Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: There are $2n$ complex numbers that satisfy both $z^{28} - z^{8} - 1 = 0$ and $\mid z \mid = 1$ . These numbers have the form $z_{m} = \cos\theta_{m} + i\sin\theta_{m}$ , where $0\leq\theta_{1} < \theta_{2} < \ldots < \theta_{2n} < 360$ and angles are measured in degrees. Find the value of $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$

Solution: $z$ can be written in the form $\text{cis\,}\theta$ . Rearranging, we find that $\text{cis\,}{28}\theta = \text{cis\,}{8}\theta+1$
Since the real part of $\text{cis\,}{28}\theta$ is one more than the real part of $\text{cis\,} {8}\theta$ and their imaginary parts are equal, it is clear that either $\text{cis\,}{28}\theta = \frac{1}{2}+\frac {\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta =  -\frac{1}{2}+\frac {\sqrt{3}}{2}i$ , or $\text{cis\,}{28}\theta = \frac{1}{2} - \frac{\sqrt{3}}{2}i$ and $\text{cis\,} {8}\theta =  -\frac{1}{2}- \frac{\sqrt{3}}{2}i$
Setting up and solving equations, $Z^{28}= \text{cis\,}{60^\circ}$ and $Z^8= \text{cis\,}{120^\circ}$ , we see that the solutions common to both equations have arguments $15^\circ , 105^\circ, 195^\circ,$ and $\ 285^\circ$ . We can figure this out by adding 360 repeatedly to the number 60 to try and see if it will satisfy what we need. We realize that it does not work in the integer values.
Again setting up equations ( $Z^{28}= \text{cis\,}{300^\circ}$ and $Z^{8} = \text{cis\,}{240^\circ}$ ) we see that the common solutions have arguments of $75^\circ, 165^\circ, 255^\circ,$ and $345^\circ$
Listing all of these values, we find that $\theta_{2} + \theta_{4} + \ldots + \theta_{2n}$ is equal to $(75 + 165 + 255 + 345) ^\circ$ which is equal to $\boxed{840}$ degrees. We only want the sum of a certain number of theta, not all of it.

Answer:  840
Problem: Let $EFGH$ $EFDC$ , and $EHBC$ be three adjacent square faces of a cube , for which $EC = 8$ , and let $A$ be the eighth vertex of the cube. Let $I$ $J$ , and $K$ , be the points on $\overline{EF}$ $\overline{EH}$ , and $\overline{EC}$ , respectively, so that $EI = EJ = EK = 2$ . A solid $S$ is obtained by drilling a tunnel through the cube. The sides of the tunnel are planes parallel to $\overline{AE}$ , and containing the edges, $\overline{IJ}$ $\overline{JK}$ , and $\overline{KI}$ . The surface area of $S$ , including the walls of the tunnel, is $m + n\sqrt {p}$ , where $m$ $n$ , and $p$ are positive integers and $p$ is not divisible by the square of any prime. Find $m + n + p$

Solution: Set the coordinate system so that vertex $E$ , where the drilling starts, is at $(8,8,8)$ . Using a little visualization (involving some similar triangles , because we have parallel lines) shows that the tunnel meets the bottom face (the xy plane one) in the line segments joining $(1,0,0)$ to $(2,2,0)$ , and $(0,1,0)$ to $(2,2,0)$ , and similarly for the other three faces meeting at the origin (by symmetry). So one face of the tunnel is the polygon with vertices (in that order), $S(1,0,0), T(2,0,2), U(8,6,8), V(8,8,6), W(2,2,0)$ , and the other two faces of the tunnel are congruent to this shape.
Observe that this shape is made up of two congruent trapezoids each with height $\sqrt {2}$ and bases $7\sqrt {3}$ and $6\sqrt {3}$ . Together they make up an area of $\sqrt {2}(7\sqrt {3} + 6\sqrt {3}) = 13\sqrt {6}$ . The total area of the tunnel is then $3\cdot13\sqrt {6} = 39\sqrt {6}$ . Around the corner $E$ we're missing an area of $6$ , the same goes for the corner opposite $E$ . So the outside area is $6\cdot 64 - 2\cdot 6 = 372$ . Thus the the total surface area is $372 + 39\sqrt {6}$ , and the answer is $372 + 39 + 6 = \boxed{417}$

Answer:  417
Problem: Find the least positive integer $n$ such that no matter how $10^{n}$ is expressed as the product of any two positive integers, at least one of these two integers contains the digit $0$

Solution: If a factor of $10^{n}$ has a $2$ and a $5$ in its prime factorization , then that factor will end in a $0$ . Therefore, we have left to consider the case when the two factors have the $2$ s and the $5$ s separated, so we need to find the first power of 2 or 5 that contains a 0.
For $n = 1:$ \[2^1 = 2 , 5^1 = 5\] $n = 2:$ \[2^2 = 4 , 5 ^ 2 =25\] $n = 3:$ \[2^3 = 8 , 5 ^3 = 125\]
and so on, until,
$n = 8:$ $2^8 = 256$ $5^8 = 390625$
We see that $5^8$ contains the first zero, so $n = \boxed{8}$

Answer:  8
Problem: Let $u$ and $v$ be integers satisfying $0 < v < u$ . Let $A = (u,v)$ , let $B$ be the reflection of $A$ across the line $y = x$ , let $C$ be the reflection of $B$ across the y-axis, let $D$ be the reflection of $C$ across the x-axis, and let $E$ be the reflection of $D$ across the y-axis. The area of pentagon $ABCDE$ is $451$ . Find $u + v$

Solution: Since $A = (u,v)$ , we can find the coordinates of the other points: $B = (v,u)$ $C = (-v,u)$ $D = (-v,-u)$ $E = (v,-u)$ . If we graph those points, we notice that since the latter four points are all reflected across the x/y-axis, they form a rectangle, and $ABE$ is a triangle. The area of $BCDE$ is $(2u)(2v) = 4uv$ and the area of $ABE$ is $\frac{1}{2}(2u)(u-v) = u^2 - uv$ . Adding these together, we get $u^2 + 3uv = u(u+3v) = 451 = 11 \cdot 41$ . Since $u,v$ are positive, $u+3v>u$ , and by matching factors we get either $(u,v) = (1,150)$ or $(11,10)$ . Since $v < u$ the latter case is the answer, and $u+v = \boxed{021}$

Answer:  021
Problem: Let $u$ and $v$ be integers satisfying $0 < v < u$ . Let $A = (u,v)$ , let $B$ be the reflection of $A$ across the line $y = x$ , let $C$ be the reflection of $B$ across the y-axis, let $D$ be the reflection of $C$ across the x-axis, and let $E$ be the reflection of $D$ across the y-axis. The area of pentagon $ABCDE$ is $451$ . Find $u + v$

Solution: We find the coordinates like in the solution above: $A = (u,v)$ $B = (v,u)$ $C = (-v,u)$ $D = (-v,-u)$ $E = (v,-u)$ . Then we apply the Shoelace Theorem \[A =  \frac{1}{2}[(u^2 + vu + vu + vu + v^2) - (v^2 - uv - uv - uv -u^2)] = 451\] \[\frac{1}{2}(2u^2 + 6uv) = 451\] \[u(u + 3v) = 451\]
This means that $(u,v) = (11, 10)$ or $(1,150)$ , but since $v < u$ , then the answer is $\boxed{021}$

Answer:  021
Problem: In the expansion of $(ax + b)^{2000},$ where $a$ and $b$ are relatively prime positive integers, the coefficients of $x^{2}$ and $x^{3}$ are equal. Find $a + b$

Solution: Using the binomial theorem $\binom{2000}{1998} b^{1998}a^2 = \binom{2000}{1997}b^{1997}a^3 \Longrightarrow b=666a$
Since $a$ and $b$ are positive relatively prime integers, $a=1$ and $b=666$ , and $a+b=\boxed{667}$

Answer:  667
Problem: The diagram shows a rectangle that has been dissected into nine non-overlapping squares . Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.
[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy]

Solution: Call the squares' side lengths from smallest to largest $a_1,\ldots,a_9$ , and let $l,w$ represent the dimensions of the rectangle.
The picture shows that \begin{align*} a_1+a_2 &= a_3\\ a_1 + a_3 &= a_4\\ a_3 + a_4 &= a_5\\ a_4 + a_5 &= a_6\\ a_2 + a_3 + a_5 &= a_7\\ a_2 + a_7 &= a_8\\ a_1 + a_4 + a_6 &= a_9\\ a_6 + a_9 &= a_7 + a_8.\end{align*}
Expressing all terms 3 to 9 in terms of $a_1$ and $a_2$ and substituting their expanded forms into the previous equation will give the expression $5a_1 = 2a_2$
We can guess that $a_1 = 2$ . (If we started with $a_1$ odd, the resulting sides would not be integers and we would need to scale up by a factor of $2$ to make them integers; if we started with $a_1 > 2$ even, the resulting dimensions would not be relatively prime and we would need to scale down.) Then solving gives $a_9 = 36$ $a_6=25$ $a_8 = 33$ , which gives us $l=61,w=69$ .  These numbers are relatively prime, as desired. The perimeter is $2(61)+2(69)=\boxed{260}$

Answer:  260
Problem: The diagram shows a rectangle that has been dissected into nine non-overlapping squares . Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.
[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy]

Solution: We can just list the equations: \begin{align*} s_3 &= s_1 + s_2 \\ s_4 &= s_3 + s_1 \\ s_5 &= s_4 + s_3 \\ s_6 &= s_5 + s_4 \\ s_7 &= s_5 + s_3 + s_2 \\ s_8 &= s_7 + s_2 \\ s_9 &= s_8 + s_2 - s_1 \\ s_9 + s_8 &= s_7 + s_6 + s_5 \end{align*} We can then write each $s_i$ in terms of $s_1$ and $s_2$ as follows \begin{align*} s_4 &= 2s_1 + s_2 \\ s_5 &= 3s_1 +2s_2 \\ s_6 &= 5s_1 + 3s_2 \\ s_7 &= 4s_1 + 4s_2 \\ s_8 &= 4s_1 + 5s_2 \\ s_9 &= 3s_1 + 6s_2 \\ \end{align*} Since $s_9 + s_8 = s_7 + s_6 + s_5 \implies (3s_1 + 6s_2) + (4s_1 + 5s_2) = (4s_1 + 4s_2) + (5s_1 + 3s_2) + (3s_1 + 2s_2),$ \[2s_2 = 5s_1 \implies \frac{2}{5}s_2 = s_1.\] Since the side lengths of the rectangle are relatively prime, we can see that $s_1 = 2$ and $s_2 = 5.$ Therefore, $2(2s_9 + s_6 + s_8) = 30s_1 + 40s_2 = \boxed{260}.$ ~peelybonehead

Answer:  260
Problem: The diagram shows a rectangle that has been dissected into nine non-overlapping squares . Given that the width and the height of the rectangle are relatively prime positive integers, find the perimeter of the rectangle.
[asy]draw((0,0)--(69,0)--(69,61)--(0,61)--(0,0));draw((36,0)--(36,36)--(0,36)); draw((36,33)--(69,33));draw((41,33)--(41,61));draw((25,36)--(25,61)); draw((34,36)--(34,45)--(25,45)); draw((36,36)--(36,38)--(34,38)); draw((36,38)--(41,38)); draw((34,45)--(41,45));[/asy]

Solution: We set the side length of the smallest square to 1, and set the side length of square $a_4$ in the previous question to a. We do some "side length chasing" and get $4a - 4 = 2a + 5$ . Solving, we get $a = 4.5$ and the side lengths are $61$ and $69$ . Thus, the perimeter of the rectangle is $2(61 + 69) = \boxed{260}.$

Answer:  260
Problem: Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is $25.$ One marble is taken out of each box randomly. The probability that both marbles are black is $27/50,$ and the probability that both marbles are white is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m + n$

Solution: If we work with the problem for a little bit, we quickly see that there is no direct combinatorics way to calculate $m/n$ . The Principle of Inclusion-Exclusion still requires us to find the individual probability of each box.
Let $a, b$ represent the number of marbles in each box, and without loss of generality let $a>b$ . Then, $a + b = 25$ , and since the $ab$ may be reduced to form $50$ on the denominator of $\frac{27}{50}$ $50|ab$ . It follows that $5|a,b$ , so there are 2 pairs of $a$ and $b: (20,5),(15,10)$
Thus, $m + n = \boxed{026}$

Answer:  026
Problem: Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is $25.$ One marble is taken out of each box randomly. The probability that both marbles are black is $27/50,$ and the probability that both marbles are white is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m + n$

Solution: Let $w_1, w_2, b_1,$ and $b_2$ represent the white and black marbles in boxes 1 and 2.
Since there are $25$ marbles in the box:
$w_1 + w_2 + b_1 + b_2 = 25$
From the fact that there is a $\frac{27}{50}$ chance of drawing one black marble from each box:
$\frac{b_1 \cdot b_2}{(b_1 + w_1)(b_2 + w_2)} = \frac{27}{50} = \frac{54}{100} = \frac{81}{150}$
Thinking of the numerator and denominator separately, if $\frac{27}{50}$ was not a reduced fraction when calculating out the probability, then $b_1 \cdot b_2 = 27$ .  Since $b_1 < 25$ , this forces the variables to be $3$ and $9$ in some permutation.  Without loss of generality, let $b_1 = 3$ and $b_2 = 9$
The denominator becomes: $(3 + w_1)(9 + w_2) = 50$
Since there have been $12$ black marbles used, there must be $13$ white marbles.  Substituting that in:
$(3 + w_1)(9 + (13 - w_1)) = 50$
$(3 + w_1)(22 - w_1) = 50$
Since the factors of $50$ that are greater than $3$ are $5, 10, 25,$ and $50$ , the quantity $3 + w_1$ must equal one of those.  However, since $w_1 < 13$ , testing $2$ and $7$ for $w_1$ does not give a correct product.  Thus, $\frac{27}{50}$ must be a reduced form of the actual fraction.
First assume that the fraction was reduced from $\frac{54}{100}$ , yielding the equations $b_1\cdot b_2 = 54$ and $(b_1 + w_1)(b_2 + w_2) = 100$ .
Factoring $b_1 \cdot b_2 = 54$ and saying WLOG that $b_1 <  b_2 < 25$ gives $(b_1, b_2) = (3, 18)$ or $(6, 9)$ .  Trying the first pair and setting the denominator equal to 100 gives: $(3 + w_1)(18 + w_2) = 100$
Since $w_1 + w_2 = 4$ , the pairs $(w_1, w_2) = (1, 3), (2,2),$ and $(3,1)$ can be tried, since each box must contain at least one white marble.  Plugging in $w_1 = w_2 = 2$ gives the true equation $(3 + 2)(18 + 2) =100$ , so the number of marbles are $(w_1, w_2, b_1, b_2) = (2, 2, 3, 18)$
Thus, the chance of drawing 2 white marbles is $\frac{w_1 \cdot w_2 }{(w_1+ b_1)(w_2 + b_2)} = \frac{4}{100} = \frac{1}{25}$ in lowest terms, and the answer to the problem is $1 + 25 = \boxed{026}.$

Answer:  026
Problem: Each of two boxes contains both black and white marbles, and the total number of marbles in the two boxes is $25.$ One marble is taken out of each box randomly. The probability that both marbles are black is $27/50,$ and the probability that both marbles are white is $m/n,$ where $m$ and $n$ are relatively prime positive integers. What is $m + n$

Solution: We know that $\frac{27}{50} = \frac{b_1}{t_1} \cdot \frac{b_2}{t_2}$ , where $b_1$ and $b_2$ are the number of black marbles in the first and the second box respectively, and $t_1$ and $t_2$ is the total number of marbles in the first and the second boxes respectively. So, $t_1 + t_2 = 25$ . Then, we can realize that $\frac{27}{50} = \frac{9}{10} \cdot \frac{3}{5} = \frac{9}{10} \cdot \frac{9}{15}$ , which means that having 9 black marbles out of 10 total in the first box and 9 marbles out of 15 total the second box is valid. Then there is 1 white marble in the first box and 6 in the second box. So, the probability of drawing two white marbles becomes $\frac{1}{10} \cdot \frac{6}{15} = \frac{1}{25}$ . The answer is $1 + 25 = \boxed{026}$

Answer:  026
Problem: For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$

Solution: \begin{eqnarray*} \frac{x+y}{2} &=& \sqrt{xy} + 2\\ x+y-4 &=& 2\sqrt{xy}\\ y - 2\sqrt{xy} + x &=& 4\\ \sqrt{y} - \sqrt{x} &=& \pm 2\end{eqnarray*}
Because $y > x$ , we only consider $+2$
For simplicity, we can count how many valid pairs of $(\sqrt{x},\sqrt{y})$ that satisfy our equation.
The maximum that $\sqrt{y}$ can be is $\sqrt{10^6} - 1 = 999$ because $\sqrt{y}$ must be an integer (this is because $\sqrt{y} - \sqrt{x} = 2$ , an integer). Then $\sqrt{x} = 997$ , and we continue this downward until $\sqrt{y} = 3$ , in which case $\sqrt{x} = 1$ . The number of pairs of $(\sqrt{x},\sqrt{y})$ , and so $(x,y)$ is then $\boxed{997}$

Answer:  997
Problem: For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$

Solution: Let $a^2$ $x$ and $b^2$ $y$ , where $a$ and $b$ are positive.
Then \[\frac{a^2 + b^2}{2} = \sqrt{{a^2}{b^2}} +2\] \[a^2 + b^2 = 2ab + 4\] \[(a-b)^2 = 4\] \[(a-b) = \pm 2\]
This makes counting a lot easier since now we just have to find all pairs $(a,b)$ that differ by 2.
Because $\sqrt{10^6} = 10^3$ , then we can use all positive integers less than 1000 for $a$ and $b$
We know that because $x < y$ , we get $a < b$
We can count even and odd pairs separately to make things easier*:
Odd: \[(1,3) , (3,5) , (5,7)  .  .  .  (997,999)\]
Even: \[(2,4) , (4,6) , (6,8)  .  .  .  (996,998)\]
This makes 499 odd pairs and 498 even pairs, for a total of $\boxed{997}$ pairs.

Answer:  997
Problem: For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$

Solution: Since the arithmetic mean is 2 more than the geometric mean, $\frac{x+y}{2} = 2 + \sqrt{xy}$ . We can multiply by 2 to get $x + y = 4 + 2\sqrt{xy}$ . Subtracting 4 and squaring gives \[((x+y)-4)^2 = 4xy\] \[((x^2 + 2xy + y^2) + 16 - 2(4)(x+y)) = 4xy\] \[x^2 - 2xy + y^2 + 16 - 8x - 8y = 0\]
Notice that $((x-y)-4)^2 = x^2 - 2xy + y^2 + 16 - 8x +8y$ , so the problem asks for solutions of \[(x-y-4)^2 = 16y\] Since the left hand side is a perfect square, and 16 is a perfect square, $y$ must also be a perfect square. Since $0 < y < (1000)^2$ $y$ must be from $1^2$ to $999^2$ , giving at most 999 options for $y$
However if $y = 1^2$ , you get $(x-5)^2 = 16$ , which has solutions $x = 9$ and $x = 1$ . Both of those solutions are not less than $y$ , so $y$ cannot be equal to 1. If $y = 2^2 = 4$ , you get $(x - 8)^2 = 64$ , which has 2 solutions, $x = 16$ , and $x = 0$ . 16 is not less than 4, and $x$ cannot be 0, so $y$ cannot be 4. However, for all other $y$ , you get exactly 1 solution for $x$ , and that gives a total of $999 - 2 = \boxed{997}$ pairs.

Answer:  997
Problem: For how many ordered pairs $(x,y)$ of integers is it true that $0 < x < y < 10^6$ and that the arithmetic mean of $x$ and $y$ is exactly $2$ more than the geometric mean of $x$ and $y$

Solution: Rearranging our conditions to
\[x^2-2xy+y^2+16-8x-8y=0 \implies\] \[(y-x)^2=8(x+y-2).\]
Thus, $4|y-x.$
Now, let $y = 4k+x.$ Plugging this back into our expression, we get
\[(k-1)^2=x-1.\]
There, a unique value of $x, y$ is formed for every value of $k$ . However, we must have
\[y<10^6 \implies (k+1)^2< 10^6-1\]
and
\[x=(k-1)^2+1>0.\]
Therefore, there are only $\boxed{997}$ pairs of $(x,y).$

Answer:  997
Problem: Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$
note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques

Solution: We can rewrite $xyz=1$ as $\frac{1}{z}=xy$
Substituting into one of the given equations, we have \[x+xy=5\] \[x(1+y)=5\] \[\frac{1}{x}=\frac{1+y}{5}.\]
We can substitute back into $y+\frac{1}{x}=29$ to obtain \[y+\frac{1+y}{5}=29\] \[5y+1+y=145\] \[y=24.\]
We can then substitute once again to get \[x=\frac15\] \[z=\frac{5}{24}.\] Thus, $z+\frac1y=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}$ , so $m+n=\boxed{005}$

Answer:  005
Problem: Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$
note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques

Solution: Let $r = \frac{m}{n} = z + \frac {1}{y}$
\begin{align*} (5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\ &=xyz + \frac{xy}{y} + \frac{xz}{x} + \frac{yz}{z} + \frac{x}{xy} + \frac{y}{yz} + \frac{z}{xz} + \frac{1}{xyz}\\ &=1 + x + z + y + \frac{1}{y} + \frac{1}{z} + \frac{1}{x} + \frac{1}{1}\\ &=2 + \left(x + \frac {1}{z}\right) + \left(y + \frac {1}{x}\right) + \left(z + \frac {1}{y}\right)\\ &=2 + 5 + 29 + r\\ &=36 + r \end{align*}
Thus $145r = 36+r \Rightarrow 144r = 36 \Rightarrow r = \frac{36}{144} = \frac{1}{4}$ . So $m + n = 1 + 4 = \boxed{5}$

Answer:  5
Problem: Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$
note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques

Solution: (Hybrid between 1/2)
Because $xyz = 1, \hspace{0.15cm} \frac{1}{x} = yz, \hspace{0.15cm} \frac{1}{y} = xz,$ and $\hspace{0.05cm}\frac{1}{z} = xy$ . Substituting and factoring, we get $x(y+1) = 5$ $\hspace{0.15cm}y(z+1) = 29$ , and $\hspace{0.05cm}z(x+1) = k$ . Multiplying them all together, we get, $xyz(x+1)(y+1)(z+1) = 145k$ , but $xyz$ is $1$ , and by the Identity property of multiplication, we can take it out. So, in the end, we get $(x+1)(y+1)(z+1) = 145k$ . And, we can expand this to get $xyz+xy+yz+xz+x+y+z+1 = 145k$ , and if we make a substitution for $xyz$ , and rearrange the terms, we get $xy+yz+xz+x+y+z = 145k-2$ This will be important.
Now, lets add the 3 equations $x(y+1) = 5, \hspace{0.15cm}y(z+1) = 29$ , and $\hspace{0.05cm}z(x+1) = k$ . We use the expand the Left hand sides, then, we add the equations to get $xy+yz+xz+x+y+z = k+34$ Notice that the LHS of this equation matches the LHS equation that I said was important. So, the RHS of both equations are equal, and thus $145k-2 = k+34$ We move all constant terms to the right, and all linear terms to the left, to get $144k = 36$ , so $k = \frac{1}{4}$ which gives an answer of $1+4 = \boxed{005}$

Answer:  005
Problem: Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$
note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques

Solution: Get rid of the denominators in the second and third equations to get $xz-5z=-1$ and $xy-29x=-1$ . Then, since $xyz=1$ , we have $\tfrac 1y-5z=-1$ and $\tfrac 1z-29x=-1$ . Then, since we know that $\tfrac 1z+x=5$ , we can subtract these two equations to get that $30x=6\implies x=5$ . The result follows that $z=\tfrac 5{24}$ and $y=24$ , so $z+\tfrac 1y=\tfrac 1{24}+\tfrac 5{24}=\tfrac 14$ , and the requested answer is $1+4=\boxed{005}.$

Answer:  005
Problem: Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$
note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques

Solution: Rewrite the equations in terms of x.
$x+\frac{1}{z}=5$ becomes $z=\frac{1}{x+5}$
$y+\frac{1}{x}=29$ becomes $y=29-\frac{1}{x}$
Now express $xyz=1$ in terms of x.
$\frac{1}{5-x}\cdot(29-\frac{1}{x})\cdot x=1$
This evaluates to $29x-1=5-x$ , giving us $x=\frac{1}{5}$ . We can now plug x into the other equations to get $y=24$ and $z=\frac{5}{24}$
Therefore, $z+\frac{1}{y}=\frac{5}{24}+\frac{1}{24}=\frac{6}{24}=\frac{1}{4}$
$1+4=\boxed{5}$ , and we are done.
~MC413551

Answer:  5