Answer: 192.0

Define a $\text{good~word}$ as a sequence of letters that consists only of the letters $A$ $B$ , and $C$ - some of these letters may not appear in the sequence - and in which $A$ is never immediately followed by $B$ $B$ is never immediately followed by $C$ , and $C$ is never immediately followed by $A$ . How many seven-letter good words are there?

Solution:
There are three letters to make the first letter in the sequence. However, after the first letter (whatever it is), only two letters can follow it, since one of the letters is restricted. Therefore, the number of seven-letter good words is $3*2^6=192$
Therefore, there are $\boxed{192}$ seven-letter good words.

Solution: There are three choices for the first letter and two choices for each subsequent letter, so there are $3\cdot2^{n-1}\ n$ -letter good words. Substitute $n=7$ to find there are $3\cdot2^6=\boxed{192}$ seven-letter good words. ~ aopsav (Credit to AoPS Alcumus)

Solution: We solve this problem using recursion. Let $f(x)$ be the number of $x$ -letter good words. Thus $f(1) = 3$ (A, B or C) and the answer is just $f(7)$ .  The recurrence relation can be found by considering the last letter of one of the valid strings of length $x - 1$ .  There are $2$ possibilities for the next letter and thus $f(x) = 2 \cdot f(x-1)$ .  Now we can find a closed form as $f(x) = 3 \cdot 2 ^{x-1}$ (easy to prove by induction) and thus $f(7) = 64 * 3 = \boxed{192}$ seven-letter good words.  ~AK2006


Answer: 68.0

Define a T-grid to be a $3\times3$ matrix which satisfies the following two properties:
Find the number of distinct T-grids

Solution:
The T-grid can be considered as a tic-tac-toe board: five $1$ 's (or X's) and four $0$ 's (or O's).
There are only $\dbinom{9}{5} = 126$ ways to fill the board with five $1$ 's and four $0$ 's. Now we just need to subtract the number of bad grids. Bad grids are ones with more than one person winning, or where someone has won twice.
Let three-in-a-row/column/diagonal be a "win" and let player $0$ be the one that fills in $0$ and player $1$ fills in $1$
Case $1$ Each player wins once.
If player X takes a diagonal, player Y cannot win. If either takes a row, all the columns are blocked, and visa versa. Therefore, they either both take a row or they both take a column.
Case $1$ $36$ cases
Case $2$ Player $1$ wins twice. (Player $0$ cannot win twice because he only has 4 moves.)
Case $2$ total: $22$
Thus, the answer is $126-22-36=\boxed{68}$

Solution:
We can use generating functions to compute the case that no row or column is completely filled with $1$ 's. Given a row, let $a$ $b$ $c$ be the events that the first, second, third respective squares are $1$ 's. Then the generating function representing the possible events that exclude a row of $1,1,1$ or $0,0,0$ from occuring is \[ab+bc+ca+a+b+c.\] Therefore, the generating function representing the possible grids where no row is filled with $0$ 's and $1$ 's is \[P(a,b,c)=((ab+bc+ca)+(a+b+c))^3.\] We expand this by the Binomial Theorem to find \[P(a,b,c)=(ab+bc+ca)^3+3(ab+bc+ca)^2(a+b+c)+3(ab+bc+ca)(a+b+c)^2+(a+b+c)^3.\] Recall that our grid has five $1$ 's, hence we only want terms where the sum of the exponents is $5$ . This is given by \[3(ab+bc+ca)^2(a+b+c).\] When we expand this, we find \[3(2abc(a+b+c)+a^2b^2+b^2c^2+c^2a^2)(a+b+c).\] We also want to make sure that each of $a$ $b$ $c$ appears at least once (so there is no column filled with $0$ 's) and the power of each of $a$ $b$ $c$ is not greater than or equal to $3$ (so there is no column filled with $1$ 's). The sum of the coefficients of the above polynomial is clearly $81$ (using $a,b,c=1$ ), and the sum of the coefficients of the terms $a^3bc$ $ab^3c$ , and $abc^3$ is $6+6+6+3+3+3+3+3+3=36$ , hence the sum of the coefficients of the desired terms is $81-36=45$ . This counts the number of grids where no column or row is filled with $0$ 's or $1$ 's. However, we could potentially have both diagonals filled with $1$ 's, but this is the only exception to our $45$ possibilities, hence the number of $T$ -grids with no row or column filled with the same digit is $44$
On the other hand, if a row (column) is filled with $0$ 's, then by the Pigeonhole Principle, another row (column) must be filled with $1$ 's. Hence this is impossible, so all other	possible $T$ -grids have a row (column) filled with $1$ 's. If the top row is filled with $1$ 's, then we have two $1$ 's left to place. Clearly they cannot go in the same row, because then the other row is filled with $0$ 's. They also cannot appear in the same column. This leaves $3\cdot 2$ arrangements--3 choices for the location of the $1$ in the second row, and 2 choices for the location of the $1$ in the last row. However, two of these arrangements will fill a diagonal with $1$ 's. Hence there are only $4$ $T$ -grids where the top row is filled with $1$ 's. The same argument applies if any other row or column is filled with $1$ 's. Hence there are $4\cdot 6=24$ such $T$ -grids.
Thus the answer is $44+24=\boxed{68}$


Answer: 761.0

Define a domino to be an ordered pair of distinct positive integers. A proper sequence of dominos is a list of distinct dominos in which the first coordinate of each pair after the first equals the second coordinate of the immediately preceding pair, and in which $(i,j)$ and $(j,i)$ do not both appear for any $i$ and $j$ .  Let $D_{40}$ be the set of all dominos whose coordinates are no larger than 40.  Find the length of the longest proper sequence of dominos that can be formed using the dominos of $D_{40}.$

Solution:
We can draw a comparison between the domino a set of 40 points (labeled 1 through 40) in which every point is connected with every other point. The connections represent the dominoes.
You need to have all even number of segments coming from each point except 0 or 2 which have an odd number of segments coming from the point. (Reasoning for this: Everytime you go to a vertex, you have to leave the vertex, so every vertex reached is equivalent to adding 2 more segments. So the degree of each vertex must be even, with the exception of endpoints) Since there are 39 segments coming from each point it is impossible to touch every segment.
But you can get up to 38 on each segment because you go in to the point then out on a different segment.  Counting going out from the starting and ending at the ending point we have:
$\frac{38\cdot 38 + 2\cdot 39}2 = \boxed{761}$


Answer: 2018.0

Define a function on the positive integers recursively by $f(1) = 2$ $f(n) = f(n-1) + 1$ if $n$ is even, and $f(n) = f(n-2) + 2$ if $n$ is odd and greater than $1$ . What is $f(2017)$
$\textbf{(A)}\ 2017 \qquad\textbf{(B)}\ 2018 \qquad\textbf{(C)}\ 4034 \qquad\textbf{(D)}\ 4035 \qquad\textbf{(E)}\ 4036$

Solution:
This is a recursive function, which means the function refers back to itself to calculate subsequent terms. To solve this, we must identify the base case, $f(1)=2$ . We also know that when $n$ is odd, $f(n)=f(n-2)+2$ . Thus we know that $f(2017)=f(2015)+2$ . Thus we know that n will always be odd in the recursion of $f(2017)$ , and we add $2$ each recursive cycle, which there are $1008$ of. Thus the answer is $1008*2+2=2018$ , which is answer $\boxed{2018}$ .
Note that when you write out a few numbers, you find that $f(n)=n+1$ for any $n$ , so $f(2017)=2018$


Answer: 396.0

Define a positive integer $n^{}_{}$ to be a factorial tail if there is some positive integer $m^{}_{}$ such that the decimal representation of $m!$ ends with exactly $n$ zeroes. How many positive integers less than $1992$ are not factorial tails?

Solution:
Let the number of zeros at the end of $m!$ be $f(m)$ . We have $f(m) = \left\lfloor \frac{m}{5} \right\rfloor + \left\lfloor \frac{m}{25} \right\rfloor + \left\lfloor \frac{m}{125} \right\rfloor + \left\lfloor \frac{m}{625} \right\rfloor + \left\lfloor \frac{m}{3125} \right\rfloor + \cdots$
Note that if $m$ is a multiple of $5$ $f(m) = f(m+1) = f(m+2) = f(m+3) = f(m+4)$
Since $f(m) \le \frac{m}{5} + \frac{m}{25} + \frac{m}{125} + \cdots  = \frac{m}{4}$ , a value of $m$ such that $f(m) = 1991$ is greater than $7964$ . Testing values greater than this yields $f(7975)=1991$
There are $\frac{7975}{5} = 1595$ distinct positive integers, $f(m)$ , less than $1992$ . Thus, there are $1991-1595 = \boxed{396}$ positive integers less than $1992$ that are not factorial tails.

Solution:
After testing various values of $m$ in $f(m)$ of solution 1 to determine $m$ for which $f(m) = 1992$ , we find that $m \in \{7980, 7981, 7982, 7983, 7984\}$ . WLOG, we select $7980$ . Furthermore, note that every time $k$ reaches a multiple of $25$ $k!$ will gain two or more additional factors of $5$ and will thus skip one or more numbers.
With this logic, we realize that the desired quantity is simply $\left \lfloor \frac{7980}{25} \right \rfloor + \left \lfloor \frac{7980}{125} \right \rfloor \cdots$ , where the first term accounts for every time $1$ number is skipped, the second term accounts for each time $2$ numbers are skipped, and so on. Evaluating this gives us $319 + 63 + 12 + 2 = \boxed{396}$ . - Spacesam(edited by srisainandan6)


Answer: 199.0

Define a regular $n$ -pointed star to be the union of $n$ line segments $P_1P_2, P_2P_3,\ldots, P_nP_1$ such that
There are no regular 3-pointed, 4-pointed, or 6-pointed stars. All regular 5-pointed stars are similar, but there are two non-similar regular 7-pointed stars. How many non-similar regular 1000-pointed stars are there?

Solution:
We use the Principle of Inclusion-Exclusion (PIE).
If we join the adjacent vertices of the  regular $n$ -star, we get a regular $n$ -gon. We number the vertices of this $n$ -gon in a counterclockwise direction: $0, 1, 2, 3, \ldots, n-1.$
A regular $n$ -star will be formed if we choose a vertex number $m$ , where $0 \le m \le n-1$ , and then form the line segments by joining the following pairs of vertex numbers: $(0 \mod{n}, m \mod{n}),$ $(m \mod{n}, 2m \mod{n}),$ $(2m \mod{n}, 3m \mod{n}),$ $\cdots,$ $((n-2)m \mod{n}, (n-1)m \mod{n}),$ $((n-1)m \mod{n}, 0 \mod{n}).$
If $\gcd(m,n) > 1$ , then the star degenerates into a regular $\frac{n}{\gcd(m,n)}$ -gon or a (2-vertex) line segment if $\frac{n}{\gcd(m,n)}= 2$ . Therefore, we need to find all $m$ such that $\gcd(m,n) = 1$
Note that $n = 1000 = 2^{3}5^{3}.$
Let $S = \{1,2,3,\ldots, 1000\}$ , and $A_{i}= \{i \in S \mid i\, \textrm{ divides }\,1000\}$ . The number of $m$ 's that are not relatively prime to $1000$ is: $\mid A_{2}\cup A_{5}\mid = \mid A_{2}\mid+\mid A_{5}\mid-\mid A_{2}\cap A_{5}\mid$ $= \left\lfloor \frac{1000}{2}\right\rfloor+\left\lfloor \frac{1000}{5}\right\rfloor-\left\lfloor \frac{1000}{2 \cdot 5}\right\rfloor$ $= 500+200-100 = 600.$
Vertex numbers $1$ and $n-1=999$ must be excluded as values for $m$ since otherwise a regular n-gon, instead of an n-star, is formed.
The cases of a 1st line segment of (0, m) and (0, n-m) give the same star. Therefore we should halve the count to get non-similar stars.
Therefore, the number of non-similar 1000-pointed stars is $\frac{1000-600-2}{2}= \boxed{199}.$


Answer: 9.0

Define a sequence recursively by $F_{0}=0,~F_{1}=1,$ and $F_{n}=$ the remainder when $F_{n-1}+F_{n-2}$ is divided by $3,$ for all $n\geq 2.$ Thus the sequence starts $0,1,1,2,0,2,\ldots$ What is $F_{2017}+F_{2018}+F_{2019}+F_{2020}+F_{2021}+F_{2022}+F_{2023}+F_{2024}?$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

Solution: A pattern starts to emerge as the function is continued. The repeating pattern is $0,1,1,2,0,2,2,1\ldots$ The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating pattern, we just need to find the sum of the numbers in the sequence, which is $\boxed{9}$


Answer: 101.0

Define a sequence recursively by $f_1(x)=|x-1|$ and $f_n(x)=f_{n-1}(|x-n|)$ for integers $n>1$ . Find the least value of $n$ such that the sum of the zeros of $f_n$ exceeds $500,000$

Solution:
First it will be shown by induction that the zeros of $f_n$ are the integers $a, {a+2,} {a+4,} \dots, {a + n(n-1)}$ , where $a = n - \frac{n(n-1)}2.$
This is certainly true for $n=1$ . Suppose that it is true for $n = m-1 \ge 1$ , and note that the zeros of $f_m$ are the solutions of $|x - m| = k$ , where $k$ is a nonnegative zero of $f_{m-1}$ . Because the zeros of $f_{m-1}$ form an arithmetic sequence with common difference $2,$ so do the zeros of $f_m$ . The greatest zero of $f_{m-1}$ is \[m-1+\frac{(m-1)(m-2)}2 =\frac{m(m-1)}2,\] so the greatest zero of $f_m$ is $m+\frac{m(m-1)}2$ and the least is $m-\frac{m(m-1)}2$
It follows that the number of zeros of $f_n$ is $\frac{n(n-1)}2+1=\frac{n^2-n+2}2$ , and their average value is $n$ . The sum of the zeros of $f_n$ is \[\frac{n^3-n^2+2n}2.\] Let $S(n)=n^3-n^2+2n$ , so the sum of the zeros exceeds $500000$ if and only if $S(n) > 1000000 = 100^3\!.$ Because $S(n)$ is increasing for $n > 2$ , the values $S(100) = 1000000 - 10000 + 200 = 990200$ and $S(101)=1030301 - 10201 + 202 = 1020302$ show that the requested value of $n$ is $\boxed{101}$

Solution:
Starting from $f_1(x)=|x-1|$ , we can track the solutions, the number of solutions, and their sum.
\[\begin{array}{c|c|c|c} n&Solutions&number&sum\\  1&1&1&1\\   2&1,3&2&4\\    3&0,2,4,6&4&12\\ 4&-2,0,2...10&7&28\\ 5&-5,-3,-1...15&11&55\\ \end{array}\]
It is clear that the solutions form arithmetic sequences with a difference of 2, and the negative solutions cancel out all but $n$ of the $1+\frac{n(n-1)}{2}$ solutions. Thus, the sum of the solutions is $n \cdot [1+\frac{n(n-1)}{2}]$ , which is a cubic function.
$n \cdot [1+\frac{n(n-1)}{2}]>500,000$
Multiplying both sides by $2$
$n \cdot [2+n(n-1)]>1,000,000$
1 million is $10^6=100^3$ , so the solution should be close to $100$
100 is slightly too small, so $\boxed{101}$ works.


Answer: 626.0

Define a sequence recursively by $t_1 = 20$ $t_2 = 21$ , and \[t_n = \frac{5t_{n-1}+1}{25t_{n-2}}\] for all $n \ge 3$ . Then $t_{2020}$ can be expressed as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$

Solution:
Let $t_n=\frac{s_n}{5}$ . Then, we have $s_n=\frac{s_{n-1}+1}{s_{n-2}}$ where $s_1 = 100$ and $s_2 = 105$ . By substitution, we find $s_3 = \frac{53}{50}$ $s_4=\frac{103}{105\cdot50}$ $s_5=\frac{101}{105}$ $s_6=100$ , and $s_7=105$ . So $s_n$ has a period of $5$ . Thus $s_{2020}=s_5=\frac{101}{105}$ . So, $\frac{101}{105\cdot 5}\implies 101+525=\boxed{626}$ .
~mn28407

Solution:
Let us examine the first few terms of this sequence and see if we can find a pattern. We are obviously given $t_1 = 20$ and $t_2 = 21$ , so now we are able to determine the numerical value of $t_3$ using this information: \[t_3 = \frac{5t_{3-1}+1}{25t_{3-2}} = \frac{5t_{2}+1}{25t_{1}} = \frac{5(21) + 1}{25(20)} = \frac{105 + 1}{500}t_3 = \frac{106}{500}  = \frac{53}{250}\] \[t_4 = \frac{5t_{4-1}+1}{25t_{4-2}} = \frac{5t_{3}+1}{25t_{2}} = \frac{5(\frac{53}{250}) + 1}{25(21)} = \frac{\frac{53}{50} + 1}{525} = \frac{\frac{103}{50}}{525} = \frac{103}{26250}\] \[t_5 = \frac{5t_{5-1}+1}{25t_{5-2}} = \frac{5t_{4}+1}{25t_{3}} = \frac{5(\frac{103}{26250}) + 1}{25(\frac{53}{250})} = \frac{\frac{103}{5250} + 1}{\frac{53}{10}} = \frac{\frac{5353}{5250}}{\frac{53}{10}} = \frac{101}{525}\] \[t_6 = \frac{5t_{6-1}+1}{25t_{6-2}} = \frac{5t_{5}+1}{25t_{4}} = \frac{5(\frac{101}{525}) + 1}{25(\frac{103}{26250})} = \frac{\frac{101}{105} + 1}{\frac{103}{1050}} = \frac{\frac{206}{105}}{\frac{103}{1050}} \implies t_6 = 20\]
Alas, we have figured this sequence is period 5! But since $2020 \equiv 5 \pmod 5$ , we can state that $t_{2020} = t_5 = \frac{101}{525}$ . According to the original problem statement, our answer is $\boxed{626}$ . ~ nikenissan


Answer: 81242.0

Define a sequence recursively by $x_0=5$ and \[x_{n+1}=\frac{x_n^2+5x_n+4}{x_n+6}\] for all nonnegative integers $n.$ Let $m$ be the least positive integer such that
\[x_m\leq 4+\frac{1}{2^{20}}.\]
In which of the following intervals does $m$ lie?
$\textbf{(A) } [9,26] \qquad\textbf{(B) } [27,80] \qquad\textbf{(C) } [81,242]\qquad\textbf{(D) } [243,728] \qquad\textbf{(E) } [729,\infty)$

Solution:
We first prove that $x_n > 4$ for all $n \ge 0$ , by induction. Observe that \[x_{n+1} - 4 = \frac{x_n^2 + 5x_n + 4 - 4(x_n+6)}{x_n+6} = \frac{(x_n - 4)(x_n+5)}{x_n+6}.\] so (since $x_n$ is clearly positive for all $n$ , from the initial definition), $x_{n+1} > 4$ if and only if $x_{n} > 4$
We similarly prove that $x_n$ is decreasing: \[x_{n+1} - x_n = \frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \frac{4-x_n}{x_n+6} < 0.\]
Now we need to estimate the value of $x_{n+1}-4$ , which we can do using the rearranged equation: \[x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6}.\] Since $x_n$ is decreasing, $\frac{x_n + 5}{x_n+6}$ is also decreasing, so we have \[\frac{9}{10} < \frac{x_n + 5}{x_n+6} \le \frac{10}{11}\] and \[\frac{9}{10}(x_n-4) < x_{n+1} - 4 \le \frac{10}{11}(x_n-4).\]
This becomes \[\left(\frac{9}{10}\right)^n = \left(\frac{9}{10}\right)^n \left(x_0-4\right) < x_{n} - 4 \le \left(\frac{10}{11}\right)^n \left(x_0-4\right) = \left(\frac{10}{11}\right)^n.\] The problem thus reduces to finding the least value of $n$ such that \[\left(\frac{9}{10}\right)^n < x_{n} - 4 \le \frac{1}{2^{20}} \text{  and  }  \left(\frac{10}{11}\right)^{n-1} > x_{n-1} - 4 > \frac{1}{2^{20}}.\]
Taking logarithms, we get $n \ln \frac{9}{10} < -20 \ln 2$ and $(n-1)\ln \frac{10}{11} > -20 \ln 2$ , i.e.
\[n > \frac{20\ln 2}{\ln\frac{10}{9}} \text{  and  }  n-1 < \frac{20\ln 2}{\ln\frac{11}{10}} .\]
As approximations, we can use $\ln\frac{10}{9} \approx \frac{1}{9}$ $\ln\frac{11}{10} \approx \frac{1}{10}$ , and $\ln 2\approx 0.7$ . These approximations allow us to estimate \[126 < n < 141,\] which gives $\boxed{81,242}$

Solution:
The condition where $x_m\leq 4+\frac{1}{2^{20}}$ gives the motivation to make a substitution to change the equilibrium from $4$ to $0$ . We can substitute $x_n = y_n + 4$ to achieve that. Now, we need to find the smallest value of $m$ such that $y_m\leq \frac{1}{2^{20}}$ given that $y_0 = 1$
Factoring the recursion $x_{n+1} = \frac{x_n^2 + 5x_n+4}{x_n + 6}$ , we get:
$x_{n+1}=\dfrac{(x_n + 4)(x_n + 1)}{x_n + 6} \Rightarrow y_{n+1}+4=\dfrac{(y_n+8)(y_n+5)}{y_n+10}$
$y_{n+1}+4=\dfrac{y_n^2+13y_n+40}{y_n+10} = \dfrac{y_n^2+9y_n +(4y_n+40)}{y_n+10}$
$y_{n+1}+4=\dfrac{y_n^2+9y_n}{y_n+10} + 4$
$y_{n+1}=\dfrac{y_n^2+9y_n}{y_n+10}$
Using wishful thinking, we can simplify the recursion as follows:
$y_{n+1} = \frac{y_n^2 + 9y_n + y_n - y_n}{y_n + 10}$
$y_{n+1} = \frac{y_n(y_n + 10) - y_n}{y_n + 10}$
$y_{n+1} = y_n - \frac{y_n}{y_n + 10}$
$y_{n+1} = y_n\left(1 - \frac{1}{y_n + 10}\right)$
The recursion looks like a geometric sequence with the ratio changing slightly after each term. Notice from the recursion that the $y_n$ sequence is strictly decreasing, so all the terms after $y_0$ will be less than 1. Also, notice that all the terms in sequence will be positive. Both of these can be proven by induction.
With both of those observations in mind, $\frac{9}{10} < 1 - \frac{1}{y_n + 10} \leq \frac{10}{11}$ . Combining this with the fact that the recursion resembles a geometric sequence, we conclude that $\left(\frac{9}{10}\right)^n < y_n \leq \left(\frac{10}{11}\right)^n.$
$\frac{9}{10}$ is approximately equal to $\frac{10}{11}$ and the ranges that the answer choices give us are generous, so we should use either $\frac{9}{10}$ or $\frac{10}{11}$ to find a rough estimate for $m$
Since $\dfrac{1}{2}=0.5$ , that means $\frac{1}{\sqrt{2}}=2^{-\frac{1}{2}} \approx 0.7$ . Additionally, $\left(\frac{9}{10}\right)^3=0.729$
Therefore, we can estimate that $2^{-\frac{1}{2}} < y_3$
Raising both sides to the 40th power, we get $2^{-20} < (y_3)^{40}$
But $y_3 = (y_0)^3$ , so $2^{-20} < (y_0)^{120}$ and therefore, $2^{-20} < y_{120}$
This tells us that $m$ is somewhere around 120, so our answer is $\boxed{81,242}$

Solution: Since the choices are rather wide ranges, we can use approximation to make it easier. Notice that \[x_{n+1} - x_n = \frac{4-x_n}{x_n+6}\] And $x_0 =5$ , we know that $x_n$ is a declining sequence, and as it get close to 4 its decline will slow, never falling below 4. So we'll use 4 to approximate $x_n$ in the denominator so that we have a solvable difference equation: \[x_{n+1} - x_n = \frac{4-x_n}{10}\] \[x_{n+1} = \frac{9}{10}x_n + \frac{2}{5}\] Solve it with $x_0 = 5$ , we have \[x_n = 4 + (\frac{9}{10})^n\] Now we wish to find $n$ so that \[(\frac{9}{10})^n \approx \frac{1}{2^{20}}\] \[n \approx \frac{\log{2^{20}}}{\log{10}-\log9} \approx \frac{20*0.3}{0.05} = 120\] Since 120 is safely within the range of [81,242], we have the answer. $\boxed{81,242}$

Solution:
We start by simplifying the recursion: $x_{n+1} = x_n + \frac{4-x_n}{x_n+6}$ .
While $x_n > 4$ $x_n$ is a decreasing sequence. Let $x_n = 4 + f_n$ . Then \[x_{n} - x_{n+1} = \frac{x_n-4}{x_n+6} = \frac{f_n}{10 + f_n} \approx \frac{f_n}{10}.\] Now notice that we want to find $m$ , such that $x_m \leq 4 + \frac{1}{2^{20}}$ , and $x_0 = 4 + \frac{1}{2^0}$ .
We are going to find an approximate number of steps to go from $4 + \frac{1}{2^i}$ to $4 + \frac{1}{2^{i+1}}$ .
If $x_j = 4 + \frac{1}{2^i}$ , then $f_j = \frac{1}{2^i}$ and if $x_k = 4 + \frac{1}{2^{i+1}}$ , then $f_k = \frac{1}{2^{i+1}}$ . Then \[x_j - x_k = \frac{1}{2^{i+1}} \approx \frac{f_j}{10} + \frac{f_{j+1}}{10} + ... + \frac{f_{k-1}}{10}.\] Furthermore, \[\frac{1}{2^i} = f_j > f_{j+1} > f_{j+2} > ... > f_{k-1} > f_k = \frac{1}{2^{i+1}}.\] Therefore, \[(k-j) \cdot \frac{\frac{1}{2^i}}{10} > \frac{f_j}{10} + \frac{f_{j+1}}{10} + ... + \frac{f_{k-1}}{10} > (k-j) \cdot \frac{\frac{1}{2^{i+1}}}{10},\] so $5 < k-j < 10$ . Therefore, to go from $f_0 = \frac{1}{2^0}$ to $f_m = \frac{1}{2^{20}}$ , we will need to do this 20 times, which means $100 < m - 0 < 200$ , which is within the range of [81,242], so we have the answer. $\boxed{81,242}$


Answer: 81242.0

Solution:
We first prove that $x_n > 4$ for all $n \ge 0$ , by induction. Observe that \[x_{n+1} - 4 = \frac{x_n^2 + 5x_n + 4 - 4(x_n+6)}{x_n+6} = \frac{(x_n - 4)(x_n+5)}{x_n+6}.\] so (since $x_n$ is clearly positive for all $n$ , from the initial definition), $x_{n+1} > 4$ if and only if $x_{n} > 4$
We similarly prove that $x_n$ is decreasing: \[x_{n+1} - x_n = \frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \frac{4-x_n}{x_n+6} < 0.\]
Now we need to estimate the value of $x_{n+1}-4$ , which we can do using the rearranged equation: \[x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6}.\] Since $x_n$ is decreasing, $\frac{x_n + 5}{x_n+6}$ is also decreasing, so we have \[\frac{9}{10} < \frac{x_n + 5}{x_n+6} \le \frac{10}{11}\] and \[\frac{9}{10}(x_n-4) < x_{n+1} - 4 \le \frac{10}{11}(x_n-4).\]
This becomes \[\left(\frac{9}{10}\right)^n = \left(\frac{9}{10}\right)^n \left(x_0-4\right) < x_{n} - 4 \le \left(\frac{10}{11}\right)^n \left(x_0-4\right) = \left(\frac{10}{11}\right)^n.\] The problem thus reduces to finding the least value of $n$ such that \[\left(\frac{9}{10}\right)^n < x_{n} - 4 \le \frac{1}{2^{20}} \text{  and  }  \left(\frac{10}{11}\right)^{n-1} > x_{n-1} - 4 > \frac{1}{2^{20}}.\]
Taking logarithms, we get $n \ln \frac{9}{10} < -20 \ln 2$ and $(n-1)\ln \frac{10}{11} > -20 \ln 2$ , i.e.
\[n > \frac{20\ln 2}{\ln\frac{10}{9}} \text{  and  }  n-1 < \frac{20\ln 2}{\ln\frac{11}{10}} .\]
As approximations, we can use $\ln\frac{10}{9} \approx \frac{1}{9}$ $\ln\frac{11}{10} \approx \frac{1}{10}$ , and $\ln 2\approx 0.7$ . These approximations allow us to estimate \[126 < n < 141,\] which gives $\boxed{81,242}$

Solution:
The condition where $x_m\leq 4+\frac{1}{2^{20}}$ gives the motivation to make a substitution to change the equilibrium from $4$ to $0$ . We can substitute $x_n = y_n + 4$ to achieve that. Now, we need to find the smallest value of $m$ such that $y_m\leq \frac{1}{2^{20}}$ given that $y_0 = 1$
Factoring the recursion $x_{n+1} = \frac{x_n^2 + 5x_n+4}{x_n + 6}$ , we get:
$x_{n+1}=\dfrac{(x_n + 4)(x_n + 1)}{x_n + 6} \Rightarrow y_{n+1}+4=\dfrac{(y_n+8)(y_n+5)}{y_n+10}$
$y_{n+1}+4=\dfrac{y_n^2+13y_n+40}{y_n+10} = \dfrac{y_n^2+9y_n +(4y_n+40)}{y_n+10}$
$y_{n+1}+4=\dfrac{y_n^2+9y_n}{y_n+10} + 4$
$y_{n+1}=\dfrac{y_n^2+9y_n}{y_n+10}$
Using wishful thinking, we can simplify the recursion as follows:
$y_{n+1} = \frac{y_n^2 + 9y_n + y_n - y_n}{y_n + 10}$
$y_{n+1} = \frac{y_n(y_n + 10) - y_n}{y_n + 10}$
$y_{n+1} = y_n - \frac{y_n}{y_n + 10}$
$y_{n+1} = y_n\left(1 - \frac{1}{y_n + 10}\right)$
The recursion looks like a geometric sequence with the ratio changing slightly after each term. Notice from the recursion that the $y_n$ sequence is strictly decreasing, so all the terms after $y_0$ will be less than 1. Also, notice that all the terms in sequence will be positive. Both of these can be proven by induction.
With both of those observations in mind, $\frac{9}{10} < 1 - \frac{1}{y_n + 10} \leq \frac{10}{11}$ . Combining this with the fact that the recursion resembles a geometric sequence, we conclude that $\left(\frac{9}{10}\right)^n < y_n \leq \left(\frac{10}{11}\right)^n.$
$\frac{9}{10}$ is approximately equal to $\frac{10}{11}$ and the ranges that the answer choices give us are generous, so we should use either $\frac{9}{10}$ or $\frac{10}{11}$ to find a rough estimate for $m$
Since $\dfrac{1}{2}=0.5$ , that means $\frac{1}{\sqrt{2}}=2^{-\frac{1}{2}} \approx 0.7$ . Additionally, $\left(\frac{9}{10}\right)^3=0.729$
Therefore, we can estimate that $2^{-\frac{1}{2}} < y_3$
Raising both sides to the 40th power, we get $2^{-20} < (y_3)^{40}$
But $y_3 = (y_0)^3$ , so $2^{-20} < (y_0)^{120}$ and therefore, $2^{-20} < y_{120}$
This tells us that $m$ is somewhere around 120, so our answer is $\boxed{81,242}$

Solution: Since the choices are rather wide ranges, we can use approximation to make it easier. Notice that \[x_{n+1} - x_n = \frac{4-x_n}{x_n+6}\] And $x_0 =5$ , we know that $x_n$ is a declining sequence, and as it get close to 4 its decline will slow, never falling below 4. So we'll use 4 to approximate $x_n$ in the denominator so that we have a solvable difference equation: \[x_{n+1} - x_n = \frac{4-x_n}{10}\] \[x_{n+1} = \frac{9}{10}x_n + \frac{2}{5}\] Solve it with $x_0 = 5$ , we have \[x_n = 4 + (\frac{9}{10})^n\] Now we wish to find $n$ so that \[(\frac{9}{10})^n \approx \frac{1}{2^{20}}\] \[n \approx \frac{\log{2^{20}}}{\log{10}-\log9} \approx \frac{20*0.3}{0.05} = 120\] Since 120 is safely within the range of [81,242], we have the answer. $\boxed{81,242}$

Solution:
We start by simplifying the recursion: $x_{n+1} = x_n + \frac{4-x_n}{x_n+6}$ .
While $x_n > 4$ $x_n$ is a decreasing sequence. Let $x_n = 4 + f_n$ . Then \[x_{n} - x_{n+1} = \frac{x_n-4}{x_n+6} = \frac{f_n}{10 + f_n} \approx \frac{f_n}{10}.\] Now notice that we want to find $m$ , such that $x_m \leq 4 + \frac{1}{2^{20}}$ , and $x_0 = 4 + \frac{1}{2^0}$ .
We are going to find an approximate number of steps to go from $4 + \frac{1}{2^i}$ to $4 + \frac{1}{2^{i+1}}$ .
If $x_j = 4 + \frac{1}{2^i}$ , then $f_j = \frac{1}{2^i}$ and if $x_k = 4 + \frac{1}{2^{i+1}}$ , then $f_k = \frac{1}{2^{i+1}}$ . Then \[x_j - x_k = \frac{1}{2^{i+1}} \approx \frac{f_j}{10} + \frac{f_{j+1}}{10} + ... + \frac{f_{k-1}}{10}.\] Furthermore, \[\frac{1}{2^i} = f_j > f_{j+1} > f_{j+2} > ... > f_{k-1} > f_k = \frac{1}{2^{i+1}}.\] Therefore, \[(k-j) \cdot \frac{\frac{1}{2^i}}{10} > \frac{f_j}{10} + \frac{f_{j+1}}{10} + ... + \frac{f_{k-1}}{10} > (k-j) \cdot \frac{\frac{1}{2^{i+1}}}{10},\] so $5 < k-j < 10$ . Therefore, to go from $f_0 = \frac{1}{2^0}$ to $f_m = \frac{1}{2^{20}}$ , we will need to do this 20 times, which means $100 < m - 0 < 200$ , which is within the range of [81,242], so we have the answer. $\boxed{81,242}$


Answer: 80.0

Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$ , and $a+d>b+c$ . How many interesting ordered quadruples are there?

Solution:
We first start out when the value of $a=1$
Doing casework, we discover that $d=5,6,7,8,9,10$ . We quickly find a pattern.
Now, doing this for the rest of the values of $a$ and $d$ , we see that the answer is simply:
$(1)+(2)+(1+3)+(2+4)+(1+3+5)+(2+4+6)+(1)+(2)+(1+3)+(2+4)$ $+(1+3+5)+(1)+(2)+(1+3)+(2+4)+(1)+(2)+(1+3)+(1)+(2)+(1)=\boxed{080}$

Solution:
Notice that if $a+d>b+c$ , then $(11-a)+(11-d)<(11-b)+(11-c)$ , so there is a bijection between the number of ordered quadruples with $a+d>b+c$ and the number of ordered quadruples with $a+d<b+c$
Quick counting gives that the number of ordered quadruples with $a+d=b+c$ is 50. To count this, consider our numbers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ . Notice that if, for example, $a+d=b+c=8$ , that the average of $a,d$ and $b,c$ must both be $4$ . In this way, there is a symmetry for this case, centered at $4$ . If instead, say, $a+d=b+c=7$ , an odd number, then there is symmetry with $(a,d);(b,c)$ about $3.5$ . Further, the number of cases for each of these centers of symmetry correspond to a triangular number. Eg centered at $2.5,3,8,8.5$ , there is $1$ case for each and so on, until centered at $5.5$ , there are $10$ possible cases. Adding these all, we have $2(1+3+6)+10=50$
Thus the answer is $\frac{\binom{10}{4}-50}{2} = \boxed{080}.$

Solution:
Think about a,b,c,and d as distinct objects, that we must place in 4 of 10 spaces. However, in only 1 of 24 of these combinations, will the placement of these objects satisfy the condition in the problem. So we know the total number of ordered quadruples is $(10*9*8*7/24)=210$
Next, intuitively, the number of quadruples where $a+d>b+c$ is equal to the number of quadruples where $a+d<b+c$ . So we need to find the number of quadruples where the two quantities are equal. To do this, all we have to do is consider the cases when $a-d$ ranges from 3 to 9. It would seem natural that a range of 3 would produce 1 option, and a range of 4 would produce 2 options. However, since b and c cannot be equal, a range of 3 or 4 produces 1 option each, a range of 5 or 6 produces 2 options each, a range of 7 or 8 produces 3 options each, and a range of 9 will produce 4 options. In addition, a range of n has 10-n options for combinations of a and d. Multiplying the number of combinations of a and d by the corresponding number of options for b and c gives us 50 total quadruplets where $a+d=b+c$
So the answer will be $\frac{210-50}{2} = \boxed{080}.$

Solution: Let $b = a+x$ and $c = a+x+y$ and $d = a+x+y+z$ for positive integers $x,y,z.$ In order to satisfy the other condition we need $z > x$ so we let $z = x+k.$ Now the only other condition we need to satisfy so $a+2x+y+k \le 10.$ This condition can be transformed into $a+2x+y+k+b  = 11$ for positive $a,x,y,k,b.$ Now we use generating functions to finish. We find the generating function of the whole expression is $(x + x^2 + \cdots)^4 \cdot (x^2+x^4 + \cdots)$ and we are looking for the $x^{11}$ coefficient. This simplifies to finding the $x^5$ coefficient of $(1+x+\cdots)^4 \cdot (1+x^2+\cdots) =\frac{1}{(1-x)^4} \cdot\frac{1}{1+x}.$ Now this expression simplifies to \[\left(\binom{4}{4}+\binom{5}{4} + \cdots +\binom{4+k}{4}x^k\right)(1-x+x^2-x^3 \cdots).\] The $x^5$ coefficient ends up to be $\binom{9}{4} -\binom{8}{4} +\binom{7}{4} -\binom{6}{4} +\binom{5}{4} -\binom{4}{4} = 126 - 70 + 35 - 15 + 5 - 1 = \boxed{080}.$

Solution: First, let $a=1$ and $d=10$ . If $b=2$ , then $c$ can be from $3$ to $8$ . If $b=3$ , then $4$ to $7$ . If $b=4$ , then $c$ is between $5$ and $6$ . We find a pattern that whenever $b$ increases by $1$ , when $a$ and $d$ are stationary, then the possible values of $c$ decrease by 2, unless it gets to zero or negative, in which case that case ends. Counting up, we have $6+4+2=12$ different possibilities when $a=1$ and $d=10$ . For $a=1$ and $d=9$ $b=2$ , then $c$ can be from $3$ to $7$ . If $b=3$ , then $c$ can be from $4$ to $6$ , and so on. Notice that the possible values for each case of $b$ gets one less than if $d$ were one greater, unless that number is zero, in which it stays zero. We then use this pattern to find all the values: $12+9+6+4+2+1+9+6+4+2+1+6+4+2+1+4+2+1+2+1+1 \Rightarrow  \\ 12\cdot1+9\cdot2+6\cdot3+4\cdot4+2\cdot5+1\cdot6= \\ 12+18+18+16+10+6=\boxed{080}.$

Solution:
Rearranging the equation obtains $b-a<d-c$ . Let $a-0=e_0$ $b-a=e_3+e_1$ $c-b=e_2$ $d-c=e_3$ $11-d=e_4$ . Add up all of these newly defined equations to obtain $e_0+e_1+e_2+2e_3+e_4=11$ . Note that since all $e_n$ were defined to be $e_n\ge1$ , to form our stars and bars argument we can let $d_n+1=e_n$ for all $n$ . Then we obtain $d_0+d_1+d_2+2d_3+d_4=5$ where $d_n$ is nonnegative. Now, we can move the $2d_3$ term to the other side and perform casework.
If $2d_3=0$ : 5 objects for 4 variables -> $\binom{8}{3}$
If $2d_3=2$ : 3 objects for 4 variables -> $\binom{6}{3}$
If $2d_3=4$ : 1 object for 4 variables -> $\binom{4}{3}$
Adding all of these cases up, we get $56+20+4=\boxed{080}$ as our requested answer.


Answer: 11.0

Define binary operations $\diamondsuit$ and $\heartsuit$ by \[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \text{and} \qquad a  \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\] for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3\, \heartsuit\, 2$ and \[a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\] for all integers $n \geq 4$ . To the nearest integer, what is $\log_{7}(a_{2019})$
$\textbf{(A) } 8 \qquad  \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$

Solution:
By definition, the recursion becomes $a_n = \left(n^{\frac1{\log_7(n-1)}}\right)^{\log_7(a_{n-1})}=n^{\frac{\log_7(a_{n-1})}{\log_7(n-1)}}$ .  By the change of base formula, this reduces to $a_n = n^{\log_{n-1}(a_{n-1})}$ .  Thus, we have $\log_n(a_n) = \log_{n-1}(a_{n-1})$ .  Thus, for each positive integer $m \geq 3$ , the value of $\log_m(a_m)$ must be some constant value $k$
We now compute $k$ from $a_3$ .  It is given that $a_3 = 3\,\heartsuit\,2 = 3^{\frac1{\log_7(2)}}$ , so $k = \log_3(a_3) = \log_3\left(3^{\frac1{\log_7(2)}}\right) = \frac1{\log_7(2)} = \log_2(7)$
Now, we must have $\log_{2019}(a_{2019}) = k = \log_2(7)$ . At this point, we simply switch some bases around. For those who are unfamiliar with logarithms, we can turn the logarithms into fractions which are less intimidating to work with.
$\frac{\log{a_{2019}}}{\log{2019}} = \frac{\log{7}}{\log{2}}\implies \frac{\log{a_{2019}}}{\log{7}} = \frac{\log{2019}}{\log{2}}\implies \log_7(a_{2019}) =\log_2(2019)$
We conclude that $\log_7(a_{2019}) = \log_2(2019) \approx \boxed{11}$

Solution:
Using the recursive definition, $a_4 = (4  \, \heartsuit \, 3) \, \diamondsuit\, (3 \, \heartsuit\, 2)$ or $a_4 = (4^{m})^{k}$ where $m = \frac{1}{\log_{7}(3)}$ and $k = \log_{7}(3^{\frac{1}{\log_{7}(2)}})$ . Using logarithm rules, we can remove the exponent of the 3 so that $k = \frac{\log_{7}(3)}{\log_{7}(2)}$ . Therefore, $a_4 = 4^{\frac{1}{\log_{7}(2)}}$ , which is $4  \, \heartsuit \, 2$
We claim that $a_n = n  \, \heartsuit \, 2$ for all $n \geq 3$ . We can prove this through induction.
Clearly, the base case where $n = 3$ holds.
$a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, ((n-1)  \, \heartsuit \, 2)$
This can be simplified as $a_n = (n^{\log_{n-1}(7)}) \, \diamondsuit \, ((n-1)^{\log_{2}(7)})$
Applying the diamond operation, we can simplify $a_n = n^h$ where $h = \log_{n-1}(7) \cdot \log_{7}(n-1)^{\log_{2}(7)}$ . By using logarithm rules to remove the exponent of $\log_{7}(n-1)$ and after cancelling, $h = \frac{1}{\log_{7}(2)}$
Therefore, $a_n = n^{\frac{1}{\log_{7}(2)}} = n  \, \heartsuit \, 2$ for all $n \geq 3$ , completing the induction.
We have $a_{2019} = 2019^{\log_{2}(7)}$ . Taking $\log_{2019}$ of both sides gives us ${\log_{2019}(a_{2019})} = {\log_{2}(7)}$ . Then, by changing to base $7$ and after cancellation, we arrive at ${\log_{7}(a_{2019})} = {\log_{2}(2019)}$ . Because $2^{11} = 2048$ and $2^{10} = 1024$ , our answer is $\boxed{11}$

Solution:
We are given that \[a_n=(n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\] \[a_n=(n^{\frac{1}{\log_7(n-1)}})^{\log_7(a_{n-1})}\] Since we are asked to find $\log_7(a_{2019})$ , we directly apply \[\log_7(a_n)=\log_7(n^{\frac{1}{\log_7(n-1)}})^{\log_7(a_{n-1})}\] Using the property that $\log_ab^c=c\log_ab$ \[\log_7(a_n)=(\log_7a_{n-1})(\log_7(n^{\frac{1}{\log_7(n-1)}}))\] Now using the property that $\frac{1}{\log_ab}=\log_ba$ \[\log_7(a_n)=(\log_7a_{n-1})(\log_7n^{\log_{n-1}7})\] Once again applying the first property yields \[\log_7(a_n)=(\log_7a_{n-1})(\log_{n-1}7)(\log_7n)\] Rearranging the expression, \[\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7a_{n-1})\]
Now expressing $\log_7a_{n-1}$ in a similar expression as $\log_7a_n$
\[\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7n-1)(\log_{n-2}7)(\log_7a_{n-2})\] \[\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7n-1)(\log_{n-2}7)(\log_7n-2)(\log_{n-3}7)...(\log_74)(\log_37)(\log_7a_3)\]
Because of the fact that $(\log_ab)(\log_ba)=1$ , we can cancel out the terms to get
\[\log_7(a_n)=(\log_7n)(\log_37)(\log_7a_3)\] \[\log_7(a_n)=(\log_7n)(\log_37)(\log_7(3^{\frac{1}{\log_72}}))\] \[\log_7(a_n)=(\log_7n)(\log_37)(\log_27)(\log_73)\] \[\log_7(a_n)=(\log_27)(\log_7n)\]
Using the Chain Rule for Logarithm, $(\log_ab)(\log_bc)=\log_ac$ , yields
\[\log_7(a_n)=(\log_2n)\] Finally, substituting in $n=2019$ , we have \[\log_7(a_{2019})=(\log_22019)\] \[\log_7(a_{2019})\approx11\boxed{11}\]


Answer: 16.0

Define the operation $\star$ by $a \star b = (a+b)b.$ What is $(3 \star 5) - (5 \star 3)?$
$\textbf{(A) } -16 \qquad\textbf{(B) } -8 \qquad\textbf{(C) } 0 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 16$

Solution: Substitute and simplify. \[(3+5)5 - (5+3)3 = (3+5)2 = 8\cdot2 = \boxed{16}\]

Solution: Note that $(a \star b) - (b \star a) = (a+b)b - (b+a)a= (a+b)(b-a)= b^2 - a^2$ . We can substitute $a=3$ and $b=5$ to get $5^2 - 3^2 = \boxed{16}$


Answer: 628.0

Define the sequence $a_1, a_2, a_3, \ldots$ by $a_n = \sum\limits_{k=1}^n \sin{k}$ , where $k$ represents radian measure. Find the index of the 100th term for which $a_n < 0$

Solution: If $n = 1$ $a_n = \sin(1) > 0$ .  Then if $n$ satisfies $a_n < 0$ $n \ge 2$ , and \[a_n = \sum_{k=1}^n \sin(k) = \cfrac{1}{\sin{1}} \sum_{k=1}^n\sin(1)\sin(k) = \cfrac{1}{2\sin{1}} \sum_{k=1}^n\cos(k - 1) - \cos(k + 1) = \cfrac{1}{2\sin(1)} [\cos(0) + \cos(1) - \cos(n) - \cos(n + 1)].\] Since $2\sin 1$ is positive, it does not affect the sign of $a_n$ .  Let $b_n = \cos(0) + \cos(1) - \cos(n) - \cos(n + 1)$ .  Now since $\cos(0) + \cos(1) = 2\cos\left(\cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)$ and $\cos(n) + \cos(n + 1) = 2\cos\left(n + \cfrac{1}{2}\right)\cos\left(\cfrac{1}{2}\right)$ $b_n$ is negative if and only if $\cos\left(\cfrac{1}{2}\right) < \cos\left(n + \cfrac{1}{2}\right)$ , or when $n \in [2k\pi - 1, 2k\pi]$ .  Since $\pi$ is irrational, there is always only one integer in the range, so there are values of $n$ such that $a_n < 0$ at $2\pi, 4\pi, \cdots$ .  Then the hundredth such value will be when $k = 100$ and $n = \lfloor 200\pi \rfloor = \lfloor 628.318 \rfloor = \boxed{628}$

Solution: Notice that $a_n$ is the imaginary part of $\sum_{k=1}^n e^{ik}$ , by Euler's formula. Using the geometric series formula, we find that this sum is equal to \[\frac{e^{i(n+1)}-e^i}{e^i-1} = \frac{\cos (n+1) - \cos 1 + i (\sin (n+1) - \sin 1) }{\cos 1 - 1 + i \sin 1}\] We multiply the fraction by the conjugate of the denominator so that we can separate out the real and imaginary parts of the above expression. Multiplying, we have \[\frac{(\cos 1 - 1)(\cos(n+1)-\cos 1) + (\sin 1)(\sin(n+1)-\sin 1) + i((\sin(n+1) - \sin 1)(\cos 1 - 1) - (\sin 1)(\cos(n+1)-\cos 1))}{\cos^2 1 - 2 \cos 1 + 1 + \sin^2 1}\] We only need to look at the imaginary part, which is \[\frac{(\sin(n+1) \cos 1 - \cos(n+1) \sin 1) - \sin 1 \cos 1 + \sin 1 - \sin (n+1) + \sin 1 \cos 1}{2-2 \cos 1} = \frac{\sin n - \sin(n+1) + \sin 1}{2-2 \cos 1}\] Since $\cos 1 < 1$ $2-2 \cos 1 > 0$ , so the denominator is positive. Thus, in order for the whole fraction to be negative, we must have $\sin (n+1) - \sin n > \sin 1 \implies 2 \cos \left(n + \frac{1}{2} \right) \sin \frac{1}{2} > \sin 1 \implies \cos \left( n + \frac{1}{2} \right) > \frac{\sin 1}{2 \sin{\frac{1}{2}}} = \cos \left(\frac{1}{2} \right),$ by sum to product. This only holds when $n$ is between $2\pi k - 1$ and $2\pi k$ for integer $k$ [continuity proof here], and since this has exactly one integer solution for every such interval, the $100$ th such $n$ is $\lfloor 200\pi \rfloor = \boxed{628}$

Solution:
Similar to solution 2, we set a complex number $z=\cos 1+i\sin 1$ . We start from $z$ instead of $1$ because $k$ starts from $1$ : be careful.
The sum of $z+z^2+z^3+z^4+z^5\dots=\frac{z-z^{n+1}}{1-z}=\frac{z}{z-1}\left(z^n-1\right)$
We are trying to make $n$ so that the imaginary part of this expression is negative.
The argument of $z$ is $1$ . The argument of $z-1$ , however, is a little more tricky. $z-1$ is on a circle centered on $(-1,0)$ with radius $1$ . The change in angle due to $z$ is $1$ with respect to the center, but the angle that $z-1$ makes with the $y$ -axis is $half$ the change, due to Circle Theorems (this intercepted arc is the argument of $z$ ), because the $y$ - axis is tangent to the circle at the origin. So $\text{arg}(z-1)=\frac{\pi+1}{2}$ . Dividing $z$ by $z-1$ subtracts the latter argument from the former, so the angle of the quotient with the $x$ -axis is $\frac{1-\pi}{2}$
We want the argument of the whole expression $-\pi<\theta<0$ . This translates into $\frac{-\pi-1}{2}<\text{arg}\left(z^n-1\right)<\frac{\pi-1}{2}$ $z^n-1$ also consists of points on the circle centered at $(-1,0)$ , so we deal with this argument similarly: the argument of $z^n$ is twice the angle $z^n-1$ makes with the $y$ -axis. Since $z^n-1$ is always negative, $\frac{-3\pi}{2}<\text{arg}\left(z^n-1\right)<\frac{-\pi}{2}$ , and the left bound is the only one that is important. Either way, the line (the line consists of both bounds) makes a $\frac{\pi}{2}-\frac{\pi-1}{2}=\frac{-1}{2}$ angle with the $y$ -axis both ways.
So the argument of $z^n$ must be in the bound $-1<\theta<0$ by doubling, namely the last $z^n$ negative before another rotation. Since there is always one $z^n$ in this category per rotation because $\pi$ is irrational, $n_{100}\equiv z^{628}$ and the answer is $\boxed{628}$


Answer: 181.0

Determine $3x_4+2x_5$ if $x_1$ $x_2$ $x_3$ $x_4$ , and $x_5$ satisfy the system of equations below.

Solution: Adding all five equations gives us $6(x_1 + x_2 + x_3 + x_4 + x_5) = 6(1 + 2 + 4 + 8 + 16)$ so $x_1 + x_2 + x_3 + x_4 + x_5 = 31$ .  Subtracting this from the fourth given equation gives $x_4 = 17$ and subtracting it from the fifth given equation gives $x_5 = 65$ , so our answer is $3\cdot17 + 2\cdot65 = \boxed{181}$

Solution: Subtracting the first equation from every one of the other equations yields \begin{align*} x_2-x_1&=6\\ x_3-x_1&=18\\ x_4-x_1&=42\\ x_5-x_1&=90 \end{align*} Thus \begin{align*} 2x_1+x_2+x_3+x_4+x_5&=6\\ 2x_1+(x_1+6)+(x_1+18)+(x_1+42)+(x_1+90)&=6\\ 6x_1+156&=6\\ x_1&=-25 \end{align*} Using the previous equations, \[3x_4+2x_5=3(x_1+42)+2(x_1+90)=\boxed{181}\]


Answer: 36.0

Determine $m$ so that $4x^2-6x+m$ is divisible by $x-3$ . The obtained value, $m$ , is an exact divisor of:
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 64$

Solution: Since the given expression is a quadratic, the factored form would be $(x-3)(4x+y)$ , where $y$ is a value such that $-12x+yx=-6x$ and $-3(y)=m$ . The only number that fits the first equation is $y=6$ , so $m=-18$ . The only choice that is a multiple of 18 is $\boxed{36}$


Answer: 36.0

Solution:
Rewrite the system of equations as \[\frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1.\] This equation is satisfied when $t \in \{4, 16, 36, 64\}$ . After clearing fractions, for each of the values $t=4,16,36,64$ , we have the equation \[x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),\] where $F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)$ and $P_k(t)=F(t)/(t-k^2)$ , for $k=1,3,5,7$
Since the polynomials on each side are equal at $t=4,16,36,64$ , we can express the difference of the two polynomials by a quartic polynomial that has roots at $t=4,16,36,64$ , so \begin{align} \tag{\dag}x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)-F(t) = -(t-4)(t-16)(t-36)(t-64) \end{align} The leading coefficient of the RHS is $-1$ because the leading coefficient of the LHS is $-1$
Plug in $t=1^2, 3^2, 5^2, 7^2$ in succession, into $(\dag)$ . In each case, most terms drop, and we end up with \begin{align*}  x^2=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}, \quad y^2=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}},\quad z^2=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}},\quad w^2=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}} \end{align*} Adding them up we get the sum as $3^2\cdot 4=\boxed{036}$

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

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


Answer: 111.0

Determine all the roots real or complex , of the system of simultaneous equations

Solution: Let $P(t)=t^3-at^2+bt-c$ have roots x, y, and z. Then \[0=P(x)+P(y)+P(z)=3-3a+3b-3c\] using our system of equations, so $P(1)=0$ . Thus, at least one of x, y, and z is equal to 1; without loss of generality, let $x=1$ . Then we can use the system of equations to find that $y=z=1$ as well, and so $\boxed{1,1,1}$ is the only solution to the system of equations.

Solution: Let $a=x-1,$ $b=y-1$ and $c=z-1.$ Then \[a+b+c=0,\] \[a^2+b^2+c^2=0,\] \[a^3+b^3+c^3=0.\] We have \begin{align*} 0&=(a+b+c)^3\\ &=(a^3+b^3+c^3)+3a^2(b+c)+3b^3(a+c)+3c^2(a+b)+6abc\\ &=0-3a^3-3b^3-3c^3+6abc\\ &=6abc. \end{align*} Then one of $a, b$ and $c$ has to be 0, and easy to prove the other two are also 0. So $\boxed{1,1,1}$ is the only solution to the system of equations.

Solution:
We are going to use Intermediate Algebra Techniques to solve this equation.
Let's start with the first one: $x+y+z=3$ . This will be referred as the FIRST equation.
We are going to use the first equation to relate to the SECOND one ( $x^2+y^2+z^2=3$ ) and the THIRD one ( $x^3+y^3+z^3=3)$
Squaring this equation: $x^2+y^2+z^2+2xy+2yz+2xz=9$
Subtracting this equation from the 2nd equation in the problem, we have $2xy+2yz+2xz=6$ , so $xy+xz+yz=3$
Now we try the same idea with the cubed terms. Cube the first equation:
$x^3+y^3+z^3+3xy^2+3xz^2+3yx^2+3yz^2+3zx^2+3zy^2=27$ . Plug in $x^3+y^3+z^3=3$ and factor partially:
$3+3(x^2(y+z)+y^2(x+z)+z^2(x+y))+6xyz=27$
Now here is the key step. Note that $z=3-x-y, y=3-x-z, x=3-y-z$ . So we are going to substitute $y+z, x+z, x+y$ for each of the expressions and we get: $-x^3+3x^2-y^3+3y^2-z^3+3z^2=8-2xyz$ (I rearranged it a bit).
Resubstituting in the second and third equation: $-3+3(3)=8-2xyz$ . So $xyz=1$
So now we have three equations for the elementary symmetric sums of $x,y,z$
Equation 4: $x+y+z=3$ (this is also equation 1)
Equation 5: $xy+yz+xz=3$
Equation 6: $xyz=1$
If we call the solutions of $t^3-3t^2+3t-1=0$ (Equation 7) $a,b,c$ , then $x,y,z$ are the three roots $a,b,c$ but in some order. Notice that Equation 7 can be factored as $(t-1)^3=0$ , which means that $t=1$ . Therefore $(x,y,z)$ are permutations of $(1,1,1)$ in some order, which can be only $(1,1,1)$ . (This step uses Vieta's formulas)
Therefore, the only solution is $\boxed{1,1,1}$


Answer: 7.0

Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132.$
$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$

Solution: We can write the two digit number in the form of $10a+b$ ; reverse of $10a+b$ is $10b+a$ . The sum of those numbers is: \[(10a+b)+(10b+a)=132\] \[11a+11b=132\] \[a+b=12\] We can use brute force to find order pairs $(a,b)$ such that $a+b=12$ . Since $a$ and $b$ are both digits, both $a$ and $b$ have to be integers less than $10$ . Thus, our ordered pairs are $(3,9); (4,8); (5,7); (6,6); (7,5); (8,4); (9,3)$ ; or $\boxed{7}$ ordered pairs.

Solution: Since the numbers are “mirror images,” their average has to be $\frac{132}{2}=66$ . The highest possible value for the tens digit is $9$ because it is a two-digit number. $9-6=3$ and $6-3=3$ , so our lowest tens digit is $3$ . The numbers between $9$ and $3$ inclusive is $9-3+1=\boxed{7}$ total possibilities.


Answer: 250.0

Determine the largest positive integer $n$ such that $1005!$ is divisible by $10^n$
$\textbf{(A) }102\qquad \textbf{(B) }112\qquad \textbf{(C) }249\qquad \textbf{(D) }502\qquad \textbf{(E) }\text{none of the above}\qquad$

Solution:
We first observe that since there will be more 2s than 5s in $1005!$ , we are looking for the largest $n$ such that $5^n$ divides $1005!$ . We will use the fact that:
\[n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor \cdots\]
(This is an application of Legendre's formula).
From $k=5$ and onwards, $\left \lfloor {\frac{1005}{5^k}}\right \rfloor = 0$ . Thus, our calculation becomes
\[n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor + \left \lfloor {\frac{1005}{5^4}}\right \rfloor\]
\[n = \left \lfloor {\frac{1005}{5}}\right \rfloor + \left \lfloor {\frac{1005}{25}}\right \rfloor + \left \lfloor {\frac{1005}{125}}\right \rfloor + \left \lfloor {\frac{1005}{625}}\right \rfloor\]
\[n = 201 + 40 + 8 + 1 = \boxed{250}\]


Answer: 54.0

Determine the number of ordered pairs $(a,b)$ of integers such that $\log_a b + 6\log_b a=5, 2 \leq a \leq 2005,$ and $2 \leq b \leq 2005.$

Solution:
The equation can be rewritten as $\frac{\log b}{\log a} + 6 \frac{\log a}{\log b} = \frac{(\log b)^2+6(\log a)^2}{\log a  \log b}=5$ Multiplying through by $\log a \log b$ and factoring yields $(\log b - 3\log a)(\log b - 2\log a)=0$ . Therefore, $\log b=3\log a$ or $\log b=2\log a$ , so either $b=a^3$ or $b=a^2$
There are $44-2+1=43$ possibilities for the square case and $12-2+1=11$ possibilities for the cube case. Thus, the answer is $43+11= \boxed{054}$

Solution:
Let $k=\log_a b$ . Then our equation becomes $k+\frac{6}{k}=5$ . Multiplying through by $k$ and solving the quadratic gives us $k=2$ or $k=3$ . Hence $a^2=b$ or $a^3=b$
For the first case $a^2=b$ $a$ can range from 2 to 44, a total of 43 values.
For the second case $a^3=b$ $a$ can range from 2 to 12, a total of 11 values.
Thus the total number of possible values is $43+11=\boxed{54}$

Solution: Using the change of base formula on the second equation to change to base $a$ , we get $\log_a(b) + \frac{6 \log_a(a)}{\log_a(b)}$ . If we substitute $x$ for $\log_a(b)$ , we get $x + \frac{6}{x}$ . Multiplying by $x$ on both sides and solving, we get $x=3,2$ . Substituting back in, we get $\log_a(b) = 3,2$ . That means $a^3 = b$ or $a^2 = b$ . Since $b \leq 2005$ , we can see that for the cubed case, the maximum $a$ can be without exceeding 2005 is 12(because $13^3 = 2197$ ) and for the squared case it can be a maximum of 44. Since $a \neq 1$ , the number of values is $(44-1)+(12-1) = \boxed{54}$


Answer: 512.0

Determine the value of $ab$ if $\log_8a+\log_4b^2=5$ and $\log_8b+\log_4a^2=7$

Solution: Use the change of base formula to see that $\frac{\log a}{\log 8} + \frac{2 \log b}{\log 4} = 5$ ; combine denominators to find that $\frac{\log ab^3}{3\log 2} = 5$ . Doing the same thing with the second equation yields that $\frac{\log a^3b}{3\log 2} = 7$ . This means that $\log ab^3 = 15\log 2 \Longrightarrow ab^3 = 2^{15}$ and that $\log a^3 b = 21\log 2 \Longrightarrow a^3 b = 2^{21}$ . If we multiply the two equations together, we get that $a^4b^4 = 2^{36}$ , so taking the fourth root of that, $ab = 2^9 = \boxed{512}$

Solution: We can simplify our expressions by changing everything to a common base and by pulling exponents out of the logarithms. The given equations then become $\frac{\ln a}{\ln 8} + \frac{2 \ln b}{\ln 4} = 5$ and $\frac{\ln b}{\ln 8} + \frac{2 \ln a}{\ln 4} = 7$ . Adding the equations and factoring, we get $(\frac{1}{\ln 8}+\frac{2}{\ln 4})(\ln a+ \ln b)=12$ . Rearranging we see that $\ln ab = \frac{12}{\frac{1}{\ln 8}+\frac{2}{\ln 4}}$ . Again, we pull exponents out of our logarithms to get $\ln ab = \frac{12}{\frac{1}{3 \ln 2} + \frac{2}{2 \ln 2}} = \frac{12 \ln 2}{\frac{1}{3} + 1} = \frac{12 \ln 2}{\frac{4}{3}} = 9 \ln 2$ . This means that $\frac{\ln ab}{\ln 2} = 9$ . The left-hand side can be interpreted as a base-2 logarithm, giving us $ab = 2^9 = \boxed{512}$

Solution: This solution is very similar to the above two, but it utilizes the well-known fact that $\log_{m^k}{n^k}= \log_m{n}.$ Thus, $\log_8a+\log_4b^2=5 \Rightarrow  \log_{2^3}{(\sqrt[3]{a})^3} + \log_{2^2}{b^2} = 5 \Rightarrow \log_2{\sqrt[3]{a}} + \log_2{b} = 5 \Rightarrow \log_2{\sqrt[3]{a}b} = 5.$ Similarly, $\log_8b+\log_4a^2=7 \Rightarrow \log_2{\sqrt[3]{b}a} = 7.$ Adding these two equations, we have $\log_2{a^{\frac{4}{3}}b^{\frac{4}{3}}} = 12 \Rightarrow ab = 2^{12\times\frac{3}{4}} = 2^9 = \boxed{512}$

Solution: We can change everything to a common base, like so: $\log_8{a} + \log_8{b^3} = 5,$ $\log_8{b} + \log_8{a^3} = 7.$ We set the value of $\log_8{a}$ to $x$ , and the value of $\log_8{b}$ to $y.$ Now we have a system of linear equations: \[x + 3y = 5,\] \[y + 3x = 7.\] Now add the two equations together then simplify, we'll get $x+y=3$ . So $\log_8{ab} = \log_8{a} + \log_8{b} = 3$ $ab = 8^3 = \boxed{512}$

Solution: Add the two equations to get $\log_8 {a}+ \log_8 {b}+ \log_{a^2}+\log_{b^2}=12$ . This can be simplified with the log property $\log_n {x}+\log_n {y}=log_n {xy}$ . Using this, we get $\log_8 {ab}+ \log_4 {a^2b^2}=12$ . Now let $\log_8 {ab}=c$ and $\log_4 {a^2b^2}=k$ . Converting to exponents, we get $8^c=ab$ and $4^k=(ab)^2$ . Sub in the $8^c$ to get $k=3c$ . So now we have that $k+c=12$ and $k=3c$ which gives $c=3$ $k=9$ . This means $\log_4 {a^2b^2}=9$ so $4^9=(ab)^2 \implies ab=(2^2)^9 \implies 2^9 \implies \boxed{512}$

Solution:
By properties of logarithms, we know that $\log_8 {a}+ \log_4 {b ^ 2} = \log_2 {a ^ {1/3}}+ \log_2 {b} = 5$
Using the fact that $\log_a {b}  + \log_a {c}  = log_a{b*c}$ , we get $\log_2 {a^{1/3} * b} = 5$
Similarly, we know that $\log_2 {a * b^{1/3}} = 7$
From these two equations, we get $a^{1/3} * b = 2^5$ and $a * b^{1/3} = 2^7$
Multiply the two equations to get $a^{4/3} * b^{4/3} = 2^{12}$ . Solving, we get that $a*b = 2^{12*3/4} = 2^9 =$ $\boxed{512}$

Solution: Adding both of the equations, we get \[\log_8{ab} +2\log_4{ab}=12\] Furthermore, we see that $\log_4 {ab}$ is $\frac{3}{2}$ times $\log_8 {ab}.$ Substituting $\log_8 {ab}$ as $x,$ we get $x+3x=12,$ so $x=3.$ Therefore, we have $\log_8 {ab} = 3,$ so $ab= 8^3=\boxed{512}$ ~ math_comb01


Answer: 4.2

Diagonal $DB$ of rectangle $ABCD$ is divided into three segments of length $1$ by parallel lines $L$ and $L'$ that pass through $A$ and $C$ and are perpendicular to $DB$ . The area of $ABCD$ , rounded to the one decimal place, is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); real x=sqrt(6), y=sqrt(3), a=0.4; pair D=origin, A=(0,y), B=(x,y), C=(x,0), E=foot(C,B,D), F=foot(A,B,D); real r=degrees(B); pair M1=F+3*dir(r)*dir(90), M2=F+3*dir(r)*dir(-90), N1=E+3*dir(r)*dir(90), N2=E+3*dir(r)*dir(-90); markscalefactor=0.02; draw(B--C--D--A--B--D^^M1--M2^^N1--N2^^rightanglemark(A,F,B)^^rightanglemark(N1,E,B)); pair W=A+a*dir(135), X=B+a*dir(45), Y=C+a*dir(-45), Z=D+a*dir(-135); label("A", A, NE); label("B", B, NE); label("C", C, dir(0)); label("D", D, dir(180)); label("$L$", (x/2,0), SW); label("$L^\prime$", C, SW); label("1", D--F, NW); label("1", F--E, SE); label("1", E--B, SE); clip(W--X--Y--Z--cycle);[/asy]
$\mathrm{(A)\ } 4.1 \qquad \mathrm{(B) \ }4.2 \qquad \mathrm{(C) \  } 4.3 \qquad \mathrm{(D) \  } 4.4 \qquad \mathrm{(E) \  }4.5$

Solution: Let $E$ be the point of intersection of $L$ and $\overline{BD}$ . Then, because $AE$ is the altitude to the hypotenuse of right triangle $ABD$ , triangles $ADE$ and $BAE$ are similar, giving \[\frac{AE}{BE} = \frac{ED}{EA},\] and so \begin{align*}AE &= \sqrt{BE \cdot ED} \\ &= \sqrt{(1+1)(1)} \\ &= \sqrt{2}.\end{align*} Thus, taking $BD$ and $AE$ as the base and perpendicular height, respectively, of triangle $ABD$ , we may compute its area as $\frac{1}{2}(3)\left(\sqrt{2}\right) = \frac{3\sqrt{2}}{2}$ . By symmetry, the area of the entire rectangle $ABCD$ is \[2\left(\frac{3\sqrt{2}}{2}\right) = 3\sqrt{2} \approx (3)(1.4) = \boxed{4.2}.\]


Answer: 3.0

Distinct points $A$ $B$ $C$ , and $D$ lie on a line, with $AB=BC=CD=1$ . Points $E$ and $F$ lie on a second line, parallel to the first, with $EF=1$ . A triangle with positive area has three of the six points as its vertices. How many possible values are there for the area of the triangle?
$\text{(A) } 3 \qquad \text{(B) } 4 \qquad \text{(C) } 5 \qquad \text{(D) } 6 \qquad \text{(E) } 7$

Solution:
Consider the classical formula for triangle area: $\frac 12 \cdot b \cdot h$ .
Each of the triangles that we can make has exactly one side lying on one of the two parallel lines. If we pick this side to be the base, the height will always be the same - it will be the distance between the two lines.
Hence each area is uniquely determined by the length of the base. And it can easily be seen, that the only possible base lengths are $1$ $2$ , and $3$ . Therefore there are only $\boxed{3}$ possible values for the area.

Solution: No matter what how we draw a triangle by selecting three non-linear points, its height will always remain the same. Therefore, we will only get different areas with different base-lengths. The possibilities are $1$ $2$ , and $3$ units for a total of $\boxed{3}$


Answer: 20.0

Distinct points $A$ and $B$ are on a semicircle with diameter $MN$ and center $C$ .
The point $P$ is on $CN$ and $\angle CAP = \angle CBP = 10^{\circ}$ . If $\stackrel{\frown}{MA} = 40^{\circ}$ , then $\stackrel{\frown}{BN}$ equals
[asy] import geometry; import graph;  unitsize(2 cm);  pair A, B, C, M, N, P;  M = (-1,0); N = (1,0); C = (0,0); A = dir(140); B = dir(20); P = extension(A, A + rotate(10)*(C - A), B, B + rotate(10)*(C - B));  draw(M--N); draw(arc(C,1,0,180)); draw(A--C--B); draw(A--P--B);  label("$A$", A, NW); label("$B$", B, E); label("$C$", C, S); label("$M$", M, SW); label("$N$", N, SE); label("$P$", P, S); [/asy]
$\textbf{(A)}\ 10^{\circ}\qquad \textbf{(B)}\ 15^{\circ}\qquad \textbf{(C)}\ 20^{\circ}\qquad \textbf{(D)}\ 25^{\circ}\qquad \textbf{(E)}\ 30^{\circ}$

Solution:
Since $\angle CAP = \angle CBP = 10^\circ$ , quadrilateral $ABPC$ is cyclic (as shown below) by the converse of the theorem "angles inscribed in the same arc are equal".

Since $\angle ACM = 40^\circ$ $\angle ACP = 140^\circ$ , so, using the fact that opposite angles in a cyclic quadrilateral sum to $180^{\circ}$ , we have $\angle ABP = 40^\circ$ . Hence $\angle ABC = \angle ABP - \angle CBP = 40^ \circ - 10^\circ = 30^\circ$
Since $CA = CB$ , triangle $ABC$ is isosceles, with $\angle BAC = \angle ABC = 30^\circ$ . Now, $\angle BAP = \angle BAC - \angle CAP = 30^\circ - 10^\circ = 20^\circ$ . Finally, again using the fact that angles inscribed in the same arc are equal, we have $\angle BCP = \angle BAP = \boxed{20}$


Answer: 7.0

Distinct points $P$ $Q$ $R$ $S$ lie on the circle $x^{2}+y^{2}=25$ and have integer coordinates. The distances $PQ$ and $RS$ are irrational numbers. What is the greatest possible value of the ratio $\frac{PQ}{RS}$
$\textbf{(A) } 3 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 3\sqrt{5} \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 5\sqrt{2}$

Solution: Because $P$ $Q$ $R$ , and $S$ are lattice points, there are only a few coordinates that actually satisfy the equation. The coordinates are $(\pm 3,\pm 4), (\pm 4, \pm 3), (0,\pm 5),$ and $(\pm 5,0).$ We want to maximize $PQ$ and minimize $RS.$ They also have to be non perfect squares, because they are both irrational. The greatest value of $PQ$ happens when $P$ and $Q$ are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be $(-4,3)$ and $(3,-4)$ because the two points are almost across from each other. Another possible pair could be $(-4,3)$ and $(5,0)$ . To find out which segment is longer, we have to compare the distances from their endpoints to a diameter (which must be the longest possible segment). The closest diameter would be from $(-4,3)$ to $(4,-3)$ . The distance between $(3,-4)$ and $(-4,3)$ is greater than the distance between $(5,0)$ and $(4,-3)$ . Therefore, the segment from $(3,-4)$ to $(-4,3)$ is the longest attainable (the other possible coordinates for $P$ and $Q$ are $(4,3)$ and $(-3, -4)$ $(3, 4)$ and $(-4, -3)$ $(-3, 4)$ and $(4, -3)$ . The least value of $RS$ is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example, $R$ is $(3,4)$ and $S$ is $(4,3).$ They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point $(3,4)$ than $(4,3)$ and vice versa. Using the distance formula, we get that $PQ$ is $\sqrt{98}$ and that $RS$ is $\sqrt{2}.$ $\frac{\sqrt{98}}{\sqrt{2}}=\sqrt{49}=\boxed{7}$

Solution: We can look at the option choices. Since we are aiming for the highest possible ratio, let's try using $7$ (though $5 \sqrt{2}$ actually is the highest ratio.) Now, looking at the problem alone, we know that to have the largest ratio possible, we have to let $RS$ be the minimum possible value while at the same time using integer coordinates. Thus, the smallest possible value of $RS$ is $\sqrt{1^{2}+1^{2}} = \sqrt{2}$ . Assuming that $\frac{PQ}{RS} = 7$ , we plug in $RS = \sqrt{2}$ and solve for $PQ$ $PQ=7\sqrt{2}$ . Remember, we don't know if this is possible yet, we are only trying to figure out if it is. But for what values of $x$ and $y$ does $\sqrt{x^{2}+y^{2}}=7\sqrt2$ ? We see that this can easily be made into a $45-45-90$ triangle. But, instead of substituting $y=x$ into the equation and then using a whole lot of algebra, we can save time and use the little trick, that if in a $45-45-90$ triangle, the two $45$ degree sides have side length $s$ , then the hypotenuse is $s\sqrt2$ . Using this, we can see that $s=7$ , and since our equation does in fact yield a sensible solution, we can be assured that our answer is $\boxed{7}$

Solution: By inspection, when $R$ is at $(3, 4)$ and $S$ is at $(4, 3),$ it makes $RS$ as small as possible with a distance of $\sqrt{2}$ . The greatest possible length of $PQ$ arises when $P$ is at $(-3, 4)$ and $Q$ is at $(4, -3).$ Using the distance formula, we find that $PQ$ has a length of $7\sqrt{2}.$ The requested fraction is then $\dfrac{PQ}{RS} = \dfrac{7\sqrt{2}}{\sqrt{2}} = \boxed{7}$


Answer: 6.0

Dots are spaced one unit apart, horizontally and vertically.  The number of square units enclosed by the polygon is
[asy] for(int a=0; a<4; ++a) { for(int b=0; b<4; ++b) { dot((a,b)); } } draw((0,0)--(0,2)--(1,2)--(2,3)--(2,2)--(3,2)--(3,0)--(2,0)--(2,1)--(1,0)--cycle); [/asy]
$\text{(A)}\ 5 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 7 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$

Solution:
We could count the area contributed by each square on the $3 \times 3$ grid:
Top-left: $0$
Top: Triangle with area $\frac{1}{2}$
Top-right: $0$
Left: Square with area $1$
Center: Square with area $1$
Right: Square with area $1$
Bottom-left: Square with area $1$
Bottom: Triangle with area $\frac{1}{2}$
Bottom-right: Square with area $1$
Adding all of these together, we get $6$ which is the same as $\boxed{6}$

Solution: By Pick's Theorem, we get the formula, $A=I+\frac{b}{2}-1$ where $I$ is the number of lattice points in the interior and $b$ being the number of lattice points on the boundary. In this problem, we can see that $I=1$ and $B=12$ . Substituting gives us $A=1+\frac{12}{2}-1=6$ Thus, the answer is $\boxed{6}$

Solution: Notice that the extra triangle on the top with area $1$ can be placed (like a jigsaw puzzle) at the bottom of the grid where there is a triangular hole, also with area $1$ . This creates a $2*3$ rectangle, with a area of $6$ . The answer is $\boxed{6}$ ~sakshamsethi


Answer: 4.0

Doug and Dave shared a pizza with $8$ equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was $8$ dollars, and there was an additional cost of $2$ dollars for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. Doug ate the remainder. Each paid for what he had eaten. How many more dollars did Dave pay than Doug?
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) }  5$

Solution: Dave and Doug paid $8+2=10$ dollars in total.  Doug paid for three slices of plain pizza, which cost $\frac{3}{8}\cdot 8=3$ .  Dave paid $10-3=7$ dollars.  Dave paid $7-3=\boxed{4}$ more dollars than Doug.


Answer: 4.0

Solution: Dave and Doug paid $8+2=10$ dollars in total.  Doug paid for three slices of plain pizza, which cost $\frac{3}{8}\cdot 8=3$ .  Dave paid $10-3=7$ dollars.  Dave paid $7-3=\boxed{4}$ more dollars than Doug.


Answer: 26.0

Solution: We note that the total length must be the same as the total height, as it is given in the problem. Calling the width of each small rectangle $2x$ , and the height $5x$ , we can see that the length is composed of $4$ widths and $5$ bars of length $2$ . This is equal to two heights of the small rectangles as well as $3$ bars of $2$ . Thus, $4(2x) + 5(2) = 2(5x) + 3(2)$ . We quickly find that $x = 2$ . The total side length is $4(4) + 5(2) = 2(10) + 3(2) = 26$ , or $\boxed{26}$


Answer: 26.0

Solution: Let the height of the panes equal $5x$ , and let the width of the panes equal $2x$ .  Now notice that the total width of the borders equals $10$ , and the total height of the borders is $6$ .  We have \[10 + 4(2x) = 6 + 2(5x)\] \[x = 2\] Now, the total side length of the window equals \[10+ 4(2x) = 10 + 16 = \boxed{26}\]


Answer: 55.0

Driving along a highway, Megan noticed that her odometer showed $15951$ (miles). This number is a palindrome-it reads the same forward and backward. Then $2$ hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this $2$ -hour period?
$\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 55 \qquad\textbf{(C)}\ 60 \qquad\textbf{(D)}\ 65 \qquad\textbf{(E)}\ 70$

Solution:
In order to get the smallest palindrome greater than $15951$ , we need to raise the middle digit. If we were to raise any of the digits after the middle, we would be forced to also raise a digit before the middle to keep it a palindrome, making it unnecessarily larger.
So we raise $9$ to the next largest value, $10$ , but obviously, that's not how place value works, so we're in the $16000$ s now. To keep this a palindrome, our number is now $16061$
So Megan drove $16061-15951=110$ miles. Since this happened over $2$ hours, she drove at $\frac{110}{2}=\boxed{55}$ mph. ~quacker88


Answer: 135.0

Driving at a constant speed, Sharon usually takes $180$ minutes to drive from her house to her mother's house. One day Sharon begins the drive at her usual speed, but after driving $\frac{1}{3}$ of the way, she hits a bad snowstorm and reduces her speed by $20$ miles per hour. This time the trip takes her a total of $276$ minutes. How many miles is the drive from Sharon's house to her mother's house?
$\textbf{(A)}\ 132 \qquad\textbf{(B)}\ 135 \qquad\textbf{(C)}\ 138 \qquad\textbf{(D)}\ 141 \qquad\textbf{(E)}\ 144$

Solution:
Let total distance be $x$ . Her speed in miles per minute is $\tfrac{x}{180}$ . Then, the distance that she drove before hitting the snowstorm is $\tfrac{x}{3}$ . Her speed in snowstorm is reduced $20$ miles per hour, or $\tfrac{1}{3}$ miles per minute. Knowing it took her $276$ minutes in total, we create equation: \[\text{Time before Storm}\, + \, \text{Time after Storm} = \text{Total Time} \Longrightarrow\] \[\frac{\text{Distance before Storm}}{\text{Speed before Storm}} + \frac{\text{Distance in Storm}}{\text{Speed in Storm}} = \text{Total Time} \Longrightarrow \frac{\tfrac{x}{3}}{\tfrac{x}{180}} + \frac{\tfrac{2x}{3}}{\tfrac{x}{180} - \tfrac{1}{3}} = 276\]
Solving equation, we get $x=135$ $\Longrightarrow \boxed{135}$


Answer: 580.0

During a recent campaign for office, a candidate made a tour of a country which we assume lies in a plane. On the first day of the tour he went east, on the second day he went north, on the third day west, on the fourth day south, on the fifth day east, etc. If the candidate went $n^{2}_{}/2$ miles on the $n^{\mbox{th}}_{}$ day of this tour, how many miles was he from his starting point at the end of the $40^{\mbox{th}}_{}$ day?

Solution: On the first day, the candidate moves $[4(0) + 1]^2/2\ \text{east},\, [4(0) + 2]^2/2\ \text{north},\, [4(0) + 3]^2/2\ \text{west},\, [4(0) + 4]^2/2\ \text{south}$ , and so on. The E/W displacement is thus $1^2 - 3^2 + 5^2 \ldots +37^2 - 39^2 = \left|\sum_{i=0}^9 \frac{(4i+1)^2}{2} - \sum_{i=0}^9 \frac{(4i+3)^2}{2}\right|$ . Applying difference of squares , we see that \begin{align*} \left|\sum_{i=0}^9 \frac{(4i+1)^2 - (4i+3)^2}{2}\right| &= \left|\sum_{i=0}^9 \frac{(4i+1+4i+3)(4i+1-(4i+3))}{2}\right|\\ &= \left|\sum_{i=0}^9 -(8i+4) \right|. \end{align*} The N/S displacement is \[\left|\sum_{i=0}^9 \frac{(4i+2)^2}{2} - \sum_{i=0}^9 \frac{(4i+4)^2}{2}\right| = \left|\sum_{i=0}^9 -(8i+6) \right|.\] Since $\sum_{i=0}^{9} i = \frac{9(10)}{2} = 45$ , the two distances evaluate to $8(45) + 10\cdot 4 = 400$ and $8(45) + 10\cdot 6 = 420$ . By the Pythagorean Theorem , the answer is $\sqrt{400^2 + 420^2} = 29 \cdot 20 = \boxed{580}$


Answer: 80.0

During the softball season, Judy had $35$ hits.  Among her hits were $1$ home run, $1$ triple and $5$ doubles.  The rest of her hits were single.  What percent of her hits were single?
$\text{(A)}\ 28\% \qquad \text{(B)}\ 35\% \qquad \text{(C)}\ 70\% \qquad \text{(D)}\ 75\% \qquad \text{(E)}\ 80\%$

Solution: From the information given, $35-5-1-1=28$ hits were single.  Thus, the percentage was \[\dfrac{28}{35}=80\% \rightarrow \boxed{80}.\]


Answer: 54.0

Solution:
The original cube has $6$ square surfaces that each have an area of $3^2 = 9$ , for a toal surface area of $6\cdot 9 = 54$
Since no two corner cubes touch, we can examine the effect of removing each corner cube individually.
Each corner cube contribues $3$ faces each of surface area $1$ to the big cube, so the surface area is decreased by $3$ when the cube is removed.
However, when the cube is removed, $3$ faces on the 3x3x3 cube will be revealed, increasing the surface area by $3$
Thus, the surface area does not change with the removal of a corner cube, and it remains $54$ , which is answer $\boxed{54}$


Answer: 36.0

Each corner of a rectangular prism is cut off.  Two (of the eight) cuts are shown.  How many edges does the new figure have?
[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((3,0)--(5,2)--(5,5)--(2,5)--(0,3)); draw((3,3)--(5,5)); draw((2,0)--(3,1.8)--(4,1)--cycle,linewidth(1)); draw((2,3)--(4,4)--(3,2)--cycle,linewidth(1)); [/asy]
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 42 \qquad \text{(E)}\ 48$
Assume that the planes cutting the prism do not intersect anywhere in or on the prism.

Solution:
In addition to the original $12$ edges, each original vertex contributes $3$ new edges.
There are $8$ original vertices, so there are $12+3\times 8=36$ edges in the new figure $\rightarrow \boxed{36}$


Answer: 3.0

Each day Walter gets $3$ dollars for doing his chores or $5$ dollars for doing them exceptionally well. After $10$ days of doing his chores daily, Walter has received a total of $36$ dollars. On how many days did Walter do them exceptionally well?
$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ 6\qquad\text{(E)}\ 7$

Solution:
If Walter had done his chores for $10$ days without doing any of them well, he would have earned $3 \cdot 10 = 30$ dollars.  He got $6$ dollars more than this.
He gets a $5 - 3 = 2$ dollar bonus every day he does his chores well.  Thus, he did his chores exceptionally well $\frac{6}{2} = 3$ days, and the answer is $\boxed{3}$

Solution:
If Walter had done his chores for $10$ days exceptionally well, he would have earned $5 \cdot 10 = 50$ dollars.  He got $50 - 36 = 14$ dollars less than this.
He gets $2$ dollars docked from his pay if he doesn't do his chores well.  Therefore, he didn't do his chores well on $\frac{14}{2} = 7$ days.  The other $10 - 7 = 3$ days, he did them exceptionally well.  Therefore, the answer is $\boxed{3}$

Solution:
Let $b$ be the number of days Walter does his chores but doesn't do them well, and let $w$ be the number of days he does his chores exceptionally well.
$b + w = 10$ since there are $10$ days Walter does chores.
$3b + 5w = 36$ since $3b$ is the amount he earns from doing his chores not well, and $5w$ is the amount he earnes from doing his chores exceptionally well, and those two sum to $36$ dollars.
Multiply the first equation by $3$ to get:
$3b + 3w = 30$
$3b + 5w = 36$
Subtract the first equation from the second equation to get:
$5w - 3w = 36 - 30$
$2w = 6$
$w = 3$
Thus, he does his chores well $3$ days, and the answer is $\boxed{3}$


Answer: 25.0

Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by $5$ minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?
$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82$

Solution: It is well known that $\text{Distance}=\text{Speed} \cdot \text{Time}$ . In the question, we want distance. From the question, we have that the time is $60$ minutes or $1$ hour.  By the equation derived from $\text{Distance}=\text{Speed} \cdot \text{Time}$ , we have $\text{Speed}=\frac{\text{Distance}}{\text{Time}}$ , so the speed is $1$ mile per $x$ minutes. Because we want the distance, we multiply the time and speed together yielding $60\text{ mins}\cdot \frac{1\text{ mile}}{x\text{ mins}}$ . The minutes cancel out, so now we have $\dfrac{60}{x}$ as our distance for the first day. The distance for the following days are: \[\dfrac{60}{x},\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}.\] We know that $x,x+5,x+10,x+15$ are all factors of $60$ , therefore, $x=5$ because the factors have to be in an arithmetic sequence with the common difference being $5$ and $x=5$ is the only solution. \[\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=12+6+4+3=\boxed{25}.\]


Answer: 50.0

Each day, Jenny ate $20\%$ of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, $32$ remained. How many jellybeans were in the jar originally?
$\textbf{(A)} \ 40 \qquad \textbf{(B)} \ 50  \qquad \textbf{(C)} \ 55 \qquad \textbf{(D)} \ 60 \qquad \textbf{(E)} \ 75$

Solution:
We can begin by labeling the number of initial jellybeans $x$ . If she ate $20\%$ of the jellybeans, then $80\%$ is remaining. Hence, after day 1, there are: $0.8 * x$
After day 2, there are: $0.8 * 0.8 * x$ or $0.64x$ jellybeans. $0.64x = 32$ , so $x = \boxed{50}$

Solution: Testing the answers choices out, we see that the answer is $\boxed{50}$


Answer: 50.0

Solution:
We can begin by labeling the number of initial jellybeans $x$ . If she ate $20\%$ of the jellybeans, then $80\%$ is remaining. Hence, after day 1, there are: $0.8 * x$
After day 2, there are: $0.8 * 0.8 * x$ or $0.64x$ jellybeans. $0.64x = 32$ , so $x = \boxed{50}$

Solution: Testing the answers choices out, we see that the answer is $\boxed{50}$


Answer: 125.0

Each edge of a cube is increased by $50$ %. The percent of increase of the surface area of the cube is: $\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 125\qquad\textbf{(C)}\ 150\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 750$

Solution: Note that increasing the length of each edge by $50$ % with result in a cube that is similar to the original cube with scale factor $1.5$ . Therefore, the surface area will increase by a factor of $1.5^2$ , or $2.25$ . Converting this back into a percent, the percent increase will be $125$ %. Therefore, the answer is $\boxed{125}$


Answer: 125.0

Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$ .
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ . A parallelepiped is a solid with six parallelogram faces
such as the one shown below.
[asy] unitsize(2cm); pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70);  draw(o--u--(u+v)); draw(o--v--(u+v), dotted); draw(shift(w)*(o--u--(u+v)--v--cycle)); draw(o--w); draw(u--(u+w)); draw(v--(v+w), dotted); draw((u+v)--(u+v+w)); [/asy]

Solution:
Denote $\alpha = \tan^{-1} \frac{\sqrt{21}}{\sqrt{31}}$ .
Denote by $d$ the length of each side of a rhombus.
Now, we put the solid to the 3-d coordinate space.
We put the bottom face on the $x-O-y$ plane.
For this bottom face, we put a vertex with an acute angle $2 \alpha$ at the origin, denoted as $O$ .
For two edges that are on the bottom face and meet at $O$ , we put one edge on the positive side of the $x$ -axis. The endpoint is denoted as $A$ . Hence, $A = \left( d , 0 , 0 \right)$ .
We put the other edge in the first quadrant of the $x-O-y$ plane. The endpoint is denoted as $B$ . Hence, $B = \left( d \cos 2 \alpha , d \sin 2 \alpha , 0 \right)$
For the third edge that has one endpoint $O$ , we denote by $C$ its second endpoint.
We denote $C = \left( u , v , w \right)$ .
Without loss of generality, we set $w > 0$ .
Hence, \[ u^2 + v^2 + w^2 = d^2 . \hspace{1cm} (1) \]
We have \begin{align*} \cos \angle AOC & = \frac{\overrightarrow{OA} \cdot \overrightarrow{OC}}{|OA| \cdot |OC|} \\ & = \frac{u}{d} , \hspace{1cm} (2) \end{align*} and \begin{align*} \cos \angle BOC & = \frac{\overrightarrow{OB} \cdot \overrightarrow{OC}}{|OB| \cdot |OC|} \\ & = \frac{u \cos 2 \alpha + v \sin 2 \alpha}{d} . \hspace{1cm} (3) \end{align*}
Case 1: $\angle AOC = \angle BOC = 2 \alpha$ or $2 \left( 90^\circ - \alpha \right)$
By solving (2) and (3), we get \begin{align*} u & = \pm d \cos 2 \alpha , \\ v & = \pm d \cos 2 \alpha \frac{1 - \cos 2 \alpha}{\sin 2 \alpha} \\ & = \pm d \cos 2 \alpha \tan \alpha . \end{align*}
Plugging these into (1), we get \begin{align*} w & = d \sqrt{1 - \cos^2 2 \alpha - \cos^2 2 \alpha \tan^2 \alpha} \\ & = d \sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \tan^2 \alpha} . \hspace{1cm} (4) \end{align*}
Case 2: $\angle AOC = 2 \alpha$ and $\angle BOC = 2 \left( 90^\circ - \alpha \right)$ , or $\angle BOC = 2 \alpha$ and $\angle AOC = 2 \left( 90^\circ - \alpha \right)$
By solving (2) and (3), we get \begin{align*} u & = \pm d \cos 2 \alpha , \\ v & = \mp d \cos 2 \alpha \frac{1 + \cos 2 \alpha}{\sin 2 \alpha} \\ & = \mp d \cos 2 \alpha \cot \alpha . \end{align*}
Plugging these into (1), we get \begin{align*} w & = d \sqrt{1 - \cos^2 2 \alpha - \cos^2 2 \alpha \cot^2 \alpha} \\ & = d \sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \cot^2 \alpha} . \hspace{1cm} (5) \end{align*}
We notice that $(4) > (5)$ . Thus, (4) (resp. (5)) is the parallelepiped with a larger (resp. smaller) height.
Therefore, the ratio of the volume of the larger parallelepiped to the smaller one is \begin{align*} \frac{(4)}{(5)} & = \frac{\sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \tan^2 \alpha}} {\sqrt{\sin^2 2 \alpha - \cos^2 2 \alpha \cot^2 \alpha}} \\ & = \sqrt{\frac{\tan^2 2 \alpha - \tan^2 \alpha}{\tan^2 2 \alpha - \cot^2 \alpha}} . \end{align*}
Recall that $\tan \alpha = \frac{\sqrt{21}}{\sqrt{31}}$ .
Thus, $\tan 2 \alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} = \frac{\sqrt{21 \cdot 31}}{5}$ .
Plugging this into the equation above, we get \begin{align*} \frac{(4)}{(5)} & = \frac{63}{62}. \end{align*}
Therefore, the answer is $63 + 62 = \boxed{125}$

Solution:
Let one of the vertices be at the origin and the three adjacent vertices be $u$ $v$ , and $w$ . For one of the parallelepipeds, the three diagonals involving the origin have length $\sqrt {21}$ . Hence, $(u+v)\cdot (u+v)=u\cdot u+v\cdot v+2u\cdot v=21$ and $(u-v)\cdot (u-v)=u\cdot u+v\cdot v-2u\cdot v=31$ . Since all of $u$ $v$ , and $w$ have equal length, $u\cdot u=13$ $v\cdot v=13$ , and $u\cdot v=-2.5$ . Symmetrically, $w\cdot w=13$ $u\cdot w=-2.5$ , and $v\cdot w=-2.5$ . Hence the volume of the parallelepiped is given by $\sqrt{\operatorname{det}\begin{pmatrix}13&-2.5&-2.5\\-2.5&13&-2.5\\-2.5&-2.5&13\end{pmatrix}}=\sqrt{\operatorname{det}\begin{pmatrix}15.5&-15.5&0\\-2.5&13&-2.5\\0&-15.5&15.5\end{pmatrix}}=\sqrt{15.5^2\operatorname\det\begin{pmatrix}1&-1&0\\-2.5&13&-2.5\\0&-1&1\end{pmatrix}}=\sqrt{15.5^2\cdot 8}$
For the other parallelepiped, the three diagonals involving the origin are of length $\sqrt{31}$ and the volume is $\sqrt{\operatorname{det}\begin{pmatrix}13&2.5&2.5\\2.5&13&2.5\\2.5&2.5&13\end{pmatrix}}=\sqrt{\operatorname{det}\begin{pmatrix}10.5&-10.5&0\\2.5&13&2.5\\0&-10.5&10.5\end{pmatrix}}=\sqrt{10.5^2\operatorname\det\begin{pmatrix}1&-1&0\\2.5&13&2.5\\0&-1&1\end{pmatrix}}=\sqrt{10.5^2\cdot 18}$
Consequently, the answer is $\sqrt\frac{10.5^2\cdot 18}{15.5^2\cdot 8}=\frac{63}{62}$ , giving $\boxed{125}$

Solution:
Observe that both parallelepipeds have two vertices (one on each base) that have three congruent angles meeting at them. Denote the parallelepiped with three acute angles meeting $P$ , and the one with three obtuse angles meeting $P'$
The area of a parallelepiped is simply the base area times the height, but because both parallelepipeds have the same base, what we want is just the ratio of the heights.
Denote the point with three acute angles meeting at it in $P$ as $A$ , and its neighbors $B$ $C$ , and $D$ . Similarly, denote the point with three obtuse angles meeting at it in $P'$ as $A'$ , and its neighbors $B'$ $C'$ , and $D'$
We have the following equations:
\[\textrm{Height of }P\textrm{ from }ACD = \frac{\textrm{Vol}(ABCD) \cdot 3}{[ACD]},\] \[\textrm{Height of }P'\textrm{ from }A'C'D' = \frac{\textrm{Vol}(A'B'C'D') \cdot 3}{[A'C'D']}.\]
However, $ACD$ and $A'C'D'$ are both half the area of a rhombus with diagonals $\sqrt{31}$ and $\sqrt{21}$ , so our ratio is really
\[\frac{P}{P'} = \frac{\textrm{Vol}(ABCD)}{\textrm{Vol}(A'B'C'D')}.\]
Because the diagonals of all of the faces are $\sqrt{31}$ and $\sqrt{21}$ , each edge of the parallelepipeds is $\sqrt{13}$ by the Pythagorean theorem.
We have $AB = AC = AD = \sqrt{13}$ , and $BC = CD = BD = \sqrt{21}$ . When we drop a perpendicular to the centroid of $BCD$ from $A$ (let's call this point $O$ ), we have $BO = \frac{\sqrt{21}}{\sqrt{3}} = \sqrt{7}$ . Thus,
\[AB^2 - BO^2 = AO^2\] \[13 - 7 = AO^2 = 6\] \[AO = \sqrt{6}.\]
The area of base $BCD$ is $\frac{21\sqrt{3}}{4}$ . Hence,
\[\textrm{Vol}(ABCD) = \frac{\sqrt{6}\cdot\frac{21\sqrt{3}}{4}}{3}\] \[= \frac{63\sqrt{2}}{12}.\]
We can apply a similar approach to $A'B'C'D'$
$A'B' = A'C' = A'D' = \sqrt{13}$ , and $B'C' = C'D' = B'D' = \sqrt{31}$ . When we drop a perpendicular to the centroid of $B'C'D'$ from $A'$ (let's call this point $O'$ ), we have $B'O' = \frac{\sqrt{31}}{\sqrt{3}} = \sqrt{\frac{31}{3}}$ . Thus,
\[A'B'^2 - B'O'^2 = A'O'^2\] \[13 - \frac{31}{3} = A'O'^2\] \[A'O' = \sqrt{8}{3} = \frac{2\sqrt{6}}{3}.\]
The area of base $B'C'D'$ is $\frac{31\sqrt{3}}{4}$ . Hence,
\[\textrm{Vol}(A'B'C'D') = \frac{\frac{2\sqrt{6}}{3}\cdot\frac{31\sqrt{3}}{4}}{3}\] \[= \frac{186\sqrt{2}}{36}\] \[= \frac{62\sqrt{2}}{12}.\]
Finally,
\[\frac{P}{P'} = \frac{\textrm{Vol}(ABCD)}{\textrm{Vol}(A'B'C'D')} = \frac{\frac{63\sqrt{2}}{12}}{\frac{62\sqrt{2}}{12}} = \frac{63}{62}.\]
Our answer is $63 + 62 = \boxed{125}$

Solution:
Since the two parallelepipeds have the same base, all we need to do is to find their respective heights.

As illustrated in the above diagram, drop a perpendicular from $D$ onto the base at $P$ . Denote the center of the base by $O$ . By symmetry, $P$ must be on $AO$ . Now we need to find $DP$
Apply Pythagorean theorem to $\triangle DPA$ we have \[DP^2 = AD^2 - AP^2.\]
Apply Pythagorean theorem to $\triangle DPC$ and then $\triangle CPO$ we have \[DP^2 = DC^2 - CP^2 = DC^2 - (CO^2 + OP^2) = DC^2 - (CO^2 + (AO-AP)^2) = DC^2 - CO^2 - (AO-AP)^2.\]
Combining the above two, we have \[AD^2 - AP^2 = DC^2 - CO^2 - (AO-AP)^2.\]
Since $AD=\sqrt{13}$ $DC=\sqrt{21}$ $CO=\frac{\sqrt{21}}{2}$ $AO=\frac{\sqrt{31}}{2}$ , plug them into the above equation and solve for the only unknown variable $AP$ , we get $AP=\frac{5}{\sqrt{31}}.$
Thus the height \[DP = \sqrt{AD^2 - AP^2} = \sqrt{13 - \frac{25}{31}} = \sqrt{\frac{378}{31}}.\]

For the other parallelepiped, using the same approach and drop a perpendicular from $E'$ onto the base at $P'$ . Similarly applying Pythagorean theorem to $\triangle E'P'C'$ $\triangle E'P'A'$ and $\triangle A'P'O'$ we have \[C'E'^2 - C'P'^2 = A'E'^2 - A'O'^2 - (C'P'-C'O')^2.\]
Plugging known values into the above equation and solve for the only unknown variable $C'P'$ , we get $C'P'=\frac{5}{\sqrt{21}}.$
Thus the height \[E'P' = \sqrt{C'E'^2 - C'P'^2} = \sqrt{13 - \frac{25}{21}} = \sqrt{\frac{248}{21}}.\]
The ratio between the two is therefore \[\frac{DP}{E'P'} = \frac{\sqrt{\frac{378}{31}}}{\sqrt{\frac{248}{21}}} = \sqrt{\frac{2\cdot3^3\cdot7}{31}\cdot\frac{3\cdot7}{2^3\cdot31}} = \frac{3^2\cdot7}{2\cdot31} = \frac{63}{62}\] , giving $\boxed{125}$


Answer: 5.0

Each half of this figure is composed of 3 red triangles, 5 blue triangles and 8 white triangles. When the upper half is folded down over the centerline, 2 pairs of red triangles coincide, as do 3 pairs of blue triangles. There are 2 red-white pairs. How many white pairs coincide?
[asy] draw((0,0)--(4,4*sqrt(3))); draw((1,-sqrt(3))--(5,3*sqrt(3))); draw((2,-2*sqrt(3))--(6,2*sqrt(3))); draw((3,-3*sqrt(3))--(7,sqrt(3))); draw((4,-4*sqrt(3))--(8,0)); draw((8,0)--(4,4*sqrt(3))); draw((7,-sqrt(3))--(3,3*sqrt(3))); draw((6,-2*sqrt(3))--(2,2*sqrt(3))); draw((5,-3*sqrt(3))--(1,sqrt(3))); draw((4,-4*sqrt(3))--(0,0)); draw((3,3*sqrt(3))--(5,3*sqrt(3))); draw((2,2*sqrt(3))--(6,2*sqrt(3))); draw((1,sqrt(3))--(7,sqrt(3))); draw((-1,0)--(9,0)); draw((1,-sqrt(3))--(7,-sqrt(3))); draw((2,-2*sqrt(3))--(6,-2*sqrt(3))); draw((3,-3*sqrt(3))--(5,-3*sqrt(3))); [/asy]
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 7 \qquad \textbf{(E)}\ 9$

Solution: Each half has $3$ red triangles, $5$ blue triangles, and $8$ white triangles. There are also $2$ pairs of red triangles, so $2$ red triangles on each side are used, leaving $1$ red triangle, $5$ blue triangles, and $8$ white triangles remaining on each half. Also, there are $3$ pairs of blue triangles, using $3$ blue triangles on each side, so there is $1$ red triangle, $2$ blue triangles, and $8$ white triangles remaining on each half. Also, we have $2$ red-white pairs. This obviously can't use $2$ red triangles on one side, since there is only $1$ on each side, so we must use $1$ red triangle and $1$ white triangle per side, leaving $2$ blue triangles and $7$ white triangles apiece. The remaining blue triangles cannot be matched with other blue triangles since that would mean there were more than $3$ blue pairs, so the remaining blue triangles must be paired with white triangles, yielding $4$ blue-white pairs, one for each of the remaining blue triangles. This uses $2$ blue triangles and $2$ white triangles on each side, leaving $5$ white triangles apiece, which must be paired with each other, so there are $5$ white-white pairs, $\boxed{5}$


Answer: 2.0

Each morning of her five-day workweek, Jane bought either a $50$ -cent muffin or a $75$ -cent bagel. Her total cost for the week was a whole number of dollars. How many bagels did she buy?
$\textbf{(A) } 1\qquad\textbf{(B) } 2\qquad\textbf{(C) } 3\qquad\textbf{(D) } 4\qquad\textbf{(E) } 5$

Solution:
If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by $25$ cents.
If Jane bought $1$ bagel, then she bought $4$ muffins. Her total cost for the week would be $75\cdot1+50\cdot4=275$ cents, or $2.75$ dollars. Clearly, she bought one more bagel but one fewer muffin at a total cost of $3.00$ dollars. Therefore, she bought $\boxed{2}$ bagels.


Answer: 2.0

Solution:
If Jane bought one more bagel but one fewer muffin, then her total cost for the week would increase by $25$ cents.
If Jane bought $1$ bagel, then she bought $4$ muffins. Her total cost for the week would be $75\cdot1+50\cdot4=275$ cents, or $2.75$ dollars. Clearly, she bought one more bagel but one fewer muffin at a total cost of $3.00$ dollars. Therefore, she bought $\boxed{2}$ bagels.


Answer: 45.0

Each of $2010$ boxes in a line contains a single red marble, and for $1 \le k \le 2010$ , the box in the $k\text{th}$ position also contains $k$ white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles. What is the smallest value of $n$ for which $P(n) < \frac{1}{2010}$
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 63 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 201 \qquad \textbf{(E)}\ 1005$

Solution:
The probability of drawing a white marble from box $k$ is $\frac{k}{k + 1}$ , and the probability of drawing a red marble from box $k$ is $\frac{1}{k+1}$
To stop after drawing $n$ marbles, we must draw a white marble from boxes $1, 2, \ldots, n-1,$ and draw a red marble from box $n.$ Thus, \[P(n) = \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac {n - 1}{n}\right) \cdot \frac{1}{n +1} = \frac{1}{n (n + 1)}.\]
So, we must have $\frac{1}{n(n + 1)} < \frac{1}{2010}$ or $n(n+1) > 2010.$
Since $n(n+1)$ increases as $n$ increases, we can simply test values of $n$ ; after some trial and error, we get that the minimum value of $n$ is $\boxed{45}$ , since $45(46) = 2070$ but $44(45) = 1980.$


Answer: 45.0

Solution:
The probability of drawing a white marble from box $k$ is $\frac{k}{k + 1}$ , and the probability of drawing a red marble from box $k$ is $\frac{1}{k+1}$
To stop after drawing $n$ marbles, we must draw a white marble from boxes $1, 2, \ldots, n-1,$ and draw a red marble from box $n.$ Thus, \[P(n) = \left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac {n - 1}{n}\right) \cdot \frac{1}{n +1} = \frac{1}{n (n + 1)}.\]
So, we must have $\frac{1}{n(n + 1)} < \frac{1}{2010}$ or $n(n+1) > 2010.$
Since $n(n+1)$ increases as $n$ increases, we can simply test values of $n$ ; after some trial and error, we get that the minimum value of $n$ is $\boxed{45}$ , since $45(46) = 2070$ but $44(45) = 1980.$


Answer: 14.0

Each of $2023$ balls is randomly placed into one of $3$ bins. Which of the following is closest to the probability that each of the bins will contain an odd number of balls?
$\textbf{(A) } \frac{2}{3} \qquad\textbf{(B) } \frac{3}{10} \qquad\textbf{(C) } \frac{1}{2} \qquad\textbf{(D) } \frac{1}{3} \qquad\textbf{(E) } \frac{1}{4}$

Solution:
We first examine the possible arrangements for parity of number of balls in each box for $2022$ balls.
If a $E$ denotes an even number and a $O$ denotes an odd number, then the distribution of balls for $2022$ balls could be $EEE,EOO,OEO,$ or $OOE$ . With the insanely overpowered magic of cheese, we assume that each case is about equally likely.
From $EEE$ , it is not possible to get to all odd by adding one ball; we could either get $OEE,EOE,$ or $EEO$ . For the other $3$ cases, though, if we add a ball to the exact right place, then it'll work.
For each of the working cases, we have $1$ possible slot the ball can go into (for $OEO$ , for example, the new ball must go in the center slot to make $OOO$ ) out of the $3$ slots, so there's a $\dfrac13$ chance. We have a $\dfrac34$ chance of getting one of these working cases, so our answer is $\dfrac34\cdot\dfrac13=\boxed{14.}$


Answer: 14.0

Solution:
We first examine the possible arrangements for parity of number of balls in each box for $2022$ balls.
If a $E$ denotes an even number and a $O$ denotes an odd number, then the distribution of balls for $2022$ balls could be $EEE,EOO,OEO,$ or $OOE$ . With the insanely overpowered magic of cheese, we assume that each case is about equally likely.
From $EEE$ , it is not possible to get to all odd by adding one ball; we could either get $OEE,EOE,$ or $EEO$ . For the other $3$ cases, though, if we add a ball to the exact right place, then it'll work.
For each of the working cases, we have $1$ possible slot the ball can go into (for $OEO$ , for example, the new ball must go in the center slot to make $OOO$ ) out of the $3$ slots, so there's a $\dfrac13$ chance. We have a $\dfrac34$ chance of getting one of these working cases, so our answer is $\dfrac34\cdot\dfrac13=\boxed{14.}$


Answer: 64.0

Each of the $100$ students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. There are $42$ students who cannot sing, $65$ students who cannot dance, and $29$ students who cannot act. How many students have two of these talents?
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 25\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 64$

Solution:
Let $a$ be the number of students that can only sing, $b$ can only dance, and $c$ can only act.
Let $ab$ be the number of students that can sing and dance, $ac$ can sing and act, and $bc$ can dance and act.
From the information given in the problem, $a + ab + b = 29, b + bc + c = 42,$ and $a + ac + c = 65$
Adding these equations together, we get $2(a + b + c) + ab + bc + ac = 136$
Since there are a total of $100$ students, $a + b + c + ab + bc + ac = 100$
Subtracting these equations, we get $a + b + c = 36$
Our answer is $ab + bc + ac = 100 - (a + b + c) = 100 - 36 = \boxed{64}$

Solution:
An easier way to solve the problem:
Since $42$ students cannot sing, there are $100-42=58$ students who can.
Similarly $65$ students cannot dance, there are $100-65=35$ students who can.
And $29$ students cannot act, there are $100-29=71$ students who can.
Therefore, there are $58+35+71=164$ students in all ignoring the overlaps between $2$ of $3$ talent categories.
There are no students who have all $3$ talents, nor those who have none $(0)$ , so only $1$ or $2$ talents are viable.
Thus, there are $164-100=\boxed{64}$ students who have $2$ of $3$ talents.