Answer: 60.0

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 $\log_zw=\boxed{060}$

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 $\boxed{060}$

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, $\log_z w = \boxed{060}$

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, $\log_z w = \boxed{060}$

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, $\log_z w$ $\boxed{060}$

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: 31.0

Let $x$ and $y$ be positive integers such that $7x^5 = 11y^{13}.$ The minimum possible value of $x$ has a prime factorization $a^cb^d.$ What is $a + b + c + d?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$

Solution:
Substitute $a^cb^d$ into $x$ . We then have $7(a^{5c}b^{5d}) = 11y^{13}$ . Divide both sides by $7$ , and it follows that:
\[(a^{5c}b^{5d}) = \frac{11y^{13}}{7}.\]
Note that because $11$ and $7$ are prime, the minimum value of $x$ must involve factors of $7$ and $11$ only. Thus, we try to look for the lowest power $p$ of $11$ such that $13p + 1 \equiv 0 \pmod{5}$ , so that we can take $11^{13p + 1}$ to the fifth root. Similarly, we want to look for the lowest power $n$ of $7$ such that $13n - 1 \equiv 0 \pmod{5}$ . Again, this allows us to take the fifth root of $7^{13n - 1}$ . Obviously, we want to add $1$ to $13p$ and subtract $1$ from $13n$ because $11^{13p}$ and $7^{13n}$ are multiplied by $11$ and divided by $7$ , respectively. With these conditions satisfied, we can simply multiply $11^{p}$ and $7^{n}$ and substitute this quantity into $y$ to attain our answer.
We can simply look for suitable values for $p$ and $n$ . We find that the lowest $p$ , in this case, would be $3$ because $13(3) + 1 \equiv 0 \pmod{5}$ . Moreover, the lowest $q$ should be $2$ because $13(2) - 1 \equiv 0 \pmod{5}$ . Hence, we can substitute the quantity $11^{3} \cdot 7^{2}$ into $y$ . Doing so gets us:
\[(a^{5c}b^{5d}) = \frac{11(11^{3} \cdot 7^{2})^{13}}{7} = 11^{40} \cdot 7^{25}.\]
Taking the fifth root of both sides, we are left with $a^cb^d = 11^{8} \cdot 7^{5}$ $a + b + c + d = 11 + 7 + 8 + 5 = \boxed{31}$

Solution:
A simpler way to tackle this problem without all that modding is to keep the equation as:
\[7*a^{5b}c^{5d} = 11y^{13}\]
As stated above, $a$ and $c$ must be the factors 7 and 11 in order to keep $x$ at a minimum. Moving all the non-y terms to the left hand side of the equation, we end up with:
\[7^{5b+1}11^{5d-1}=y^{13}\]
The above equation means that $y$ must also contain only the factors 7 and 11 (again, in order to keep $x$ at a minimum), so we end up with:
\[7^{5b+1}11^{5d-1}=7^{13h}11^{13j}\]
$h$ and $j$ are arbitrary variables placed in order to show that $y$ could have more than just one 7 or one 11 as factors)
Since 7 and 11 are prime, we know that $5b+1=13h$ and $5d-1=13j$ . The smallest positive combinations that would work are $b=5,h=2$ and $d=8,j=3$ . Therefore, $a+b+c+d=7+5+11+8=31$ $\boxed{31}$ is correct.

Solution:
Another way to solve this problem solve for x. First, we can divide both sides by 7 to get:
\[x^5 = \frac{11y^{13}}{7}\]
Next, we take the fifth root on both sides, which gets us:
\[x = \sqrt[5]{\frac{11y^{13}}{7}}\]
\[x = y^2 \cdot \sqrt[5]{\frac{11y^{3}}{7}}\]
Since we know x is a positive integer that we are trying to minimize, we can let y equal the smallest number that will make x an integer. In this case, we let $y = 11^3 \cdot 7^2$ (Make sure you see why this makes x the smallest integer possible!), which when plugged in, results in:
\[x = (11^3 \cdot 7^2)^2 \cdot \sqrt[5]{\frac{11}{7}\cdot (11^3 \cdot 7^2)^3}\]
\[x = (11^3 \cdot 7^2)^2 \cdot \sqrt[5]{\frac{11}{7}\cdot 11^9 \cdot 7^6}\]
\[x = (11^3 \cdot 7^2)^2 \cdot \sqrt[5]{11^{10} \cdot 7^5}\]
\[x = (11^3 \cdot 7^2)^2 \cdot 11^2 \cdot 7^1\]
\[x = 7^5 \cdot 11^ 8\]
This gets us $a^cb^d = 7^5\cdot 11^8$ , so $a + b + c + d = 7 + 5 + 11 + 8 = \boxed{31}$ ~lucaswujc, $\LaTeX$ help from Technodoggo


Answer: 31.0

Solution:
According to the problem, we have that $x^5$ and $y^{13}$ must be a multiple of both $7$ and $11$ . Thus, in their prime factorisation, there must be $7$ and $11$ . Thus, we have $a=7$ and $b=11$ . Now, let $x=7^c11^d\implies x^5=7^{5c}11^{5d}\implies7x^5=7^{5c+1}11^{5d}$ .
We can now divide both sides by 11 in our original equation $7x^5=11y^{13}$ to get $y^{13}=7^{5c+1}11^{5d-1}$ . As we are only considering integers, we have \[5c+1\equiv0~(\text{mod}~13)\implies5c\equiv12~(\text{mod}~13)\] and \[5d-1\equiv0~(\text{mod}~13)\implies5d\equiv1~(\text{mod}~13).\]
We can apply brute force to solve for $c$ and $d$ as the numbers aren't big. For the first congruence, we find that $25$ is the smallest number that satisfies, thus $c=5$ . For the second congruence, we find that $40$ is the smallest number that satisfies, thus $d=8$ . Summarising, we get $a+b+c+d=7+11+5+8=\boxed{31}$


Answer: 89.0

Let $x$ and $y$ be real numbers satisfying $x^4y^5+y^4x^5=810$ and $x^3y^6+y^3x^6=945$ . Evaluate $2x^3+(xy)^3+2y^3$

Solution:
The expression we want to find is $2(x^3+y^3) + x^3y^3$
Factor the given equations as $x^4y^4(x+y) = 810$ and $x^3y^3(x^3+y^3)=945$ , respectively. Dividing the latter by the former equation yields $\frac{x^2-xy+y^2}{xy} = \frac{945}{810}$ . Adding 3 to both sides and simplifying yields $\frac{(x+y)^2}{xy} = \frac{25}{6}$ . Solving for $x+y$ and substituting this expression into the first equation yields $\frac{5\sqrt{6}}{6}(xy)^{\frac{9}{2}} = 810$ . Solving for $xy$ , we find that $xy = 3\sqrt[3]{2}$ , so $x^3y^3 = 54$ . Substituting this into the second equation and solving for $x^3+y^3$ yields $x^3+y^3=\frac{35}{2}$ . So, the expression to evaluate is equal to $2 \times \frac{35}{2} + 54 = \boxed{089}$

Solution: Factor the given equations as $x^4y^4(x+y) = 810$ and $x^3y^3(x+y)(x^2-xy+y^2)=945$ , respectively. By the first equation, $x+y=\frac{810}{x^4y^4}$ . Plugging this in to the second equation and simplifying yields $(\frac{x}{y}-1+\frac{y}{x})=\frac{7}{6}$ . Now substitute $\frac{x}{y}=a$ . Solving the quadratic in $a$ , we get $a=\frac{x}{y}=\frac{2}{3}$ or $\frac{3}{2}$ As both of the original equations were symmetric in $x$ and $y$ , WLOG, let $\frac{x}{y}=\frac{2}{3}$ , so $x=\frac{2}{3}y$ . Now plugging this in to either one of the equations, we get the solutions $y=\frac{3(2^{\frac{2}{3}})}{2}$ $x=2^{\frac{2}{3}}$ . Now plugging into what we want, we get $8+54+27=\boxed{089}$

Solution: Add three times the first equation to the second equation and factor to get $(xy)^3(x^3+3x^2y+3xy^2+y^3)=(xy)^3(x+y)^3=3375$ . Taking the cube root yields $xy(x+y)=15$ . Noting that the first equation is $(xy)^3\cdot(xy(x+y))=810$ , we find that $(xy)^3=\frac{810}{15}=54$ . Plugging this into the second equation and dividing yields $x^3+y^3 = \frac{945}{54} = \frac{35}{2}$ . Thus the sum required, as noted in Solution 1, is $54+\frac{35}{2}\cdot2 = \boxed{089}$

Solution: As with the other solutions, factor. But this time, let $a=xy$ and $b=x+y$ . Then $a^4b=810$ . Notice that $x^3+y^3 = (x+y)(x^2-xy+y^2) = b(b^2-3a)$ , so we are looking for $2b(b^2-3a)+a^3$ . Now, if we divide the second equation by the first one, we get $7/6 = \frac{b^2-3a}{a}$ ; then $\frac{b^2}{a}=\frac{25}{6}$ . Therefore, $a = \frac{6}{25}b^2$ . Substituting $\frac{6}{25}b^2$ for $a$ in equation 1, simplifying, and then taking the cube root gives us $b^3 = \frac{5^3}{2}.$ Finding $a^3$ by cubing $a = \frac{6}{25}b^2$ on both sides and simplifying using our previous substitution, we get $a^3 = 54$ . Substituting this into the first equation and then dividing by $27$ , we get $2b(b^2 - 3a) = 35$ . Our final answer is $35+54=\boxed{089}$

Solution:
Factor the given equations as: \[x^4y^4(x+y)=810\] \[x^3y^3(x^3+y^3)=x^3y^3(x+y)(x^2-xy+y^2)=945\] We note that these expressions (as well as the desired expression) can be written exclusively in terms of $x+y$ and $xy$ . We make the substitution $s=x+y$ and $p=xy$ (for sum and product, respectively).
\[x^4y^4(x+y)=p^4s=810\] \[x^3y^3(x+y)(x^2-xy+y^2)=p^3(s)(s^2-3p)=s^3p^3-3p^4s=945\]
We see that $p^4s$ shows up in both equations, so we can eliminate it and find $sp$ , after which we can get $p^3$ from the first equation. If you rewrite the desired expression using $s$ and $p$ , it becomes clear that you don't need to actually find the values of $s$ and $p$ , but I will do so for the sake of completion.
\[s^3p^3=945+3p^4s\] \[s^3p^3=945+3(810)=3375\] \[sp=15\]
\[p^3=\frac{810}{sp}=54\] \[p=3\cdot2^{1/3}\] \[s=\frac{15}{p}=5\cdot2^{-1/3}\]
The desired expression can be written as: \[2(x^3+y^3)+(xy)^3=2(x+y)(x^2-xy+y^2)+(xy)^3\] \[2(s)(s^2-3p)+p^3=2s^3-6sp+p^3\]
Plugging in $s$ and $p$ , we get: \[2(5\cdot2^{-1/3})^3-6(15)+54=125-90+54=\boxed{089}\]

Solution:
Factor the first and second equations as $(xy)^4(x+y)=810$ and $(xy)^3(x+y)(x^2-xy+y^2)=945$ , respectively. Dividing them (allowed, since neither are $0$ ), we have \[\frac{xy}{x^2-xy+y^2}=\frac67\] or \[x^2-\frac{13}{6}xy+y^2=0.\] Plugging into the quadratic formula and solving for $x$ in terms of $y,$ we have \[x=\frac{\frac{13y}{6}\pm \sqrt{\frac{169y^2}{36}-4y^2}}{2}=\frac{2y}3 , \frac{3y}2 .\] WLOG, let $x=\frac{3y}2 .$ Plugging into our first equation, we have \[\left(\frac{3}{2}y\right)^4\left(\frac52 y\right)=810 \implies y^3 = 4 .\] Plugging this result (and the one for $x$ in terms of $y$ ) into our desired expression, we have
\begin{align*} 2x^3+(xy)^3+2y^3 &= \frac{27}{4}y^3 + \left(\frac{3}{2} y^2\right)^3 +2y^3 \\ &= \frac{35}{4}y^3 +\frac{27}{8}(y^3)^2 \\ &= 35+54 \\ &= \boxed{089} ~ASAB

Solution: Take $w=x+y$ and $z=xy$ . Remark that \begin{align*} ~&2x^{3}+(xy)^{3}+2y^{3} \\ =~&2(x^{3}+y^{3})+(xy)^{3} \\ =~&2\left[(x+y)^{3}-3xy(x+y)\right]+(xy)^{3} \\ =~&2(w^{3}-3wz)+z^{3} \\ =~& 2w^{3}-6wz+z^{3}.\end{align*} The given equations imply that \[wz^{4}=810~~~\text{and}~~~(wz)^{3}-3wz^{4}=945.\] Substituting the first equation into the second, we have that $(wz)^{3}=945+3\cdot 810=3375$ , thus $wz=\sqrt[3]{3375}=15$ . Now \[z^{3}=\frac{wz^{4}}{wz}=\frac{810}{15}=54\] and \[w^{3}=\frac{3375}{54}=\frac{125}{2}.\] Thus \begin{align*}2w^{3}-6wz+z^{3}&=2\left(\frac{125}{2}\right)-6(15)+54 \\ &=125-90+54 \\ &=\boxed{89}

Solution:
$x^4y^4(x+y)=810; x^3y^3(x^3+y^3)=945, \frac{x^2-xy+y^2}{xy}=\frac{7}{6}, \frac{x^2+y^2}{xy}=\frac{13}{6}$
Let $x^2+y^2=13k; xy=6k$ , then we can see $(x+y)^2-12k=13k, x+y=5\sqrt{k}$ , now, we see $x^4y^4\cdot (x+y)=1296k^4\cdot 5\sqrt{k}, k=\frac{1}{\sqrt[3]{4}}$
The rest is easy, $2(x^3+y^3)+x^3y^3=216k^3+2[(x+y)^3-3xy(x+y)]=216k^3+2\cdot 35k^{\frac{3}{2}}=\boxed{89}$

Solution: Let's first put the left hand sides of the equations in factored forms. Doing this we obtain $(xy)^4 \cdot (x + y) = 810$ and $(xy)(x + y)(x^2 - xy + y^2) = 945$ . Now, we will subtract and add the equations to gather information on $x$ and $y$ . When we subtract the equations and clean it up via factoring, we yield $(xy)^3 \cdot (x + y) \cdot (x - y)^2 = 115$ , and when we add them, we yield $(xy)^3 \cdot (x + y) \cdot (x^2 + y^2) = 1755$ . Now with some intuition, you should divide the equations to obtain $\frac{(x^2 + y^2)}{(x - y)^2} = 13$ . Now, we clean this up to obtain the following factoring of $0 = 2 \cdot (2x - 3y) \cdot (3x - 2y)$ . This implies that $x = \frac{3y}{2}$ . We plug that into the target expression to reduce it down to one variable, and get that target expression is $2x^3 + (xy)^3 + 2y^3 = \frac{27}{4} \cdot y^3 + \frac{27}{8} \cdot y^6 + 2y^3$ . This means that if we can find a way to get $y^3$ , then the rest is trivial. We get $y^3$ by plugging in $x = \frac{3y}{2}$ into $x^3 \cdot y^6 + y^3 \cdot x^6 = 945$ . However, this time we only factor as $(xy)^3 \cdot (x^3 + y^3)$ because we particularly want a cubic degree on $y$ . Plugging in $x = \frac{3y}{2}$ we get $y^3 = 4$ . Now lets plug this into our target expression to get $\frac{27}{4} \cdot 4 + \frac{27}{8} \cdot 4^2 + 2 \cdot 4 = \boxed{89}$

Solution:
Let $x, y$ be roots of some polynomial $P(\alpha)$ . Then we have that $P(\alpha) = (\alpha - x)(\alpha - y) \implies P(\alpha) = \alpha^2 - (x + y) \alpha + xy$ . Now from the first equation we have that $x + y = 810/(xy)^4$ . For convenience, denote $z = xy$
Define $P_k = x^k + y^k$ , where $x, y$ are roots of the above polynomial. Then we have that by Newton's formulas $P_k = (x + y)P_{k - 1} - xyP_{k - 2} = 810/z^4P_{k - 1} - zP_{k - 2}$ , where $P_0 = 2$ and $P_1 = x + y = 810/z^4$ . We desire $2P_3 + z^3$ . Now building up this recurrence;
\[P_2 = \frac{810}{z^4}P_1 - zP_0 = \frac{810^2}{z^8} - 2z.\] Then we have that \[P_3 = \frac{810}{z^4} \left(\frac{810^2}{z^8} - 3z \right).\] Plugging this value of $P_3$ into the second equation $z^3P_3 = 810$ , yields $z^9 = 164025$ . Now $z^3 = 54$ . To compute $P_3$ is now trivial, as returning this new value of $z$ into the second equation yields $P_3 = 35/2$ . Hence $z^3 + 2P_3 = 54 + 35 = \boxed{089}$


Answer: 107.0

Let $x$ and $y$ be real numbers such that $\frac{\sin x}{\sin y} = 3$ and $\frac{\cos x}{\cos y} = \frac12$ . The value of $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y}$ can be expressed in the form $\frac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$

Solution:
Let the equation $\frac{\sin x}{\sin y} = 3$ be equation 1, and let the equation $\frac{\cos x}{\cos y} = \frac12$ be equation 2.
Hungry for the widely-used identity $\sin^2\theta + \cos^2\theta = 1$ , we cross multiply equation 1 by $\sin y$ and multiply equation 2 by $\cos y$
Equation 1 then becomes:
$\sin x = 3\sin y$
Equation 2 then becomes:
$\cos x = \frac{1}{2} \cos y$
Aha! We can square both of the resulting equations and match up the resulting LHS with the resulting RHS:
$1 = 9\sin^2 y + \frac{1}{4} \cos^2 y$
Applying the identity $\cos^2 y = 1 - \sin^2 y$ (which is similar to $\sin^2\theta + \cos^2\theta = 1$ but a bit different), we can change $1 = 9\sin^2 y + \frac{1}{4} \cos^2 y$ into:
$1 = 9\sin^2 y + \frac{1}{4} - \frac{1}{4} \sin^2 y$
Rearranging, we get $\frac{3}{4} = \frac{35}{4} \sin^2 y$
So, $\sin^2 y = \frac{3}{35}$
Squaring Equation 1 (leading to $\sin^2 x = 9\sin^2 y$ ), we can solve for $\sin^2 x$
$\sin^2 x = 9\left(\frac{3}{35}\right) = \frac{27}{35}$
Using the identity $\cos 2\theta = 1 - 2\sin^2\theta$ , we can solve for $\frac{\cos 2x}{\cos 2y}$
$\cos 2x = 1 - 2\sin^2 x = 1 - 2\cdot\frac{27}{35} = 1 - \frac{54}{35} = -\frac{19}{35}$
$\cos 2y = 1 - 2\sin^2 y = 1 - 2\cdot\frac{3}{35} = 1 - \frac{6}{35} = \frac{29}{35}$
Thus, $\frac{\cos 2x}{\cos 2y} = \frac{-19/35}{29/35} = -\frac{19}{29}$
Plugging in the numbers we got back into the original equation :
We get $\frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} = \frac32 + \left(-\frac{19}{29} \right) = \frac{49}{58}$
So, the answer is $49+58=\boxed{107}$

Solution: As mentioned above, the first term is clearly $\frac{3}{2}.$ For the second term, we first wish to find $\frac{\cos 2x}{\cos 2y} =\frac{2\cos^2 x - 1}{2 \cos^2y -1}.$ Now we first square the first equation getting $\frac{\sin^2x}{\sin^2y} =\frac{1-\cos^2x}{1 - \cos^2y} =9.$ Squaring the second equation yields $\frac{\cos^2x}{\cos^2y} =\frac{1}{4}.$ Let $\cos^2x = a$ and $\cos^2y = b.$ We have the system of equations \begin{align*} 1-a &= 9-9b \\ 4a &= b \\ \end{align*} Multiplying the first equation by $4$ yields $4-4a = 36 - 36b$ and so $4-b =36 - 36b \implies b =\frac{32}{35}.$ We then find $a =\frac{8}{35}.$ Therefore the second fraction ends up being $\frac{\frac{64}{35}-1}{\frac{16}{35}-1} = -\frac{19}{29}$ so that means our desired sum is $\frac{49}{58}$ so the desired sum is $\boxed{107}.$

Solution:
Dgxje.png
We draw 2 right triangles with angles x and y that have the same hypotenuse.
We get $b^2 + 9a^2 = 4b^2 + a^2$ . Then, we find $8a^2 = 3b^2$
Now, we can scale the triangle such that $a = \sqrt{3}$ $b = \sqrt{8}$ . We find all the side lengths, and we find the hypotenuse of both these triangles to equal $\sqrt{35}$ This allows us to find sin and cos easily.
The first term is $\frac{3}{2}$ , refer to solution 1 for how to find it.
The second term is $\frac{\cos^2(x) - \sin^2(x)}{\cos^2(y) - \sin^2(y)}$ . Using the diagram, we can easily compute this as $\frac{\frac{8}{35} - \frac{27}{35}}{\frac{32}{35} - \frac{3}{35}} = \frac{-19}{29}$
Summing these you get $\frac{3}{2} + \frac{-19}{29} = \frac{49}{58} \implies \boxed{107}$

Solution:
Let $a = \sin(x), b = \sin(y)$ The first equation yields $\frac{a}{b} = 3.$ Using $sin^2(x) + cos^2(x) = 1$ the second equation yields \[\frac{\sqrt{1-a^2}}{\sqrt{1-b^2}} = \frac{1}{2} \rightarrow \frac{1-a^2}{1-b^2} = \frac{1}{4}\]
Solving this yields $\left(a, b\right) = \left(3\sqrt{\frac{3}{35}},\sqrt{\frac{3}{35}}\right).$ Finding the first via double angle for sin yields \[\frac{\sin(2x)}{\sin(2y)} = \frac{2\sin{x}\cos{x}}{2\sin{y}\cos{y}} = 3 \cdot \frac{1}{2} = \frac{3}{2}\] Double angle for cosine is \[\cos(2x) = 1-2\sin^2{x}\] so $\frac{\cos(2x)}{\sin(2x)} = \frac{1-2a^2}{1-2b^2} = -\frac{19}{29}.$ Adding yields $\frac{49}{58} \rightarrow 49 + 58 = \boxed{107}$


Answer: 154.0

Let $x$ and $y$ be two-digit integers such that $y$ is obtained by reversing the digits
of $x$ . The integers $x$ and $y$ satisfy $x^2 - y^2 = m^2$ for some positive integer $m$ .
What is $x + y + m$
$\textbf{(A) } 88 \qquad \textbf{(B) } 112 \qquad \textbf{(C) } 116 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 154$

Solution:
Let $x = 10a+b, y = 10b+a$ . The given conditions imply $x>y$ , which implies $a>b$ , and they also imply that both $a$ and $b$ are nonzero.
Then, $x^2 - y^2 = (x-y)(x+y) = (9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$
Since this must be a perfect square, all the exponents in its prime factorization must be even. $99$ factorizes into $3^2 \cdot 11$ , so $11|(a-b)(a+b)$ . However, the maximum value of $a-b$ is $9-1=8$ , so $11|a+b$ . The maximum value of $a+b$ is $9+8=17$ , so $a+b=11$
Then, we have $33^2(a-b) = m^2$ , so $a-b$ is a perfect square, but the only perfect squares that are within our bound on $a-b$ are $1$ and $4$ . We know $a+b=11$ , and, for $a-b=1$ , adding equations to eliminate $b$ gives us $2a=12 \Longrightarrow a=6, b=5$ . Testing $a-b=4$ gives us $2a=15 \Longrightarrow a=\frac{15}{2}, b=\frac{7}{2}$ , which is impossible, as $a$ and $b$ must be digits. Therefore, $(a,b) = (6,5)$ , and $x+y+m=\boxed{154}$

Solution:
The first steps are the same as Solution 1.  Let $x = 10a+b, y = 10b+a$ , where we know that a and b are digits (whole numbers less than $10$ ).
Like Solution 1, we end up getting $(9a - 9b)(11a + 11b) = 99(a-b)(a+b) = m^2$ .  This is where the solution diverges.
We know that the left side of the equation is a perfect square because $m$ is an integer.  If we factor $99$ into its prime factors, we get $3^2\cdot 11$ .  In order to get a perfect square on the left side, $(a-b)(a+b)$ must make both prime exponents even.  Because the a and b are digits, a simple guess would be that $(a+b)$ (the bigger number) equals $11$ while $(a-b)$ is a factor of nine (1 or 9).  The correct guesses are $a = 6, b = 5$ causing $x = 65, y = 56,$ and $m = 33$ .  The sum of the numbers is $\boxed{154}$

Solution:
Once again, the solution is quite similar as the above solutions. Since $x$ and $y$ are two digit integers, we can write $x = 10a+b, y = 10b+a$ and because $x^2 - y^2 = (x-y)(x+y)$ , substituting and factoring, we get $99(a+b)(a-b) = m^2$
Therefore, $(a+b)(a-b) = \frac{m^2}{99}$ and $\frac{m^2}{99}$ must be an integer. A quick strategy is to find the smallest such integer $m$ such that $\frac{m^2}{99}$ is an integer. We notice that 99 has a prime factorization of $3^2 \cdot 11.$
Let $m^2 = n.$ Since we need a perfect square and 3 is already squared, we just need to square 11. So $3^2 \cdot 11^2$ gives us 1089 as $n$ and $m = \sqrt{1089} = 33.$ We now get the equation $(x-y)(x+y) = 33^2$ , which we can also write as $(x-y)(x+y) = 11^2 \cdot 3^2$
A very simple guess assumes that $x-y=3^2$ and $x+y=11^2$ since $x$ and $y$ are positive. Finally, we come to the conclusion that $x=65$ and $y=56$ , so $x+y+m$ $=$ $\boxed{154}$

Solution:
Continue the same as Solution $3$ until we get $33$ . Knowing that $33^2 = 1089$ , we have narrowed down our Pythagorean triples. We know that the $2$ other squares should be larger than $33^2$ , so we can start testing.
If we start testing the $40$ s, it is fruitless since the closest to $33^2$ would be $33 - 45 - 54$ which is not a Pythagorean triple. We can start by testing out the $50$ s, and it turns our that $33 - 56 - 65$ is a Pythagorean triple. Therefore, our answer is $33+56+65$ $\boxed{154}$


Answer: 0.5

Let $x$ be a real number such that $\sec x - \tan x = 2$ . Then $\sec x + \tan x =$
$\textbf{(A)}\ 0.1 \qquad  \textbf{(B)}\ 0.2 \qquad  \textbf{(C)}\ 0.3 \qquad  \textbf{(D)}\ 0.4 \qquad  \textbf{(E)}\ 0.5$

Solution: $(\sec x - \tan x)(\sec x + \tan x) = \sec^{2} x - \tan^{2} x = 1$ , so $\sec x + \tan x = \boxed{0.5}$

Solution:
Note that $\sec x - \tan x = (1-\sin x)/\cos x$ and $\sec x + \tan x = (1+\sin x)/\cos x$ . Let $(1+\sin x)/\cos x = y$ . Multiplying, we get $(1-\sin^{2}x)/\cos^{2}x = 1$ .Then, $2y = 1$ $\sec x + \tan x =  \boxed{0.5}$ . ~songmath20
Edited 5.1.2023


Answer: 67.0

Let $x$ be a real number such that $\sin^{10}x+\cos^{10} x = \tfrac{11}{36}$ . Then $\sin^{12}x+\cos^{12} x = \tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
We can substitute $y = \sin^2{x}$ . Since we know that $\cos^2{x}=1-\sin^2{x}$ , we can do some simplification.
This yields $y^5+(1-y)^5=\frac{11}{36}$ . From this, we can substitute again to get some cancellation through binomials. If we let $z=\frac{1}{2}-y$ , we can simplify the equation to: \[\left(\frac{1}{2}+z\right)^5+\left(\frac{1}{2}-z\right)^5=\frac{11}{36}.\] After using binomial theorem, this simplifies to: \[\frac{1}{16}(80z^4+40z^2+1)=\frac{11}{36}.\] If we use the quadratic formula, we obtain $z^2=\frac{1}{12}$ , so $z=\pm\frac{1}{2\sqrt{3}}$ (observe that either choice of $z$ doesn't matter). Substituting $z,$ we get:
\[\sin^{12}{x}+\cos^{12}{x}=\left(\frac{1}{2}-z\right)^6+\left(\frac{1}{2}+z\right)^6=2z^6 + \frac{15z^4}{2} + \frac{15z^2}{8} + \frac{1}{32}=\frac{13}{54}.\]
Therefore, the answer is $13+54=\boxed{067}$

Solution: First, for simplicity, let $a=\sin{x}$ and $b=\cos{x}$ . Note that $a^2+b^2=1$ . We then bash the rest of the problem out. Take the fifth power of this expression and get $a^{10}+b^{10}+5a^2b^2(a^6+b^6)+10a^4b^4(a^2+b^2)=\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1$ . Note that we also have $\frac{11}{36}=a^{10}+b^{10}=(a^{10}+b^{10})(a^2+b^2)=a^{12}+b^{12}+a^2b^2(a^8+b^8)$ . So, it suffices to compute $a^2b^2(a^8+b^8)$ . Let $y=a^2b^2$ . We have from cubing $a^2+b^2=1$ that $a^6+b^6+3a^2b^2(a^2+b^2)=1$ or $a^6+b^6=1-3y$ . Next, using $\frac{11}{36}+5a^2b^2(a^6+b^6)+10a^4b^4=1$ , we get $a^2b^2(a^6+b^6)+2a^4b^4=\frac{5}{36}$ or $y(1-3y)+2y^2=y-y^2=\frac{5}{36}$ . Solving gives $y=\frac{5}{6}$ or $y=\frac{1}{6}$ . Clearly $y=\frac{5}{6}$ is extraneous, so $y=\frac{1}{6}$ . Now note that $a^4+b^4=(a^2+b^2)^2-2a^2b^2=\frac{2}{3}$ , and $a^8+b^8=(a^4+b^4)^2-2a^4b^4=\frac{4}{9}-\frac{1}{18}=\frac{7}{18}$ . Thus we finally get $a^{12}+b^{12}=\frac{11}{36}-\frac{7}{18}\cdot\frac{1}{6}=\frac{13}{54}$ , giving $\boxed{067}$

Solution:
Newton sums is basically constructing the powers of the roots of the polynomials instead of deconstructing them which was done in Solution $2$ . Let $\sin^2x$ and $\cos^2x$ be the roots of some polynomial $F(a)$ . Then, by Vieta, $F(a)=a^2-a+b$ for some $b=\sin^2x\cdot\cos^2x$
Let $S_k=\left(\sin^2x\right)^k+\left(\cos^2x\right)^k$ . We want to find $S_6$ . Clearly $S_1=1$ and $S_2=1-2b$ . Newton sums tells us that $S_k-S_{k-1}+bS_{k-2}=0\Rightarrow S_k=S_{k-1}-bS_{k-2}$ where $k\ge 3$ for our polynomial $F(a)$
Bashing, we have \[S_3=S_2-bS_1\Rightarrow S_3=(1-2b)-b(1)=1-3b\] \[S_4=S_3-bS_2\Rightarrow S_4=(1-3b)-b(1-2b)=2b^2-4b+1\] \[S_5=S_4-bS_3\Rightarrow S_5=(2b^2-4b+1)-b(1-3b)=5b^2-5b+1=\frac{11}{36}\]
Thus \[5b^2-5b+1=\frac{11}{36}\Rightarrow 5b^2-5b+\frac{25}{36}=0, 36b^2-36b+5=0, (6b-1)(6b-5)=0\] $b=\frac{1}{6} \text{ or } \frac{5}{6}$ . Clearly, $\sin^2x\cdot\cos^2x\not=\frac{5}{6}$ so $\sin^2x\cdot\cos^2x=b=\frac{1}{6}$
Note $S_4=\frac{7}{18}$ . Solving for $S_6$ , we get $S_6=S_5-\frac{1}{6}S_4=\frac{13}{54}$ . Finally, $13+54=\boxed{067}$

Solution:
Factor the first equation. \[\sin^{10}x + \cos^{10}x = (\sin^2x+\cos^2x)(\sin^8x-\sin^6x\cos^2x+\sin^4x\cos^4x-\sin^2x\cos^6x+\cos^8x)\] First of all, $\sin^4x+\cos^4x = 1-2\sin^2x\cos^2x$ because $\sin^4x+\cos^4x=(\sin^2x + \cos^2x)^2 -2\sin^2x\cos^2x = 1 - 2\sin^2x\cos^2x$ We group the first, third, and fifth term and second and fourth term. The first group: \begin{align*} \sin^8+\sin^4x\cos^4x+\cos^8x &= (\sin^4x+\cos^4x)^2-\sin^4x\cos^4x)\\ &= (1 - 2\sin^2x\cos^2x)^2-\sin^4x\cos^4x)\\ &= 1+4\sin^4x\cos^4x-4\sin^2x\cos^2x \end{align*} The second group: \begin{align*} -\sin^6x\cos^2x-\sin^2x\cos^6x  &= -\sin^2x\cos^2x(\sin^4x+\cos^4x)\\ &= -\sin^2x\cos^2x(1-2\sin^2x\cos^2x)\\ &= -\sin^2x\cos^2x+2\sin^4x\cos^4x \end{align*} Add the two together to make \[1+4\sin^4x\cos^4x-4\sin^2x\cos^2x-\sin^2x\cos^2x+2\sin^4x\cos^4x = 1 - 5\sin^2x\cos^2x+5\sin^4x\cos^4x\] Because this equals $\frac{11}{36}$ , we have \[5\sin^4x\cos^4x- 5\sin^2x\cos^2x+\frac{25}{36}=0\] Let $\sin^2x\cos^2x = a$ so we get \[5a^2- 5a+\frac{25}{36}=0 \Rightarrow a^2-a+\frac{5}{36}\] Solving the quadratic gives us \[a = \frac{1 \pm \frac{2}{3}}{2}\] Because $\sin^2x\cos^2x \le \frac{1}{4}$ , we finally get $a = \frac{1 - \frac{2}{3}}{2} = \frac{1}{6}$
Now from the second equation, \begin{align*} \sin^{12}x + \cos^{12}x &= (\sin^4x+\cos^4x)(\sin^8x-\sin^4x\cos^4x+\cos^8x)\\ &= (1-2\sin^2x\cos^2x)((\sin^4x+\cos^4x)^2-3\sin^4x\cos^4x)\\ &= (1-2\sin^2x\cos^2x)((1-2\sin^2x\cos^2x)^2-3\sin^4x\cos^4x) \end{align*} Plug in $\sin^2x\cos^2x = \frac{1}{6}$ to get \[(1-2(\frac{1}{6}))(1-2(\frac{1}{6})^2-3(\frac{1}{6})^2) = \frac{13}{54}\] which yields the answer $\boxed{067}$

Solution:
Define the recursion $a_n=(\sin^2 x)^n+(\cos^2 x)^n$ We know that the characteristic equation of $a_n$ must have 2 roots, so we can recursively define $a_n$ as $a_n=p*a_{n-1}+q*a_{n-2}$ $p$ is simply the sum of the roots of the characteristic equation, which is $\sin^2 x+\cos^2 x=1$ $q$ is the product of the roots, which is $-(\sin^2 x)(\cos^2 x)$ . This value is not trivial and we have to solve for it.
We know that $a_0=2$ $a_1=1$ $a_5=\frac{11}{36}$ .
Solving the rest of the recursion gives
\[a_2=1+2q\] \[a_3=1+3q\] \[a_4=1+4q+2q^2\] \[a_5=1+5q+5q^2=\frac{11}{36}\] \[a_6=1+6q+9q^2+2q^3\]
Solving for $q$ in the expression for $a_5$ gives us $q^2+q+\frac{5}{36}=0$ , so $q=-\frac{5}{6}, -\frac{1}{6}$ . Since $q=-(\sin^2 x)(\cos^2 x)$ , we know that the minimum value it can attain is $-\frac{1}{4}$ by AM-GM, so $q$ cannot be $-\frac{5}{6}$ .
Plugging in the value of $q$ into the expression for $a_6$ , we get $a_6=1-1+\frac{1}{4}-\frac{1}{108}=\frac{26}{108}=\frac{13}{54}$ . Our final answer is then $13+54=\boxed{067}$

Solution:
Let $m=\sin^2 x$ and $n=\cos^2 x$ , then $m+n=1$ and $m^5+n^5=\frac{11}{36}$
$m^6+n^6=(m^5+n^5)(m+n)-mn(m^4+n^4)=(m^5+n^5)-mn(m^4+n^4)$
Now factoring $m^5+n^5$ as solution 4 yields $m^5+n^5=(m+n)(m^4-m^3n+m^2n^2-mn^3+n^4)$ $=m^4+n^4-mn(m^2-mn+n^2)=m^4+n^4-mn[(m+n)^2-3mn]=m^4+n^4-mn(1-3mn)$
Since $(m+n)^4=m^4+4m^3n+6m^2n^2+4mn^3+n^4$ $m^4+n^4=(m+n)^4-2mn(2m^2+3mn+2n^2)=1-2mn(2m^2+3mn+2n^2)$
Notice that $2m^2+3mn+2n^2$ can be rewritten as $[\sqrt{2}(a+b)]^2-mn=2-mn$ . Thus, $m^4+n^4=1-2mn(2-mn)$ and $m^5+n^5=1-2mn(2-mn)-mn(1-3mn)=1-5mn+5(mn)^2=\frac{11}{36}$ . As in solution 4, we get $mn=\frac{1}{6}$ and $m^4+n^4=1-2*\frac{1}{6}(2-\frac{1}{6})=\frac{7}{18}$
Substitute $m^4+n^4=\frac{7}{18}$ and $mn=\frac{1}{6}$ , then $m^6+n^6=\frac{11}{36}-\frac{1}{6}*\frac{7}{18}=\frac{13}{54}$ , and the desired answer is $\boxed{067}$

Solution:
Let $a_n=\sin^nx+\cos^nx$ for non-negative integers $n$ . Then $a_0=2$ and $a_2=1$ . In addition, \[a_n=\sin^nx+\cos^nx=\left(\sin^{n-2}x+\cos^{n-2}x\right)\left(\sin^2x+\cos^2x\right)-\sin^2x\cos^2x\left(\sin^{n-4}x+\cos^{n-4}x\right)=a_{n-2}-Xa_{n-4},\] where $X=\sin^2x\cos^2x$ . So we can compute
\begin{align*}
a_4&=1-2X\\
a_6&=1-3X\\
a_8&=1-4X+2X^2\\
a_{10}&=1-5X+5X^2=\frac{11}{36}
\end{align*}so $X=\frac{1}{6},\frac{5}{6}$ . But by the sin double angle formula, $\sin^2x\cos^2x=\frac{1}{4}\sin^22x\leq\frac{1}{4}$ , so $X=\frac{1}{6}$ . Then \[a_{12}=a_{10}-Xa_8=\frac{11}{36}-\frac{1}{6}\cdot\frac{7}{18}=\frac{13}{54}\] so the answer is $\boxed{067}$ as desired.


Answer: 16.0

Let $x$ be chosen at random from the interval $(0,1)$ . What is the probability that $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$ ?
Here $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$
$\mathrm{(A)}\ \frac 18 \qquad \mathrm{(B)}\ \frac 3{20} \qquad \mathrm{(C)}\ \frac 16 \qquad \mathrm{(D)}\ \frac 15  \qquad \mathrm{(E)}\ \frac 14$

Solution:
Let $k$ be an arbitrary integer. For which $x$ do we have $\lfloor\log_{10}4x\rfloor = \lfloor\log_{10}x\rfloor = k$
The equation $\lfloor\log_{10}x\rfloor = k$ can be rewritten as $10^k \leq x < 10^{k+1}$ . The second one gives us $10^k \leq 4x < 10^{k+1}$ . Combining these, we get that both hold at the same time if and only if $10^k \leq x < \frac{10^{k+1}}4$
Hence for each integer $k$ we get an interval of values for which $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$ . These intervals are obviously pairwise disjoint.
For any $k\geq 0$ the corresponding interval is disjoint with $(0,1)$ , so it does not contribute to our answer. On the other hand, for any $k<0$ the entire interval is inside $(0,1)$ . Hence our answer is the sum of the lengths of the intervals for $k<0$
For a fixed $k$ the length of the interval $\left[ 10^k, \frac{10^{k+1}}4 \right)$ is $\frac 32\cdot 10^k$
This means that our result is $\frac 32 \left( 10^{-1} + 10^{-2} + \cdots \right) = \frac 32 \cdot \frac 19 = \boxed{16}$

Solution:
The largest value for $x$ is $10^{0}$ . If $x > 10^{-1}$ , then $\lfloor\log_{10}4x\rfloor$ doesn't fulfill the condition unless $10^{-2} \leq x < 0.25 * 10^{-1}$ . The same holds when you get smaller, because $x = 0.25 * 10^{n}$ for $n \leq 0$ is the lowest value such that $4x$ becomes a higher power of 10.
Recognize that this is a geometric sequence. The probability of choosing $x$ such that $\lfloor\log_{10}4x\rfloor$ and $\lfloor\log_{10}x\rfloor$ both equal $-1$ is $(9/10)* (15/90) =15/100$ , because there is a 90 percent chance of choosing $x > 10^{-1}$ , and only values of $x$ between $10^{-1}$ and $0.25*10^{0}$ work in this case. Then, for $x$ such that $\lfloor\log_{10}4x\rfloor$ and $\lfloor\log_{10}x\rfloor$ both equal $-2$ , you have $(1/10) * ((9/10) *(15/90))$ . This is a geometric series with ratio $1/10$ . Using $a/(1-r)$ for the sum of an infinite geometric sequence, we get $(15/100)/(1-(1/10)) =  \boxed{16}$


Answer: 13.0

Let $x$ be the least real number greater than $1$ such that $\sin(x)= \sin(x^2)$ , where the arguments are in degrees. What is $x$ rounded up to the closest integer?
$\textbf{(A) } 10 \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 14 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 20$

Solution:
The smallest $x$ to make $\sin(x) = \sin(x^2)$ would require $x=x^2$ , but since $x$ needs to be greater than $1$ , these solutions are not valid.
The next smallest $x$ would require $x=180-x^2$ , or $x^2+x=180$
After a bit of guessing and checking, we find that $12^2+12=156$ , and $13^2+13=182$ , so the solution lies between $12{ }$ and $13$ , making our answer $\boxed{13}.$

Solution: For choice $\textbf{(A)},$ we have \begin{align*} \left| \sin x - \sin \left( x^2 \right) \right| & = \left| \sin 10^\circ - \sin \left( \left( 10^2 \right)^\circ \right) \right| \\ & = \left| \sin 10^\circ - \sin 100^\circ \right| \\ & = \left| \sin 10^\circ - \sin 80^\circ \right| \\ & = \sin 80^\circ - \sin 10^\circ . \end{align*} For choice $\textbf{(B)},$ we have \begin{align*} \left| \sin x - \sin \left( x^2 \right) \right| & = \left| \sin 13^\circ - \sin \left( \left( 13^2 \right)^\circ \right) \right| \\ & = \left| \sin 13^\circ - \sin 169^\circ \right| \\ & = \left| \sin 10^\circ - \sin 11^\circ \right| \\ & = \sin 11^\circ - \sin 10^\circ . \end{align*} For choice $\textbf{(C)},$ we have \begin{align*} \left| \sin x - \sin \left( x^2 \right) \right| & = \left| \sin 14^\circ - \sin \left( \left( 14^2 \right)^\circ \right) \right| \\ & = \left| \sin 14^\circ - \sin 196^\circ \right| \\ & = \left| \sin 14^\circ + \sin 16^\circ \right| \\ & = \sin 14^\circ + \sin 16^\circ . \end{align*} For choice $\textbf{(D)},$ we have \begin{align*} \left| \sin x - \sin \left( x^2 \right) \right| & = \left| \sin 19^\circ - \sin \left( \left( 19^2 \right)^\circ \right) \right| \\ & = \left| \sin 19^\circ - \sin 361^\circ \right| \\ & = \left| \sin 19^\circ - \sin 1^\circ \right| \\ & = \sin 19^\circ - \sin 1^\circ . \end{align*} For choice $\textbf{(E)},$ we have \begin{align*} \left| \sin x - \sin \left( x^2 \right) \right| & = \left| \sin 20^\circ - \sin \left( \left( 20^2 \right)^\circ \right) \right| \\ & = \left| \sin 20^\circ - \sin 400^\circ \right| \\ & = \left| \sin 20^\circ - \sin 40^\circ \right| \\ & = \sin 40^\circ - \sin 20^\circ . \end{align*} Therefore, the answer is $\boxed{13},$ as $\sin 11^\circ - \sin 10^\circ$ is the closest to $0.$


Answer: 3.0

Let $x$ be the number \[0.\underbrace{0000...0000}_{1996\text{ zeros}}1,\] where there are 1996 zeros after the decimal point.  Which of the following expressions represents the largest number?
$\text{(A)}\ 3+x \qquad \text{(B)}\ 3-x \qquad \text{(C)}\ 3\cdot x \qquad \text{(D)}\ 3/x \qquad \text{(E)}\ x/3$

Solution:
Estimate each of the options.
$A$ will give a number that is just over $3$
$B$ will give a number that is just under $3$ .  This eliminates $B$ , because $A$ is bigger.
$C$ will give a number that is barely over $0$ , since it is three times a tiny number.  This eliminates $C$ , because $A$ is bigger.
$D$ will give a huge number. $\frac{1}{x}$ will get very, very large in magnitude when $x$ gets close to zero.  You can see this by examining the sequence $x=0.1$ , which gives $10$ as the reciprocal, $x=0.01$ , which gives $100$ as the reciprocal, and $x=0.001$ , which gives $1000$ as the reciprocal.  Thus, $D$ will be huge, and this eliminates $A$
$E$ will give a small number, since you're dividing a tiny number into thirds.  This eliminates $E$ , and gives $\boxed{3}$ as the answer.


Answer: 33.0

Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution:
First, let define a triangle with side lengths $\sqrt{2x}$ $\sqrt{2z}$ , and $l$ , with altitude from $l$ 's equal to $\sqrt{xz}$ $l = \sqrt{2x - xz} + \sqrt{2z - xz}$ , the left side of one equation in the problem.
Let $\theta$ be angle opposite the side with length $\sqrt{2x}$ . Then the altitude has length $\sqrt{2z} \cdot \sin(\theta) = \sqrt{xz}$ and thus $\sin(\theta) = \sqrt{\frac{x}{2}}$ , so $x=2\sin^2(\theta)$ and the side length $\sqrt{2x}$ is equal to $2\sin(\theta)$
We can symmetrically apply this to the two other equations/triangles.
By law of sines, we have $\frac{2\sin(\theta)}{\sin(\theta)} = 2R$ , with $R=1$ as the circumradius, same for all 3 triangles.
The circumcircle's central angle to a side is $2 \arcsin(l/2)$ , so the 3 triangles' $l=1, \sqrt{2}, \sqrt{3}$ ,  have angles $120^{\circ}, 90^{\circ}, 60^{\circ}$ , respectively.
This means that by half angle arcs, we see that we have in some order, $x=2\sin^2(\alpha)$ $y=2\sin^2(\beta)$ , and $z=2\sin^2(\gamma)$ (not necessarily this order, but here it does not matter due to symmetry), satisfying that $\alpha+\beta=180^{\circ}-\frac{120^{\circ}}{2}$ $\beta+\gamma=180^{\circ}-\frac{90^{\circ}}{2}$ , and $\gamma+\alpha=180^{\circ}-\frac{60^{\circ}}{2}$ . Solving, we get $\alpha=\frac{135^{\circ}}{2}$ $\beta=\frac{105^{\circ}}{2}$ , and $\gamma=\frac{165^{\circ}}{2}$
We notice that \[[(1-x)(1-y)(1-z)]^2=[\sin(2\alpha)\sin(2\beta)\sin(2\gamma)]^2=[\sin(135^{\circ})\sin(105^{\circ})\sin(165^{\circ})]^2\] \[=\left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{6}-\sqrt{2}}{4} \cdot \frac{\sqrt{6}+\sqrt{2}}{4}\right)^2 = \left(\frac{\sqrt{2}}{8}\right)^2=\frac{1}{32} \to \boxed{033}. \blacksquare\]

Solution:
(This eventually whittles down to the same concept as Solution 1)
Note that in each equation in this system, it is possible to factor $\sqrt{x}$ $\sqrt{y}$ , or $\sqrt{z}$ from each term (on the left sides), since each of $x$ $y$ , and $z$ are positive real numbers. After factoring out accordingly from each terms one of $\sqrt{x}$ $\sqrt{y}$ , or $\sqrt{z}$ , the system should look like this: \begin{align*} \sqrt{x}\cdot\sqrt{2-y} + \sqrt{y}\cdot\sqrt{2-x} &= 1 \\ \sqrt{y}\cdot\sqrt{2-z} + \sqrt{z}\cdot\sqrt{2-y} &= \sqrt2 \\ \sqrt{z}\cdot\sqrt{2-x} + \sqrt{x}\cdot\sqrt{2-z} &= \sqrt3. \end{align*}
This should give off tons of trigonometry vibes. To make the connection clear, $x = 2\cos^2 \alpha$ $y = 2\cos^2 \beta$ , and $z = 2\cos^2 \theta$ is a helpful substitution:
\begin{align*} \sqrt{2\cos^2 \alpha}\cdot\sqrt{2-2\cos^2 \beta} + \sqrt{2\cos^2 \beta}\cdot\sqrt{2-2\cos^2 \alpha} &= 1 \\ \sqrt{2\cos^2 \beta}\cdot\sqrt{2-2\cos^2 \theta} + \sqrt{2\cos^2 \theta}\cdot\sqrt{2-2\cos^2 \beta} &= \sqrt2 \\ \sqrt{2\cos^2 \theta}\cdot\sqrt{2-2\cos^2 \alpha} + \sqrt{2\cos^2 \alpha}\cdot\sqrt{2-2\cos^2 \theta} &= \sqrt3. \end{align*}
From each equation $\sqrt{2}^2$ can be factored out, and when every equation is divided by 2, we get:
\begin{align*} \sqrt{\cos^2 \alpha}\cdot\sqrt{1-\cos^2 \beta} + \sqrt{\cos^2 \beta}\cdot\sqrt{1-\cos^2 \alpha} &= \frac{1}{2} \\ \sqrt{\cos^2 \beta}\cdot\sqrt{1-\cos^2 \theta} + \sqrt{\cos^2 \theta}\cdot\sqrt{1-\cos^2 \beta} &= \frac{\sqrt2}{2} \\ \sqrt{\cos^2 \theta}\cdot\sqrt{1-\cos^2 \alpha} + \sqrt{\cos^2 \alpha}\cdot\sqrt{1-\cos^2 \theta} &= \frac{\sqrt3}{2}. \end{align*}
which simplifies to (using the Pythagorean identity $\sin^2 \phi + \cos^2 \phi = 1 \; \forall \; \phi \in \mathbb{C}$ ):
\begin{align*} \cos \alpha\cdot\sin \beta + \cos \beta\cdot\sin \alpha &= \frac{1}{2} \\ \cos \beta\cdot\sin \theta + \cos \theta\cdot\sin \beta &= \frac{\sqrt2}{2} \\ \cos \theta\cdot\sin \alpha + \cos \alpha\cdot\sin \theta &= \frac{\sqrt3}{2}. \end{align*}
which further simplifies to (using sine addition formula $\sin(a + b) = \sin a \cos b + \cos a \sin b$ ):
\begin{align*} \sin(\alpha + \beta) &= \frac{1}{2} \\ \sin(\beta + \theta) &= \frac{\sqrt2}{2} \\ \sin(\alpha + \theta) &= \frac{\sqrt3}{2}. \end{align*}
Taking the inverse sine ( $0\leq\theta\frac{\pi}{2}$ ) of each equation yields a simple system:
\begin{align*} \alpha + \beta &= \frac{\pi}{6} \\ \beta + \theta &= \frac{\pi}{4} \\ \alpha + \theta &= \frac{\pi}{3} \end{align*}
giving solutions:
\begin{align*} \alpha &= \frac{\pi}{8} \\ \beta &= \frac{\pi}{24} \\ \theta &= \frac{5\pi}{24} \end{align*}
Since these unknowns are directly related to our original unknowns, there are consequent solutions for those:
\begin{align*} x &= 2\cos^2\left(\frac{\pi}{8}\right) \\ y &= 2\cos^2\left(\frac{\pi}{24}\right) \\ z &= 2\cos^2\left(\frac{5\pi}{24}\right) \end{align*}
When plugging into the expression $\left[ (1-x)(1-y)(1-z) \right]^2$ , noting that $-\cos 2\phi = 1 - 2\cos^2 \phi\; \forall \; \phi \in \mathbb{C}$ helps to simplify this expression into:
\begin{align*} \left[ (-1)^3\left(\cos \left(2\cdot\frac{\pi}{8}\right)\cos \left(2\cdot\frac{\pi}{24}\right)\cos \left(2\cdot\frac{5\pi}{24}\right)\right)\right]^2 \\ = \left[ (-1)\left(\cos \left(\frac{\pi}{4}\right)\cos \left(\frac{\pi}{12}\right)\cos \left(\frac{5\pi}{12}\right)\right)\right]^2 \end{align*}
Now, all the cosines in here are fairly standard:
\begin{align*} \cos \frac{\pi}{4} &= \frac{\sqrt{2}}{2} \\ \cos \frac{\pi}{12} &=\frac{\sqrt{6} + \sqrt{2}}{4} & (= \cos{\frac{\frac{\pi}{6}}{2}} ) \\ \cos \frac{5\pi}{12} &= \frac{\sqrt{6} - \sqrt{2}}{4} & (=  \cos\left({\frac{\pi}{6} + \frac{\pi}{4}} \right) ) \end{align*}
With some final calculations:
\begin{align*} &(-1)^2\left(\frac{\sqrt{2}}{2}\right)^2\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)^2\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)^2 \\ =& \left(\frac{1}{2}\right) \left(\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)\left(\frac{\sqrt{6} - \sqrt{2}}{4}\right)\right)^2 \\ =&\frac{1}{2} \frac{4^2}{16^2} = \frac{1}{32} \end{align*}
This is our answer in simplest form $\frac{m}{n}$ , so $m + n = 1 + 32 = \boxed{033}$

Solution:
Let $1-x=a;1-y=b;1-z=c$ , rewrite those equations
$\sqrt{(1-a)(1+b)}+\sqrt{(1+a)(1-b)}=1$
$\sqrt{(1-b)(1+c)}+\sqrt{(1+b)(1-c)}=\sqrt{2}$
$\sqrt{(1-a)(1+c)}+\sqrt{(1-c)(1+a)}=\sqrt{3}$
and solve for $m/n = (abc)^2 = a^2b^2c^2$
Square both sides and simplify, to get three equations:
$2ab-1=2\sqrt{(1-a^2)(1-b^2)}$
$2bc~ ~ ~ ~  ~ ~=2\sqrt{(1-b^2)(1-c^2)}$
$2ac+1=2\sqrt{(1-c^2)(1-a^2)}$
Square both sides again, and simplify to get three equations:
$a^2+b^2-ab=\frac{3}{4}$
$b^2+c^2~ ~ ~ ~ ~ ~=1$
$a^2+c^2+ac=\frac{3}{4}$
Subtract first and third equation, getting $(b+c)(b-c)=a(b+c)$ $a=b-c$
Put it in first equation, getting $b^2-2bc+c^2+b^2-b(b-c)=b^2+c^2-bc=\frac{3}{4}$ $bc=\frac{1}{4}$
Since $a^2=b^2+c^2-2bc=\frac{1}{2}$ $m/n = a^2b^2c^2 = a^2(bc)^2 = \frac{1}{2}\left(\frac{1}{4}\right)^2=\frac{1}{32}$ and so the final answer is $\boxed{033}$

Solution:
Denote $u = 1 - x$ $v = 1 - y$ $w = 1 - z$ .
Hence, the system of equations given in the problem can be written as \begin{align*} \sqrt{(1-u)(1+v)} + \sqrt{(1+u)(1-v)} & = 1 \hspace{1cm} (1) \\ \sqrt{(1-v)(1+w)} + \sqrt{(1+v)(1-w)} & = \sqrt{2} \hspace{1cm} (2) \\ \sqrt{(1-w)(1+u)} + \sqrt{(1+w)(1-u)} & = \sqrt{3} . \hspace{1cm} (3)  \end{align*}
Each equation above takes the following form: \[ \sqrt{(1-a)(1+b)} + \sqrt{(1+a)(1-b)} = k . \]
Now, we simplify this equation by removing radicals.
Denote $p = \sqrt{(1-a)(1+b)}$ and $q = \sqrt{(1+a)(1-b)}$
Hence, the equation above implies \[ \left\{ \begin{array}{l} p + q = k \\ p^2 = (1-a)(1+b) \\ q^2 = (1+a)(1-b) \end{array} \right.. \]
Hence, $q^2 - p^2 = (1+a)(1-b) - (1-a)(1+b) = 2 (a-b)$ .
Hence, $q - p = \frac{q^2 - p^2}{p+q} = \frac{2}{k} (a-b)$
Because $p + q = k$ and $q - p = \frac{2}{k} (a-b)$ , we get $q = \frac{a-b}{k} + \frac{k}{2}$ .
Plugging this into the equation $q^2 = (1+a)(1-b)$ and simplifying it, we get \[ a^2 + \left( k^2 - 2 \right) ab + b^2 = k^2 - \frac{k^4}{4} . \]
Therefore, the system of equations above can be simplified as \begin{align*} u^2 - uv + v^2 & = \frac{3}{4} \\ v^2 + w^2 & = 1 \\ w^2 + wu + u^2 & = \frac{3}{4}  . \end{align*}
Denote $w' = - w$ .
The system of equations above can be equivalently written as \begin{align*} u^2 - uv + v^2 & = \frac{3}{4} \hspace{1cm} (1') \\ v^2 + w'^2 & = 1 \hspace{1cm} (2') \\ w'^2 - w'u + u^2 & = \frac{3}{4} \hspace{1cm} (3') . \end{align*}
Taking $(1') - (3')$ , we get \[ (v - w') (v + w' - u) = 0 . \]
Thus, we have either $v - w' = 0$ or $v + w' - u = 0$
$\textbf{Case 1}$ $v - w' = 0$
Equation (2') implies $v = w' = \pm \frac{1}{\sqrt{2}}$
Plugging $v$ and $w'$ into Equation (2), we get contradiction. Therefore, this case is infeasible.
$\textbf{Case 2}$ $v + w' - u = 0$
Plugging this condition into (1') to substitute $u$ , we get \[ v^2 + v w' + w'^2 = \frac{3}{4} \hspace{1cm} (4) . \]
Taking $(4) - (2')$ , we get \[ v w' = - \frac{1}{4} . \hspace{1cm} (5) . \]
Taking (4) + (5), we get \[ \left( v + w' \right)^2 = \frac{1}{2} . \]
Hence, $u^2 = \left( v + w' \right)^2 = \frac{1}{2}$
Therefore, \begin{align*} \left[ (1-x)(1-y)(1-z) \right]^2 & = u^2 (vw)^2 \\ & = u^2 (vw')^2 \\ & = \frac{1}{2} \left( - \frac{1}{4} \right)^2 \\ & = \frac{1}{32} . \end{align*}
Therefore, the answer is $1 + 32 = \boxed{033}$

Solution:
In given equations, $0 \leq x,y,z \leq 2,$ so we define some points: \[\bar {O} = (0, 0), \bar {A} = (1, 0), \bar{M} = \left(\frac {1}{\sqrt{2}},\frac {1}{\sqrt{2}}\right),\] \[\bar {X} = \left(\sqrt {\frac {x}{2}}, \sqrt{1 – \frac{x}{2}}\right), \bar {Y'} = \left(\sqrt {\frac {y}{2}}, \sqrt{1 – \frac{y}{2}}\right),\] \[\bar {Y} = \left(\sqrt {1 – \frac{y}{2}},\sqrt{\frac {y}{2}}\right), \bar {Z} = \left(\sqrt {1 – \frac{z}{2}},\sqrt{\frac {z}{2}}\right).\] Notice, that \[\mid \vec {AO} \mid = \mid \vec {MO} \mid = \mid \vec {XO} \mid =\mid \vec {YO} \mid = \mid \vec {Y'O} \mid =\mid \vec {ZO} \mid = 1\] and each points lies in the first quadrant.
We use given equations and get some scalar products: \[(\vec {XO} \cdot \vec {YO}) = \frac {1}{2} = \cos \angle XOY \implies  \angle XOY = 60 ^\circ,\] \[(\vec {XO} \cdot \vec {ZO}) = \frac {\sqrt{3}}{2} = \cos \angle XOZ \implies  \angle XOZ = 30^\circ,\] \[(\vec {Y'O} \cdot \vec {ZO}) = \frac {1}{\sqrt{2}} = \cos \angle Y'OZ \implies  \angle Y'OZ = 45^\circ.\] So $\angle YOZ =  \angle XOY – \angle XOZ =  60 ^\circ – 30 ^\circ = 30 ^\circ,   \angle Y'OY =  \angle Y'OZ +  \angle YOZ = 45^\circ + 30 ^\circ = 75^\circ.$
Points $Y$ and $Y'$ are symmetric with respect to $OM.$
Case 1 \[\angle YOA = \frac{90^\circ – 75^\circ}{2} = 7.5^\circ, \angle ZOA = 30^\circ + 7.5^\circ = 37.5^\circ, \angle XOA = 60^\circ + 7.5^\circ = 67.5^\circ .\] \[1 – x = \left(\sqrt{1 – \frac{x}{2}} \right)^2– \left(\sqrt{\frac {x}{2}}\right)^2  = \sin^2 \angle XOA – \cos^2 \angle XOA = –\cos 2 \angle XOA = –\cos 135^\circ,\] \[1 – y = \cos 15^\circ, 1 – z = \cos 75^\circ \implies  \left[ (1–x)(1–y)(1–z) \right]^2 = \left[ \sin 45^\circ \cdot \cos 15^\circ \cdot \sin 15^\circ \right]^2 =\] \[=\left[ \frac {\sin 45^\circ \cdot \sin 30^\circ}{2} \right]^2  = \frac {1}{32} \implies \boxed{033}.\] Case 2


Answer: 49.0

Let $x,$ $y,$ and $z$ be positive real numbers that satisfy \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) \ne 0.\] The value of $xy^5z$ can be expressed in the form $\frac{1}{2^{p/q}},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$

Solution: Since there are only two dependent equations given and three unknowns, the three expressions given can equate to any common value (that isn't 0, of course), so to simplify the problem let us assume without loss of generality that \[2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) = 2.\] Then \begin{align*} 2\log_{x}(2y) = 2 &\implies x=2y\\ 2\log_{2x}(4z) = 2 &\implies 2x=4z\\ \log_{2x^4}(8yz) = 2 &\implies 4x^8 = 8yz \end{align*} Solving these equations, we quickly see that $4x^8 = (2y)(4z) = x(2x) \rightarrow x=2^{-1/6}$ and then $y=z=2^{-1/6 - 1} = 2^{-7/6}.$ Finally, our desired value is $2^{-1/6} \cdot (2^{-7/6})^5 \cdot 2^{-7/6} = 2^{-43/6}$ and thus $p+q = 43 + 6 = \boxed{049.}$

Solution:
Notice that $2y\cdot 4z=8yz$ $2\log_2(2y)=\log_2\left(4y^2\right)$ and $2\log_2(4z)=\log_2\left(16z^2\right)$
From this, we see that $8yz$ is the geometric mean of $4y^2$ and $16z^2$ . So, for constant $C\ne 0$ \[\frac{\log 4y^2}{\log x}=\frac{\log 8yz}{\log 2x^4}=\frac{\log 16z^2}{\log 2x}=C\] Since $\log 4y^2,\log 8yz,\log 16z^2$ are in an arithmetic progression, so are $\log x,\log 2x^4,\log 2x$
Therefore, $2x^4$ is the geometric mean of $x$ and $2x$ \[2x^4=\sqrt{x\cdot 2x}\implies 4x^8=2x^2\implies 2x^6=1\implies x=2^{-1/6}\] We can plug $x$ in to any of the two equal fractions aforementioned. So, without loss of generality: \[\frac{\log 4y^2}{\log x}=\frac{\log 16z^2}{\log 2x}\implies \log\left(4y^2\right)\log\left(2x\right)=\log\left(16z^2\right)\log\left(x\right)\] \[\implies \log\left(4y^2\right)\cdot \frac{5}{6}\log 2=\log\left(16z^2\right)\cdot \frac{-1}{6}\log 2\] \[\implies 5\log\left(4y^2\right)=-\log\left(16z^2\right)\implies 5\log\left(4y^2\right)+\log\left(16z^2\right)=0\] \[\implies \left(4y^2\right)^5\cdot 16z^2=1\implies 16384y^{10}z^2=1\implies y^{10}z^2=\frac{1}{16384}\implies y^5z=\frac{1}{128}\]
Thus $xy^5z=2^{-\frac{1}{6}-7}=2^{-\frac{43}{6}}$ and $43+6=\boxed{049}$

Solution:
Since we are given that $xy^5z = 2^{-p/q}$ , we may assume that $x, y$ , and $z$ are all powers of two.  We shall thus let $x = 2^X$ $y = 2^Y$ , and $z = 2^Z$ .  Let $a = \log_{2^X}(2^{Y+1})$ .  From this we get the system of equations: \[\] $(1)$ \[a = \log_{2^X}(2^{Y+1}) \Rightarrow aX = Y + 1\] $(2)$ \[a = \log_{2^{X + 1}}(2^{Z + 2}) \Rightarrow aX + a = Z + 2\] $(3)$ \[2a = \log_{2^{4X + 1}}(2^{Y + Z + 3}) \Rightarrow 8aX + 2a = Y + Z + 3\]
Plugging equation $(1)$ into equation $(2)$ yields $Y + a = Z + 1$ .  Plugging equation $(1)$ into equation $(3)$ and simplifying yields $7Y + 2a + 6 = Z + 1$ , and substituting $Y + a$ for $Z + 1$ and simplifying yields $Y + 1 = \frac{-a}{6}$ .  But $Y + 1 = aX$ , so $aX =  \frac{-a}{6}$ , so $X = \frac{-1}{6}$
Knowing this, we may substitute $\frac{-1}{6}$ for $X$ in equations $(1)$ and $(2)$ , yielding $\frac{-a}{6} = Y + 1$ and $\frac{5a}{6} = Z + 2$ .  Thus, we have that $-5(Y + 1) = Z + 2 \rightarrow 5Y + Z = -7$ .  We are looking for $xy^5z = 2^{X+ 5Y + Z}$ $X = \frac{-1}{6}$ and $5Y + Z = -7$ , so $xy^5z = 2^{-43/6} = \frac{1}{2^{43/6}}$ .  The answer is $43+6=\boxed{049}$

Solution:
We know that \[\frac{\log (4y^2)}{\log (x)} = \frac{\log (16z^2)}{\log (2x)} = \frac{2\log (8yz)}{\log (2x^2)}\] By the Mediant theorem
Substituting into the original equation yields us $\frac{2\log (8yz)}{\log (2x^2)} = \frac{\log (8yz)}{\log (2x^4)} \Rightarrow 2\log (2x^4) = \log (2x^2) \Rightarrow x=2^{-1/6}.$ For some constant $C\not= 0,$ Let $2\log_{x}(2y) = 2\log_{2x}(4z) = \log_{2x^4}(8yz) = C.$ Then, we obtain the system of equations \[y=2^{-13C/12}\] \[z=2^{-19C/12}\] \[8yz=2^{C/3}.\]
Adding the first two equations and subtracting the third, we find $C=1.$ Thus, \[xy^5z=2^{-1/6} \cdot 2^{-65/12} \cdot 2^{-19/12}=2^{-43/6} \Rightarrow p+q=\boxed{049}.\]


Answer: 33.0

Let $x,y$ and $z$ be positive real numbers that satisfy the following system of equations: \[\log_2\left({x \over yz}\right) = {1 \over 2}\] \[\log_2\left({y \over xz}\right) = {1 \over 3}\] \[\log_2\left({z \over xy}\right) = {1 \over 4}\] Then the value of $\left|\log_2(x^4y^3z^2)\right|$ is $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
Denote $\log_2(x) = a$ $\log_2(y) = b$ , and $\log_2(z) = c$
Then, we have: $a-b-c = \frac{1}{2}$ $-a+b-c = \frac{1}{3}$ $-a-b+c = \frac{1}{4}$
Now, we can solve to get $a = \frac{-7}{24}, b = \frac{-9}{24}, c = \frac{-5}{12}$ . Plugging these values in, we obtain $|4a + 3b + 2c|  = \frac{25}{8} \implies \boxed{033}$ . ~akliu

Solution:
$\log_2(y/xz) + \log_2(z/xy) = \log_2(1/x^2) = -2\log_2(x) = \frac{7}{12}$
$\log_2(x/yz) + \log_2(z/xy) = \log_2(1/y^2) = -2\log_2(y) = \frac{3}{4}$
$\log_2(x/yz) + \log_2(y/xz) = \log_2(1/z^2) = -2\log_2(z) = \frac{5}{6}$
$\log_2(x) = -\frac{7}{24}$
$\log_2(y) = -\frac{3}{8}$
$\log_2(z) = -\frac{5}{12}$
$4\log_2(x) + 3\log_2(y) + 2\log_2(z) = -25/8$
$25 + 8 = \boxed{033}$

Solution:
Adding all three equations, $\log_2(\frac{1}{xyz}) = \frac{1}{2}+\frac{1}{3}+\frac{1}{4} = \frac{13}{12}$ . Subtracting this from every equation, we have: \[2\log_2x = -\frac{7}{12},\] \[2\log_2y = -\frac{3}{4},\] \[2\log_2z = -\frac{5}{6}\] Our desired quantity is the absolute value of $4\log_2x+3\log_2y+2\log_2z = 2(\frac{7}{12})+3/2(\frac{3}{4})+\frac{5}{6} = \frac{25}{8}$ , so our answer is $25+8 = \boxed{033}$ .
~Spoirvfimidf


Answer: 265.0

Let $x,y,$ and $z$ be real numbers satisfying the system \begin{align*} \log_2(xyz-3+\log_5 x)&=5,\\ \log_3(xyz-3+\log_5 y)&=4,\\ \log_4(xyz-3+\log_5 z)&=4.\\ \end{align*} Find the value of $|\log_5 x|+|\log_5 y|+|\log_5 z|$

Solution: First, we get rid of logs by taking powers: $xyz-3+\log_5 x=2^{5}=32$ $xyz-3+\log_5 y=3^{4}=81$ , and $(xyz-3+\log_5 z)=4^{4}=256$ . Adding all the equations up and using the $\log {xy}=\log {x}+\log{y}$ property, we have $3xyz+\log_5{xyz} = 378$ , so we have $xyz=125$ . Solving for $x,y,z$ by substituting $125$ for $xyz$ in each equation, we get $\log_5 x=-90, \log_5 y=-41, \log_5 z=134$ , so adding all the absolute values we have $90+41+134=\boxed{265}$


Answer: 273.0

Let $x,y,$ and $z$ be real numbers satisfying the system of equations \begin{align*} xy + 4z &= 60 \\ yz + 4x &= 60 \\ zx + 4y &= 60. \end{align*} Let $S$ be the set of possible values of $x.$ Find the sum of the squares of the elements of $S.$

Solution:
We first subtract the second equation from the first, noting that they both equal $60$ \begin{align*} xy+4z-yz-4x&=0 \\ 4(z-x)-y(z-x)&=0 \\ (z-x)(4-y)&=0 \end{align*}
Case 1: Let $y=4$
The first and third equations simplify to: \begin{align*} x+z&=15 \\ xz&=44 \end{align*} from which it is apparent that $x=4$ and $x=11$ are solutions.
Case 2: Let $x=z$
The first and third equations simplify to: \begin{align*} xy+4x&=60 \\ x^2+4y&=60 \end{align*}
We subtract the following equations, yielding: \begin{align*} x^2+4y-xy-4x&=0 \\ x(x-4)-y(x-4)&=0 \\ (x-4)(x-y)&=0 \end{align*}
We thus have $x=4$ and $x=y$ , substituting in $x=y=z$ and solving yields $x=6$ and $x=-10$
Then, we just add the squares of the solutions (make sure not to double count the $4$ ), and get \[4^2+11^2+(-6)^2+10^2=16+121+36+100=\boxed{273}.\] ~SAHANWIJETUNGA

Solution:
We index these equations as (1), (2), and (3), respectively.
Taking $(1)-(2)$ , we get \[ \left( x - z \right) \left( y - 4 \right) = 0 . \]
Denote $x' = x - 4$ $y' = y - 4$ $z' = z - 4$ .
Thus, the above equation can be equivalently written as \[ \left( x' - z' \right) y' = 0 . \hspace{1cm} (1') \]
Similarly, by taking $(2)-(3)$ , we get \[ \left( y' - x' \right) z' = 0 . \hspace{1cm} (2') \]
By taking $(3) - (1)$ , we get \[ \left( z' - y' \right) x' = 0 . \hspace{1cm} (3') \]
From $(3')$ , we have the following two cases.
Case 1: $x' = 0$
Plugging this into $(1')$ and $(2')$ , we get $y'z' = 0$ .
Thus, $y' = 0$ or $z' = 0$
Because we only need to compute all possible values of $x$ , without loss of generality, we only need to analyze one case that $y' = 0$
Plugging $x' = 0$ and $y' = 0$ into (1), we get a feasible solution $x = 4$ $y = 4$ $z = 11$
Case 2: $x' \neq 0$ and $z' - y' = 0$
Plugging this into $(1')$ and $(2')$ , we get $\left( x' - y' \right) y' = 0$
Case 2.1: $y' = 0$
Thus, $z' = 0$ . Plugging $y' = 0$ and $z' = 0$ into (1), we get a feasible solution $x = 11$ $y = 4$ $z = 4$
Case 2.2: $y' \neq 0$ and $x' = y'$
Thus, $x' = y' = z'$ . Plugging these into (1), we get $\left( x, y, z \right) = \left( -10, -10, -10 \right)$ or $\left( 6, 6, 6 \right)$
Putting all cases together, $S = \left\{ 4, 11, -10, 6 \right\}$ .
Therefore, the sum of the squares of the elements of $S$ is \begin{align*} 4^2 + 11^2 + \left( -10 \right)^2 + 6^2 = \boxed{273}

Solution:
We index these equations as (1), (2), and (3), respectively.
Using equation (1), we get $z = \frac{60 - xy}{4} = 15 - \frac{xy}{4}$ We need to solve for x, so we plug this value of z into equation (3) to get:
\[15x - \frac{x^2y}{4} - 4y = 60\] \[\frac{y}{4} * x^2 - 15x + (60 + 4y) = 0\]
We use the quadratic formula to get possible values of x:
\[x = \frac{15 \pm \sqrt{15^2 - 4(\frac{y}{4})(60 + 4y)}}{\frac{y}{2}}\] \[x = \frac{30 \pm 2\sqrt{225 - 60y + 4y^2}}{y}\] \[x = \frac{30 \pm 2\sqrt{(2y-15)^2}}{y}\] \[x = \frac{30 \pm 2(2y - 15)}{y}\] \[x = \frac{30 \pm (4y - 30)}{y}\]
Here, we have two cases, (plus) and (minus)
In the plus case, we have: \[x = \frac{(30 + 4y - 30)}{y}\] \[x = 4\]
So, our first case gives us one value of x, which is 4.
In the minus case, we have: \[x = \frac{30 - (4y - 30)}{y}\] \[x = \frac{60 - 4y}{y}\] \[x = \frac{60}{y} - 4\]
For this case, we now have values of x in terms of y.
Plugging this expression for x in equation (1), we get \[60 - 4y + 4z = 60\] \[4z = 4y\] \[z = y\]
So we know that for this case, z = y.
Using this information in equation (2), we get \[y^2 + \frac{240}{y} - 16 = 60\] \[y^2 + \frac{240}{y} - 76 = 0\] Multiplying both sides by y, we get a cubic expression: \[y^3 + 0y^2 - 76y + 240 = 0\] Here we just have to figure out the values of y that make this equation true. I used Vieta's Formulas to get a possible list, but you could also use the rational root theorem and synthetic division to find these. We call the three values of y that solve this equation: $y_{1},y_{2},y_{3}$ Using Vieta's Formulas, you get these three expressions: \[y_{1} + y_{2} + y_{3} = 0\] \[y_{1} * y_{2} * y_{3} = -240\] \[(y_{1} * y_{2}) + (y_{2} * y_{3}) + (y_{1} * y_{3}) = -76\]
In addition, we know that $y \vert 60$ , because of our expression for x.
Since the three values of y multiply to a negative number but also add to 0, we know that one value is negative and the other two are positive, and that the absolute value of the negative value is greater than both of the positive values.
List of possible values for y are $\{-60,-30,-20,-15,-12,-10,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6,10,12,15,20,30,60\}$ From a list of these values, the only values that work are $y_1 = -10, y_2 = 6, y_3 = 4$ because \[-10 + 6 + 4 = 0\] \[-10 * 6 * 4 = -240\] \[(-10 * 6) + (-10 * 4) + (6 * 4) = -60 - 40 + 24 = -76\]
Plugging in these values for y into our expression for x, we get: \[x = \frac{60}{-10} - 4 = -10\] \[x = \frac{60}{6} - 4 = 6\] \[x = \frac{60}{4} - 4 = 11\]
So, now we have accounted for both cases, and we have 4 values of x = $\{-10,4,6,11\}$ Squaring all these terms we get: 100 + 16 + 36 + 121 = 273, so our answer is $\boxed{273}$

Solution:
Since all three equations are in the form $\frac{K}{a} + 4a = 60$ where $K = xyz$ , we can rearrange this to see that $x$ $y$ , and $z$ all satisfy
\[ 4a^2 - 60a + K = 0. \]
Let this quadratic have roots $a_1$ and $a_2$ . Then, there are two cases to consider: two of $x$ $y$ $z$ are equal to $a_1$ and the third is equal to $a_2$ , or all of $x$ $y$ $z$ are equal to $a_1$
Case 1: WLOG let $x = y = a_1$ and $z = a_2$
Then by Vieta's,
\[x + z = 15 \hspace{1cm} (1)\] \[xz = \frac{x^2z}{4} \hspace{1cm} (2)\]
which gives $x = y = 4$ and $z = 11$ . But we can swap $x$ $y$ , and $z$ however we like, so this also gives $x = 11$ as a solution. In total, this case yields $x = 4$ and $x = 11$ as possible values of $x$
Case 2: WLOG let $x = y = z = a_1$
Again, by Vieta's,
\[x + a_2 = 15 \hspace{1cm} (1)\] \[xa_2 = \frac{x^3}{4} \hspace{1cm} (2)\]
We can use $(2)$ to isolate $a_2$ in terms of $x$ , then plug that into $(1)$ to get that $x^2 + 4x - 60 = 0$ . This yields $x = -10$ and $x = 6$ as additional possible values of $x$
In all, $x$ can be any of $\{ 4, 11, -10, 6 \}$ , so the requested answer is $4^2 + 11^2 + (-10)^2 + 6^2 = \boxed{273}.$

Solution:
We index these equations as (1), (2), and (3) respectively (same as solution 2). There are two possible cases:
Case 1: $x = \pm 4$
In this case, we simply plug in $x = 4$ and $x = -4$ . We note that $x=4$ is a valid case.
Case 2: $x \neq \pm 4$
In this case, using equation (3), we get $y = 15 - \frac{xz}{4}$ . Plugging that into equation (1), we get $z = \frac{240-60x}{16-x^2}$ . Plugging that expression back into the original expression for $y$ we obtain $y = 15 - \frac{60x-15x^2}{16-x^2}$ .
Now we plug these two expressions into equation (2): \[\left(15 - \frac{60x-15x^2}{16-x^2}\right)\left(\frac{240-60x}{16-x^2}\right) = 60-4x\] multiplying both sides by $(16-x^2)$ and factoring, we get: \[\left(15(4-x)(4+x)-60x+15x^2\right)\left(\frac{60(4-x)}{(4+x)(4-x)}\right) = 4(15 - x)(4 + x)(4 - x)\] which simplifies to: \[x^3 - 7x^2 - 104x + 660 = 0\] we note that $x = 6$ is a root. Factoring, we get the other roots, -10 and 11.
Our desired answer is the sum of the square of all these roots: \[4^2 + 6^2 + (-10)^2 + 11^2 = \boxed{273}\]


Answer: 4032.0

Let $x=-2016$ . What is the value of $\Bigg\vert\Big\vert |x|-x\Big\vert-|x|\Bigg\vert-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$

Solution:
Substituting carefully, $\Bigg\vert\Big\vert 2016-(-2016)\Big\vert-2016\Bigg\vert-(-2016)$
becomes $|4032-2016|+2016=2016+2016=4032$ which is $\boxed{4032}$

Solution:
Solution by e_power_pi_times_i
Substitute $-y = x = -2016$ into the equation. Now, it is $\Bigg\vert\Big\vert |y|+y\Big\vert-|y|\Bigg\vert+y$ . Since $y = 2016$ , it is a positive number, so $|y| = y$ . Now the equation is $\Bigg\vert\Big\vert y+y\Big\vert-y\Bigg\vert+y$ . This further simplifies to $2y-y+y = 2y$ , so the answer is $\boxed{4032}$


Answer: 4032.0

Let $x=-2016$ . What is the value of $\bigg|$ $||x|-x|-|x|$ $\bigg|$ $-x$
$\textbf{(A)}\ -2016\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2016\qquad\textbf{(D)}\ 4032\qquad\textbf{(E)}\ 6048$

Solution:
By: dragonfly
First of all, lets plug in all of the $x$ 's into the equation.
$\bigg|$ $||-2016|-(-2016)|-|-2016|$ $\bigg|$ $-(-2016)$
Then we simplify to get
$\bigg|$ $|2016+2016|-2016$ $\bigg|$ $+2016$
which simplifies into
$\bigg|$ $2016$ $\bigg|$ $+2016$
and finally we get $\boxed{4032}$

Solution:
Consider $x$ is negative.
We replace all instances of $x$ with $|x|$
$\bigg|$ $||x|+|x||-|x|$ $\bigg|$ $+|x|$
$=$ $\bigg|$ $|2x|-|x|$ $\bigg|$ $+|x|$
$=$ $|x|$ $+|x|$
$=|2x|$
$=4032 \implies \boxed{4032}$


Answer: 125.0

Solution:
We note that in general,
It now becomes apparent that if we multiply the numerator and denominator of $\frac{4}{ (\sqrt{5}+1) (\sqrt[4]{5}+1) (\sqrt[8]{5}+1) (\sqrt[16]{5}+1) }$ by $(\sqrt[16]{5} - 1)$ , the denominator will telescope to $\sqrt[1]{5} - 1 = 4$ , so
It follows that $(x + 1)^{48} = (\sqrt[16]5)^{48} = 5^3 = \boxed{125}$

Solution:
Like Solution $2$ , let $z=\sqrt[16]{5}$ Then, the expression becomes
$x=\frac{4}{(z+1)(z^2+1)(z^4+1)(z^8+1)}$ Now, multiplying by the conjugate of each binomial in the denominator, we obtain...
$x=\frac{4(z-1)(z^2-1)(z^4-1)(z^8-1)}{(z^2-1)(z^4-1)(z^8-1)(z^{16}-1)}\implies x=\frac{4(z-1)}{z^{16}-1}$ Plugging back in,
$x=\frac{4({\sqrt[16]{5}-1)}}{5-1}\implies x=\sqrt[16]{5}-1$
Hence, after some basic exponent rules, we find the answer is $\boxed{125}$


Answer: 241.0

Let $x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}$ . What is the greatest integer that does not exceed $100x$

Solution:
Note that $\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{\sum_{n=1}^{44} \cos n}{\sum_{n=46}^{89} \cos n} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\cos 89 + \cos 88 + \dots + \cos 46}$ by the cofunction identities.(We could have also written it as $\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{\sum_{n=46}^{89} \sin n}{\sum_{n=1}^{44} \sin n} = \frac {\sin 89 + \sin 88 + \dots + \sin 46}{\sin 1 + \sin 2 + \dots + \sin 44}$ .)
Now use the sum-product formula $\cos x + \cos y = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$ We want to pair up $[1, 44]$ $[2, 43]$ $[3, 42]$ , etc. from the numerator and $[46, 89]$ $[47, 88]$ $[48, 87]$ etc. from the denominator. Then we get: \[\frac{\sum_{n=1}^{44} \cos n}{\sum_{n=1}^{44} \sin n} = \frac{\sum_{n=1}^{44} \cos n}{\sum_{n=46}^{89} \cos n} = \frac{2\cos(\frac{45}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})]}{2\cos(\frac{135}{2})[\cos(\frac{43}{2})+\cos(\frac{41}{2})+\dots+\cos(\frac{1}{2})]} \Rightarrow \frac{\cos(\frac{45}{2})}{\cos(\frac{135}{2})}\]
To calculate this number, use the half angle formula. Since $\cos\left(\frac{x}{2}\right) = \pm \sqrt{\frac{\cos x + 1}{2}}$ , then our number becomes: \[\frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}}\] in which we drop the negative roots (as it is clear cosine of $22.5$ and $67.5$ are positive). We can easily simplify this:
\begin{eqnarray*} \frac{\sqrt{\frac{\frac{\sqrt{2}}{2} + 1}{2}}}{\sqrt{\frac{\frac{-\sqrt{2}}{2} + 1}{2}}} &=& \sqrt{\frac{\frac{2+\sqrt{2}}{4}}{\frac{2-\sqrt{2}}{4}}} \\ &=& \sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}} \cdot \sqrt{\frac{2+\sqrt{2}}{2+\sqrt{2}}} \\ &=& \sqrt{\frac{(2+\sqrt{2})^2}{2}} \\ &=& \frac{2+\sqrt{2}}{\sqrt{2}} \cdot \sqrt{2} \\ &=& \sqrt{2}+1 \end{eqnarray*}
And hence our answer is $\lfloor 100x \rfloor =  \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}$

Solution:
\begin{eqnarray*} x &=& \frac {\sum_{n = 1}^{44} \cos n^\circ}{\sum_{n = 1}^{44} \sin n^\circ} = \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\\ &=& \frac {\cos (45 - 1) + \cos(45 - 2) + \dots + \cos(45 - 44)}{\sin 1 + \sin 2 + \dots + \sin 44} \end{eqnarray*}
Using the identity $\sin a + \sin b = 2\sin \frac{a+b}2 \cos \frac{a-b}{2}$ $\Longrightarrow \sin x + \cos x$ $= \sin x + \sin (90-x)$ $= 2 \sin 45 \cos (45-x)$ $= \sqrt{2} \cos (45-x)$ , that summation reduces to
\begin{eqnarray*}x &=& \left(\frac {1}{\sqrt {2}}\right)\left(\frac {(\cos 1 + \cos2 + \dots + \cos44) + (\sin1 + \sin2 + \dots + \sin44)}{\sin1 + \sin2 + \dots + \sin44}\right)\\ &=& \left(\frac {1}{\sqrt {2}}\right)\left(1 + \frac {\cos 1 + \cos 2 + \dots + \cos 44}{\sin 1 + \sin 2 + \dots + \sin 44}\right) \end{eqnarray*}
This fraction is equivalent to $x$ .  Therefore, \begin{eqnarray*} x &=& \left(\frac {1}{\sqrt {2}}\right)\left(1 + x\right)\\ \frac {1}{\sqrt {2}} &=& x\left(\frac {\sqrt {2} - 1}{\sqrt {2}}\right)\\ x &=& \frac {1}{\sqrt {2} - 1} = 1 + \sqrt {2}\\ \lfloor 100x \rfloor &=& \lfloor 100(1 + \sqrt {2}) \rfloor = \boxed{241}

Solution:
A slight variant of the above solution, note that
\begin{eqnarray*} \sum_{n=1}^{44} \cos n + \sum_{n=1}^{44} \sin n &=& \sum_{n=1}^{44} \sin n + \sin(90-n)\\ &=& \sqrt{2}\sum_{n=1}^{44} \cos(45-n) = \sqrt{2}\sum_{n=1}^{44} \cos n\\ \sum_{n=1}^{44} \sin n &=& (\sqrt{2}-1)\sum_{n=1}^{44} \cos n \end{eqnarray*}
This is the ratio we are looking for. $x$ reduces to $\frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1$ , and $\lfloor 100(\sqrt{2} + 1)\rfloor = \boxed{241}$

Solution:
Consider the sum $\sum_{n = 1}^{44} \text{cis } n^\circ$ . The fraction is given by the real part divided by the imaginary part.
The sum can be written $- 1 + \sum_{n = 0}^{44} \text{cis } n^\circ = - 1 + \frac {\text{cis } 45^\circ - 1}{\text{cis } 1^\circ - 1}$ (by De Moivre's Theorem with geometric series)
$= - 1 + \frac {\frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2}}{\text{cis } 1^\circ - 1} = - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 + \frac {i \sqrt {2}}{2} \right) (\text{cis } ( - 1^\circ) - 1)}{(\cos 1^\circ - 1)^2 + \sin^2 1^\circ}$ (after multiplying by complex conjugate
$= - 1 + \frac {\left( \frac {\sqrt {2}}{2} - 1 \right) (\cos 1^\circ - 1) + \frac {\sqrt {2}}{2}\sin 1^\circ + i\left( \left(1 - \frac {\sqrt {2}}{2} \right) \sin 1^\circ + \frac {\sqrt {2}}{2} (\cos 1^\circ - 1)\right)}{2(1 - \cos 1^\circ)}$
$= - \frac {1}{2} - \frac {\sqrt {2}}{4} - \frac {i\sqrt {2}}{4} + \frac {\sin 1^\circ \left( \frac {\sqrt {2}}{2} + i\left( 1 - \frac {\sqrt {2}}{2} \right) \right)}{2(1 - \cos 1^\circ)}$
Using the tangent half-angle formula , this becomes $\left( - \frac {1}{2} + \frac {\sqrt {2}}{4}[\cot (1/2^\circ) - 1] \right) + i\left( \frac {1}{2}\cot (1/2^\circ) - \frac {\sqrt {2}}{4}[\cot (1/2^\circ) + 1] \right)$
Dividing the two parts and multiplying each part by 4, the fraction is $\frac { - 2 + \sqrt {2}[\cot (1/2^\circ) - 1]}{2\cot (1/2^\circ) - \sqrt {2}[\cot (1/2^\circ) + 1]}$
Although an exact value for $\cot (1/2^\circ)$ in terms of radicals will be difficult, this is easily known: it is really large!
So treat it as though it were $\infty$ . The fraction is approximated by $\frac {\sqrt {2}}{2 - \sqrt {2}} = \frac {\sqrt {2}(2 + \sqrt {2})}{2} = 1 + \sqrt {2}\Rightarrow \lfloor 100(1+\sqrt2)\rfloor=\boxed{241}$

Solution:
Consider the sum $\sum_{n = 1}^{44} \text{cis } n^\circ$ . The fraction is given by the real part divided by the imaginary part.
The sum can be written as $\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)$ .
Consider the rhombus $OABC$ on the complex plane such that $O$ is the origin, $A$ represents $\text{cis } n^\circ$ $B$ represents $\text{cis } n^\circ + \text{cis } 45-n^\circ$ and $C$ represents $\text{cis } n^\circ$ . Simple geometry shows that $\angle BOA = 22.5-k^\circ$ , so the angle that $\text{cis } n^\circ + \text{cis } 45-n^\circ$ makes with the real axis is simply $22.5^\circ$ . So $\sum_{n=1}^{22} (\text{cis } n^\circ + \text{cis } 45-n^\circ)$ is the sum of collinear complex numbers, so the angle the sum makes with the real axis is $22.5^\circ$ . So our answer is $\lfloor 100 \cot(22.5^\circ) \rfloor = \boxed{241}$

Solution: We write $x =\frac{\sum_{n=46}^{89} \sin n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}$ since $\cos x = \sin (90^{\circ}-x).$ Now we by the sine angle sum we know that $\sin (x+45^{\circ}) = \sin 45^{\circ}(\sin x + \cos x).$ So the expression simplifies to $\sin 45^{\circ}\left(\frac{\sum_{n=1}^{44} (\sin n^{\circ}+\cos n^{\circ})}{\sum_{n=1}^{44} \sin n^{\circ}}\right) = \sin 45^{\circ}\left(1+\frac{\sum_{n=1}^{44} \cos n^{\circ}}{\sum_{n=1}^{44} \sin n^{\circ}}\right)=\sin 45^{\circ}(1+x).$ Therefore we have the equation $x = \sin 45^{\circ}(1+x) \implies x = \sqrt{2}+1.$ Finishing, we have $\lfloor 100x \rfloor = \boxed{241}.$

Solution:
We can pair the terms of the summations as below.
\[\dfrac{(\cos{1} + \cos{44}) + (\cos{2} + \cos{43}) + (\cos{3} + \cos{42}) + \cdots + (\cos{22} + \cos{23})}{(\sin{1} + \sin{44}) + (\sin{2} + \sin{43}) + (\sin{3} + \sin{42}) + \cdots + (\sin{22} + \sin{23})}.\]
From here, we use the cosine and sine subtraction formulas as shown.
\begin{align*}
&\dfrac{(\cos(45-44) + \cos(45-1)) + (\cos(45-43) + \cos(45-2))+ \cdots (\cos(45-23) + \cos(45-22))}{(\sin(45-44) + \sin(45-1)) + (\sin(45-43) + \sin(45-2))+ \cdots (\sin(45-23) + \sin(45-22))} \\
&= \dfrac{(\cos{45}\cos{44} +\sin{45}\sin{44} + \cos{45}\cos{1} + \sin{45}\sin{1})+ \cdots + (\cos{45}\cos{23} + \sin{45}\sin{23} + \cos{45}\cos{22} + \sin{45}\sin{22})}{(\sin{45}\cos{44}-\cos{45}\sin{44} + \sin{45}\cos{1} - \cos{45}\sin{1}) + \cdots + (\sin{45}\cos{23} -\cos{45}\sin{23} + \sin{45}\cos{22} - \cos{45}\sin{22})} \\
&= \dfrac{\sqrt{2}/2(\cos{44} + \sin{44} + \cos{1}+\sin{1} +\cos{43} + \sin{43} + \cos{2} + \sin{2} + \cdots + \cos{23} + \sin{23} + \cos{22} + \sin{22})}{\sqrt{2}/2(\cos{44} -\sin{44} +\cos{1} - \sin{1} + \cos{43}-\sin{43} + \cos{2} -\sin{2} +\cdots + \cos{23}-\sin{23} + \cos{22} - \sin{22})} \\
&=\dfrac{\sum\limits_{n=1}^{44} \cos n^\circ + \sum\limits_{n=1}^{44} \sin n^\circ}{\sum\limits_{n=1}^{44} \cos n^\circ - \sum\limits_{n=1}^{44} \sin n^\circ}.
\end{align*}
Since all of these operations have been performed to ensure equality, we can set the final line of the above mess equal to our original expression.
\[\dfrac{\sum\limits_{n=1}^{44} \cos n^\circ + \sum\limits_{n=1}^{44} \sin n^\circ}{\sum\limits_{n=1}^{44} \cos n^\circ - \sum\limits_{n=1}^{44} \sin n^\circ} = \frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ}.\]
For the sake of clarity, let $\sum\limits_{n=1}^{44} \cos n^\circ = C$ and $\sum\limits_{n=1}^{44} \sin n^\circ = S$ . Then, we have
\[\dfrac{C+S}{C-S} = \dfrac{C}{S} \implies CS+S^2 = C^2-CS.\]
Finishing, we have $S^2+2CS=C^2$ . Adding $C^2$ to both sides gives $(C+S)^2 = 2C^2$ , or $C+S = \pm C\sqrt{2}$ . Taking the positive case gives $S= C(\sqrt{2}-1)$ . Finally,
\[x=\frac{\sum\limits_{n=1}^{44} \cos n^\circ}{\sum\limits_{n=1}^{44} \sin n^\circ} = \dfrac{C}{S} = \dfrac{1}{\sqrt{2}-1} = \sqrt{2} +1 \implies \lfloor{100x\rfloor} = \boxed{241}.\]


Answer: 6.0

Let $x_0,x_1,x_2,\dotsc$ be a sequence of numbers, where each $x_k$ is either $0$ or $1$ . For each positive integer $n$ , define \[S_n = \sum_{k=0}^{n-1} x_k 2^k\] Suppose $7S_n \equiv 1 \pmod{2^n}$ for all $n \geq 1$ . What is the value of the sum \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}?\] $\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) }12\qquad \textbf{(D) } 14\qquad \textbf{(E) }15$

Solution:
In binary numbers, we have \[S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0})_2.\] It follows that \[8S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0}000)_2.\] We obtain $7S_n$ by subtracting the equations: \[\begin{array}{clccrccccccr}   & (x_{n-1} & x_{n-2} & x_{n-3} & x_{n-4} & \ldots & x_2 & x_1 & x_0 & 0 & 0 & 0 \ )_2 \\ -\quad & & & & (x_{n-1} & \ldots & x_5 & x_4 & x_3 & x_2 & x_1 & x_0)_2 \\ \hline   & & & & & & & & & & &  \\ [-2.5ex]   & ( \ \ ?& ? & ? & 0 \ \ \ & \ldots & 0 & 0 & 0 & 0 & 0 & 1 \ )_2 \\ \end{array}\] We work from right to left: \begin{alignat*}{6} x_0=x_1=x_2=1  \quad &\implies \quad &x_3 &= 0& \\  \quad &\implies \quad &x_4 &= 1& \\  \quad &\implies \quad &x_5 &= 1& \\ \quad &\implies \quad &x_6 &= 0& \\  \quad &\implies \quad &x_7 &= 1& \\  \quad &\implies \quad &x_8 &= 1& \\ \quad &\quad \vdots & & & \end{alignat*} For all $n\geq3,$ we conclude that
Finally, we get $(x_{2019},x_{2020},x_{2021},x_{2022})=(0,1,1,0),$ from which \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \boxed{6}.\] ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution:
First, notice that \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] Then since $S_n$ is the modular inverse of $7$ in $\mathbb{Z}_{2^n}$ , we can perform the Euclidean Algorithm to find it for $n = 2019,2023$
Starting with $2019$ \begin{align*} 7S_{2019} &\equiv 1 \pmod{2^{2019}} \\ 7S_{2019} &= 2^{2019}k + 1. \end{align*} Now, take both sides $\operatorname{mod} \ 7$ \[0 \equiv 2^{2019}k + 1 \pmod{7}.\] Using Fermat's Little Theorem, \[2^{2019} = (2^{336})^6 \cdot 2^3 \equiv 2^3 \equiv 1 \pmod{7}.\] Thus, \[0 \equiv k + 1 \pmod{7} \implies k \equiv 6 \pmod{7} \implies k = 7j + 6.\] Therefore, \[7S_{2019} = 2^{2019} (7j + 6) + 1 \implies S_{2019} = \frac{2^{2019} (7j + 6) + 1}{7}.\]
We may repeat this same calculation with $S_{2023}$ to yield \[S_{2023} = \frac{2^{2023} (7h + 3) + 1}{7}.\] Now, we notice that $S_n$ is basically an integer expressed in binary form with $n$ bits.
This gives rise to a simple inequality, \[0 \leqslant S_n \leqslant 2^n.\] Since the maximum possible number that can be generated with $n$ bits is \[\underbrace{{11111\dotsc1}_2}_{n} = \sum_{k=0}^{n-1} 2^k = 2^n - 1 \leqslant 2^n.\] Looking at our calculations for $S_{2019}$ and $S_{2023}$ , we see that the only valid integers that satisfy that constraint are $j = h = 0$ \[\frac{S_{2023} - S_{2019}}{2^{2019}} = \frac{\tfrac{2^{2023} \cdot 3 + 1}{7} - \tfrac{2^{2019} \cdot 6 + 1}{7}}{2^{2019}} = \frac{2^4 \cdot 3 - 6}{7} = \boxed{6}.\] ~zoomanTV

Solution:
As in Solution 2, we note that \[x_{2019}+2x_{2020}+4x_{2021}+8x_{2022}=\frac{S_{2023}-S_{2019}}{2^{2019}}.\] We also know that $7S_{2023} \equiv 1 \pmod{2^{2023}}$ and $7S_{2019} \equiv 1 \pmod{2^{2019}}$ , this implies: \[\textbf{(1) } 7S_{2023}=2^{2023}\cdot{x} + 1,\] \[\textbf{(2) } 7S_{2019}=2^{2019}\cdot{y} + 1.\] Dividing by $7$ , we can isolate the previous sums: \[\textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7},\] \[\textbf{(4) } S_{2019}=\frac{2^{2019}\cdot{y} + 1}{7}.\] The maximum value of $S_n$ occurs when every $x_i$ is equal to $1$ . Even when this happens, the value of $S_n$ is less than $2^n$ . Therefore, we can construct the following inequalities: \[\textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7} < 2^{2023},\] \[\textbf{(4) } S_{2019}=\frac{2^{2019}\cdot{y} + 1}{7} < 2^{2019}.\] From these two equations, we can deduce that both $x$ and $y$ are less than $7$
Reducing $\textbf{1}$ and $\textbf{2}$ $\pmod{7},$ we see that \[2^{2023}\cdot{x}\equiv 6\pmod{7},\] and \[2^{2019}\cdot{y}\equiv 6\pmod{7}.\]
The powers of $2$ repeat every $3, \pmod{7}.$
Therefore, $2^{2023}\equiv 2 \pmod 7$ and $2^{2019} \equiv 1 \pmod {7}.$ Substituing this back into the above equations, \[2x\equiv{6}\pmod{7}\] and \[y\equiv{6}\pmod{7}.\]
Since $x$ and $y$ are integers less than $7$ , the only values of $x$ and $y$ are $3$ and $6$ respectively.
The requested sum is \begin{align*} \frac{S_{2023}-S_{2019}}{2^{2019}} &= \frac{\frac{2^{2023}\cdot{x} + 1}{7} - \frac{2^{2019}\cdot{y} + 1}{7}}{2^{2019}} \\ &= \frac{1}{2^{2019}}\left(\frac{2^{2023}\cdot{3} + 1}{7} -\left(\frac{2^{2019}\cdot{6} + 1}{7}  \right)\right) \\ &= \frac{3\cdot{2^4}-6}{7} \\ &= \boxed{6} -Benedict T (countmath1)

Solution:
Note that, as in Solution 2, we have \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] This is because \[S_{2023} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2019}2^{2019} + \cdots + x_{2022}2^{2022}\] and \[S_{2019} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2018}2^{2018}.\] Note that \[S_{2023} - S_{2019} = x_{2019}2^{2019} + \cdots + x_{2022}2^{2022} = 2^{2019}(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}).\] Therefore, \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] Multiplying both sides by 7 gives us \[7(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}) = \frac{7S_{2023} - 7S_{2019}}{2^{2019}}.\] We can write \[7S_{2023} = 1\pmod{2^{2023}} = 1 + 2^{2023}a = 1 + 2^{2019}*16a\] and \[7S_{2019} = 1\pmod{2^{2019}} = 1 + 2^{2019}b\] for some a and b. Substituting, we get \[7(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}) = \frac{(1 + 2^{2019} * 16a) - (1 + 2^{2019}b)}{2^{2019}} = 16a - b.\] Therefore, our answer can be written as \[\frac{16a - b}{7}.\] Another thing to notice is that a and b are integers between 0 and 6. This is because \[7(1 + 2 + 4 + 8 + \cdots + 2^{2022}) \geqslant 7S_{2023} = 1 + 2^{2023}a\] which is \[7(2^{2023}) - 7 \geqslant 1 + 2^{2023}a\] \[(7-a) \geqslant \frac{8}{2^{2023}}\] which only holds when a is less than 7 because the right is very small positive number, so the left must be positive, too. Clearly, a is also non-negative, because otherwise, \[7S_{2023} = 1 + 2^{2023}a < 0\] which would mean \[S_{2023} < 0\] which cannot happen, so a is greater than 0.  A similar explanation for b shows that b is an integer between 0 and 6 inclusive.
Going back to the solution, if our answer to the problem is n, then \[16a - b = 7n\] and \[16a = 7n + b,\] so we can try the five option choices and see which one, when multiplied by 7 and added to some whole number between 0 and 6 results in a multiple of 16. Trying all the option choices, we see that you need to add 7n to something more than 6 to equal a multiple of 16 other than for option A. Therefore, the answer is $\boxed{6}.$


Answer: 6.0

Solution:
In binary numbers, we have \[S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0})_2.\] It follows that \[8S_n = (x_{n-1} x_{n-2} x_{n-3} x_{n-4} \ldots x_{2} x_{1} x_{0}000)_2.\] We obtain $7S_n$ by subtracting the equations: \[\begin{array}{clccrccccccr}   & (x_{n-1} & x_{n-2} & x_{n-3} & x_{n-4} & \ldots & x_2 & x_1 & x_0 & 0 & 0 & 0 \ )_2 \\ -\quad & & & & (x_{n-1} & \ldots & x_5 & x_4 & x_3 & x_2 & x_1 & x_0)_2 \\ \hline   & & & & & & & & & & &  \\ [-2.5ex]   & ( \ \ ?& ? & ? & 0 \ \ \ & \ldots & 0 & 0 & 0 & 0 & 0 & 1 \ )_2 \\ \end{array}\] We work from right to left: \begin{alignat*}{6} x_0=x_1=x_2=1  \quad &\implies \quad &x_3 &= 0& \\  \quad &\implies \quad &x_4 &= 1& \\  \quad &\implies \quad &x_5 &= 1& \\ \quad &\implies \quad &x_6 &= 0& \\  \quad &\implies \quad &x_7 &= 1& \\  \quad &\implies \quad &x_8 &= 1& \\ \quad &\quad \vdots & & & \end{alignat*} For all $n\geq3,$ we conclude that
Finally, we get $(x_{2019},x_{2020},x_{2021},x_{2022})=(0,1,1,0),$ from which \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \boxed{6}.\] ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution:
First, notice that \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] Then since $S_n$ is the modular inverse of $7$ in $\mathbb{Z}_{2^n}$ , we can perform the Euclidean Algorithm to find it for $n = 2019,2023$
Starting with $2019$ \begin{align*} 7S_{2019} &\equiv 1 \pmod{2^{2019}} \\ 7S_{2019} &= 2^{2019}k + 1. \end{align*} Now, take both sides $\operatorname{mod} \ 7$ \[0 \equiv 2^{2019}k + 1 \pmod{7}.\] Using Fermat's Little Theorem, \[2^{2019} = (2^{336})^6 \cdot 2^3 \equiv 2^3 \equiv 1 \pmod{7}.\] Thus, \[0 \equiv k + 1 \pmod{7} \implies k \equiv 6 \pmod{7} \implies k = 7j + 6.\] Therefore, \[7S_{2019} = 2^{2019} (7j + 6) + 1 \implies S_{2019} = \frac{2^{2019} (7j + 6) + 1}{7}.\]
We may repeat this same calculation with $S_{2023}$ to yield \[S_{2023} = \frac{2^{2023} (7h + 3) + 1}{7}.\] Now, we notice that $S_n$ is basically an integer expressed in binary form with $n$ bits.
This gives rise to a simple inequality, \[0 \leqslant S_n \leqslant 2^n.\] Since the maximum possible number that can be generated with $n$ bits is \[\underbrace{{11111\dotsc1}_2}_{n} = \sum_{k=0}^{n-1} 2^k = 2^n - 1 \leqslant 2^n.\] Looking at our calculations for $S_{2019}$ and $S_{2023}$ , we see that the only valid integers that satisfy that constraint are $j = h = 0$ \[\frac{S_{2023} - S_{2019}}{2^{2019}} = \frac{\tfrac{2^{2023} \cdot 3 + 1}{7} - \tfrac{2^{2019} \cdot 6 + 1}{7}}{2^{2019}} = \frac{2^4 \cdot 3 - 6}{7} = \boxed{6}.\] ~zoomanTV

Solution:
As in Solution 2, we note that \[x_{2019}+2x_{2020}+4x_{2021}+8x_{2022}=\frac{S_{2023}-S_{2019}}{2^{2019}}.\] We also know that $7S_{2023} \equiv 1 \pmod{2^{2023}}$ and $7S_{2019} \equiv 1 \pmod{2^{2019}}$ , this implies: \[\textbf{(1) } 7S_{2023}=2^{2023}\cdot{x} + 1,\] \[\textbf{(2) } 7S_{2019}=2^{2019}\cdot{y} + 1.\] Dividing by $7$ , we can isolate the previous sums: \[\textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7},\] \[\textbf{(4) } S_{2019}=\frac{2^{2019}\cdot{y} + 1}{7}.\] The maximum value of $S_n$ occurs when every $x_i$ is equal to $1$ . Even when this happens, the value of $S_n$ is less than $2^n$ . Therefore, we can construct the following inequalities: \[\textbf{(3) } S_{2023}=\frac{2^{2023}\cdot{x} + 1}{7} < 2^{2023},\] \[\textbf{(4) } S_{2019}=\frac{2^{2019}\cdot{y} + 1}{7} < 2^{2019}.\] From these two equations, we can deduce that both $x$ and $y$ are less than $7$
Reducing $\textbf{1}$ and $\textbf{2}$ $\pmod{7},$ we see that \[2^{2023}\cdot{x}\equiv 6\pmod{7},\] and \[2^{2019}\cdot{y}\equiv 6\pmod{7}.\]
The powers of $2$ repeat every $3, \pmod{7}.$
Therefore, $2^{2023}\equiv 2 \pmod 7$ and $2^{2019} \equiv 1 \pmod {7}.$ Substituing this back into the above equations, \[2x\equiv{6}\pmod{7}\] and \[y\equiv{6}\pmod{7}.\]
Since $x$ and $y$ are integers less than $7$ , the only values of $x$ and $y$ are $3$ and $6$ respectively.
The requested sum is \begin{align*} \frac{S_{2023}-S_{2019}}{2^{2019}} &= \frac{\frac{2^{2023}\cdot{x} + 1}{7} - \frac{2^{2019}\cdot{y} + 1}{7}}{2^{2019}} \\ &= \frac{1}{2^{2019}}\left(\frac{2^{2023}\cdot{3} + 1}{7} -\left(\frac{2^{2019}\cdot{6} + 1}{7}  \right)\right) \\ &= \frac{3\cdot{2^4}-6}{7} \\ &= \boxed{6} -Benedict T (countmath1)

Solution:
Note that, as in Solution 2, we have \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] This is because \[S_{2023} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2019}2^{2019} + \cdots + x_{2022}2^{2022}\] and \[S_{2019} = x_{0}2^{0} + x_{1}2^{1} + \cdots + x_{2018}2^{2018}.\] Note that \[S_{2023} - S_{2019} = x_{2019}2^{2019} + \cdots + x_{2022}2^{2022} = 2^{2019}(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}).\] Therefore, \[x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022} = \frac{S_{2023} - S_{2019}}{2^{2019}}.\] Multiplying both sides by 7 gives us \[7(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}) = \frac{7S_{2023} - 7S_{2019}}{2^{2019}}.\] We can write \[7S_{2023} = 1\pmod{2^{2023}} = 1 + 2^{2023}a = 1 + 2^{2019}*16a\] and \[7S_{2019} = 1\pmod{2^{2019}} = 1 + 2^{2019}b\] for some a and b. Substituting, we get \[7(x_{2019} + 2x_{2020} + 4x_{2021} + 8x_{2022}) = \frac{(1 + 2^{2019} * 16a) - (1 + 2^{2019}b)}{2^{2019}} = 16a - b.\] Therefore, our answer can be written as \[\frac{16a - b}{7}.\] Another thing to notice is that a and b are integers between 0 and 6. This is because \[7(1 + 2 + 4 + 8 + \cdots + 2^{2022}) \geqslant 7S_{2023} = 1 + 2^{2023}a\] which is \[7(2^{2023}) - 7 \geqslant 1 + 2^{2023}a\] \[(7-a) \geqslant \frac{8}{2^{2023}}\] which only holds when a is less than 7 because the right is very small positive number, so the left must be positive, too. Clearly, a is also non-negative, because otherwise, \[7S_{2023} = 1 + 2^{2023}a < 0\] which would mean \[S_{2023} < 0\] which cannot happen, so a is greater than 0.  A similar explanation for b shows that b is an integer between 0 and 6 inclusive.
Going back to the solution, if our answer to the problem is n, then \[16a - b = 7n\] and \[16a = 7n + b,\] so we can try the five option choices and see which one, when multiplied by 7 and added to some whole number between 0 and 6 results in a multiple of 16. Trying all the option choices, we see that you need to add 7n to something more than 6 to equal a multiple of 16 other than for option A. Therefore, the answer is $\boxed{6}.$


Answer: 3.0

Let $x_1$ and $x_2$ be such that $x_1\not=x_2$ and $3x_i^2-hx_i=b$ $i=1, 2$ . Then $x_1+x_2$ equals
$\mathrm{(A)\ } -\frac{h}{3} \qquad \mathrm{(B) \ }\frac{h}{3} \qquad \mathrm{(C) \  } \frac{b}{3} \qquad \mathrm{(D) \  } 2b \qquad \mathrm{(E) \  }-\frac{b}{3}$

Solution: Notice that $x_1$ and $x_2$ are the distinct solutions to the quadratic $3x^2-hx-b=0$ . By Vieta, the sum of the roots of this quadratic is the negation of the coefficient of the linear term divided by the coefficient of the quadratic term, so in this case $-\frac{-h}{3}=\frac{h}{3}, \boxed{3}$


Answer: 7.0

Let $x_1, x_2, \ldots , x_n$ be a sequence of integers such that
(i) $-1 \le x_i \le 2$ for $i = 1,2, \ldots n$ (ii) $x_1 + \cdots + x_n = 19$ ; and
(iii) $x_1^2 + x_2^2 + \cdots + x_n^2 = 99$ .
Let $m$ and $M$ be the minimal and maximal possible values of $x_1^3 + \cdots + x_n^3$ , respectively. Then $\frac Mm =$
$\mathrm{(A) \ }3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ }7$

Solution:
Let $a =$ number of $2$ s, $b =$ number of $1$ s, $c =$ number of $-1$
\[4a+b+c=99\] \[2a+b-c=19\]
Multiplying the second equation by $2$ gives $4a+2b-2c=38$ . By subtracting this equation from the first equation we get $3c-b=61$ $3c=61+b$ . As we need to minimize the value of $c$ $c = 21$ $b = 2$ $a = 19$
$x_1^3 + \cdots + x_n^3 = 8 \cdot 19 +2 - 21 = 133$
Therefore, $\frac Mm = \frac{133}{19}=\boxed{7}$


Answer: 7.0

Solution:
Clearly, we can ignore the possibility that some $x_i$ are zero, as adding/removing such variables does not change the truth value of any condition, nor does it change the value of the sum of cubes. Thus we'll only consider $x_i\in\{-1,1,2\}$
Also, order of the $x_i$ does not matter, so we are only interested in the counts of the variables of each type. Let $a$ of the $x_i$ be equal to $-1$ $b$ equal to $1$ , and $c$ equal to $2$
The conditions (ii) and (iii) simplify to: (ii) $2c + b - a = 19$ (iii) $4c + a + b = 99$ and we want to find the maximum and minimum of $2^3c + 1^3b + (-1)^3a = 8c + b - a$ over all non-negative solutions of the above two equations.
Subtracting twice (ii) from (iii) we get $b = 3a-61$ . By entering that into one of the two equations and simplifying we get $c=40-a$
Thus all the solutions of our system of equations have the form $(a,3a-61,40-a)$
As all three variables must be non-negative integers, we have $3a-61 \geq 0 \Rightarrow a \geq 21$ and $40-a\geq 0 \Rightarrow a \leq 40$
For $(a,b,c)$ of the form $(a,3a-61,40-a)$ the expression we are maximizing/minimizing simplifies to $320-8a+3a-61-a = 259 - 6a$ . Clearly, the maximum is achieved for $a=21$ and the minimum for $a=40$ . Their values are $M=133$ and $m=19$ , and therefore $\frac Mm = \boxed{7}$


Answer: 2.0

Let $x_1<x_2<x_3$ be the three real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$ . Find $x_2(x_1+x_3)$

Solution: Substituting $n$ for $2014$ , we get \[\sqrt{n}x^3 - (1+2n)x^2 + 2 = \sqrt{n}x^3 - x^2 - 2nx^2 + 2\] \[= x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0\] Noting that $nx^2 - 1$ factors as a difference of squares to \[(\sqrt{n}x - 1)(\sqrt{n}x+1)\] we can factor the left side as \[(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))\] This means that $\frac{1}{\sqrt{n}}$ is a root, and the other two roots are the roots of $x^2 - 2\sqrt{n}x - 2$ . Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to $2\sqrt{n}$ , so the positive root must be greater than $2\sqrt{n}$ in order to produce this sum when added to a negative value. Since $0 < \frac{1}{\sqrt{2014}} < 2\sqrt{2014}$ is clearly true, $x_2 = \frac{1}{\sqrt{2014}}$ and $x_1 + x_3 = 2\sqrt{2014}$ . Multiplying these values together, we find that $x_2(x_1+x_3) = \boxed{002}$

Solution:
From Vieta's formulae, we know that \[x_1x_2x_3 = \dfrac{-2}{\sqrt{2014}}\] \[x_1 + x_2 + x_3 = \dfrac{4029}{\sqrt{2014}}\] and \[x_1x_2 + x_2x_3 + x_1x_3 = 0\] Thus, we know that \[x_2(x_1 + x_3) = -x_1x_3\]
Now consider the polynomial with roots $x_1x_2, x_2x_3,$ and $x_1x_3$ . Expanding the polynomial \[(x - x_1x_2)(x - x_2x_3)(x - x_1x_3)\] we get the polynomial \[x^3 - (x_1x_2 + x_2x_3 + x_1x_3)x^2 + (x_1x_2x_3)(x_1 + x_2 + x_3)x - (x_1x_2x_3)^2\] Substituting the values obtained from Vieta's formulae, we find that this polynomial is \[x^3 - \dfrac{8058}{2014}x - \dfrac{4}{2014}\] We know $x_1x_3$ is a root of this polynomial, so we set it equal to 0 and simplify the resulting expression to \[1007x^3 - 4029x - 2 = 0\]
Given the problem conditions, we know there must be at least 1 integer solution, and that it can't be very large (because the $x^3$ term quickly gets much larger/smaller than the other 2). Trying out some numbers, we quickly find that $x = -2$ is a solution. Factoring it out, we get that \[1007x^3 - 4029x - 2 = (x+2)(1007x^2 - 2014x - 1)\] Since the other quadratic factor clearly does not have any integer solutions and since the AIME has only positive integer answers, we know that this must be the answer they are looking for. Thus, \[x_1x_3 = -2\] so \[-x_1x_3 = \boxed{002}\] and we're done.

Solution:
Observing the equation, we notice that the coefficient for the middle term $-4029$ is equal to \[-2{\sqrt{2014}}^2-1\]
Also notice that the coefficient for the ${x^3}$ term is $\sqrt{2014}$ . Therefore, if the original expression was to be factored into a linear binomial and a quadratic trinomial, the $x$ term of the binomial would have a coefficient of $\sqrt{2014}$ . Similarly, the $x$ term of the trinomial would also have a coefficient of $\sqrt{2014}$ . The factored form of the expression would look something like the following: \[({\sqrt{2014}}x-a)(x^2-n{\sqrt{2014}}x-b)\] where ${a, b,c}$ are all positive integers (because the ${x^2}$ term of the original expression is negative, and the constant term is positive), and \[{ab=2}\]
Multiplying this expression out gives \[{{\sqrt{2014}x^3-(2014n+a)x^2+(an{\sqrt{2014}}-b{\sqrt{2014}})x+ab}}\] Equating this with the original expression gives \[{2014n+a}=-4029\] The only positive integer solutions of this expression is $(n, a)=(1, 2015)$ or $(2, 1)$ . If $(n, a)=(1, 2015)$ then setting ${an{\sqrt{2014}}-b{\sqrt{2014}}}=0$ yields ${b=2015}$ and therefore ${ab=2015^2}$ which clearly isn't equal to $2$ as the constant term. Therefore, $(n, a)=(2, 1)$ and the factored form of the expression is: \[({\sqrt{2014}}x-1)(x^2-2{\sqrt{2014}}x-2)\] Therefore, one of the three roots of the original expression is \[{x=\dfrac{1}{\sqrt{2014}}}\] Using the quadratic formula yields the other two roots as \[{x={\sqrt{2014}}+{\sqrt{2016}}}\] and \[{x={\sqrt{2014}}-{\sqrt{2016}}}\] Arranging the roots in ascending order (in the order $x_1<x_2<x_3$ ), \[{\sqrt{2014}}-{\sqrt{2016}}<\dfrac{1}{\sqrt{2014}}<{\sqrt{2014}}+{\sqrt{2016}}\] Therefore, \[x_2(x_1+x_3)=\dfrac{1}{\sqrt{2014}}{2\sqrt{2014}}=\boxed{002}\]

Solution:
By Vieta's, we are seeking to find $x_2(x_1+x_3)=x_1x_2+x_2x_3=-x_1x_3=\frac{2}{\sqrt{2014}x_2}$ . Substitute $n=-x_1x_3$ and $x_2=\frac{2}{\sqrt{2014}n}$ . Substituting this back into the original equation, we have $\frac{4}{1007n^3}-\frac{8058}{1007n^2}+2=0$ , so $2n^3-\frac{8058}{1007}n+\frac{4}{1007}=2n^3-\frac{8058n-4}{1007}=0$ . Hence, $8058n-4\equiv 2n-4 \equiv 0 \pmod{1007}$ , and $n\equiv 2\pmod{1007}$ . But since $n\le 999$ because it is our desired answer, the only possible value for $n$ is $\boxed{002}$ BEST PROOOFFFF
Stormersyle & mathleticguyyy

Solution: Let $x =\frac{y}{\sqrt{2014}}.$ The original equation simplifies to $\frac{y^3}{2014} -\frac{4029y^2}{2014}+2 = 0 \implies y^3 - 4029y^2 + 4028=0.$ Here we clearly see that $y=1$ is a root. Dividing $y-1$ from the sum we find that $(y-1)(y^2-4028y-4028)=0.$ From simple bounding we see that $y=1$ is the middle root. Therefore $x_{2}(x_{1}+x_{3}) =\frac{1}{\sqrt{2014}} \cdot\frac{4028}{\sqrt{2014}} = \boxed{002}.$

Solution:
We will estimate the roots of the polynomial. (This strategy normally doesn't work on AIME #9, but playing around with a function is often good strategy for getting an intuition for the problem. In this problem, estimation happens to be a valid solution path. It isn't a proof, but given the constraint that the answer is an integer, we can be certain that our answer is correct.)
Let $p(x) = \sqrt{2014}x^3-4029x^2+2=0$ . We start by estimating $p(-1)$ $p(0)$ , and $p(1)$ (A natural first step for function analysis.):
$p(-1)\approx -45-4029+2 \approx -4000$
$p(0) = 2$
$p(1) \approx 45-4029+2 \approx -4000$
We conclude by Intermediate Value Theorem (or just common sense), that there is a root on $(-1, 0)$ and another root on $(0, 1)$
We know that $p(1) < 0$ and that $\lim_{x\to\infty} p(x) = \infty$ . We conclude that the third root is on $(1, \infty)$ . Therefore, $x_1 \in (-1, 0)$ $x_2 \in (0, 1)$ , and $x_3 \in (1, \infty)$
We will estimate $x_3$ . For $x > 1$ , the constant term of $p(x)$ is negligible. We simplify and get $p(x_3) \approx \sqrt{2014}x_{3}^3 - 4029x_{3}^2 = 0$ . Solving for $x_3$ (We can divide by $x_{3}^2$ because we know $x_3 \neq 0$ ), we get $x_3 \approx \frac{4029}{\sqrt{2014}} \approx 2\sqrt{2014} \approx 90$ . We can intuitively bound $x_3$ between $88$ and $92$
We will now estimate $x_1$ and $x_2$ $x_1$ and $x_2$ are close to $0$ . As a result, the $\sqrt{2014}x^3$ term is negligible. We simplify and get $p(x) \approx -4029x^2 + 2 = 0$ . Solving for $x$ , we get $x \approx \pm \sqrt{\frac{2}{4029}} \approx \pm \sqrt{\frac{1}{2014.5}} \approx \pm \frac{1}{45}$ . We can intuitively bound $x_1$ between $-\frac{1}{43}$ and $-\frac{1}{47}$ . Similarly, we can intuitively bound $x_2$ between $\frac{1}{47}$ and $\frac{1}{43}$
We calculate that the minimum possible value of $x_2(x_1 + x_3)$ is $\frac{88-\frac{1}{43}}{47}$ and the maximum possible value is $\frac{92-\frac{1}{47}}{43}$ . The only integer that falls is this range is $\boxed{002}$


Answer: 384.0

Let $x_1=97$ , and for $n>1$ , let $x_n=\frac{n}{x_{n-1}}$ . Calculate the product $x_1x_2x_3x_4x_5x_6x_7x_8$

Solution: Since $x_n=\frac{n}{x_{n-1}}$ $x_n \cdot x_{n - 1} = n$ .  Setting $n = 2, 4, 6$ and $8$ in this equation gives us respectively $x_1x_2 = 2$ $x_3x_4 = 4$ $x_5x_6 = 6$ and $x_7x_8 = 8$ so \[x_1x_2x_3x_4x_5x_6x_7x_8 = 2\cdot4\cdot6\cdot8 = \boxed{384}.\] Notice that the value of $x_1$ was completely unneeded!

Solution:
Another way to do this is to realize that most of our numbers will be canceled out in the multiplication in the end, and to just list out the terms of our product and cancel:
\[x_1x_2x_3x_4x_5x_6x_7x_8=x_1\cdot\dfrac{2}{x_1}\cdot\dfrac{3}{\dfrac{2}{x_1}}\cdot\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}\cdot\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}\cdot\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}\cdot\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}\cdot\dfrac{8}{\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}}\] \[=\left (x_1\cdot\dfrac{2}{x_1} \right )\cdot \left (\dfrac{3}{\dfrac{2}{x_1}}\cdot\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}} \right )\cdot \left (\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}\cdot\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}} \right )\cdot \left (\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}\cdot\dfrac{8}{\dfrac{7}{\dfrac{6}{\dfrac{5}{\dfrac{4}{\dfrac{3}{\dfrac{2}{x_1}}}}}}} \right )\] \[=(2)\cdot (4)\cdot (6)\cdot (8)=\boxed{384}\]