Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Determine $x^2+y^2+z^2+w^2$ if

Solution: As in Solution 1, we have \[x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),\] where $F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)$ and $P_k(t)=F(t)/(t-k^2)$ , for $k=1,3,5,7$
Now the coefficient of $t^3$ on both sides must be equal. So instead of expanding it fully, we will find what the coefficients of the $t^4$ and $t^3$ terms are, so we can eventually apply Vieta's. We can write the long equation as \[(x^2 + y^2 + z^2 + w^2)t^3 + \dots = t^4 - (1^2 + 3^2 + 5^2 + 7^2)t^3 + \dots\] Rearranging gives us \[t^4 - (1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2)t^3 \dots = 0.\] By Vieta's, we know that the sum of the roots of this equation is \[1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2.\] (recall that the roots of the original and this manipulated form of it had roots $2^2, 4^2, 6^2,$ and $8^2$ ). Thus, \[x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2 - 1^2 - 3^2 - 5^2 - 7^2 = \boxed{36}.\]

Answer:  36
Problem: Determine $x^2+y^2+z^2+w^2$ if

Solution: Notice how on each line, we have equations of the form $\frac{x^2}{a-1^2}+\frac{y^2}{a-3^2}+\frac{z^2}{a-5^2}+\frac{w^2}{a-7^2}=1$ . We can let this be a polynomial, with respect to $a$ . We can say that $w^2$ $x^2$ $y^2$ , and $z^2$ are coefficients, and not variables. So, we can now expand the fractions to get $(a-1)(a-9)(a-25)(a-49)=x^2(a-9)(a-25)(a-49)$ $+ y^2(a-1)(a-25)(a-49)$ $+ z^2(a-1)(a-3)(a-7)$ $+ w^2(a-1)(a-9)(a-25)$
Now, we have arrived at this huge expression, but what do we do with it?
Well, we can look at what we want to find - $x^2+y^2+z^2+w^2$ . So, we want the sum of $x^2$ $y^2$ $z^2$ , and $w^2$ . Looking back to our expression, we can note how on the right hand side, the $a^3$ terms add to $x^2+y^2+z^2+w^2$ . Also, on the left hand side, the $a^3$ coefficient is $84$ (which is achievable by Vieta's formulas rather than expanding if you want to save a few seconds). So, moving all the $a^3$ terms to the left hand side, then we have that by Vieta's formulas, the sum of the roots is $-84-x^2-y^2-z^2-w^2=-(2^2+4^2+6^2+8^2)$ . Then, we can solve to find that $x^2+y^2+z^2+w^2=120-84=\boxed{036}$

Answer:  036
Problem: Let $x$ $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$ $\log_y w = 40$ and $\log_{xyz} w = 12$ . Find $\log_z w$

Solution: The logarithmic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.
$x^{24}=w$ $y^{40}=w$ , and $(xyz)^{12}=w$ . If we now convert everything to a power of $120$ , it will be easy to isolate $z$ and $w$
$x^{120}=w^5$ $y^{120}=w^3$ , and $(xyz)^{120}=w^{10}$
With some substitution, we get $w^5w^3z^{120}=w^{10}$ and $\log_zw=\boxed{060}$

Answer:  060
Problem: Let $x$ $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$ $\log_y w = 40$ and $\log_{xyz} w = 12$ . Find $\log_z w$

Solution: First we'll convert everything to exponential form. $x^{24}=w$ $y^{40}=w$ , and $(xyz)^{12}=w$ . The only expression containing $z$ is $(xyz)^{12}=w$ . It now becomes clear that one way to find $\log_z w$ is to find what $x^{12}$ and $y^{12}$ are in terms of $w$
Taking the square root of the equation $x^{24}=w$ results in $x^{12}=w^{\frac{1}{2}}$ . Raising both sides of $y^{40}=w$ to the $\frac{12}{40}$ th power gives $y^{12}=w^{\frac{3}{10}}$
Going back to $(xyz)^{12}=w$ , we can substitute the $x^{12}$ and $y^{12}$ with $w^{1/2}$ and $w^{3/10}$ , respectively. We now have $w^{1/2}w^{3/10}z^{12}=w$ . Simplifying, we get $z^{60}=w$ .
So our answer is $\boxed{060}$

Answer:  060
Problem: Let $x$ $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$ $\log_y w = 40$ and $\log_{xyz} w = 12$ . Find $\log_z w$

Solution: Applying the change of base formula, \begin{align*} \log_x w = 24 &\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\ \log_y w = 40 &\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \\ \log_{xyz} w = 12 &\implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} \end{align*} Therefore, $\frac {\log z}{\log w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}$
Hence, $\log_z w = \boxed{060}$

Answer:  060
Problem: Let $x$ $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$ $\log_y w = 40$ and $\log_{xyz} w = 12$ . Find $\log_z w$

Solution: Since $\log_a b = \frac{1}{\log_b a}$ , the given conditions can be rewritten as $\log_w x = \frac{1}{24}$ $\log_w y = \frac{1}{40}$ , and $\log_w xyz = \frac{1}{12}$ . Since $\log_a \frac{b}{c} = \log_a b - \log_a c$ $\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}$ . Therefore, $\log_z w = \boxed{060}$

Answer:  060
Problem: Let $x$ $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$ $\log_y w = 40$ and $\log_{xyz} w = 12$ . Find $\log_z w$

Solution: If we convert all of the equations into exponential form, we receive $x^{24}=w$ $y^{40}=w$ , and $(xyz)^{12}=w$ . The last equation can also be written as $x^{12}y^{12}z^{12}=w$ . Also note that by multiplying the first two equations, we get, $x^{24}y^{40}= w^{2}$ . Taking the square root of this, we find that $x^{12}y^{20}=w$ . Recall, $x^{12}y^{12}z^{12}=w$ . Thus, $z^{12}= y^{8}$ . Also recall, $y^{40}=w$ . Therefore, $z^{60}$ $y^{40}$ $w$ . So, $\log_z w$ $\boxed{060}$

Answer:  060
Problem: Let $x$ $y$ and $z$ all exceed $1$ and let $w$ be a positive number such that $\log_x w = 24$ $\log_y w = 40$ and $\log_{xyz} w = 12$ . Find $\log_z w$

Solution: Converting all of the logarithms to exponentials gives $x^{24} = w, y^{40} =w,$ and $x^{12}y^{12}z^{12}=w.$ Thus, we have $y^{40} = x^{24} \Rightarrow z^3=y^2.$ We are looking for $\log_z w,$ which by substitution, is $\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} =\boxed{60}.$

Answer:  60
Problem: Let $f(x)=|x-p|+|x-15|+|x-p-15|$ , where $0 < p < 15$ . Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \leq x\leq15$

Solution: It is best to get rid of the absolute values first.
Under the given circumstances, we notice that $|x-p|=x-p$ $|x-15|=15-x$ , and $|x-p-15|=15+p-x$
Adding these together, we find that the sum is equal to $30-x$ , which attains its minimum value (on the given interval $p \leq x \leq 15$ ) when $x=15$ , giving a minimum of $\boxed{015}$

Answer:  015
Problem: Let $f(x)=|x-p|+|x-15|+|x-p-15|$ , where $0 < p < 15$ . Determine the minimum value taken by $f(x)$ for $x$ in the interval $p \leq x\leq15$

Solution: Let $p$ be equal to $15 - \varepsilon$ , where $\varepsilon$ is an almost neglectable value. Because of the small value $\varepsilon$ , the domain of $f(x)$ is basically the set ${15}$ . plugging in $15$ gives $\varepsilon + 0 + 15 - \varepsilon$ , or $15$ , so the answer is $\boxed{15}$

Answer:  15
Problem: What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$

Solution: If we were to expand by squaring, we would get a quartic polynomial , which isn't always the easiest thing to deal with.
Instead, we substitute $y$ for $x^2+18x+30$ , so that the equation becomes $y=2\sqrt{y+15}$
Now we can square; solving for $y$ , we get $y=10$ or $y=-6$ . The second root is extraneous since $2\sqrt{y+15}$ is always non-negative (and moreover, plugging in $y=-6$ , we get $-6=6$ , which is obviously false). Hence we have $y=10$ as the only solution for $y$ . Substituting $x^2+18x+30$ back in for $y$
Both of the roots of this equation are real, since its discriminant is $18^2 - 4 \cdot 1 \cdot 20 = 244$ , which is positive. Thus by Vieta's formulas , the product of the real roots is simply $\boxed{020}$

Answer:  020
Problem: What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$

Solution: We begin by noticing that the polynomial on the left is $15$ less than the polynomial under the radical sign. Thus: \[(x^2+ 18x + 45) - 2\sqrt{x^2+18x+45} - 15 = 0.\] Letting $n = \sqrt{x^2+18x+45}$ , we have $n^2-2n-15 = 0 \Longrightarrow (n-5)(n+3) = 0$ . Because the square root of a real number can't be negative, the only possible $n$ is $5$
Substituting that in, we have \[\sqrt{x^2+18x+45} = 5 \Longrightarrow x^2 + 18x + 45 = 25 \Longrightarrow x^2+18x+20=0.\]
Reasoning as in Solution 1, the product of the roots is $\boxed{020}$

Answer:  020
Problem: What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$

Solution: Begin by completing the square on both sides of the equation, which gives \[(x+9)^2-51=2\sqrt{(x+3)(x+15)}\] Now by substituting $y=x+9$ , we get $y^2-51=2\sqrt{(y-6)(y+6)}$ , or \[y^4-106y^2+2745=0\] The solutions in $y$ are then \[y=x+9=\pm3\sqrt{5},\pm\sqrt{61}\] Turns out, $\pm3\sqrt{5}$ are a pair of extraneous solutions. Thus, our answer is then \[\left(\sqrt{61}-9\right)\left(-\sqrt{61}-9\right)=81-61=\boxed{020}\] By difference of squares.

Answer:  020
Problem: What is the product of the real roots of the equation $x^2 + 18x + 30 = 2 \sqrt{x^2 + 18x + 45}$

Solution: We are given the equation \[x^2+18x+30=2\sqrt{x^2+18x+45}\] Squaring both sides yields \[(x^2+18x+30)^2=4(x^2+18x+45)\] \[(x^2+18x+30)^2=4(x^2+18x+30+15)\] \[(x^2+18x+30)^2=4(x^2+18x+30)+60\] \[(x^2+18x+30)^2-4(x^2+18x+30)-60=0\] Substituting $y=x^2+18x+30$ yields \[y^2-4y-60=0\] \[(y+6)(y-10)=0\] Thus $y=x^2+18x+30=-6,10$ . However if $y=-6$ , the left side of the equation \[x^2+18x+30=2\sqrt{x^2+18x+45}\] would be negative while the right side is negative. Thus $y=10$ is the only possible value and we have \[x^2+18x+30=10\] \[x^2+18x+20=0\] Since the discriminant $\sqrt{18^2-4\cdot20}$ is real, both the roots are real. Thus by Vieta's Formulas, the product of the roots is the constant, $\boxed{20}$

Answer:  20
Problem: A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle.
[asy] size(150);  defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]

Solution: Because we are given a right angle, we look for ways to apply the Pythagorean Theorem .  Let the foot of the perpendicular from $O$ to $AB$ be $D$ and let the foot of the perpendicular from $O$ to the line $BC$ be $E$ . Let $OE=x$ and $OD=y$ . We're trying to find $x^2+y^2$
Applying the Pythagorean Theorem, $OA^2 = OD^2 + AD^2$ and $OC^2 = EC^2 + EO^2$
Thus, $\left(\sqrt{50}\right)^2 = y^2 + (6-x)^2$ , and $\left(\sqrt{50}\right)^2 = x^2 + (y+2)^2$ . We solve this system to get $x = 1$ and $y = 5$ , such that the answer is $1^2 + 5^2 = \boxed{026}$

Answer:  026
Problem: A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle.
[asy] size(150);  defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]

Solution: Drop perpendiculars from $O$ to $AB$ (with foot $T_1$ ), $M$ to $OT_1$ (with foot $T_2$ ), and $M$ to $AB$ (with foot $T_3$ ).
Also, mark the midpoint $M$ of $AC$
First notice that by computation, $OAC$ is a $\sqrt {50} - \sqrt {40} - \sqrt {50}$ isosceles triangle, so $AC = MO$ .
Then, notice that $\angle MOT_2 = \angle T_3MO = \angle BAC$ . Therefore, the two blue triangles are congruent, from which we deduce $MT_2 = 2$ and $OT_2 = 6$ . As $T_3B = 3$ and $MT_3 = 1$ , we subtract and get $OT_1 = 5,T_1B = 1$ . Then the Pythagorean Theorem tells us that $OB^2 = \boxed{026}$

Answer:  026
Problem: A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle.
[asy] size(150);  defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]

Solution: Draw segment $OB$ with length $x$ , and draw radius $OQ$ such that $OQ$ bisects chord $AC$ at point $M$ . This also means that $OQ$ is perpendicular to $AC$ . By the Pythagorean Theorem, we get that $AC=\sqrt{(BC)^2+(AB)^2}=2\sqrt{10}$ , and therefore $AM=\sqrt{10}$ . Also by the Pythagorean theorem, we can find that $OM=\sqrt{50-10}=2\sqrt{10}$
Next, find $\angle BAC=\arctan{\left(\frac{2}{6}\right)}$ and $\angle OAM=\arctan{\left(\frac{2\sqrt{10}}{\sqrt{10}}\right)}$ . Since $\angle OAB=\angle OAM-\angle BAC$ , we get \[\angle OAB=\arctan{2}-\arctan{\frac{1}{3}}\] \[\tan{(\angle OAB)}=\tan{(\arctan{2}-\arctan{\frac{1}{3}})}\] By the subtraction formula for $\tan$ , we get \[\tan{(\angle OAB)}=\frac{2-\frac{1}{3}}{1+2\cdot \frac{1}{3}}\] \[\tan{(\angle OAB)}=1\] \[\cos{(\angle OAB)}=\frac{1}{\sqrt{2}}\] Finally, by the Law of Cosines on $\triangle OAB$ , we get \[x^2=50+36-2(6)\sqrt{50}\frac{1}{\sqrt{2}}\] \[x^2=\boxed{026}.\]

Answer:  026
Problem: A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle.
[asy] size(150);  defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]

Solution: We use coordinates. Let the circle have center $(0,0)$ and radius $\sqrt{50}$ ; this circle has equation $x^2 + y^2 = 50$ . Let the coordinates of $B$ be $(a,b)$ . We want to find $a^2 + b^2$ $A$ and $C$ with coordinates $(a,b+6)$ and $(a+2,b)$ , respectively, both lie on the circle. From this we obtain the system of equations
$a^2 + (b+6)^2 = 50$
$(a+2)^2 + b^2 = 50$
After expanding these terms, we notice by subtracting the first and second equations, we can cancel out $a^2$ and $b^2$ . after substituting $a=3b+8$ and plugging back in, we realize that $(a,b)=(-7,-5)$ or $(5,-1)$ . Since the first point is out of the circle, we find that $(5,-1)$ is the only relevant answer. This paragraph is written by ~hastapasta.
Solving, we get $a=5$ and $b=-1$ , so the distance is $a^2 + b^2 = \boxed{026}$

Answer:  026
Problem: A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle.
[asy] size(150);  defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]

Solution: I will use the law of cosines in triangle $\triangle OAC$ and $\triangle OBC$
$AC = \sqrt{AB^2 + BC^2} = \sqrt{6^2 + 2^2} = 2 \sqrt{10}$
$\cos \angle ACB = \frac{2}{2\sqrt{10}} = \frac{1}{\sqrt{10}}$
$\cos \angle ACO = \frac{AC^2+OC^2-OA^2}{2 \cdot AC \cdot OC} = \frac{(2\sqrt{10})^2+(\sqrt{50})^2-(\sqrt{50})^2}{2 \cdot 2\sqrt{10} \cdot \sqrt{50}} = \frac{1}{\sqrt{5}}$
$\sin \angle ACB = \sqrt{1-\cos^2 \angle ACB} = \sqrt{1-(\frac{1}{\sqrt{10}})^2} = \frac{3}{\sqrt{10}}$
$\sin \angle ACO = \sqrt{1-\cos^2 \angle ACO} = \sqrt{1-(\frac{1}{\sqrt{5}})^2} = \frac{2}{\sqrt{5}}$
$\cos \angle OCB = \cos (\angle ACB - \angle ACO) = \cos \angle ACB \cdot \cos \angle ACO + \sin \angle ACB \cdot \sin \angle ACO = \frac{1}{\sqrt{10}} \cdot \frac{1}{\sqrt{5}} + \frac{3}{\sqrt{10}} \cdot \frac{2}{\sqrt{5}} = \frac{7}{5\sqrt{2}}$
$OB^2 = OC^2 + BC^2 - 2 \cdot OC \cdot BC \cdot \cos \angle OCB = (\sqrt{50})^2 + 2^2 - 2 \cdot \sqrt{50} \cdot 2 \cdot \frac{7}{5\sqrt{2}} = 50 + 4 - 28 = \boxed{026}$

Answer:  026
Problem: A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle.
[asy] size(150);  defaultpen(linewidth(0.6)+fontsize(11)); real r=10; pair O=(0,0), A=r*dir(45),B=(A.x,A.y-r); path P=circle(O,r); pair C=intersectionpoint(B--(B.x+r,B.y),P); // Drawing arc instead of full circle //draw(P); draw(arc(O, r, degrees(A), degrees(C))); draw(C--B--A--B); dot(A); dot(B); dot(C); label("$A$",A,NE); label("$B$",B,S); label("$C$",C,SE); [/asy]

Solution: Notice that $50=5^2+5^2=7^2+1^2$ , and by the size of the diagram, it seems reasonable that $OA$ represents $5^2+5^2$ , and $OC$ means the $7^1+1^2$ , and indeed, the values work ( $7-5=2$ and $5+1=6$ ), so $OB^2=5^2+1^2=\boxed{026}$

Answer:  026
Problem: Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have?

Solution: One way to solve this problem is by substitution . We have
$x^2+y^2=(x+y)^2-2xy=7$ and $x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10$
Hence observe that we can write $w=x+y$ and $z=xy$
This reduces the equations to $w^2-2z=7$ and $w(7-z)=10$
Because we want the largest possible $w$ , let's find an expression for $z$ in terms of $w$
$w^2-7=2z \implies z=\frac{w^2-7}{2}$
Substituting, $w^3-21w+20=0$ , which factorizes as $(w-1)(w+5)(w-4)=0$ (the Rational Root Theorem may be used here, along with synthetic division).
The largest possible solution is therefore $x+y=w=\boxed{004}$

Answer:  004
Problem: Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have?

Solution: An alternate way to solve this is to let $x=a+bi$ and $y=c+di$
Because we are looking for a value of $x+y$ that is real, we know that $d=-b$ , and thus $y=c-bi$
Expanding $x^2+y^2=7+0i$ will give two equations, since the real and imaginary parts must match up.
$(a+bi)^2+(c-bi)^2=7+0i$
$(a^2+c^2-2b^2)+(2ab-2cb)i=7+0i$
Looking at the imaginary part of that equation, $2ab-2cb=0$ , so $a=c$ , and $x$ and $y$ are actually complex conjugates.
Looking at the real part of the equation and plugging in $a=c$ $2a^2-2b^2=7$ , or $2b^2=2a^2-7$
Now, evaluating the real part of $(a+bi)^3+(a-bi)^3$ , which equals $10$ (ignoring the odd powers of $i$ , since they would not result in something in the form of $10+0i$ ):
$a^3+3a(bi)^2+a^3+3a(-bi)^2=10$
$2a^3-6ab^2=10$
Since we know that $2b^2=2a^2-7$ , it can be plugged in for $b^2$ in the above equation to yield:
$2a^3-3a(2a^2-7)=10$
$-4a^3+21a=10$
$4a^3-21a+10=0$
Since the problem is looking for $x+y=2a$ to be a positive integer, only positive half-integers (and whole-integers) need to be tested.  From the Rational Roots theorem, $a=10, a=5, a=\frac{5}{2}$ all fail, but $a=2$ does work.  Thus, the real part of both numbers is $2$ , and their sum is $\boxed{004}$

Answer:  004
Problem: Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have?

Solution: Begin by assuming that $x$ and $y$ are roots of some polynomial of the form $w^2+bw+c$ , such that by Vieta's Formulæ and some algebra (left as an exercise to the reader), $b^2-2c=7$ and $3bc-b^3=10$ .
Substituting $c=\frac{b^2-7}{2}$ , we deduce that $b^3-21b-20=0$ , whose roots are $-4$ $-1$ , and $5$ .
Since $-b$ is the sum of the roots and is maximized when $b=-4$ , the answer is $-(-4)=\boxed{004}$

Answer:  004
Problem: Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have?

Solution: $x^3 + y^3 = 10 = (x+y)(x^2-xy+y^2) = (x+y)(7-xy) \implies xy = 7 - \frac{10}{x+y}.$ Also, $(x+y)^3 = x^3 + 3x^2y+3xy^2+y^3 = 10 + 3xy(x+y).$ Substituting our above into this, we get $10 + 3(7-\frac{10}{x+y})(x+y) = 21x+21y-20 = (x+y)^3$ .  Letting $p = x+y$ , we have that $p^3 - 21p + 20 = 0$ .  Testing $p = 1$ , we find that this is a root, to get $(p-1)(p^2+p-20) = 0 \implies p = -5, 1, 4 \implies \boxed{4}$

Answer:  4
Problem: Let $a_n=6^{n}+8^{n}$ . Determine the remainder upon dividing $a_ {83}$ by $49$

Solution: Firstly, we try to find a relationship between the numbers we're provided with and $49$ . We notice that $49=7^2$ , and both $6$ and $8$ are greater or less than $7$ by $1$
Thus, expressing the numbers in terms of $7$ , we get $a_{83} = (7-1)^{83}+(7+1)^{83}$
Applying the Binomial Theorem , half of our terms cancel out and we are left with $2(7^{83}+3403\cdot7^{81}+\cdots + 83\cdot7)$ . We realize that all of these terms are divisible by $49$ except the final term.
After some quick division, our answer is $\boxed{035}$

Answer:  035
Problem: Let $a_n=6^{n}+8^{n}$ . Determine the remainder upon dividing $a_ {83}$ by $49$

Solution: As the value of $a$ is obviously $6^{83}+8^{83}$ we look for a pattern with others. With a bit of digging, we discover that $6^n+6^m$ where $m$ and $n$ are odd is equal to $\boxed{35}\:49$

Answer:  35
Problem: Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent off to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and denominator?

Solution: We can use complementary counting , by finding the probability that none of the three knights are sitting next to each other and subtracting it from $1$
Imagine that the $22$ other (indistinguishable) people are already seated, and fixed into place.
We will place $A$ $B$ , and $C$ with and without the restriction.
There are $22$ places to put $A$ , followed by $21$ places to put $B$ , and $20$ places to put $C$ after $A$ and $B$ . Hence, there are $22\cdot21\cdot20$ ways to place $A, B, C$ in between these people with restrictions.
Without restrictions, there are $22$ places to put $A$ , followed by $23$ places to put $B$ , and $24$ places to put $C$ after $A$ and $B$ . Hence, there are $22\cdot23\cdot24$ ways to place $A,B,C$ in between these people without restrictions.
Thus, the desired probability is $1-\frac{22\cdot21\cdot20}{22\cdot23\cdot24}=1-\frac{420}{552}=1-\frac{35}{46}=\frac{11}{46}$ , and the answer is $11+46=\boxed{057}$

Answer:  057