated team, a winless team, or both.
The number of ways that we can have an undefeated team is $5 \cdot 2^6,$ as there are five ways to choose the team itself and $2^6$ outcomes for the games that are not concerning the undefeated team. Reversing all the wings to losses yields a winless team, so this situation is symmetrical to the first one. Therefore, there are $5 \cdot 2^6 + 5 \cdot 2^6$ ways for an undefeated or winless team.
In the process, however, we have accidentally overcounted the scenario of both an undefeated and winless team. There are $5 \cdot 4 \cdot 2^3$ ways for this scenario to happen, because there are five ways to choose the undefeated team, four ways for the winless, and three games that don't concern either of the teams. Therefore, there are $5 \cdot 2^6 + 5 \cdot 2^6 - 5 \cdot 4 \cdot 2^3 = 480$ ways to have an undefeated and/or winless team, so there are $1024 - 480 = 544$ to not have any.
Our probability is thus $\dfrac{544}{1024},$ which simplifies to $\dfrac{17}{32},$ for our answer of $17 + 32 = \boxed{049}.$

Answer:  049
Problem: Two squares of a $7\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible?

Solution: There are ${49 \choose 2}$ possible ways to select two squares to be painted yellow. There are four possible ways to rotate each board. Given an arbitrary pair of yellow squares, these four rotations will either yield two or four equivalent but distinct boards.
Note that a pair of yellow squares will only yield $2$ distinct boards upon rotation iff the yellow squares are rotationally symmetric about the center square; there are $\frac{49-1}{2}=24$ such pairs. There are then ${49 \choose 2}-24$ pairs that yield $4$ distinct boards upon rotation; in other words, for each of the ${49 \choose 2}-24$ pairs, there are three other pairs that yield an equivalent board.
Thus, the number of inequivalent boards is $\frac{{49 \choose 2} - 24}{4} + \frac{24}{2} = \boxed{300}$ . For a $(2n+1) \times (2n+1)$ board, this argument generalizes to $n(n+1)(2n^2+2n+1)$ inequivalent configurations.

Answer:  300
Problem: Two squares of a $7\times 7$ checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane board. How many inequivalent color schemes are possible?

Solution: There are 4 cases: 
1. The center square is occupied, in which there are $12$ cases.
2. The center square isn't occupied and the two squares that are opposite to each other with respect to the center square, in which there are $12$ cases.
3. The center square isn't occupied and the two squares can rotate to each other with a $90^{\circ}$ rotation with each other and with respect to the center square, in which case there are $12$ cases.
4. And finally, the two squares can't rotate to each other with respect to the center square in which case there are $\dbinom{12}{2} \cdot \frac{16}{4} = 264$ cases. 
Add up all the values for each case to get $\boxed{300}$ as your answer.

Answer:  300
Problem: The harmonic mean of two positive integers is the reciprocal of the arithmetic mean of their reciprocals. For how many ordered pairs of positive integers $(x,y)$ with $x<y$ is the harmonic mean of $x$ and $y$ equal to $6^{20}$

Solution: The harmonic mean of $x$ and $y$ is equal to $\frac{1}{\frac{\frac{1}{x}+\frac{1}{y}}2} = \frac{2xy}{x+y}$ , so we have $xy=(x+y)(3^{20}\cdot2^{19})$ , and by SFFT $(x-3^{20}\cdot2^{19})(y-3^{20}\cdot2^{19})=3^{40}\cdot2^{38}$ . Now, $3^{40}\cdot2^{38}$ has $41\cdot39=1599$ factors, one of which is the square root ( $3^{20}2^{19}$ ). Since $x<y$ , the answer is half of the remaining number of factors, which is $\frac{1599-1}{2}= \boxed{799}$

Answer:  799