Answer: 697.0

There is a complex number $z$ with imaginary part $164$ and a positive integer $n$ such that
\[\frac {z}{z + n} = 4i.\]
Find $n$

Solution:
Let $z = a + 164i$
Then \[\frac {a + 164i}{a + 164i + n} = 4i\] and \[a + 164i = \left (4i \right ) \left (a + n + 164i \right ) = 4i \left (a + n \right ) - 656.\]
By comparing coefficients, equating the real terms on the leftmost and rightmost side of the equation,
we conclude that \[a = -656.\]
By equating the imaginary terms on each side of the equation,
we conclude that \[164i = 4i \left (a + n \right ) = 4i \left (-656 + n \right ).\]
We now have an equation for $n$ \[4i \left (-656 + n \right ) = 164i,\]
and this equation shows that $n = \boxed{697}.$

Solution:
\[\frac {z}{z+n}=4i\]
\[1-\frac {n}{z+n}=4i\]
\[1-4i=\frac {n}{z+n}\]
\[\frac {1}{1-4i}=\frac {z+n}{n}\]
\[\frac {1+4i}{17}=\frac {z}{n}+1\]
Since their imaginary part has to be equal,
\[\frac {4i}{17}=\frac {164i}{n}\]
\[n=\frac {(164)(17)}{4}=697\]
\[n = \boxed{697}.\]

Solution:
Below is an image of the complex plane. Let $\operatorname{Im}(z)$ denote the imaginary part of a complex number $z$  $z$ must lie on the line $\operatorname{Im}(z)=164$ $z+n$ must also lie on the same line, since $n$ is real and does not affect the imaginary part of $z$
Consider $z$ and $z+n$ in terms of their magnitude (distance from the origin) and phase (angle formed by the point, the origin, and the positive real axis, measured counterclockwise from said axis). When multiplying/dividing two complex numbers, you can multiply/divide their magnitudes and add/subtract their phases to get the magnitude and phase of the product/quotient. Expressed in a formula, we have $z_1z_2 = r_1\angle\theta_1 \cdot r_2\angle\theta_2 = r_1r_2\angle(\theta_1+\theta_2)$ and $\frac{z_1}{z_2} = \frac{r_1\angle\theta_1}{r_2\angle\theta_2} = \frac{r_1}{r_2}\angle(\theta_1-\theta_2)$ , where $r$ is the magnitude and $\theta$ is the phase, and $z_n=r_n\angle\theta_n$
Since $4i$ has magnitude $4$ and phase $90^\circ$ (since the positive imaginary axis points in a direction $90^\circ$ counterclockwise from the positive real axis), $z$ must have a magnitude $4$ times that of $z+n$ . We denote the length from the origin to $z+n$ with the value $x$ and the length from the origin to $z$ with the value $4x$ . Additionally, $z$ , the origin, and $z+n$ must form a right angle, with $z$ counterclockwise from $z+n$
This means that $z$ , the origin, and $z+n$ form a right triangle. The hypotenuse is the length from $z$ to $z+n$ and has length $n$ , since $n$ is defined to be a positive integer. The area of the triangle can be expressed using the two legs, as $\frac{x \cdot 4x}{2}$ , or using the hypotenuse and its corresponding altitude, as $\frac{164n}{2}$ , so $\frac{x \cdot 4x}{2} = \frac{164n}{2} \implies x^2 = 41n$ . By Pythagorean Theorem, $x^2+(4x)^2 = n^2 \implies 17x^2 = n^2$ . Substituting out $x^2$ using the earlier equation, we get $17\cdot41n = n^2 \implies n = \boxed{697}$ . ~ emerald_block

Solution: Taking the reciprocal of our equation gives us $1 + \frac{n}{z} = \frac{1}{4i}.$ Therefore, \[\frac{n}{z} = \frac{1-4i}{4i} = \frac{17}{-16+4i}.\] Since $z$ has an imaginary part of $164$ , we must multiply both sides of our RHS fraction by $\frac{164}{4} = 41$ so that its denominator's imaginary part matches the LHS fraction's denominator's imaginary part. That looks like this: \[\frac{n}{z} = \frac{697}{-656 + 164i}.\] Therefore, we can conclude the the real part of $z$ is $-656$ and $n = \boxed{697}.$ (it wasn't necessary to find the real part)


Answer: 6.0

There is a list of seven numbers. The average of the first four numbers is $5$ , and the average of the last four numbers is $8$ . If the average of all seven numbers is $6\frac{4}{7}$ , then the number common to both sets of four numbers is
$\text{(A)}\ 5\frac{3}{7}\qquad\text{(B)}\ 6\qquad\text{(C)}\ 6\frac{4}{7}\qquad\text{(D)}\ 7\qquad\text{(E)}\ 7\frac{3}{7}$

Solution:
Remember that if a list of $n$ numbers has an average of $k$ , then the sum $S$ of all the numbers on the list is $S = nk$
So if the average of the first $4$ numbers is $5$ , then the first four numbers total $4 \cdot 5 = 20$
If the average of the last $4$ numbers is $8$ , then the last four numbers total $4 \cdot 8 = 32$
If the average of all $7$ numbers is $6\frac{4}{7}$ , then the total of all seven numbers is $7 \cdot 6\frac{4}{7} = 7\cdot 6 + 4 =  46$
If the first four numbers are $20$ , and the last four numbers are $32$ , then all "eight" numbers are $20 + 32 = 52$ .  But that's counting one number twice.  Since the sum of all seven numbers is $46$ , then the number that was counted twice is $52 - 46 = 6$ , and the answer is $\boxed{6}$


Answer: 220.0

There is a polynomial $P(x)$ with integer coefficients such that \[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\] holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$

Solution:
Because $0 < x < 1$ , we have \begin{align*} P \left( x \right) & = \sum_{a=0}^6  \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \binom{6}{a} x^{2310a} \left( - 1 \right)^{6-a} x^{105b} x^{70c} x^{42d} x^{30e} \\ & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} x^{2310 a + 105 b + 70 c + 42 d + 30 e} . \end{align*}
Denote by $c_{2022}$ the coefficient of $P \left( x \right)$ .
Thus, \begin{align*} c_{2022} & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} \Bbb I \left\{ 2310 a + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & =  \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-0} \Bbb I \left\{ 2310 \cdot 0 + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & = \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \Bbb I \left\{ 105 b + 70 c + 42 d + 30 e = 2022 \right\} . \end{align*}
Now, we need to find the number of nonnegative integer tuples $\left( b , c , d , e \right)$ that satisfy \[ 105 b + 70 c + 42 d + 30 e = 2022 . \hspace{1cm} (1) \]
Modulo 2 on Equation (1), we have $b \equiv 0 \pmod{2}$ .
Hence, we can write $b = 2 b'$ . Plugging this into (1), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c , d , e \right)$ that satisfy \[ 105 b' + 35 c + 21 d + 15 e = 1011 . \hspace{1cm} (2) \]
Modulo 3 on Equation (2), we have $2 c \equiv 0 \pmod{3}$ .
Hence, we can write $c = 3 c'$ . Plugging this into (2), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d , e \right)$ that satisfy \[ 35 b' + 35 c' + 7 d + 5 e = 337 . \hspace{1cm} (3) \]
Modulo 5 on Equation (3), we have $2 d \equiv 2 \pmod{5}$ .
Hence, we can write $d = 5 d' + 1$ . Plugging this into (3), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d' , e \right)$ that satisfy \[ 7 b' + 7 c' + 7 d' + e = 66 . \hspace{1cm} (4) \]
Modulo 7 on Equation (4), we have $e \equiv 3 \pmod{7}$ .
Hence, we can write $e = 7 e' + 3$ . Plugging this into (4), the problem reduces to finding the number of
nonnegative integer tuples $\left( b' , c' , d' , e' \right)$ that satisfy \[ b' + c' + d' + e' = 9 . \hspace{1cm} (5) \]
The number of nonnegative integer solutions to Equation (5) is $\binom{9 + 4 - 1}{4 - 1} = \binom{12}{3} =  \boxed{220}$

Solution: Note that $2022 = 210\cdot 9 +132$ . Since the only way to express $132$ in terms of $105$ $70$ $42$ , or $30$ is $132 = 30+30+30+42$ , we are essentially just counting the number of ways to express $210*9$ in terms of these numbers. Since $210 = 2*105=3*70=5*42=7*30$ , it can only be expressed as a sum in terms of only one of the numbers ( $105$ $70$ $42$ , or $30$ ). Thus, the answer is (by sticks and stones) \[\binom{12}{3} = \boxed{220}\]


Answer: 10.0

There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$

Solution:
\[\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ \Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\ \Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\ \Rightarrow \ &n^2 + 4n - 437 = 0\end{split}\]
Solving by the quadratic formula, $n = \frac{-4\pm \sqrt{16+437\cdot4}}{2} = \frac{-4\pm 42}{2} = \frac{38}{2} = 19$ (since clearly $n \geq 0$ ). The answer is therefore $1 + 9 = \boxed{10}$

Solution: Dividing both sides by $n!$ gives \[(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.\] Since $n$ is non-negative, $n=19$ . The answer is $1 + 9 = \boxed{10}$

Solution: Dividing both sides by $n!$ as before gives $(n+1)+(n+1)(n+2)=440$ . Now factor out $(n+1)$ , giving $(n+1)(n+3)=440$ . By considering the prime factorization of $440$ , a bit of experimentation gives us $n+1=20$ and $n+3=22$ , so $n=19$ , so the answer is $1 + 9 = \boxed{10}$

Solution: Since $(n+1)! + (n+2)! = (n+1)n! + (n+2)(n+1)n! = 440 \cdot n!$ , the result can be factored into $(n+1)(n+3)n!=440 \cdot n!$ and divided by $n!$ on both sides to get $(n+1)(n+3)=440$ . From there, it is easier to complete the square with the quadratic $(n+1)(n+3) = n^2 + 4n + 3$ , so $n^2+4n+4=441 \Rightarrow (n+2)^2=441$ . Solving for $n$ results in $n=19,-23$ , and since $n>0$ $n=19$ and the answer is $1 + 9 = \boxed{10}$

Solution: Rewrite $(n+1)! + (n+2)! = 440 \cdot n!$ as $(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!.$ Factoring out the $n!$ we get $n!(n + 1 + (n+1)(n+2)) = 440 \cdot n!.$ Expand this to get $n!(n^2 + 4n + 3) = 440 \cdot n!.$ Factor this and divide by $n!$ to get $(n + 1)(n + 3) = 440.$ If we take the prime factorization of $440$ we see that it is $2^3 * 5 * 11.$ Intuitively, we can find that $n + 1 = 20$ and $n + 3 = 22.$ Therefore, $n = 19.$ Since the problem asks for the sum of the didgits of $n$ , we finally calculate $1 + 9 = 10$ and get answer choice $\boxed{10}$


Answer: 10.0

Solution:
\[\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ \Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\ \Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\ \Rightarrow \ &n^2 + 4n - 437 = 0\end{split}\]
Solving by the quadratic formula, $n = \frac{-4\pm \sqrt{16+437\cdot4}}{2} = \frac{-4\pm 42}{2} = \frac{38}{2} = 19$ (since clearly $n \geq 0$ ). The answer is therefore $1 + 9 = \boxed{10}$

Solution: Dividing both sides by $n!$ gives \[(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.\] Since $n$ is non-negative, $n=19$ . The answer is $1 + 9 = \boxed{10}$

Solution: Dividing both sides by $n!$ as before gives $(n+1)+(n+1)(n+2)=440$ . Now factor out $(n+1)$ , giving $(n+1)(n+3)=440$ . By considering the prime factorization of $440$ , a bit of experimentation gives us $n+1=20$ and $n+3=22$ , so $n=19$ , so the answer is $1 + 9 = \boxed{10}$

Solution: Since $(n+1)! + (n+2)! = (n+1)n! + (n+2)(n+1)n! = 440 \cdot n!$ , the result can be factored into $(n+1)(n+3)n!=440 \cdot n!$ and divided by $n!$ on both sides to get $(n+1)(n+3)=440$ . From there, it is easier to complete the square with the quadratic $(n+1)(n+3) = n^2 + 4n + 3$ , so $n^2+4n+4=441 \Rightarrow (n+2)^2=441$ . Solving for $n$ results in $n=19,-23$ , and since $n>0$ $n=19$ and the answer is $1 + 9 = \boxed{10}$

Solution: Rewrite $(n+1)! + (n+2)! = 440 \cdot n!$ as $(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!.$ Factoring out the $n!$ we get $n!(n + 1 + (n+1)(n+2)) = 440 \cdot n!.$ Expand this to get $n!(n^2 + 4n + 3) = 440 \cdot n!.$ Factor this and divide by $n!$ to get $(n + 1)(n + 3) = 440.$ If we take the prime factorization of $440$ we see that it is $2^3 * 5 * 11.$ Intuitively, we can find that $n + 1 = 20$ and $n + 3 = 22.$ Therefore, $n = 19.$ Since the problem asks for the sum of the didgits of $n$ , we finally calculate $1 + 9 = 10$ and get answer choice $\boxed{10}$


Answer: 112.0

There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Define $a$ to be $\log_{20x} (22x) = \log_{2x} (202x)$ , what we are looking for. Then, by the definition of the logarithm, \[\begin{cases}  (20x)^{a} &= 22x \\ (2x)^{a} &= 202x.  \end{cases}\] Dividing the first equation by the second equation gives us $10^a = \frac{11}{101}$ , so by the definition of logs, $a = \log_{10} \frac{11}{101}$ . This is what the problem asked for, so the fraction $\frac{11}{101}$ gives us $m+n = \boxed{112}$

Solution:
We could assume a variable $v$ which equals to both $\log_{20x} (22x)$ and $\log_{2x} (202x)$
So that $(20x)^v=22x \textcircled{1}$ and $(2x)^v=202x \textcircled{2}$
Express $\textcircled{1}$ as: $(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot \left(10^v\right)=22x \textcircled{3}$
Substitute $\textcircled{{2}}$ to $\textcircled{3}$ $202x \cdot (10^v)=22x$
Thus, $v=\log_{10} \left(\frac{22x}{202x}\right)= \log_{10} \left(\frac{11}{101}\right)$ , where $m=11$ and $n=101$
Therefore, $m+n = \boxed{112}$

Solution:
We have \begin{align*} \log_{20x} (22x) & = \frac{\log_k 22x}{\log_k 20x} \\ & = \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} . \end{align*}
We have \begin{align*} \log_{2x} (202x) & = \frac{\log_k 202x}{\log_k 2x} \\ & = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} . \end{align*}
Because $\log_{20x} (22x)=\log_{2x} (202x)$ , we get \[ \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} . \]
We denote this common value as $\lambda$
By solving the equality $\frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} = \lambda$ , we get $\log_k x = \frac{\log_k 22 - \lambda \log_k 20}{\lambda - 1}$
By solving the equality $\frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} = \lambda$ , we get $\log_k x = \frac{\log_k 202 - \lambda \log_k 2}{\lambda - 1}$
By equating these two equations, we get \[ \frac{\log_k 22 - \lambda \log_k 20}{\lambda - 1} = \frac{\log_k 202 - \lambda \log_k 2}{\lambda - 1} . \]
Therefore, \begin{align*} \log_{20x} (22x) & = \lambda \\ & = \frac{\log_k 22 - \log_k 202}{\log_k 20 - \log_k 2} \\ & = \frac{\log_k \frac{11}{101}}{\log_k 10} \\ & = \log_{10} \frac{11}{101} . \end{align*}
Therefore, the answer is $11 + 101 = \boxed{112}$

Solution: Let $a$ be the exponent such that $(20x)^a = 22x$ and $(2x)^a = 202x$ . Dividing, we get \begin{align*} \dfrac{(20x)^a}{(2x)^a} &= \dfrac{22x}{202x}. \\ \left(\dfrac{20x}{2x}\right)^a &= \dfrac{22x}{202x}. \\ 10^a &= \dfrac{11}{101}. \\ \end{align*} Thus, we see that $\log_{10} \left(\dfrac{11}{101}\right) = a = \log_{20x} 22x$ , so the answer is $11 + 101 = \boxed{112}$

Solution: By the change of base rule, we have $\frac{\log 22x}{\log 20x}=\frac{\log 202x}{\log 2x}$ , or $\frac{\log 22 +\log x}{\log 20 +\log x}=\frac{\log 202 +\log x}{\log 2 +\log x}=k$ . We also know that if $a/b=c/d$ , then this also equals $\frac{a-c}{b-d}$ . We use this identity and find that $k=\frac{\log 202 -\log 22}{\log 2 -\log 20}=-\log\frac{202}{22}=\log\frac{11}{101}$ . The requested sum is $11+101=\boxed{112}.$

Solution: By change of base formula, \[\frac{\log_{2x} 22x}{\log_{2x} 20x} = \frac{{\log_{2x} 11} + 1}{{\log_{2x} 10} + 1} = {\log_{2x} 101} + 1\] \[\log_{2x} 11 + 1 = (\log_{2x} 10)(\log_{2x} 101) + \log{2x} 1010 + 1\] \[\frac{\log_{2x} \frac{11}{1010}}{\log_{2x} 10} = \log_{2x} 101\] \[\log_{10} {\frac{11}{1010}} = \log_{2x} 101\] \[\log_{10} {\frac{11}{1010}} + 1 = \log_{2x} 101 + 1 = \log_{2x} 202x = \log_{20x} {22x}\] Thus, \[\log_{20x} 22x = \log_{10} \left( \frac{11}{1010} \times 10 \right) = \log_{10} \frac{11}{101}\] The requested answer is $11 + 101 = \boxed{112}$


Answer: 650.0

There is a set of 1000 switches, each of which has four positions, called $A, B, C$ , and $D$ .  When the position of any switch changes, it is only from $A$ to $B$ , from $B$ to $C$ , from $C$ to $D$ , or from $D$ to $A$ .  Initially each switch is in position $A$ .  The switches are labeled with the 1000 different integers $(2^{x})(3^{y})(5^{z})$ , where $x, y$ , and $z$ take on the values $0, 1, \ldots, 9$ .  At step i of a 1000-step process, the $i$ -th switch is advanced one step, and so are all the other switches whose labels divide the label on the $i$ -th switch.  After step 1000 has been completed, how many switches will be in position $A$

Solution:
For each $i$ th switch (designated by $x_{i},y_{i},z_{i}$ ), it advances itself only one time at the $i$ th step; thereafter, only a switch with larger $x_{j},y_{j},z_{j}$ values will advance the $i$ th switch by one step provided $d_{i}= 2^{x_{i}}3^{y_{i}}5^{z_{i}}$ divides $d_{j}= 2^{x_{j}}3^{y_{j}}5^{z_{j}}$ . Let $N = 2^{9}3^{9}5^{9}$ be the max switch label. To find the divisor multiples in the range of $d_{i}$ to $N$ , we consider the exponents of the number $\frac{N}{d_{i}}= 2^{9-x_{i}}3^{9-y_{i}}5^{9-z_{i}}$ . In general, the divisor-count of $\frac{N}{d}$ must be a multiple of 4 to ensure that a switch is in position A:
We consider the cases where the 3 factors above do not contribute multiples of 4.
The number of switches in position A is $1000-125-225 = \boxed{650}$


Answer: 7.0

There is a set of five positive integers whose average (mean) is 5, whose median is 5, and whose only mode is 8.  What is the difference between the largest and smallest integers in the set?
$\text{(A)}\ 3 \qquad \text{(B)}\ 5 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 8$

Solution: When these numbers are ordered in ascending order, 5, the median, falls right in the middle, which is the third integer from the left. Since there is a unique mode of 8, both integers to the right of 5 must be 8s. Since the mean is 5, the sum of the integers is 25, which means the 2 leftmost integers have to sum to 4. 2 and 2 does not work because that would result in two modes. However, 1 and 3 does, and so our answer is 8-1=7. $\boxed{7}$


Answer: 9.0

There is a smallest positive real number $a$ such that there exists a positive real number $b$ such that all the roots of the polynomial $x^3-ax^2+bx-a$ are real. In fact, for this value of $a$ the value of $b$ is unique. What is this value of $b$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12$

Solution:
The acceleration must be zero at the $x$ -intercept; this intercept must be an inflection point for the minimum $a$ value.
Derive $f(x)$ so that the acceleration $f''(x)=0$ . Using the power rule, \begin{align*} f(x) &= x^3-ax^2+bx-a \\ f’(x) &= 3x^2-2ax+b \\  f’’(x) &= 6x-2a \end{align*} So $x=\frac{a}{3}$ for the inflection point/root. Furthermore, the slope of the function must be zero - maximum - at the intercept, thus having a triple root at $x=a/3$ (if the slope is greater than zero, there will be two complex roots and we do not want that).
The function with the minimum $a$
\[f(x)=\left(x-\frac{a}{3}\right)^3\] \[x^3-ax^2+\left(\frac{a^2}{3}\right)x-\frac{a^3}{27}\] Since this is equal to the original equation $x^3-ax^2+bx-a$ , equating the coefficients, we get that
\[\frac{a^3}{27}=a\rightarrow a^2=27\rightarrow a=3\sqrt{3}\] \[b=\frac{a^2}{3}=\frac{27}{3}=\boxed{9}\]

Solution: Let the roots of the polynomial be $r, s, t$ . By Vieta's formulas we have $r+s+t=a$ $rs+st+rt=b$ , and $rst=a$ . Since both $a$ and $b$ are positive, it follows that all 3 roots $r, s, t$ are positive as well, and so we can apply AM-GM to get \[\tfrac 13 (r+s+t) \ge \sqrt[3]{rst} \quad \Rightarrow \quad a \ge 3\sqrt[3]{a}.\] Cubing both sides and then dividing by $a$ (since $a$ is positive we can divide by $a$ and not change the sign of the inequality) yields \[a^2 \ge 27 \quad \Rightarrow \quad a \ge  3\sqrt{3}.\] Thus, the smallest possible value of $a$ is $3\sqrt{3}$ which is achieved when all the roots are equal to $\sqrt{3}$ . For this value of $a$ , we can use Vieta's to get $b=\boxed{9}$

Solution: We see that with cubics, the number $3$ comes up a lot, and as $9=3\cdot3$ has the most relation to $3$ , we can assume $b=\boxed{9}$


Answer: 547.0

There is a unique angle $\theta$ between $0^{\circ}$ and $90^{\circ}$ such that for nonnegative integers $n$ , the value of $\tan{\left(2^{n}\theta\right)}$ is positive when $n$ is a multiple of $3$ , and negative otherwise. The degree measure of $\theta$ is $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime integers. Find $p+q$

Solution:
Note that if $\tan \theta$ is positive, then $\theta$ is in the first or third quadrant, so $0^{\circ} < \theta < 90^{\circ} \pmod{180^{\circ}}$
Furthermore, the only way $\tan{\left(2^{n}\theta\right)}$ can be positive for all $n$ that are multiples of $3$ is when: \[2^0\theta \equiv 2^3\theta \equiv 2^6\theta \equiv ... \pmod{180^{\circ}}.\] (This is because if it isn't the same value, the terminal angle will gradually shift from the first quadrant into different quadrants, making the condition for positive tan untrue. This must also be true in order for $\theta$ to be unique.)
This is the case if $2^3\theta \equiv 2^0\theta \pmod{180^{\circ}}$ , so $7\theta \equiv 0^{\circ} \pmod{180^{\circ}}$ . Therefore, recalling that $0^{\circ}<\theta<90^{\circ},$ the possible $\theta$ are: \[\frac{180}{7}^{\circ}, \frac{360}{7}^{\circ}, \frac{540}{7}^{\circ}.\]
$\frac{180}{7}^{\circ}$ does not work since $\tan\left(2 \cdot \frac{180}{7}^{\circ}\right)$ is positive.
$\frac{360}{7}^{\circ}$ does not work because $\tan\left(4 \cdot \frac{360}{7}^{\circ}\right)$ is positive.
Thus, $\theta = \frac{540}{7}^{\circ}$ , and a quick check verifies that it does work. Our desired answer is $540 + 7 = \boxed{547}$

Solution:
As in the previous solution, we note that $\tan \theta$ is positive when $\theta$ is in the first or third quadrant. In order for $\tan\left(2^n\theta\right)$ to be positive for all $n$ divisible by $3$ , we must have $\theta$ $2^3\theta$ $2^6\theta$ , etc to lie in the first or third quadrants. We already know that $\theta\in(0,90)$ . We can keep track of the range of $2^n\theta$ for each $n$ by considering the portion in the desired quadrants, which gives \[n=1 \implies (90,180)\] \[n=2\implies (270,360)\] \[n=3 \implies (180,270)\] \[n=4 \implies (90,180)\] \[n=5\implies(270,360)\] \[n=6 \implies (180,270)\] \[\cdots\] at which point we realize a pattern emerging. Specifically, the intervals repeat every $3$ after $n=1$ . We can use these repeating intervals to determine the desired value of $\theta$ since the upper and lower bounds will converge to such a value (since it is unique, as indicated in the problem). Let's keep track of the lower bound.
Initially, the lower bound is $0$ (at $n=0$ ), then increases to $\frac{90}{2}=45$ at $n=1$ . This then becomes $45+\frac{45}{2}$ at $n=2$ $45+\frac{45}{2}$ at $n=3$ $45+\frac{45}{2}+\frac{45}{2^3}$ at $n=4$ $45+\frac{45}{2}+\frac{45}{2^3}+\frac{45}{2^4}$ at $n=5$ , etc. Due to the observed pattern of the intervals, the lower bound follows a partial geometric series. Hence, as $n$ approaches infinity, the lower bound converges to \[\sum_{k=0}^{\infty}\left(45+\frac{45}{2}\right)\cdot \left(\frac{1}{8}\right)^k=\frac{45+\frac{45}{2}}{1-\frac{1}{8}}=\frac{\frac{135}{2}}{\frac{7}{8}}=\frac{540}{7}\implies p+q=540+7=\boxed{547}\] -ktong

Solution: Since $\tan\left(\theta\right) > 0$ $0 < \theta < 90$ .  Since $\tan\left(2\theta\right) < 0$ $\theta$ has to be in the second half of the interval (0, 90) ie (45, 90).  Since $\tan\left(4\theta\right) < 0$ $\theta$ has to be in the second half of that interval ie (67.5, 90).  And since $\tan\left(8\theta\right) > 0$ $\theta$ has to be in the first half of (67.5, 90).  Inductively, the pattern repeats: $\theta$ is in the second half of the second half of the first half of the second half of the second half... of the interval (0, 90).  Consider the binary representation of numbers in the interval (0, 1).  Numbers in the first half of the interval start with 0.0... and numbers in the second half start with 0.1... .  Similarly, numbers in the second half of the second half start with 0.11... etc.  So if we want a number in the first half of the second half of the second half... of the interval, we want its binary representation to be $0.11011011011..._2 = \frac{6}{7}_{10}$ .  So we want the number which is 6/7 of the way through the interval (0, 90) so $\theta = \frac{6}{7}\cdot 90 = \frac{540}{7}$ and $p+q = 540 + 7 = \boxed{547}$

Solution:
With some simple arithmetic and guess and check, we can set the lower bound and upper bounds for the "first round of $3$ powers of two", which are $\frac{540}{8} = 67.5$ and $\frac{630}{8} = 78.75$ . Going on to the "second round of $3$ powers of two, we set the new lower and upper bounds as $\frac{360 \times 12.5}{64} = 70.3125$ and $\frac{360 \times 13.75}{64} = 77.34375$ using some guess and check and bashing. Now, it is obvious that the bounds for the "zeroth round of $3$ powers of two" are $0$ and $90$ , and notice that $90 - 78.75 = 11.25$ and $78.75 - 77.34375 = 1.40625$ and $\frac{11.25}{1.40625} = 8$ .
This is obviously a geometric series, so setting $11.25$ as $u$ , we obtain $90 - (u + \frac{u}{8} + \frac{u}{64} + ...)$ $90 - \frac{u}{1-\frac{1}{8}}$ $\frac{u}{\frac{7}{8}}$ $\frac{45}{4} \times \frac{8}{7}$ which simplifies to $\frac{90}{7}$ . We can now finally subtract $\frac{90}{7}$ from $\frac{630}{7}$ and then we get $\frac{540}{7}$ as the unique angle, so $\boxed{547}$ is our answer.
-fidgetboss_4000

Solution: Since $7$ is the only number n such that f(x) = $2^{\lfloor x\rfloor}$ $\text{(mod 7)}$ has a period of 3, we find that $\theta$ is a multiple of ${\frac{180}{7}}^\circ$ . Note that the tangents of ${\frac{180}{7}}^\circ$ ${\frac{360}{7}}^\circ$ ${\frac{540}{7}}^\circ$ are positive while those of ${\frac{720}{7}}^\circ$ ${\frac{900}{7}}^\circ$ , and ${\frac{1080}{7}}^\circ$ are negative. With a bit of trial and error, we find $\theta$ is ${\frac{540}{7}}^\circ$ and $540+7$ is $\boxed{547}$


Answer: 13.0

There is a unique positive integer $n$ such that \[\log_2{(\log_{16}{n})} = \log_4{(\log_4{n})}.\] What is the sum of the digits of $n?$
$\textbf{(A) } 4 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 13$

Solution:
We can use the fact that $\log_{a^b} c = \frac{1}{b} \log_a c.$ This can be proved by using change of base formula to base $a.$
So, the original equation $\log_2{(\log_{2^4}{n})} = \log_{2^2}{(\log_{2^2}{n})}$ becomes \[\log_2\left({\frac{1}{4}\log_{2}{n}}\right) = \frac{1}{2}\log_2\left({\frac{1}{2}\log_2{n}}\right).\] Using log property of addition, we expand both sides and then simplify: \begin{align*} \log_2{\frac{1}{4}}+\log_2{(\log_{2}{n}}) &= \frac{1}{2}\left[\log_2{\frac{1}{2}} +\log_{2}{(\log_2{n})}\right] \\ \log_2{\frac{1}{4}}+\log_2{(\log_{2}{n}}) &= \frac{1}{2}\left[-1 +\log_{2}{(\log_2{n})}\right] \\ -2+\log_2{(\log_{2}{n}}) &= -\frac{1}{2}+ \frac{1}{2}(\log_{2}{(\log_2{n})}). \end{align*} Subtracting $\frac{1}{2}(\log_{2}{(\log_2{n})})$ from both sides and adding $2$ to both sides gives us \[\frac{1}{2}(\log_{2}{(\log_2{n})}) = \frac{3}{2}.\] Multiplying by $2,$ exponentiating, and simplifying gives us \begin{align*} \log_{2}{(\log_2{n})} &= 3 \\ \log_2{n}&=8 \\ n&=256. \end{align*} Adding the digits together, we have $2+5+6=\boxed{13}.$

Solution: We will apply the following logarithmic identity: \[\log_{p^k}{q^k}=\log_{p}{q},\] which can be proven by the Change of Base Formula \[\log_{p^k}{q^k}=\frac{\log_{p}{q^k}}{\log_{p}{p^k}}=\frac{k\log_{p}{q}}{k}=\log_{p}{q}.\] Note that $\log_{16}{n}\neq0,$ so we rewrite the original equation as follows: \begin{align*} \log_4{(\log_{16}{n})^2}&=\log_4{(\log_4{n})} \\ (\log_{16}{n})^2&=\log_4{n} \\ (\log_{16}{n})^2&=\log_{16}{n^2} \\ (\log_{16}{n})^2&=2\log_{16}{n} \\ \log_{16}{n}&=2, \end{align*} from which $n=16^2=256.$ The sum of its digits is $2+5+6=\boxed{13}.$

Solution: Using the change of base formula on the RHS of the initial equation yields \[\log_2{(\log_{16}{n})} = \frac{\log_2{(\log_4{n})}}{\log_2{4}}.\] This means we can multiply each side by $2$ for \[\log_2{(\log_{16}{n})^2} = \log_2{(\log_4{n})}.\] Canceling out the logs gives \[(\log_{16}{n})^2=\log_4{n}.\] We use change of base on the RHS to see that \begin{align*} (\log_{16}{n})^2&=\frac{ \log_{16}{n}}{\log_{16}{4}} \\ (\log_{16}{n})^2&=2 \log_{16}{n}. \end{align*} Substituting in $m = \log_{16}{n}$ gives $m^2=2m,$ so $m$ is either $0$ or $2.$ Since $m=0$ yields no solution for $n$ (since this would lead to use taking the log of $0$ ), we get $2 = \log_{16}{n},$ or $n=16^2=256,$ for the digit-sum of $2 + 5 + 6 = \boxed{13}.$

Solution: Suppose $\log_2(\log_{16}n)=k\implies\log_{16}n=2^k\implies n=16^{2^k}.$ Similarly, we have $\log_4(\log_4 n)=k\implies \log_4 n=4^k\implies n=4^{4^k}.$ Thus, we have \[16^{2^k}=(4^2)^{2^k}=4^{2^{k+1}}\] and \[4^{4^k}=4^{2^{2k}},\] so $k+1=2k\implies k=1.$ Plugging this in to either one of the expressions for $n$ gives $256$ , and the requested answer is $2+5+6=\boxed{13}.$

Solution: We know that, as the answer is an integer, $n$ must be some power of $16.$ Testing $16$ yields \begin{align*} \log_2{(\log_{16}{16})} &= \log_4{(\log_4{16})} \\ \log_2{1} &= \log_4{2} \\ 0 &= \frac{1}{2}, \end{align*} which does not work. We then try $256,$ giving us \begin{align*} \log_2{(\log_{16}{256})} &= \log_4{(\log_4{256})} \\ \log_2{2} &= \log_4{4} \\ 1 &= 1, \end{align*} which holds true. Thus, $n = 256,$ so the answer is $2 + 5 + 6 = \boxed{13}.$


Answer: 17.0

There is a unique positive real number $x$ such that the three numbers $\log_8{2x}$ $\log_4{x}$ , and $\log_2{x}$ , in that order, form a geometric progression with positive common ratio.  The number $x$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$

Solution:
Since these form a geometric series, $\frac{\log_2{x}}{\log_4{x}}$ is the common ratio. Rewriting this, we get $\frac{\log_x{4}}{\log_x{2}} = \log_2{4} = 2$ by base change formula. Therefore, the common ratio is 2. Now $\frac{\log_4{x}}{\log_8{2x}} = 2 \implies \log_4{x} = 2\log_8{2} + 2\log_8{x} \implies \frac{1}{2}\log_2{x} = \frac{2}{3} + \frac{2}{3}\log_2{x}$
$\implies -\frac{1}{6}\log_2{x} = \frac{2}{3} \implies \log_2{x} = -4 \implies x = \frac{1}{16}$ . Therefore, $1 + 16 = \boxed{017}$

Solution: If we set $x=2^y$ , we can obtain three terms of a geometric sequence through logarithm properties. The three terms are \[\frac{y+1}{3}, \frac{y}{2}, y.\] In a three-term geometric sequence, the middle term squared is equal to the product of the other two terms, so we obtain the following: \[\frac{y^2+y}{3} = \frac{y^2}{4},\] which can be solved to reveal $y = -4$ . Therefore, $x = 2^{-4} = \frac{1}{16}$ , so our answer is $\boxed{017}$

Solution: Let $r$ be the common ratio. We have \[r = \frac{\log_4{(x)}}{\log_8{(2x)}} = \frac{\log_2{(x)}}{\log_4{(x)}}\] Hence we obtain \[(\log_4{(x)})(\log_4{(x)}) = (\log_8{(2x)})(\log_2{(x)})\] Ideally we change everything to base $64$ and we can get: \[(\log_{64}{(x^3)})(\log_{64}{(x^3)}) = (\log_{64}{(x^6)})(\log_{64}{(4x^2)})\] Now divide to get: \[\frac{\log_{64}{(x^3)}}{\log_{64}{(4x^2)}} = \frac{\log_{64}{(x^6)}}{\log_{64}{(x^3)}}\] By change-of-base we obtain: \[\log_{(4x^2)}{(x^3)} = \log_{(x^3)}{(x^6)} = 2\] Hence $(4x^2)^2 = x^3 \rightarrow 16x^4 = x^3 \rightarrow x = \frac{1}{16}$ and we have $1+16 = \boxed{017}$ as desired.

Solution: Let $r$ be the common ratio, and let $a$ be the starting term ( $a=\log_{8}{(2x)}$ ). We then have: \[\log_{8}{(2x)}=a, \log_{4}{(x)}=ar, \log_{2}{(x)}=ar^2\] Rearranging these equations gives: \[8^a=2x, 4^{ar}=x, 2^{ar^2}=x\] Deal with the last two equations first: Setting them equal gives: \[4^{ar}=2^{ar^2} \implies 2^{2ar}=2^{ar^2} \implies 2ar=ar^2 \implies r=2\] Using this value of $r$ , substitute into the first and second equations (or the first and third, it doesn't really matter) to get: \[8^a=2x, 4^{2a}=x\] Changing these to a common base gives: \[2^{3a}=2x, 2^{4a}=x\] Dividing the first equation by 2 on both sides yields: \[2^{3a-1}=x\] Setting these equations equal to each other and removing the exponent again gives: \[3a-1=4a \implies a=-1\] Substituting this back into the first equation gives: \[8^{-1}=2x \implies 2x=\frac{1}{8} \implies x=\frac{1}{16}\] Therefore, $m+n=1+16=\boxed{017}$

Solution:
We can relate the logarithms as follows:
\[\frac{\log_4{x}}{\log_8{(2x)}}=\frac{\log_2{x}}{\log_4{x}}\] \[\log_8{(2x)}\log_2{x}=\log_4{x}\log_4{x}\]
Now we can convert all logarithm bases to $2$ using the identity $\log_a{b}=\log_{a^c}{b^c}$
\[\log_2{\sqrt[3]{2x}}\log_2{x}=\log_2{\sqrt{x}}\log_2{\sqrt{x}}\]
We can solve for $x$ as follows:
\[\frac{1}{3}\log_2{(2x)}\log_2{x}=\frac{1}{4}\log_2{x}\log_2{x}\] \[\frac{1}{3}\log_2{(2x)}=\frac{1}{4}\log_2{x}\] \[\frac{1}{3}\log_2{2}+\frac{1}{3}\log_2{x}=\frac{1}{4}\log_2{x}\] We get $x=\frac{1}{16}$ . Verifying that the common ratio is positive, we find the answer of $\boxed{017}$

Solution: If the numbers are in a geometric sequence, the middle term must be the geometric mean of the surrounding terms. We can rewrite the first two logarithmic expressions as $\frac{1+\log_2{x}}{3}$ and $\frac{1}{2}\log_2{x}$ , respectively. Therefore: \[\frac{1}{2}\log_2{x}=\sqrt{\left(\frac{1+\log_2{x}}{3}\right)\left(\log_2{x}\right)}\] Let $n=\log_2{x}$ . We can rewrite the expression as: \[\frac{n}{2}=\sqrt{\frac{n(n+1)}{3}}\] \[\frac{n^2}{4}=\frac{n(n+1)}{3}\] \[4n(n+1)=3n^2\] \[4n^2+4n=3n^2\] \[n^2+4n=0\] \[n(n+4)=0\] \[n=0 \text{ and } -4\] Zero does not work in this case, so we consider $n=-4$ $\log_2{x}=-4 \rightarrow x=\frac{1}{16}$ . Therefore, $1+16=\boxed{017}$

Solution: Again, by the Change of Base Formula, obtain that the common ratio is 2. If we let $y$ be the exponent of $\log_8 (2x)$ , then we have $8^y=2x;\:4^{2y}=x;\:2^{4y}=x.$ Wee can then divide the first equation by two to have the right side be $x$ . Also, $2^{4y}=\left(2^{4}\right)^y=16^y$ . Setting this equal to $\frac{8^y}{2}$ , we can divide the two equations to get $2^y=\frac12$ . Therefore, $y=-1$ . After that, we can raise $16$ to the $-1$ th power to get $x=16^{-1}\Rightarrow x=\frac1{16}$ . We then get our sum of $1+16=\boxed{017}$


Answer: 1.0

There is a unique sequence of integers $a_1, a_2, \cdots a_{2023}$ such that \[\tan2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}\] whenever $\tan 2023x$ is defined. What is $a_{2023}?$
$\textbf{(A) } -2023 \qquad\textbf{(B) } -2022 \qquad\textbf{(C) } -1 \qquad\textbf{(D) } 1 \qquad\textbf{(E) } 2023$

Solution:
\begin{align*} \cos 2023 x + i \sin 2023 x &= (\cos x + i \sin x)^{2023}\\ &= \cos^{2023} x + \binom{2023}{1} \cos^{2022} x (i\sin x) + \binom{2023}{2} \cos^{2021} x (i \sin x)^{2} +\binom{2023}{3} \cos^{2023} x (i \sin x)^{3}\\ &+ \dots + \binom{2023}{2022} \cos x (i \sin x)^{2022} + (i \sin x)^{2023}\\ &= \cos^{2023} x + i \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{2} \cos^{2021} x \sin^{2} x - i\binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots\\ &- \binom{2023}{2022} \cos x \sin^{2022} x - i \sin^{2023} x\\ \end{align*}
By equating real and imaginary parts:
\[\cos 2023 x = \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x\]
\[\sin 2023 x = \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x\]
\begin{align*} \tan2023x  &= \frac{ \sin2023x }{ \cos2023x } = \frac{ \binom{2023}{1} \cos^{2022} x \sin x - \binom{2023}{3} \cos^{2020} x \sin^{3} x + \dots - \sin^{2023} x }{ \cos^{2023} x - \binom{2023}{2} \cos^{2021} x \sin^{2} x + \dots - \binom{2023}{2022} \cos x \sin^{2022} x }\\ &= \frac{ \binom{2023}{1} \frac{\cos^{2022} x \sin x}{\cos^{2023} x} - \binom{2023}{3} \frac{\cos^{2020} x \sin^{3} x}{\cos^{2023} x} + \dots - \frac{\sin^{2023} x}{\cos^{2023} x} }{ \frac{\cos^{2023} x}{\cos^{2023} x} - \binom{2023}{2} \frac{\cos^{2021} x \sin^{2} x}{\cos^{2023} x} + \dots - \binom{2023}{2022} \frac{\cos x \sin^{2022} x}{\cos^{2023} x}  }\\ &= \frac{ \binom{2023}{1} \tan x - \binom{2023}{3} \tan^{3}x + \dots - \tan^{2023}x }{ 1 - \binom{2023}{2} \tan^{2}x + \dots - \binom{2023}{2022} \tan^{2022} x }\\ \end{align*}
\[a_{2023} = \boxed{1}\]

Solution: Note that $\tan{kx} = \frac{\binom{k}{1}\tan{x} - \binom{k}{3}\tan^{3}{x} + \cdots \pm \binom{k}{k}\tan^{k}{x}}{\binom{k}{0}\tan^{0}{x} - \binom{k}{2}\tan^{2}{x} + \cdots + \binom{k}{k-1}\tan^{k-1}{x}}$ , where k is odd and the sign of each term alternates between positive and negative. To realize this during the test, you should know the formulas of $\tan{2x}, \tan{3x},$ and $\tan{4x}$ , and can notice the pattern from that. The expression given essentially matches the formula of $\tan{kx}$ exactly. $a_{2023}$ is evidently equivalent to $\pm\binom{2023}{2023}$ , or 1. However, it could be positive or negative. Notice that in the numerator, whenever the exponent of the tangent term is congruent to 1 mod 4, the term is positive. Whenever the exponent of the tangent term is 3 mod 4, the term is negative. 2023, which is assigned to k, is congruent to 3 mod 4. This means that the term of $\binom{k}{k}\tan^{k}{x}$ is $\boxed{1}$

Solution:
For odd $n$ , we have \begin{align*} \tan nx & = \frac{\sin nx}{\cos nx} \\ & = \frac{\frac{1}{2i} \left( e^{i n x} - e^{-i n x} \right)} {\frac{1}{2} \left( e^{i n x} + e^{-i n x} \right)} \\ & = - i \frac{e^{i n x} - e^{-i n x}}{e^{i n x} + e^{-i n x}} \\ & = - i  \frac{\left( \cos x + i \sin x \right)^n - \left( \cos x - i \sin x \right)^n} {\left( \cos x + i \sin x \right)^n + \left( \cos x - i \sin x \right)^n} \\ & = \frac{ - 2 i \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \cos x \right)^{n - 2m - 1} \left( i \sin x \right)^{2m + 1}} {2 \sum_{m=0}^{(n-1)/2} \binom{n}{2m} \left( \cos x \right)^{n - 2m} \left( i \sin x \right)^{2m}} \\ & = \frac{ \frac{1}{\left( \cos x \right)^n} \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \cos x \right)^{n - 2m - 1} \left( i \sin x \right)^{2m + 1}} {i \frac{1}{\left( \cos x \right)^n} \sum_{m=0}^{(n-1)/2} \binom{n}{2m} \left( \cos x \right)^{n - 2m} \left( i \sin x \right)^{2m}} \\ & = \frac{ \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( i \tan x \right)^{2m + 1}} {i \sum_{m=0}^{(n-1)/2} \binom{n}{2m} \left( i \tan x \right)^{2m}} \\ & = \frac{ \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \tan x \right)^{2m + 1} i^{2m + 1}} {\sum_{m=0}^{(n-1)/2} \binom{n}{2m} \left( \tan x \right)^{2m} i^{2m + 1}} \\ & = \frac{ \sum_{m=0}^{(n-1)/2} \binom{n}{2m + 1} \left( \tan x \right)^{2m + 1} \left( -1 \right)^m} {\sum_{m=0}^{(n-1)/2} \binom{n}{2m} \left( \tan x \right)^{2m} \left( -1 \right)^m}  . \end{align*}
Thus, for $n = 2023$ , we have \begin{align*} a_{2023} & = \binom{2023}{2023} \left( -1 \right)^{(2023-1)/2} \\ & = \left( -1 \right)^{1011} \\ & = \boxed{1}


Answer: 336.0

There is an unlimited supply of congruent equilateral triangles made of colored paper. Each triangle is a solid color with the same color on both sides of the paper. A large equilateral triangle is constructed from four of these paper triangles. Two large triangles are considered distinguishable if it is not possible to place one on the other, using translations, rotations, and/or reflections, so that their corresponding small triangles are of the same color.
Given that there are six different colors of triangles from which to choose, how many distinguishable large equilateral triangles may be formed?
[asy] pair A,B; A=(0,0); B=(2,0); pair C=rotate(60,A)*B; pair D, E, F; D = (1,0); E=rotate(60,A)*D; F=rotate(60,C)*E; draw(C--A--B--cycle); draw(D--E--F--cycle); [/asy]

Solution:
If two of our big equilateral triangles have the same color for their center triangle and the same multiset of colors for their outer three triangles, we can carry one onto the other by a combination of rotation and reflection.  Thus, to make two triangles distinct, they must differ either in their center triangle or in the collection of colors which make up their outer three triangles.
There are 6 possible colors for the center triangle.
Thus, in total we have $6\cdot(20 + 30 + 6) = \boxed{336}$ total possibilities.

Solution:
We apply Burnside's Lemma and consider 3 rotations of 120 degrees, 240 degrees, and 0 degrees. We also consider three reflections from the three lines of symmetry in the triangle. Thus, we have to divide by $3+3=6$ for our final count.
Case 1: 0 degree rotation. This is known as the identity rotation, and there are $6^4=1296$ choices because we don't have any restrictions.
Case 2: 120 degree rotation. Note that the three "outer" sides of the triangle have to be the same during this, and the middle one can be anything. We have $6*6=36$ choices from this.
Case 3: 240 degree rotation. Similar to the 120 degree rotation, each must be the same except for the middle. We have $6*6=36$ choices from this.
Case 4: symmetry about lines. We multiply by 3 for these because the amount of colorings fixed under symmetry are the exact same each time. Two triangles do not change under this, and they must be the same. The other two triangles (1 middle and 1 outer) can be anything because they stay the same during the reflection. We have $6*6*6=216$ ways for one symmetry. There are 3 symmetries, so there are $216*3=648$ combinations in all.
Now, we add our cases up: $1296+36+36+648=2016$ . We have to divide by 6, so $2016/6=\boxed{336}$ distinguishable ways to color the triangle.

Solution: There are $6$ choices for the center triangle. Note that given any $3$ colors, there is a unique way to assign them to the corner triangles. We have $6$ different colors to choose from, so the number of ways to color the corner triangles is the same as the number of ways to arrange $6 - 1 = 5$ dividers and $3$ identical items. Therefore, our answer is \[6 \binom{5 + 3}{3} = 6\binom{8}{3} = \boxed{336}.\]


Answer: 12.0

Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of $10$ hours per week helping around the house for $6$ weeks. For the first $5$ weeks she helps around the house for $8$ $11$ $7$ $12$ and $10$ hours. How many hours must she work for the final week to earn the tickets?
$\mathrm{(A)}\ 9 \qquad\mathrm{(B)}\ 10 \qquad\mathrm{(C)}\ 11 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 13$

Solution:
Let $x$ be the number of hours she must work for the final week. We are looking for the average, so \[\frac{8 + 11 + 7 + 12 + 10 + x}{6} = 10\] Solving gives: \[\frac{48 + x}{6} = 10\]
\[48 + x = 60\]
\[x = 12\]
So, the answer is $\boxed{12}$

Solution:
Use average deviation:
The average is 10 hour per day. If work 8 hours then it is 2 hours short; if work 11 hours then there is 1 hour surplus, the last day need to cancel out the collective deviation from the previous 5 days.
So we got
\[-2+1-3+2+0=-2\]
The last day need to have +2 deviation to cancel out the -2 collective deviation to get 10 as average value, so $\boxed{12}$


Answer: 11.0

Thirteen black and six white hexagonal tiles were used to create the figure below. If a new figure is created by attaching a border of white tiles with the same size and shape as the others, what will be the difference between the total number of white tiles and the total number of black tiles in the new figure?
[asy] defaultpen(linewidth(1)); real x=sqrt(3)/2; path p=rotate(30)*polygon(6); filldraw(p^^shift(0,3)*p^^shift(4x,0)*p^^shift(3x,1.5)*p^^shift(2x,3)*p^^shift(-4x,0)*p^^shift(-3x,1.5)*p^^shift(-2x,3)*p^^shift(3x,-1.5)*p^^shift(-3x,-1.5)*p^^shift(2x,-3)*p^^shift(-2x,-3)*p^^shift(0,-3)*p, black, black); draw(shift(2x,0)*p^^shift(-2x,0)*p^^shift(x,1.5)*p^^shift(-x,1.5)*p^^shift(x,-1.5)*p^^shift(-x,-1.5)*p); [/asy]
$\textbf{(A)}\ 5\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 11\qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$

Solution: The first ring around the middle tile has $6$ tiles, and the second has $12$ . From this pattern, the third ring has $18$ tiles. Of these, $6+18=24$ are white and $1+12=13$ are black, with a difference of $24-13 = \boxed{11}$


Answer: 3.0

Three $\Delta$ 's and a $\diamondsuit$ will balance nine $\bullet$ 's.  One $\Delta$ will balance a $\diamondsuit$ and a $\bullet$
[asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,2)--(-12,0)--(12,0)--(12,2)); draw(ellipse((-12,5),8,3)); draw(ellipse((12,5),8,3)); label("$\Delta \hspace{2 mm}\Delta \hspace{2 mm}\Delta \hspace{2 mm}\diamondsuit $",(-12,6.5),S); label("$\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm} \bullet $",(12,5.2),N); label("$\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet \hspace{2 mm}\bullet $",(12,5.2),S); fill((44,0)--(40,-2)--(48,-2)--cycle,black); draw((34,2)--(34,0)--(54,0)--(54,2)); draw(ellipse((34,5),6,3)); draw(ellipse((54,5),6,3)); label("$\Delta $",(34,6.5),S);  label("$\bullet \hspace{2 mm}\diamondsuit $",(54,6.5),S); [/asy]
How many $\bullet$ 's will balance the two $\diamondsuit$ 's in this balance?
[asy] unitsize(5.5); fill((0,0)--(-4,-2)--(4,-2)--cycle,black); draw((-12,4)--(-12,2)--(12,-2)--(12,0)); draw(ellipse((-12,7),6.5,3)); draw(ellipse((12,3),6.5,3)); label("$?$",(-12,8.5),S); label("$\diamondsuit \hspace{2 mm}\diamondsuit $",(12,4.5),S); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$

Solution:
For simplicity, suppose $\Delta = a$ $\diamondsuit = b$ and $\bullet = c$ .  Then, \[3a+b=9c\] \[a=b+c\] and we want to know what $2b$ is in terms of $c$ .  Substituting the second equation into the first, we have \[4b=6c\Rightarrow 2b=3c\]
Thus, we need $3$ $\bullet$ 's $\rightarrow \boxed{3}$


Answer: 4.0

Three $\text{A's}$ , three $\text{B's}$ , and three $\text{C's}$ are placed in the nine spaces so that each row and column contains one of each letter. If $\text{A}$ is placed in the upper left corner, how many arrangements are possible?
[asy] size((80)); draw((0,0)--(9,0)--(9,9)--(0,9)--(0,0)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); draw((0,3)--(9,3)); draw((0,6)--(9,6)); label("A", (1.5,7.5)); [/asy]
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

Solution: There are $2$ ways to place the remaining $\text{As}$ $2$ ways to place the remaining $\text{Bs}$ , and $1$ way to place the remaining $\text{Cs}$ for a total of $(2)(2)(1) = \boxed{4}$


Answer: 864.0

Three 12 cm $\times$ 12 cm squares are each cut into two pieces $A$ and $B$ , as shown in the first figure below, by joining the midpoints of two adjacent sides. These six pieces are then attached to a regular hexagon , as shown in the second figure, so as to fold into a polyhedron . What is the volume (in $\mathrm{cm}^3$ ) of this polyhedron?
AIME 1985 Problem 15.png

Solution: Note that gluing two of the given polyhedra together along a hexagonal face (rotated $60^\circ$ from each other) yields a cube , so the volume is $\frac12 \cdot 12^3 = 864$ , so our answer is $\boxed{864}$


Answer: 31.0

Three bags of jelly beans contain 26, 28, and 30 beans.  The ratios of yellow beans to all beans in each of these bags are $50\%$ $25\%$ , and $20\%$ , respectively.  All three bags of candy are dumped into one bowl.  Which of the following is closest to the ratio of yellow jelly beans to all beans in the bowl?
$\text{(A)}\ 31\% \qquad \text{(B)}\ 32\% \qquad \text{(C)}\ 33\% \qquad \text{(D)}\ 35\% \qquad \text{(E)}\ 95\%$

Solution:
In bag $A$ , there are $26$ jellybeans, and $50\%$ are yellow.  That means there are $26\times 50\% = 26\times 0.50 = 13$ yellow jelly beans in this bag.
In bag $B$ , there are $28$ jellybeans, and $25\%$ are yellow.  That means there are $28\times 25\% = 28\times 0.25 = 7$ yellow jelly beans in this bag.
In bag $C$ , there are $30$ jellybeans, and $20\%$ are yellow.  That means there are $30\times 20\% = 30\times 0.20 = 6$ yellow jelly beans in this bag.
In all three bags, there are $13 + 7 + 6 = 26$ yellow jelly beans in total , and $26 + 28 + 30 = 84$ jelly beans of all types in total.
Thus, $\frac{26}{84} = \frac{26}{84}\cdot 100\% = \frac{13 }{ 42} \cdot 100\% = 30.9\%$ of all jellybeans are yellow.  Thus, the correct answer is $\boxed{31}$


Answer: 55.0

Three balls are randomly and independantly tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $i$ is $2^{-i}$ for $i=1,2,3,....$ More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is $\frac pq,$ where $p$ and $q$ are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins $3,17,$ and $10.$ ) What is $p+q?$
$\textbf{(A) }55 \qquad \textbf{(B) }56 \qquad \textbf{(C) }57\qquad \textbf{(D) }58 \qquad \textbf{(E) }59$

Solution:
"Evenly spaced" just means the bins form an arithmetic sequence.
Suppose the middle bin in the sequence is $x$ . There are $x-1$ different possibilities for the first bin, and these two bins uniquely determine the final bin. Now, the probability that these $3$ bins are chosen is $6\cdot 2^{-3x} = 6\cdot \frac{1}{8^x}$ , so the probability $x$ is the middle bin is $6\cdot\frac{x-1}{8^x}$ . Then, we want the sum \begin{align*} 6\sum_{x=2}^{\infty}\frac{x-1}{8^x} &= \frac{6}{8}\left[\frac{1}{8} + \frac{2}{8^2} + \frac{3}{8^3}\cdots\right]\\ &= \frac34\left[\left(\frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3}+\cdots \right) + \left(\frac{1}{8^2} + \frac{1}{8^3} + \frac{1}{8^4} + \cdots \right) + \cdots\right]\\ &= \frac34\left[\frac17\cdot \left(1 + \frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3} + \cdots \right)\right]\\ &= \frac34\cdot \frac{8}{49}\\ &= \frac{6}{49} \end{align*} The answer is $6+49=\boxed{55}.$

Solution:
As in solution 1, note that "evenly spaced" means the bins are in arithmetic sequence. We let the first bin be $a$ and the common difference be $d$ . Further note that each $(a, d)$ pair uniquely determines a set of $3$ bins.
We have $a\geq1$ because the leftmost bin in the sequence can be any bin, and $d\geq1$ , because the bins must be distinct.
This gives us the following sum for the probability: \begin{align*} 6 \sum_{a=1}^{\infty} \sum_{d=1}^{\infty} 2^{-3a-3d} &= 6 \sum_{a=1}^{\infty} \sum_{d=1}^{\infty} 2^{-3a} \cdot 2^{-3d} \\ &= 6 \left( \sum_{a=1}^{\infty} 2^{-3a} \right) \left( \sum_{d=1}^{\infty} 2^{-3d} \right) \\ &= 6 \left( \sum_{a=1}^{\infty} 8^{-a} \right) \left( \sum_{d=1}^{\infty} 8^{-d} \right) \\ &= 6 \left( \frac{1}{7} \right) \left( \frac{1}{7} \right) \\ &= \frac{6}{49} .\end{align*} Therefore the answer is $6 + 49 = \boxed{55}$

Solution: This is a slightly messier variant of solution 2. If the first ball is in bin $i$ and the second ball is in bin $j>i$ , then the third ball is in bin $2j-i$ . Thus the probability is \begin{align*} 6\sum_{i=1}^{\infty}\sum_{j=i+1}^\infty2^{-i}2^{-j}2^{-2j+i}&=6\sum_{i=1}^{\infty}\sum_{j=i+1}^\infty2^{-3j}\\ &=6\sum_{i=1}^{\infty}\left(\frac{2^{-3(i+1)}}{1-\tfrac{1}{8}}\right)\\ &=6\sum_{i=1}^\infty\frac{8}{7}\cdot2^{-3}\cdot2^{-3i}\\ &=\frac{6}{7}\sum_{i=1}^\infty2^{-3i}\\ &=\frac{6}{7}\cdot\frac{2^{-3}}{1-\tfrac18}\\ &=\frac{6}{49}. \end{align*} Therefore the answer is $6 + 49 = \boxed{55}$

Solution: Based on the value of $n,$ we construct the following table: \[\begin{array}{c|c|c|c}  & & & \\ [-1.5ex] \textbf{Exactly }\boldsymbol{n}\textbf{ Spaces Apart} & \textbf{Bin \#s} & \textbf{Expression} & \textbf{Prob. of One Such Perm.} \\ [1ex]  \hline\hline   & & & \\ [-1.5ex]  n=1 & 1,2,3 & 2^{-1}\cdot2^{-2}\cdot2^{-3} & 2^{-6} \\ [1ex]   & 2,3,4 & 2^{-2}\cdot2^{-3}\cdot2^{-4} & 2^{-9} \\ [1ex]   & 3,4,5 & 2^{-3}\cdot2^{-4}\cdot2^{-5} & 2^{-12} \\ [1ex]   & 4,5,6 & 2^{-4}\cdot2^{-5}\cdot2^{-6} & 2^{-15} \\ [1ex]   & \cdots & \cdots & \cdots \\ [1ex]  \hline  & & & \\ [-1.5ex]  n=2 & 1,3,5 & 2^{-1}\cdot2^{-3}\cdot2^{-5} & 2^{-9} \\ [1ex]   & 2,4,6 & 2^{-2}\cdot2^{-4}\cdot2^{-6} & 2^{-12} \\ [1ex]   & 3,5,7 & 2^{-3}\cdot2^{-5}\cdot2^{-7} & 2^{-15} \\ [1ex]   & 4,6,8 & 2^{-4}\cdot2^{-6}\cdot2^{-8} & 2^{-18} \\ [1ex]   & \cdots & \cdots & \cdots \\ [1ex]  \hline  & & & \\ [-1.5ex]  n=3 & 1,4,7 & 2^{-1}\cdot2^{-4}\cdot2^{-7} & 2^{-12} \\ [1ex]   & 2,5,8 & 2^{-2}\cdot2^{-5}\cdot2^{-8} & 2^{-15} \\ [1ex]   & 3,6,9 & 2^{-3}\cdot2^{-6}\cdot2^{-9} & 2^{-18} \\ [1ex]  & 4,7,10 & 2^{-4}\cdot2^{-7}\cdot2^{-10} & 2^{-21} \\ [1ex]  & \cdots & \cdots & \cdots \\ [1ex]  \hline   & & & \\ [-1.5ex]    \cdots & \cdots & \cdots & \cdots \\ [1ex] \end{array}\] Since three balls have $3!=6$ permutations, the requested probability is \begin{align*} 6\left(\sum_{k=0}^{\infty}2^{-6-3k}+\sum_{k=0}^{\infty}2^{-9-3k}+\sum_{k=0}^{\infty}2^{-12-3k}+\cdots\right)&=6\left(2^{-6}\sum_{k=0}^{\infty}2^{-3k}+2^{-9}\sum_{k=0}^{\infty}2^{-3k}+2^{-12}\sum_{k=0}^{\infty}2^{-3k}+\cdots\right) \\ &=\left(6\sum_{k=0}^{\infty}2^{-3k}\right)\cdot\left(2^{-6}+2^{-9}+2^{-12}+\cdots\right) \\ &=\left(6\sum_{k=0}^{\infty}2^{-3k}\right)\cdot\left(\sum_{k=0}^{\infty}2^{-6-3k}\right) \\ &=\left(6\sum_{k=0}^{\infty}2^{-3k}\right)\cdot\left(2^{-6}\sum_{k=0}^{\infty}2^{-3k}\right) \\ &=\frac{6}{1-2^{-3}}\cdot\frac{2^{-6}}{1-2^{-3}} \\ &=\frac{6}{49} \end{align*} by infinite geometric series, from which the answer is $6+49=\boxed{55}.$


Answer: 4.0

Three cards, each with a positive integer written on it, are lying face-down on a table. Casey, Stacy, and Tracy are told that
First, Casey looks at the number on the leftmost card and says, "I don't have enough information to determine the other two numbers."  Then Tracy looks at the number on the rightmost card and says, "I don't have enough information to determine the other two numbers."  Finally, Stacy looks at the number on the middle card and says, "I don't have enough information to determine the other two numbers."  Assume that each person knows that the other two reason perfectly and hears their comments.  What number is on the middle card?
$\textrm{(A)}\ 2 \qquad \textrm{(B)}\ 3 \qquad \textrm{(C)}\ 4 \qquad \textrm{(D)}\ 5 \qquad \textrm{(E)}\ \text{There is not enough information to determine the number.}$

Solution:
Initially, there are the following possibilities for the numbers on the cards: $(1,2,10)$ $(1,3,9)$ $(1,4,8)$ $(1,5,7)$ $(2,3,8)$ $(2,4,7)$ $(2,5,6)$ , and $(3,4,6)$
If Casey saw the number $3$ , she would have known the other two numbers. As she does not, we eliminated the possibility $(3,4,6)$
At this moment, if the last card contained a $10$ $9$ , or a $6$ , Tracy would know the other two numbers. (Note that Tracy is aware of the fact that $(3,4,6)$ was eliminated. If she saw the number $6$ , she would know that the other two are $2$ and $5$ .) This eliminates three more possibilities.
Thus before Stacy took her look, we are left with four possible cases: $(2,4,7)$ $(1,4,8)$ $(1,5,7)$ , and $(2,3,8)$ . As Stacy could not find out the exact combination, the middle number must be $\boxed{4}$


Answer: 24.0

Three circles, each of radius $3$ , are drawn with centers at $(14, 92)$ $(17, 76)$ , and $(19, 84)$ . A line passing through $(17,76)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line?

Solution:
1984 AIME-6.png
The line passes through the center of the bottom circle; hence it is the circle's diameter and splits the circle into two equal areas. For the rest of the problem, we do not have to worry about that circle.
Draw the midpoint of $\overline{AC}$ (the centers of the other two circles), and call it $M$ . If we draw the feet of the perpendiculars from $A,C$ to the line (call $E,F$ ), we see that $\triangle AEM\cong \triangle CFM$ by HA congruency ; hence $M$ lies on the line. The coordinates of $M$ are $\left(\frac{19+14}{2},\frac{84+92}{2}\right) = \left(\frac{33}{2},88\right)$
Thus, the slope of the line is $\frac{88 - 76}{\frac{33}{2} - 17} = -24$ , and the answer is $\boxed{024}$

Solution:
First of all, we can translate everything downwards by $76$ and to the left by $14$ . Then, note that a line passing through a given point intersecting a circle with a center as that given point will always cut the circle in half, so we can re-phrase the problem:
Two circles, each of radius $3$ , are drawn with centers at $(0, 16)$ , and $(5, 8)$ . A line passing through $(3,0)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the absolute value of the slope of this line?
Note that this is equivalent to finding a line such that the distance from $(0,16)$ to the line is the same as the distance from $(5,8)$ to the line. Let the line be $y - ax - b = 0$ . Then, we have that: \[\frac{|-5a + 8 - b|}{\sqrt{a^2+1}}= \frac{|16 - b|}{\sqrt{a^2+1}} \Longleftrightarrow |-5a+8-b| = |16-b|\] We can split this into two cases.
Case 1: $16-b = -5a + 8 - b \Longleftrightarrow a = -\frac{8}{5}$
In this case, the absolute value of the slope of the line won’t be an integer, and since this is an AIME problem, we know it’s not possible.
Case 2: $b-16 = -5a + 8 - b \Longleftrightarrow 2b + 5a = 24$
But we also know that it passes through the point $(3,0)$ , so $-3a-b = 0 \Longleftrightarrow b = -3a$ . Plugging this in, we see that $2b + 5a = 24 \Longleftrightarrow a = -24$ $\boxed{24}$

Solution: Consider the region of the plane between $x=16$ and $x=17$ . The parts of the circles centered at $(14,92)$ and $(19,84)$ in this region have equal area. This is by symmetry- the lines defining the region are 2 units away from the centers of each circle and therefore cut off congruent segments. We will draw the line in a way that uses this symmetry and makes identical cuts on the circles. Since $(17,76)$ is $8$ units below the center of the lower circle, we will have the line exit the region $8$ units above the center of the upper circle, at $(16,100)$ . We then find that the slope of the line is $-24$ and our answer is $\boxed{024}$

Solution: We can redefine the coordinate system so that the center of the center circle is the origin, for easier calculations, as the slope of the line and the congruence of the circles do not depend on it. $O_1=(-3, 16)$ $O_2=(0,0)$ , and $O_3=(2,8)$ . A line bisects a circle iff it passes through the center. Therefore, we can ignore the bottom circle because it contributes an equal area with any line. A line passing through the centroid of any plane system with two perpendicular lines of reflectional symmetry bisects it. We have defined two points of the line, which are $(0,0)$ and $(-\frac{1}{2},12)$ . We use the slope formula to calculate the slope, which is $-24$ , leading to an answer of $\boxed{024}$ $QED \blacksquare$

Solution: Notice that any line that passes through the bottom circle's center cuts it in half, so all we really care about are the top two circles. Suppose $\ell$ is the desired line. Draw lines $\ell_1$ and $\ell_2$ both parallel to $\ell$ such that $\ell_1$ passes through $(14,92)$ and $\ell_2$ passes through $(19,84)$ . Clearly, $\ell$ must be the "average" of $\ell_1$ and $\ell_2$ . Suppose $\ell:=y=mx+b, \ell_1:=y=mx+c, \ell_2:=y=mx+d$ . Then $b=76-17m, c=92-14m, d=84-19m$ . So we have that \[76-17m=\frac{92-14m+84-19m}{2},\] which yields $m=-24$ for an answer of $\boxed{024}$


Answer: 50.0

Three circular arcs of radius $5$ units bound the region shown. Arcs $AB$ and $AD$ are quarter-circles, and arc $BCD$ is a semicircle. What is the area, in square units, of the region?
[asy] pair A,B,C,D; A = (0,0); B = (-5,5); C = (0,10); D = (5,5); draw(arc((-5,0),A,B,CCW)); draw(arc((0,5),B,D,CW)); draw(arc((5,0),D,A,CCW)); label("$A$",A,S); label("$B$",B,W); label("$C$",C,N); label("$D$",D,E);[/asy]
$\text{(A)}\ 25\qquad\text{(B)}\ 10+5\pi\qquad\text{(C)}\ 50\qquad\text{(D)}\ 50+5\pi\qquad\text{(E)}\ 25\pi$

Solution:
Draw two squares:  one that has opposing corners at $A$ and $B$ , and one that has opposing corners at $A$ and $D$ .  These squares share side $\overline{AO}$ , where $O$ is the center of the large semicircle.
These two squares have a total area of $2 \cdot 5^2$ , but have two quarter circle "bites" of radius $5$ that must be removed.  Thus, the bottom part of the figure has area
$2\cdot 25 - 2 \cdot \frac{1}{4}\pi \cdot 5^2$
$50 - \frac{25\pi}{2}$
This is the area of the part of the figure underneath $\overline{BD}$ .  The part of the figure over $\overline{BD}$ is just a semicircle with radius $5$ , which has area of $\frac{1}{2}\pi\cdot 5^2 = \frac{25\pi}{2}$
Adding the two areas gives a total area of $\boxed{50}$


Answer: 408.0

Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio $3: 2: 1,$ what is the least possible total for the number of bananas?

Solution:
Denote the number of bananas the first monkey took from the pile as $b_1$ , the second $b_2$ , and the third $b_3$ ; the total is $b_1 + b_2 + b_3$ . Thus, the first monkey got $\frac{3}{4}b_1 + \frac{3}{8}b_2 + \frac{11}{24}b_3$ , the second monkey got $\frac{1}{8}b_1 + \frac{1}{4}b_2 + \frac{11}{24}b_3$ , and the third monkey got $\frac{1}{8}b_1 + \frac{3}{8}b_2 + \frac{1}{12}b_3$
Taking into account the ratio aspect, say that the third monkey took $x$ bananas in total. Then,
$x = \frac{1}{4}b_1 + \frac{1}{8}b_2 + \frac{11}{72}b_3 = \frac{1}{16}b_1 + \frac{1}{8}b_2 + \frac{11}{48}b_3 = \frac{1}{8}b_1 + \frac{3}{8}b_2 + \frac{1}{12}b_3$
Solve this to find that $\frac{b_1}{11} = \frac{b_2}{13} = \frac{b_3}{27}$ . All three fractions must be integral. Also note some other conditions we have picked up in the course of the problem, namely that $b_1$ is divisible by $8$ $b_2$ is divisible by $8$ , and $b_3$ is divisible by $72$ (however, since the denominator contains a $27$ , the factors of $3$ cancel, and it only really needs to be divisible by $8$ ). Thus, the minimal value is when each fraction is equal to $8$ , and the solution is $8(11 + 13 + 27) = \boxed{408}$

Solution:
Let the first monkey take $8x$ bananas, the second monkey take $8y$ , and the third monkey take $24z$ . I chose these numbers to make it so, when each monkey splits his bananas, they will get an integer amount of each variable. So, when the first monkey distributes his bananas, he gets $6x$ , and the other monkeys get $1x$ . So, we can make expressions for how many bananas each monkey gets. Also, the variables have to be integers too.
Monkey 1: $6x+3y+11z$
Monkey 2: $1x+2y+11z$
Monkey 3: $1x+3y+2z$
So, they are in a ratio $3:2:1$ . But, we can turn it into an equation by multiplying the amount of bananas each monkey has by 2, 3, 6. Now, the ratio is $6:6:6$ , so, $12x+6y+22z = 3x+6y+33z = 6x+18y+12z$ . Subtracting $3x+6y+12z$ from all, we get $9x+10z = 21z = 3x+12y$ . Let's split this into 3 equations.
\[9x+10z = 21z\] \[21z = 3x+12y\] \[9x+10z = 3x+12y\]
Let's look at the first equation. Rearranging, it gets us $9x = 11z$
We can rearrange the third equation, then divide by 2, then subtract the second equation. \[21z-6y = 12y-5z\] \[26z = 18y\] \[13z = 9y\]
It is clear $z$ is a multiple of 9, so let $z = 9$ . Then we get the $x = 11$ , and $y = 13$ . Testing, we confirm this will get the first monkey 204 bananas, the second 136, and the third, 68. Adding them up, we get that there were $\boxed{408}$ bananas originally in the pile.

Solution:
In this solution, you start out the same as solution one. Convert everything into the fractions of largest denominator terms (this is necessary) until you get
$27x=11z$ $27y=13z$
While solving, make sure to leave a list of numbers that must divide $x$ $y$ , and $z$ . For example, just by looking at the basic fractions you receive from writing the starting equations, 24 divides $z$ , 8 divides $y$ and $x$ . In the expression above, it's also clear that 27 divides $z$ , 13 divides $y$ , and 11 divides $z$ . You might be wondering why I wrote the expression in terms of z. That's because $z$ has the largest divisor. The LCM of all it's divisors shows that $z$ must be divisible by 216. The total amount of bananas can be found to equal to $17z/9$ . This means there are two possible solutions under 1000: 408 and 816. Trial and error can be done quickly to find the smallest possible solution to be $\boxed{408}$

Solution: Let $A,B,C$ be the fraction of bananas taken by the first, second, and third monkeys respectively. Then we have the system of equations \[\frac{3}{4}A+\frac{3}{8}B+\frac{11}{24}C=\frac{1}{2}\] \[\frac{1}{8}A+\frac{1}{4}B+\frac{11}{24}C=\frac{1}{3}\] \[\frac{1}{8}A+\frac{3}{8}B+\frac{2}{24}C=\frac{1}{6}.\] Solve this your favorite way to get that \[(A,B,C)=\left( \frac{11}{51}, \frac{13}{51}, \frac{9}{17} \right).\] We need the amount taken by the first and second monkeys to be divisible by 8 and the third by 24 (but for the third, we already have divisibility by 9). Thus our minimum is $8 \cdot 51 = \boxed{408}.$


Answer: 108.0

Three concentric circles centered at $O$ have radii of $1$ $2$ , and $3$ . Points $B$ and $C$ lie on the largest circle. The region between the two smaller circles is shaded, as is the portion of the region between the two larger circles bounded by central angle $BOC$ , as shown in the figure below. Suppose the shaded and unshaded regions are equal in area. What is the measure of $\angle{BOC}$ in degrees?
[asy] size(150); import graph;  draw(circle((0,0),3)); real radius = 3; real angleStart = -54;  // starting angle of the sector real angleEnd = 54;  // ending angle of the sector label("$O$",(0,0),W); pair O = (0, 0); filldraw(arc(O, radius, angleStart, angleEnd)--O--cycle, gray); filldraw(circle((0,0),2),gray); filldraw(circle((0,0),1),white); draw((1.763,2.427)--(0,0)--(1.763,-2.427)); label("$B$",(1.763,2.427),NE); label("$C$",(1.763,-2.427),SE);  [/asy] $\textbf{(A) } 108\qquad\textbf{(B) } 120\qquad\textbf{(C) } 135\qquad\textbf{(D) } 144\qquad\textbf{(E) } 150$

Solution:
Let $x=\angle{BOC}$
We see that the shaded region is the inner ring plus a sector $x^\circ$ of the outer ring. The area of this in terms of $x$ is $\left( 4 \pi - \pi \right)+\frac{x}{360} \left( 9 \pi - 4 \pi \right)$ . This simplifies to $3 \pi + \frac{x}{360}(5 \pi)$
Also, the unshaded portion is comprised of the smallest circle plus the sector $(360-x)^\circ$ of the outer ring. The area of this is $\pi + \frac{360-x}{360}(5 \pi)$
We are told these are equal, therefore $\pi + \frac{x}{360}(5 \pi) = 3 \pi + \frac{360-x}{360}(5 \pi)$ . Solving for $x$ reveals $x=\boxed{108}$

Solution:
Notice for the 3rd most outer ring of the circle, the ratio of the shaded region to non-shaded region is the ratio of $\angle{BOC}$ to $360-\angle{BOC}$ . With that, all we need to do is solve for the shaded region.
The inner most circle has radius $1$ , and the second circle has radius 2. Therefore, the first shaded area has $4 \pi - \pi = 3 \pi$ area. The circle has total area $9 \pi$ , so the other shaded region must have $1.5 \pi$ area, as the non-shaded and shaded area is equivalent. So for the 3rd outer ring, the total area is $9 \pi - 4 \pi = 5 \pi$ , so the non-shaded part of the outer ring is $5 \pi - 1.5 \pi = 3.5 \pi$
Now as said before, the ratio of these two areas is the ratio of $\angle{BOC}$ and $360 - \angle{BOC}$ . So, $\frac{3.5}{1.5} = \frac{7}{3}$ . We have $7x:3x$ where $7x+3x = 360$ $x = 36$ , so our answer is $3x = 108, \boxed{108}$

Solution: The AMC 8s allow a ruler tool that you can rotate and drag. You can use the tool to make a straight segment (which we know is $180$ degrees), and we let the angle of desire be $x$ . We can estimate that $180-x$ is just about $30$ degrees short of $x$ itself, so $x-30=180-x$ , solving gives $x=105$ , therefore the closest answer is $\boxed{108}$

Solution: Suppose the desired angle is some fraction $x$ of the total degree measure of the circle. We now compile a list of the shaded and unshaded areas. The inner circle of radius $1$ is completely unshaded, so it contributes $1$ to the unshaded area. (Everything will be a multiple of $\pi$ , so we omit it.) The inner annulus has area $2^2 - 1^2 = 3$ , which it contributes to the shaded area. The outer annulus has a total area of $3^2 - 2^2 = 5$ ; the fraction $x$ is shaded, so the shaded portion of the outer annulus contributes $5x$ to the shaded area, while the other $1 - x$ fraction is unshaded, so the unshaded portion contributes $5(1-x)$ to the unshaded area. We now equate and solve. \[1 + 5(1-x) = 3 + 5x\] Upon solving, we find that $x = \frac{3}{10}$ , so the degree measure is $360 \cdot \frac{3}{10} = \boxed{108}$


Answer: 41.0

Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$

Solution:
Reinterpret the problem in the following manner. Equilateral triangle $ABC$ has a point $X$ on the interior such that $AX = 5,$ $BX = 4,$ and $CX = 3.$ $60^\circ$ counter-clockwise rotation about vertex $A$ maps $X$ to $X'$ and $B$ to $C.$
Note that angle $XAX'$ is $60$ and $XA = X'A = 5$ which tells us that triangle $XAX'$ is equilateral and that $XX' = 5.$ We now notice that $XC = 3$ and $X'C = 4$ which tells us that angle $XCX'$ is $90$ because there is a $3$ $4$ $5$ Pythagorean triple. Now note that $\angle ABC + \angle ACB = 120^\circ$ and $\angle XCA + \angle XBA = 90^\circ,$ so $\angle XCB+\angle XBC = 30^\circ$ and $\angle BXC = 150^\circ.$ Applying the law of cosines on triangle $BXC$ yields
\[BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos150^\circ = 4^2+3^2-24 \cdot \frac{-\sqrt{3}}{2} = 25+12\sqrt{3}\]
and thus the area of $ABC$ equals \[\frac{\sqrt{3}}{4}\cdot BC^2 = 25\frac{\sqrt{3}}{4}+9.\]
so our final answer is $3+4+25+9 = \boxed{041}.$

Solution: Let's call the circle center $X$ . It has a distance of 3, 4, 5 to an equilateral triangle $LMN$ . Consider $X$ ’s pedal triangle $ABC$ . Since $X$ ’s antipedal triangle is equilateral, $X$ must be the one of the isogonic centers of $\triangle{ABC}$ . We’ll take the one inside $ABC$ , i.e., the Fermat point, because it leads to larger $\triangle LMN$ . Now we construct the three equilateral triangles $ABD$ $ACE$ , and $BCF$ , the same way the Fermat point is constructed. Then we have $\angle DXE = \angle EXF = \angle FXE = 120$ . Since $AEMCX$ is concyclic with $XM$ =4 as diameter, we have $AC=4\sin(60)$ . Similarly, $AB=3\sin(60)$ , and $BC=5\sin(60)$ . So $\triangle ABC$ is a 3-4-5 right triangle with $\angle BAC=90$ . With some more angle chasing we get \[\angle MXC+\angle LXB = \angle MAC + \angle LAB = 180 – \angle BAC = 90\] \[\angle LXM = 360 – (\angle MXC + \angle LXB + \angle BXC) = 360 –(90+120)=150\] By Law of Cosines, we have \[LM^2 = 3^2+4^2-2*3*4\cos(150)=25+12\sqrt 3\] And the area follows. \[[LMN] = \frac{25}{4}\sqrt{3} + 9; \boxed{041}.\] By Mathdummy

Solution: We have $x=3$ $y=4$ , and $z=5$ . Because $AD=m$ is the median of $\triangle AOB$ , by Stewart's Theorem we have \[m^2=\frac 12 (x^2+y^2)-\frac 14 \cdot z^2\quad \Rightarrow \quad m = \frac 52.\] Because $CD$ is the altitude of equilateral triangle $OBC$ , we have $CD=\frac{\sqrt{3}}2\cdot z$ . Then in $\triangle ADC$ , we have $\angle ADC=\varphi+90^\circ$ , so $\cos(\angle ADC)=-\sin\varphi$ , and the Law of Cosines gives \[s^2=m^2+\frac 34\cdot z^2 + mz\sqrt{3}\sin\varphi = 25 \left(1+\frac {\sqrt{3}}{2}\cdot\sin\varphi  \right)\] To calculate $\sin\phi$ we apply the Law of Cosines to $\triangle ADO$ to get \[mz\cos\varphi = m^2+\frac 14\cdot z^2 - x^2 \quad \Rightarrow \quad \cos\varphi = \frac 7{25}\quad \Rightarrow \quad \sin\varphi = \frac {24}{25}.\] Finally, we get $s^2=25+12\sqrt{3}$ and and thus the area of $\triangle ABC$ equals \[\frac{\sqrt{3}}{4}\cdot s^2 = 25\frac{\sqrt{3}}{4}+9.\] so our final answer is $3+4+25+9 = \boxed{041}.$


Answer: 32.0

Three congruent circles with centers $P$ $Q$ , and $R$ are tangent to the sides of rectangle $ABCD$ as shown.  The circle centered at $Q$ has diameter $4$ and passes through points $P$ and $R$ .  The area of the rectangle is
[asy] pair A,B,C,D,P,Q,R; A = (0,4); B = (8,4); C = (8,0); D = (0,0); P = (2,2); Q = (4,2); R = (6,2); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(R); draw(A--B--C--D--cycle); draw(circle(P,2)); draw(circle(Q,2)); draw(circle(R,2)); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$P$",P,W); label("$Q$",Q,W); label("$R$",R,W); [/asy]
$\text{(A)}\ 16 \qquad \text{(B)}\ 24 \qquad \text{(C)}\ 32 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 128$

Solution:
If circle $Q$ has diameter $4$ , then so do congruent circles $P$ and $R$ .  Draw a diameter through $P$ parallel to $AD$ .  The diameter will be congruent to $AD$ , and thus $AD = 4$ , which is the height of the rectangle.
Draw a horizontal line $PQR$ that extends to the sides of the rectangle.  This line is $2$ diameters long, so it has length $4\cdot 2 = 8$ .  It is parallel and congruent to $AB$ , so the width of the rectangle is $8$
Thus, the area of the rectangle is $lw = 4\cdot 8 = 32$ , and the answer is $\boxed{32}$

Solution: The area of circle $P$ and circle $R$ almost fill the rectangle.  Circle $P$ has radius $\frac{4}{2} = 2$ , and so does circle $R$ .  Thus, the sum of their areas is $\pi \cdot 2^2 + \pi \cdot 2^2 = 8\pi$ .  Since $\pi \approx 3.14$ , the area of the circles is just over $24$ .  The area of the rectangle is greater than the area of the circles. $64$ would be too high, as the two circles appear to take up much more than half the area of the rectangle.  But $24$ is too low, as it is less than the area of the two circles.  Thus, the only reasonable answer is $\boxed{32}$ , which is $32$


Answer: 164.0

Three cubes are each formed from the pattern shown. They are then stacked on a table one on top of another so that the $13$ visible numbers have the greatest possible sum. What is that sum?
[asy] unitsize(.8cm);  pen p = linewidth(1); draw(shift(-2,0)*unitsquare,p); label("1",(-1.5,0.5)); draw(shift(-1,0)*unitsquare,p); label("2",(-0.5,0.5)); draw(unitsquare,p); label("32",(0.5,0.5)); draw(shift(1,0)*unitsquare,p); label("16",(1.5,0.5)); draw(shift(0,1)*unitsquare,p); label("4",(0.5,1.5)); draw(shift(0,-1)*unitsquare,p); label("8",(0.5,-0.5)); [/asy]
$\mathrm{(A)}\ 154\qquad\mathrm{(B)}\ 159\qquad\mathrm{(C)}\ 164\qquad\mathrm{(D)}\ 167\qquad\mathrm{(E)}\ 189$

Solution: Conversely, maximize the sum. Two cubes have 4 exposed faces. Since $32>16+8+4+2$ , 32 must be on the side. There are two distinct (asymmetrical) configurations with 32 on the side, but $(32,16,2,1)$ is the greatest at 51. There are 2 such cubes, so 51*2. The top cube has one unexposed face, so use 1 as the unexposed face. $2(51)+32+16+8+4+2=164 \implies \boxed{164}$


Answer: 26.0

Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?
[asy] path cell=((0,0)--(1,0)--(1,1)--(0,1)--cycle); path sw=((0,0)--(1,sqrt(3))); path se=((5,0)--(4,sqrt(3))); draw(cell, linewidth(1)); draw(shift(2,0)*cell, linewidth(1)); draw(shift(4,0)*cell, linewidth(1)); draw(shift(1,3)*cell, linewidth(1)); draw(shift(3,3)*cell, linewidth(1)); draw(shift(2,6)*cell, linewidth(1)); draw(shift(0.45,1.125)*sw, EndArrow); draw(shift(2.45,1.125)*sw, EndArrow); draw(shift(1.45,4.125)*sw, EndArrow); draw(shift(-0.45,1.125)*se, EndArrow); draw(shift(-2.45,1.125)*se, EndArrow); draw(shift(-1.45,4.125)*se, EndArrow); label("$+$", (1.5,1.5)); label("$+$", (3.5,1.5)); label("$+$", (2.5,4.5));[/asy]
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 24\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 35$

Solution:
If the lower cells contain $A, B$ and $C$ , then the second row will contain $A + B$ and $B + C$ , and the top cell will contain $A + 2B + C$ . To obtain the
smallest sum, place $1$ in the center cell and $2$ and $3$ in the outer ones. The top
number will be $7$ . For the largest sum, place $9$ in the center cell and $7$ and $8$ in
the outer ones. This top number will be $33$ . The difference is $33 - 7 = \boxed{26}$


Answer: 47.0

Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?
$\mathrm{(A)}\ \frac{1}{4} \qquad \mathrm{(B)}\ \frac{3}{8} \qquad \mathrm{(C)}\ \frac{4}{7} \qquad \mathrm{(D)}\ \frac{5}{7} \qquad \mathrm{(E)}\ \frac{3}{4}$

Solution:
We will try to use symmetry as much as possible.
Pick the first vertex $A$ , its choice clearly does not influence anything.
Pick the second vertex $B$ . With probability $3/7$ vertices $A$ and $B$ have a common edge, with probability $3/7$ they are in opposite corners of the same face, and with probability $1/7$ they are in opposite corners of the cube. We will handle each of the cases separately.
In the first case, there are $2$ faces that contain the edge $AB$ . In each of these faces there are $2$ other vertices. If one of these $4$ vertices is the third vertex $C$ , the entire triangle $ABC$ will be on a face. On the other hand, if $C$ is one of the two remaining vertices, the triangle will contain points inside the cube. Hence in this case the probability of choosing a good $C$ is $2/6 = 1/3$
In the second case, the triangle $ABC$ will not intersect the cube if point $C$ is one of the two points on the side that contains $AB$ . Hence the probability of $ABC$ intersecting the inside of the cube is $2/3$
In the third case, already the diagonal $AB$ contains points inside the cube, hence this case will be good regardless of the choice of $C$
Summing up all cases, the resulting probability is: \[\frac 37\cdot\frac 13 + \frac 37\cdot \frac 23 + \frac 17\cdot 1 = \boxed{47}\]

Solution:
There are $\binom{8}{3}=56$ ways to pick three vertices from eight total vertices; this is our denominator. In order to have three points inside the cube, they cannot be on the surface. Thus, we can use complementary probability.
There are $\binom{4}{3}=4$ ways to choose three points from the vertices of a single face. Since there are six faces, $4 \times 6 = 24$
Thus, the probability of what we don't want is $\frac{24}{56} = \frac{3}{7}$ . Using complementary probability,
\[1- \frac 37 = \boxed{47}\]


Answer: 6.0

Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34$ . What is the distance between two adjacent parallel lines?
$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$

Solution:
Since two parallel chords have the same length ( $38$ ), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$ . Thus, the distance from the center of the circle to the chord of length $34$ is
\[2d + d = 3d\]
and the distance between each of the chords is just $2d$ . Let the radius of the circle be $r$ . Drawing radii to the points where the lines intersect the circle, we create two different right triangles:
By the Pythagorean theorem, we can create the following system of equations:
\[19^2 + d^2 = r^2\]
\[17^2 + (2d + d)^2 = r^2\]
Solving, we find $d = 3$ , so $2d = \boxed{6}$

Solution: If $d$ is the requested distance, and $r$ is the radius of the circle, Stewart's Theorem applied to $\triangle OCD$ with cevian $\overleftrightarrow{OP}$ gives \[19\cdot 38\cdot 19 + \tfrac{1}{2}d\cdot 38\cdot\tfrac{1}{2}d=19r^{2}+19r^{2}.\] This simplifies to $13718+\tfrac{19}{2}d^{2}=38r^{2}$ . Similarly, another round of Stewart's Theorem applied to $\triangle OEF$ with cevian $\overleftrightarrow{OQ}$ gives \[17\cdot 34\cdot 17 + \tfrac{3}{2}d\cdot 34\cdot\tfrac{3}{2}d=17r^{2}+17r^{2}.\] This simplifies to $9826+\tfrac{153}{2}d^{2}=34r^{2}$ . Dividing the top equation by $38$ and the bottom equation by $34$ results in the system of equations \begin{align*} 361+\tfrac{1}{4}d^{2} &= r^{2} \\ 289+\tfrac{9}{4}d^{2} &= r^{2} \\ \end{align*} By transitive, $361+\tfrac{1}{4}d^{2}=289+\tfrac{9}{4}d^{2}$ . Therefore $(\tfrac{9}{4}-\tfrac{1}{4})d^{2}=361-289\rightarrow 2d^{2}=72\rightarrow d^{2}=36\rightarrow d=\boxed{6}.$


Answer: 6.0

Solution:
Since two parallel chords have the same length ( $38$ ), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$ . Thus, the distance from the center of the circle to the chord of length $34$ is
\[2d + d = 3d\]
and the distance between each of the chords is just $2d$ . Let the radius of the circle be $r$ . Drawing radii to the points where the lines intersect the circle, we create two different right triangles:
By the Pythagorean theorem, we can create the following system of equations:
\[19^2 + d^2 = r^2\]
\[17^2 + (2d + d)^2 = r^2\]
Solving, we find $d = 3$ , so $2d = \boxed{6}$

Solution: If $d$ is the requested distance, and $r$ is the radius of the circle, Stewart's Theorem applied to $\triangle OCD$ with cevian $\overleftrightarrow{OP}$ gives \[19\cdot 38\cdot 19 + \tfrac{1}{2}d\cdot 38\cdot\tfrac{1}{2}d=19r^{2}+19r^{2}.\] This simplifies to $13718+\tfrac{19}{2}d^{2}=38r^{2}$ . Similarly, another round of Stewart's Theorem applied to $\triangle OEF$ with cevian $\overleftrightarrow{OQ}$ gives \[17\cdot 34\cdot 17 + \tfrac{3}{2}d\cdot 34\cdot\tfrac{3}{2}d=17r^{2}+17r^{2}.\] This simplifies to $9826+\tfrac{153}{2}d^{2}=34r^{2}$ . Dividing the top equation by $38$ and the bottom equation by $34$ results in the system of equations \begin{align*} 361+\tfrac{1}{4}d^{2} &= r^{2} \\ 289+\tfrac{9}{4}d^{2} &= r^{2} \\ \end{align*} By transitive, $361+\tfrac{1}{4}d^{2}=289+\tfrac{9}{4}d^{2}$ . Therefore $(\tfrac{9}{4}-\tfrac{1}{4})d^{2}=361-289\rightarrow 2d^{2}=72\rightarrow d^{2}=36\rightarrow d=\boxed{6}.$


Answer: 1150.0

Three flower beds overlap as shown. Bed A has 500 plants, bed B has 450 plants, and bed C has 350 plants. Beds A and B share 50 plants, while beds A and C share 100. The total number of plants is
[asy] draw((0,0)--(3,0)--(3,1)--(0,1)--cycle); draw(circle((.3,-.1),.7)); draw(circle((2.8,-.2),.8)); label("A",(1.3,.5),N); label("B",(3.1,-.2),S); label("C",(.6,-.2),S); [/asy]
$\text{(A)}\ 850 \qquad \text{(B)}\ 1000 \qquad \text{(C)}\ 1150 \qquad \text{(D)}\ 1300 \qquad \text{(E)}\ 1450$

Solution:
Plants shared by two beds have been counted
twice, so the total is $500 + 450 + 350 - 50 - 100 = \boxed{1150}$

Solution:
Bed A has $350$ plants it doesn't
share with B or C. Bed B has $400$ plants it doesn't
share with A or C. And C has $250$ it doesn't share
with A or B. The total is $350 + 400 + 250 + 50 + 100 = \boxed{1150}$ plants.


Answer: 15.0

Three fourths of a pitcher is filled with pineapple juice. The pitcher is emptied by pouring an equal amount of juice into each of $5$ cups. What percent of the total capacity of the pitcher did each cup receive?
$\textbf{(A) }5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }15 \qquad \textbf{(D) }20 \qquad \textbf{(E) }25$

Solution: Each cup is filled with $\frac{3}{4} \cdot \frac{1}{5} = /frac{3}{20}$ of the amount of juice in the pitcher, so the percentage is $\frac{3}{20} \cdot 100 = \boxed{15}$

Solution: The pitcher is $\frac{3}{4}$ full, i.e. $75\%$ full. Therefore each cup receives $\frac{75}{5}=\boxed{15}$ percent of the total capacity.

Solution: Assume that the pitcher has a total capacity of $100$ ounces. Since it is filled three fourths with pineapple juice, it contains $75$ ounces of pineapple juice, which means that each cup will contain $\frac{75}{5}=15$ ounces of pineapple juice. Since the total capacity of the pitcher was $100$ ounces, it follows that each cup received $15\%$ of the total capacity of the pitcher, yielding $\boxed{15}$ as the answer.


Answer: 4.0

Three friends have a total of $6$ identical pencils, and each one has at least one pencil. In how many ways can this happen?
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12$

Solution:
Like in solution 1 and solution 2, assign one pencil to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups use casework.
Let the three friends be $a$ $b$ $c$ repectively.
$a + b + c = 3$
Case $1:a=0$
$b + c = 3$
$b = 0,1,2,3$
$c = 3,2,1,0$
$\boxed{4}$ solutions.


Answer: 10.0

Solution: For each person to have at least one pencil, assign one pencil to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups, use Ball-and-urn to find the number of possibilities is $\binom{3+3-1}{3-1} = \binom{5}{2} = \boxed{10}$

Solution:
Like in solution 1, for each person to have at least one pencil, assign one of the pencils to each of the three friends so that you have $3$ left. In partitioning the remaining $3$ pencils into $3$ distinct groups, use number of non-negative integral solutions.
Let the three friends be $a, b, c$ respectively.
$a + b + c = 3$ The total being 3 and 2 plus signs, which implies $\binom{3+2}{2} = \binom{5}{2} = \boxed{10}$

Solution:
For each of the 3 People to have at least one pencil when distributing 6 pencil amongst them, we can use another formula from the Ball-and-urn counting technique, shown below:
for n = number of items, and s = slots:
Now we can plug in our values,
number of items = 6, and slots = 3:
$\binom{6-1}{3-1}     =     \binom{5}{2}     =    \boxed{10}$


Answer: 252.0

Three generous friends, each with some money, redistribute the money as followed:
Amy gives enough money to Jan and Toy to double each amount has.
Jan then gives enough to Amy and Toy to double their amounts.
Finally, Toy gives enough to Amy and Jan to double their amounts.
If Toy had 36 dollars at the beginning and 36 dollars at the end, what is the total amount that all three friends have?
$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 180\qquad\textbf{(C)}\ 216\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 288$

Solution:
If Toy had $36$ dollars at the beginning, then after Amy doubles his money, he has $36 \times 2 = 72$ dollars after the first step.
Then Jan doubles his money, and Toy has $72 \times 2 = 144$ dollars after the second step.
Then Toy doubles whatever Amy and Jan have.  Since Toy ended up with $36$ , he spent $144 - 36 = 108$ to double their money.  Therefore, just before this third step, Amy and Jan must have had $108$ dollars in total.  And, just before this step, Toy had $144$ dollars.  Altogether, the three had $144 + 108 = 252$ dollars, and the correct answer is $\boxed{252}$


Answer: 37.0

Three hexagons of increasing size are shown below. Suppose the dot pattern continues so that each successive hexagon contains one more band of dots. How many dots are in the next hexagon?
[asy] // diagram by SirCalcsALot, edited by MRENTHUSIASM size(250); path p = scale(0.8)*unitcircle; pair[] A; pen grey1 = rgb(100/256, 100/256, 100/256); pen grey2 = rgb(183/256, 183/256, 183/256); for (int i=0; i<7; ++i) { A[i] = rotate(60*i)*(1,0);} path hex = A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--cycle; fill(p,grey1); draw(scale(1.25)*hex,black+linewidth(1.25)); pair S = 6A[0]+2A[1]; fill(shift(S)*p,grey1); for (int i=0; i<6; ++i) { fill(shift(S+2*A[i])*p,grey2);} draw(shift(S)*scale(3.25)*hex,black+linewidth(1.25)); pair T = 16A[0]+4A[1]; fill(shift(T)*p,grey1); for (int i=0; i<6; ++i) {   fill(shift(T+2*A[i])*p,grey2);  fill(shift(T+4*A[i])*p,grey1);  fill(shift(T+2*A[i]+2*A[i+1])*p,grey1); } draw(shift(T)*scale(5.25)*hex,black+linewidth(1.25)); [/asy]
$\textbf{(A) }35 \qquad \textbf{(B) }37 \qquad \textbf{(C) }39 \qquad \textbf{(D) }43 \qquad \textbf{(E) }49$

Solution: Looking at the rows of each hexagon, we see that the first hexagon has $1$ dot, the second has $2+3+2$ dots, and the third has $3+4+5+4+3$ dots. Given the way the hexagons are constructed, it is clear that this pattern continues. Hence, the fourth hexagon has $4+5+6+7+6+5+4=\boxed{37}$ dots.

Solution:
The dots in the next hexagon have four bands. From innermost to outermost:
Together, the answer is $1+6+12+18=\boxed{37}.$

Solution: The first hexagon has $1$ dot, the second hexagon has $1+6$ dots, the third hexagon has $1+6+12$ dots, and so on. The pattern continues since to go from hexagon $n$ to hexagon $(n+1),$ we add a new band of dots around the outside of the existing ones, with each side of the band having side length $(n+1).$ Thus, the number of dots added is $6(n+1)-6 = 6n$ (we subtract $6$ as each of the corner hexagons in the band is counted as part of two sides.). We therefore predict that the fourth hexagon has $1+6+12+18=\boxed{37}$ dots.

Solution: From the full diagram below, the answer is $\boxed{37} draw(shift(R)*scale(7.25)*hex,black+linewidth(1.25)); [/asy] ~MRENTHUSIASM


Answer: 150.0

Three identical rectangles are put together to form rectangle $ABCD$ , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is  5 feet, what is the area in square feet of rectangle $ABCD$
[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]
$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$

Solution: We can see that there are $2$ rectangles lying on top of the other and that is the same as the length of one rectangle. Now we know that the shorter side is $5$ , so the bigger side is $10$ , if we do $5 \cdot 2 = 10$ . Now we get the sides of the big rectangle being $15$ and $10$ , so the area is $\boxed{150}$ . ~avamarora

Solution: Using the diagram we find that the larger side of the small rectangle is $2$ times the length of the smaller side. Therefore, the longer side is $5 \cdot 2 = 10$ . So the area of the identical rectangles is $5 \cdot 10 = 50$ . We have 3 identical rectangles that form the large rectangle. Therefore the area of the large rectangle is $50 \cdot 3 = \boxed{150}$ .    ~~fath2012

Solution:
We see that if the short sides are 5, the long side has to be $5\cdot2=10$ because the long side is equal to the 2 short sides and because the rectangles are congruent. If that is to be, then the long side of the BIG rectangle(rectangle $ABCD$ )
is $10+5=15$ because long side + short side of the small rectangle is $15$ . The short side of rectangle $ABCD$ is $10$ because it is the long side of the short rectangle. Multiplying $15$ and $10$ together gets us $15\cdot10$ which is $\boxed{150}$ .
~~mathboy282

Solution:
We see that the $2$ rectangles lying on top of each other give us the height of the rectangle. Using what we know, we can find out that the $2$ rectangles put together is a square. So, we can infer that the length of the rectangles is $10$ . Adding that to the width of the third rectangle which is $5$ , we get that the length of the rectangle is $15$ . Multiplying $10$ and $15$ gives us $15\cdot10$ which is $\boxed{150}$ .
~~awesomepony566

Solution:
There are two rectangles lying on the side and one standing up. Given that one small side is 5, we can determine that two of the small sides make up a big side which means that the long side is equal to 10. The top side of the rectangle is made up of one small side and one long side, therefore the dimensions for the rectangle is 10x15. 10 multiplied by 15 is 150, hence the answer $\boxed{150}$ .
~elenafan


Answer: 147.0

Three identical square sheets of paper each with side length $6$ are stacked on top of each other. The middle sheet is rotated clockwise $30^\circ$ about its center and the top sheet is rotated clockwise $60^\circ$ about its center, resulting in the $24$ -sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a-b\sqrt{c}$ , where $a$ $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a+b+c$
$(\textbf{A})\: 75\qquad(\textbf{B}) \: 93\qquad(\textbf{C}) \: 96\qquad(\textbf{D}) \: 129\qquad(\textbf{E}) \: 147$

Solution:
 The $24$ -sided polygon is made out of $24$ shapes like $\triangle ABC$ . Then $\angle BAC=360^\circ/24=15^\circ$ , and $\angle EAC = 45^\circ$ , so $\angle{EAB} = 30^{\circ}$ . Then $EB=AE\tan 30^\circ = \sqrt{3}$ ; therefore $BC=EC-EB=3-\sqrt{3}$ . Thus \[[ABC] = \frac{BC}{EC}\cdot [ACE] = \frac{3-\sqrt{3}}{3}\cdot \frac 92\] and the required area is $24\cdot[ABC] =108-36\sqrt{3}$ . Finally $108+36+3=\boxed{147}$ .
~lopkiloinm

Solution:
As shown in Image:2021_AMC_12B_(Nov)_Problem_15,_sol.png , all 12 vertices of three squares form a regular dodecagon (12-gon).
Denote by $O$ the center of this dodecagon.
Hence, $\angle AOB = \frac{360^\circ}{12} = 30^\circ$
Because the length of a side of a square is 6, $AO = 3 \sqrt{2}$
Hence, $AB = 2 AO \sin \frac{\angle AOB}{2} = 3 \left( \sqrt{3} - 1 \right)$
We notice that $\angle MAB = \angle MBA = 30^\circ$ .
Hence, $AM = \frac{AB}{2\cos \angle MAB} = 3 - \sqrt{3}$
Therefore, the area of the region that three squares cover is \begin{align*} & {\rm Area} \ ABCDEFGHIJKL - 12 {\rm Area} \ \triangle MAB \\ & = 12 {\rm Area} \ \triangle OAB - 12 {\rm Area} \ \triangle MAB \\ & = 12 \cdot \frac{1}{2} OA \cdot OB \sin \angle AOB - 12 \cdot \frac{1}{2} MA \cdot MB \sin \angle AMB \\ & = 6 OA^2 \sin \angle AOB - 6 MA^2 \sin \angle AMB \\ & = 108 - 36 \sqrt{3} . \end{align*}
Therefore, the answer is $\boxed{147}$