problem: Every morning Aya goes for a $9$-kilometer-long walk and stops at a coffee shop afterwards. When she walks at a constant speed of $s$ kilometers per hour, the walk takes her 4 hours, including $t$ minutes spent in the coffee shop. When she walks $s+2$ kilometers per hour, the walk takes her 2 hours and 24 minutes, including $t$ minutes spent in the coffee shop. Suppose Aya walks at $s+\frac{1}{2}$ kilometers per hour. Find the number of minutes the walk takes her, including the $t$ minutes spent in the coffee shop.

solution:
$\frac{9}{s} + t = 4$ in hours and $\frac{9}{s+2} + t = 2.4$ in hours.
Subtracting the second equation from the first, we get,
$\frac{9}{s} - \frac{9}{s+2} = 1.6$
Multiplying by $(s)(s+2)$, we get
$9s+18-9s=18=1.6s^{2} + 3.2s$
Multiplying by 5/2 on both sides, we get
$0 = 4s^{2} + 8s - 45$
Factoring gives us
$(2s-5)(2s+9) = 0$, of which the solution we want is $s=2.5$.
Substituting this back to the first equation, we can find that $t = 0.4$ hours.
Lastly, $s + \frac{1}{2} = 3$ kilometers per hour, so
$\frac{9}{3} + 0.4 = 3.4$ hours, or $\framebox{204}$ minutes
-Failure.net
The amount of hours spent while walking on the first travel is $\frac{240-t}{6}$. Thus, we have the equation $(240-t)(s) = 540$, and by the same logic, the second equation yields $(144-t)(s+2) = 540$. We have $240s-st = 540$, and $288+144s-2t-st = 540$. We subtract the two equations to get $96s+2t-288 = 0$, so we have $48s+t = 144$, so $t = 144-48s$, and now we have $(96+48s)(s) = 540$. The numerator of $s$ must evenly divide 540, however, $s$ must be less than 3. We can guess that $s = 2.5$. Now, $2.5+0.5 = 3$. Taking $\frac{9}{3} = 3$, we find that it will take three hours for the 9 kilometers to be traveled. The t minutes spent at the coffeeshop can be written as $144-48(2.5)$, so t = 24. $180 + 24 = 204$. -sepehr2010

answer: 204

question: Every morning Aya goes for a $9$-kilometer-long walk and stops at a coffee shop afterwards. When she walks at a constant speed of $s$ kilometers per hour, the walk takes her 4 hours, including $t$ minutes spent in the coffee shop. When she walks $s+2$ kilometers per hour, the walk takes her 2 hours and 24 minutes, including $t$ minutes spent in the coffee shop. Suppose Aya walks at $s+\frac{1}{2}$ kilometers per hour. Find the number of minutes the walk takes her, including the $t$ minutes spent in the coffee shop.


problem: There exist real numbers $x$ and $y$, both greater than 1, such that $\log_x\left(y^x\right)=\log_y\left(x^{4y}\right)=10$. Find $xy$.

solution:
By properties of logarithms, we can simplify the given equation to $x\log_xy=4y\log_yx=10$. Let us break this into two separate equations:
\[x\log_xy=10\]
\[4y\log_yx=10.\]
We multiply the two equations to get:
\[4xy\left(\log_xy\log_yx\right)=100.\]
Also by properties of logarithms, we know that $\log_ab\cdot\log_ba=1$; thus, $\log_xy\cdot\log_yx=1$. Therefore, our equation simplifies to:
\[4xy=100\implies xy=\boxed{025}.\]
~Technodoggo
Convert the two equations into exponents:
\[x^{10}=y^x~(1)\]
\[y^{10}=x^{4y}~(2).\]
Take $(1)$ to the power of $\frac{1}{x}$:
\[x^{\frac{10}{x}}=y.\]
Plug this into $(2)$:
\[x^{(\frac{10}{x})(10)}=x^{4(x^{\frac{10}{x}})}\]
\[{\frac{100}{x}}={4x^{\frac{10}{x}}}\]
\[{\frac{25}{x}}={x^{\frac{10}{x}}}=y,\]
So $xy=\boxed{025}$
~alexanderruan
Similar to solution 2, we have:
$x^{10}=y^x$ and $y^{10}=x^{4y}$
Take the tenth root of the first equation to get
$x=y^{\frac{x}{10}}$
Substitute into the second equation to get
$y^{10}=y^{\frac{4xy}{10}}$
This means that $10=\frac{4xy}{10}$, or $100=4xy$, meaning that $xy=\boxed{25}$.
~MC413551

answer: 025

question: There exist real numbers $x$ and $y$, both greater than 1, such that $\log_x\left(y^x\right)=\log_y\left(x^{4y}\right)=10$. Find $xy$.


problem: Alice and Bob play the following game. A stack of $n$ tokens lies before them. The players take turns with Alice going first. On each turn, the player removes either $1$ token or $4$ tokens from the stack. Whoever removes the last token wins. Find the number of positive integers $n$ less than or equal to $2024$ for which there exists a strategy for Bob that guarantees that Bob will win the game regardless of Alice's play.

solution:
Let's first try some experimentation. Alice obviously wins if there is one coin. She will just take it and win. If there are 2 remaining, then Alice will take one and then Bob will take one, so Bob wins. If there are $3$, Alice will take $1$, Bob will take one, and Alice will take the final one. If there are $4$, Alice will just remove all $4$ at once. If there are $5$, no matter what Alice does, Bob can take the final coins in one try. Notice that Alice wins if there are $1$, $3$, or $4$ coins left. Bob wins if there are $2$ or $5$ coins left.
After some thought, you may realize that there is a strategy for Bob. If there is n is a multiple of $5$, then Bob will win. The reason for this is the following: Let's say there are a multiple of $5$ coins remaining in the stack. If Alice takes $1$, Bob will take $4$, and there will still be a multiple of $5$. If Alice takes $4$, Bob will take $1$, and there will still be a multiple of $5$. This process will continue until you get $0$ coins left. For example, let's say there are $205$ coins. No matter what Alice does, Bob can simply just do the complement. After each of them make a turn, there will always be a multiple of $5$ left. This will continue until there are $5$ coins left, and Bob will end up winning.
After some more experimentation, you'll realize that any number that is congruent to $2$ mod $5$ will also work. This is because Bob can do the same strategy, and when there are $2$ coins left, Alice is forced to take $1$ and Bob takes the final coin. For example, let's say there are $72$ coins. If Alice takes $1$, Bob will take $4$. If Alice takes $4$, Bob will take $1$. So after they each make a turn, the number will always be equal to $2$ mod $5$. Eventually, there will be only $2$ coins remaining, and we've established that Alice will simply take $1$ and Bob will take the final coin.
So we have to find the number of numbers less than or equal to $2024$ that are either congruent to $0$ mod $5$ or $2$ mod $5$. There are $404$ numbers in the first category: $5, 10, 15, \dots, 2020$. For the second category, there are $405$ numbers. $2, 7, 12, 17, \dots, 2022$. So the answer is $404 + 405 = \boxed{809}$
~lprado
We will use winning and losing positions, where a $W$ marks when Alice wins and an $L$ marks when Bob wins.
$1$ coin: $W$
$2$ coins: $L$
$3$ coins: $W$
$4$ coins: $W$
$5$ coins: $L$
$6$ coin: $W$
$7$ coins: $L$
$8$ coins: $W$
$9$ coins: $W$
$10$ coins: $L$
$11$ coin: $W$
$12$ coins: $L$
$13$ coins: $W$
$14$ coins: $W$
$15$ coins: $L$
We can see that losing positions occur when $n$ is congruent to $0, 2 \mod{5}$ and winning positions occur otherwise. In other words, there will be $2$ losing positions out of every $5$ consecutive values of n. As $n$ ranges from $1$ to $2020$, $\frac{2}{5}$ of these values are losing positions where Bob will win. As $n$ ranges from $2021$ to $2024$, $2022$ is the only value where Bob will win. Thus, the answer is $2020\times\frac{2}{5}+1=\boxed{809}$
~alexanderruan
Denote by $A_i$ and $B_i$ Alice's or Bob's $i$th moves, respectively.
Case 1: $n \equiv 0 \pmod{5}$.
Bob can always take the strategy that $B_i = 5 - A_i$.
This guarantees him to win.
In this case, the number of $n$ is $\left\lfloor \frac{2024}{5} \right\rfloor = 404$.
Case 2: $n \equiv 1 \pmod{5}$.
In this case, consider Alice's following strategy: $A_1 = 1$ and $A_i = 5 - B_{i-1}$ for $i \geq 2$.
Thus, under Alice's this strategy, Bob has no way to win.
Case 3: $n \equiv 4 \pmod{5}$.
In this case, consider Alice's following strategy: $A_1 = 4$ and $A_i = 5 - B_{i-1}$ for $i \geq 2$.
Thus, under Alice's this strategy, Bob has no way to win.
Case 4: $n \equiv 2 \pmod{5}$.
Bob can always take the strategy that $B_i = 5 - A_i$.
Therefore, after the $\left\lfloor \frac{n}{5} \right\rfloor$th turn, there are two tokens leftover.
Therefore, Alice must take 1 in the next turn that leaves the last token on the table.
Therefore, Bob can take the last token to win the game.
This guarantees him to win.
In this case, the number of $n$ is $\left\lfloor \frac{2024 - 2}{5} \right\rfloor +1 = 405$.
Case 5: $n \equiv 3 \pmod{5}$.
Consider Alice's following strategy: $A_1 = 1$ and $A_i = 5 - B_{i-1}$ for $i \geq 2$.
By doing so, there will finally be 2 tokens on the table and Bob moves first. Because Bob has the only choice of taking 1 token, Alice can take the last token and win the game.
Therefore, in this case, under Alice's this strategy, Bob has no way to win.
Putting all cases together, the answer is $404 + 405 = \boxed{\textbf{(809) }}$.
Since the game Alice and Bob play is impartial (the only difference between player 1 and player 2 is that player 1 goes first (note that games like chess are not impartial because each player can only move their own pieces)), we can use the Sprague-Grundy Theorem to solve this problem. We will use induction to calculate the Grundy Values for this game.
We claim that heaps of size congruent to $0,2 \bmod{5}$ will be in outcome class $\mathcal{P}$ (win for player 2 = Bob), and heaps of size equivalent to $1,3,4 \bmod{5}$ will be in outcome class $\mathcal{N}$ (win for player 1 = Alice). Note that the mex (minimal excludant) of a set of nonnegative integers is the least nonnegative integer not in the set. e.g. mex$(1, 2, 3) = 0$ and mex$(0, 1, 2, 4) = 3$.

$\text{heap}(0) = \{\} = *\text{mex}(\emptyset) = 0$

$\text{heap}(1) = \{0\} = *\text{mex}(0) = *$

$\text{heap}(2) = \{*\} = *\text{mex}(1) = 0$

$\text{heap}(3) = \{0\} = *\text{mex}(0) = *$

$\text{heap}(4) = \{0, *\} = *\text{mex}(0, 1) = *2$

$\text{heap}(5) = \{*, *2\} = *\text{mex}(1, 2) = 0$

$\text{heap}(6) = \{0, 0\} = *\text{mex}(0, 0) = *$

$\text{heap}(7) = \{*, *\} = *\text{mex}(1, 1) = 0$

$\text{heap}(8) = \{*2, 0\} = *\text{mex}(0, 2) = *$

$\text{heap}(9) = \{0, *\} = *\text{mex}(0, 1) = *2$

$\text{heap}(10) = \{*, *2\} = *\text{mex}(1, 2) = 0$

We have proven the base case. We will now prove the inductive hypothesis: If $n \equiv 0 \bmod{5}$, $\text{heap}(n) = 0$, $\text{heap}(n+1) = *$, $\text{heap}(n+2) = 0$, $\text{heap}(n+3) = *$, and $\text{heap}(n+4) = *2$, then $\text{heap}(n+5) = 0$, $\text{heap}(n+6) = *$, $\text{heap}(n+7) = 0$, $\text{heap}(n+8) = *$, and $\text{heap}(n+9) = *2$.

$\text{heap}(n+5) = \{\text{heap}(n+1), \text{heap}(n+4)\} = \{*, *2\} = *\text{mex}(1, 2) = 0$

$\text{heap}(n+6) = \{\text{heap}(n+2), \text{heap}(n+5)\} = \{0, 0\} = *\text{mex}(0, 0) = *$

$\text{heap}(n+7) = \{\text{heap}(n+3), \text{heap}(n+6)\} = \{*, *\} = *\text{mex}(1, 1) = 0$

$\text{heap}(n+8) = \{\text{heap}(n+4), \text{heap}(n+7)\} = \{*2, 0\} = *\text{mex}(2, 1) = *$

$\text{heap}(n+9) = \{\text{heap}(n+5), \text{heap}(n+8)\} = \{0, *\} = *\text{mex}(0, 1) = *2$

We have proven the inductive hypothesis. QED.
There are $2020*\frac{2}{5}=808$ positive integers congruent to $0,2 \bmod{5}$ between 1 and 2020, and 1 such integer between 2021 and 2024. $808 + 1 = \boxed{809}$.

~numerophile
We start with $n$ as some of the smaller values. After seeing the first 4 where Bob wins automatically, with trial and error we see that $2, 5, 7,$ and $10$ are spaced alternating in between 2 and 3 apart. This can also be proven with modular arithmetic, but  this is an easier solution for some people. We split them into 2 different sets with common difference 5: {2,7,12 ...} and {5,10,15...}. Counting up all the numbers in each set can be done as follows:
Set 1 ${2,7,12...}$
$2024-2=2022$ (because the first term is two)
$\lfloor \frac{2024}{5} \rfloor = 404$
Set 2 ${5,10,15}$
$\lfloor \frac{2024}{5} \rfloor = 404$

And because we forgot 2022 we add 1 more.
$404+404+1=809$
-Multpi12
(Edits would be appreciated)
LaTexed by BossLu99

answer: 809

question: Alice and Bob play the following game. A stack of $n$ tokens lies before them. The players take turns with Alice going first. On each turn, the player removes either $1$ token or $4$ tokens from the stack. Whoever removes the last token wins. Find the number of positive integers $n$ less than or equal to $2024$ for which there exists a strategy for Bob that guarantees that Bob will win the game regardless of Alice's play.


problem: Jen enters a lottery by picking $4$ distinct numbers from $S=\{1,2,3,\cdots,9,10\}.$ $4$ numbers are randomly chosen from $S.$ She wins a prize if at least two of her numbers were $2$ of the randomly chosen numbers, and wins the grand prize if all four of her numbers were the randomly chosen numbers. The probability of her winning the grand prize given that she won a prize is $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

solution:
This is a conditional probability problem. Bayes' Theorem states that
\[P(A|B)=\dfrac{P(B|A)\cdot P(A)}{P(B)}\]

in other words, the probability of $A$ given $B$ is equal to the probability of $B$ given $A$ times the probability of $A$ divided by the probability of $B$. In our case, $A$ represents the probability of winning the grand prize, and $B$ represents the probability of winning a prize. Clearly, $P(B|A)=1$, since by winning the grand prize you automatically win a prize. Thus, we want to find $\dfrac{P(A)}{P(B)}$.
Let us calculate the probability of winning a prize. We do this through casework: how many of Jen's drawn numbers match the lottery's drawn numbers?
To win a prize, Jen must draw at least $2$ numbers identical to the lottery. Thus, our cases are drawing $2$, $3$, or $4$ numbers identical.
Let us first calculate the number of ways to draw exactly $2$ identical numbers to the lottery. Let Jen choose the numbers $a$, $b$, $c$, and $d$; we have $\dbinom42$ ways to choose which $2$ of these $4$ numbers are identical to the lottery. We have now determined $2$ of the $4$ numbers drawn in the lottery; since the other $2$ numbers Jen chose can not be chosen by the lottery, the lottery now has $10-2-2=6$ numbers to choose the last $2$ numbers from. Thus, this case is $\dbinom62$, so this case yields $\dbinom42\dbinom62=6\cdot15=90$ possibilities.
Next, let us calculate the number of ways to draw exactly $3$ identical numbers to the lottery. Again, let Jen choose $a$, $b$, $c$, and $d$. This time, we have $\dbinom43$ ways to choose the identical numbers and again $6$ numbers left for the lottery to choose from; however, since $3$ of the lottery's numbers have already been determined, the lottery only needs to choose $1$ more number, so this is $\dbinom61$. This case yields $\dbinom43\dbinom61=4\cdot6=24$.
Finally, let us calculate the number of ways to all $4$ numbers matching. There is actually just one way for this to happen.
In total, we have $90+24+1=115$ ways to win a prize. The lottery has $\dbinom{10}4=210$ possible combinations to draw, so the probability of winning a prize is $\dfrac{115}{210}$. There is actually no need to simplify it or even evaluate $\dbinom{10}4$ or actually even know that it has to be $\dbinom{10}4$; it suffices to call it $a$ or some other variable, as it will cancel out later. However, let us just go through with this. The probability of winning a prize is $\dfrac{115}{210}$. Note that the probability of winning a grand prize is just matching all $4$ numbers, which we already calculated to have $1$ possibility and thus have probability $\dfrac1{210}$. Thus, our answer is $\dfrac{\frac1{210}}{\frac{115}{210}}=\dfrac1{115}$. Therefore, our answer is $1+115=\boxed{116}$.
~Technodoggo
For getting all $4$ right, there is only $1$ way.
For getting $3$ right, there is $\dbinom43$ multiplied by $\dbinom61$ = $24$ ways.
For getting $2$ right, there is $\dbinom42$ multiplied by $\dbinom62$ = $90$ ways.
$\frac{1}{1+24+90}$  = $\frac{1}{115}$
Therefore, the answer is $1+115 = \boxed{116}$
~e___

answer: 116

question: Jen enters a lottery by picking $4$ distinct numbers from $S=\{1,2,3,\cdots,9,10\}.$ $4$ numbers are randomly chosen from $S.$ She wins a prize if at least two of her numbers were $2$ of the randomly chosen numbers, and wins the grand prize if all four of her numbers were the randomly chosen numbers. The probability of her winning the grand prize given that she won a prize is $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.


problem: Consider the paths of length $16$ that follow the lines from the lower left corner to the upper right corner on an $8\times 8$ grid. Find the number of such paths that change direction exactly four times, as in the examples shown below.

solution:
We divide the path into eight “$R$” movements and eight “$U$” movements. Five sections of alternative $RURUR$ or $URURU$ are necessary in order to make four “turns.” We use the first case and multiply by $2$.

For $U$, we have seven ordered pairs of positive integers $(a,b)$ such that $a+b=8$.
For $R$, we subtract $1$ from each section (to make the minimum stars of each section $0$) and we use Stars and Bars to get ${7 \choose 5}=21$.

Thus our answer is $7\cdot21\cdot2=\boxed{294}$.
~eevee9406
Draw a few examples of the path. However, notice one thing in common - if the path starts going up, there will be 3 "segments" where the path goes up, and two horizontal "segments." Similarly, if the path starts going horizontally, we will have three horizontal segments and two vertical segments. Those two cases are symmetric, so we only need to consider one. If our path starts going up, by stars and bars, we can have $\binom{7}{2}$ ways to split the 8 up's into 3 lengths, and have $\binom{7}{1}$ to split the 8-horizontals into 2 lengths. We multiply them together, and multiply by 2 for symmetry, giving us $2*\binom{7}{2}*\binom{7}{1}=294.$
~nathan27 (original by alexanderruan)
Notice that the $RURUR$ case and the $URURU$ case is symmetrical. WLOG, let's consider the RURUR case.
Now notice that there is a one-to-one correspondence between this problem and the number of ways to distribute 8 balls into 3 boxes and also 8 other balls into 2 other boxes, such that each box has a nonzero amount of balls.
There are ${8+2-3 \choose 2}$ ways for the first part, and ${8+1-2 \choose 1}$ ways for the second part, by stars and bars.
The answer is $2\cdot {7 \choose 2} \cdot {7 \choose 1} = \boxed{294}$.
~northstar47
Feel free to edit this solution
Starting at the origin, you can either first go up or to the right. If you go up first, you will end on the side opposite to it (the right side) and if you go right first, you will end up on the top. It can then be observed that if you choose the turning points in the middle $7 \times 7$ grid, that will automatically determine your start and ending points. For example, in the diagram if you choose the point $(3,2)$ and $(5,3)$, you must first move three up or two right, determining your first point, and move 5 up or 3 right, determining your final point. Knowing this is helpful because if we first move anywhere horizontally, we have $7$ points on each column to choose from and starting from left to right, we have $6,5,4,3,2,1$ points on that row to choose from. This gives us $7(6)+7(5)+7(4)+7(3)+7(2)+7(1)$ which simplifies to $7\cdot21$. The vertical case is symmetrical so we have $7\cdot21\cdot2 = \boxed{294}$
~KEVIN_LIU
As in Solution 1, there are two cases: $RURUR$ or $URURU$. We will work with the first case and multiply by $2$ at the end. We use stars and bars; we can treat the $R$s as the stars and the $U$s as the bars. However, we must also use stars and bars on the $U$s to see how many different patterns of bars we can create for the reds. We must have $1$ bar in $8$ blacks, so we use stars and bars on the equation \[x + y = 8\]. However, each divider must have at least one black in it, so we do the change of variable $x' = x-1$ and $y' = x-1$. Our equation becomes \[x' + y' = 6\]. By stars and bars, this equation has $\binom{6 + 2 - 1}{1} = 7$ valid solutions. Now, we use stars and bars on the reds. We must distribute two bars amongst the reds, so we apply stars and bars to \[x + y + z = 8\]. Since each group must have one red, we again do a change of variables with $x' = x-1$, $y' = y-1$, and $z' = z-1$. We are now working on the equation \[x' + y' + z' = 5\]. By stars and bars, this has $\binom{5 + 3 - 1}{2} = 21$ solutions. The number of valid paths in this case is the number of ways to create the bars times the number of valid arrangements of the stars given fixed bars, which equals $21 \cdot 7 = 147$. We must multiply by two to account for both cases, so our final answer is $147 \cdot 2 = \boxed{294}$.
~ [cxsmi](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Cxsmi)

answer: 294

question: Consider the paths of length $16$ that follow the lines from the lower left corner to the upper right corner on an $8\times 8$ grid. Find the number of such paths that change direction exactly four times, as in the examples shown below.


problem: Find the largest possible real part of \[(75+117i)z+\frac{96+144i}{z}\]where $z$ is a complex number with $|z|=4$.

solution:
Let $z=a+bi$ such that $a^2+b^2=4^2=16$. The expression becomes:
\[(75+117i)(a+bi)+\dfrac{96+144i}{a+bi}.\]
Call this complex number $w$. We simplify this expression.
\begin{align*}
w&=(75+117i)(a+bi)+\dfrac{96+144i}{a+bi} \\
&=(75a-117b)+(117a+75b)i+48\left(\dfrac{2+3i}{a+bi}\right) \\
&=(75a-117b)+(116a+75b)i+48\left(\dfrac{(2+3i)(a-bi)}{(a+bi)(a-bi)}\right) \\
&=(75a-117b)+(116a+75b)i+48\left(\dfrac{2a+3b+(3a-2b)i}{a^2+b^2}\right) \\
&=(75a-117b)+(116a+75b)i+48\left(\dfrac{2a+3b+(3a-2b)i}{16}\right) \\
&=(75a-117b)+(116a+75b)i+3\left(2a+3b+(3a-2b)i\right) \\
&=(75a-117b)+(116a+75b)i+6a+9b+(9a-6b)i \\
&=(81a-108b)+(125a+69b)i. \\
\end{align*}
We want to maximize $\text{Re}(w)=81a-108b$. We can use elementary calculus for this, but to do so, we must put the expression in terms of one variable. Recall that $a^2+b^2=16$; thus, $b=\pm\sqrt{16-a^2}$. Notice that we have a $-108b$ in the expression; to maximize the expression, we want $b$ to be negative so that $-108b$ is positive and thus contributes more to the expression. We thus let $b=-\sqrt{16-a^2}$. Let $f(a)=81a-108b$. We now know that $f(a)=81a+108\sqrt{16-a^2}$, and can proceed with normal calculus.
\begin{align*}
f(a)&=81a+108\sqrt{16-a^2} \\
&=27\left(3a+4\sqrt{16-a^2}\right) \\
f'(a)&=27\left(3a+4\sqrt{16-a^2}\right)' \\
&=27\left(3+4\left(\sqrt{16-a^2}\right)'\right) \\
&=27\left(3+4\left(\dfrac{-2a}{2\sqrt{16-a^2}}\right)\right) \\
&=27\left(3-4\left(\dfrac a{\sqrt{16-a^2}}\right)\right) \\
&=27\left(3-\dfrac{4a}{\sqrt{16-a^2}}\right). \\
\end{align*}
We want $f'(a)$ to be $0$ to find the maximum.
\begin{align*}
0&=27\left(3-\dfrac{4a}{\sqrt{16-a^2}}\right) \\
&=3-\dfrac{4a}{\sqrt{16-a^2}} \\
3&=\dfrac{4a}{\sqrt{16-a^2}} \\
4a&=3\sqrt{16-a^2} \\
16a^2&=9\left(16-a^2\right) \\
16a^2&=144-9a^2 \\
25a^2&=144 \\
a^2&=\dfrac{144}{25} \\
a&=\dfrac{12}5 \\
&=2.4. \\
\end{align*}
We also find that $b=-\sqrt{16-2.4^2}=-\sqrt{16-5.76}=-\sqrt{10.24}=-3.2$.
Thus, the expression we wanted to maximize becomes $81\cdot2.4-108(-3.2)=81\cdot2.4+108\cdot3.2=\boxed{540}$.
~Technodoggo
Same steps as solution one until we get $\text{Re}(w)=81a-108b$. We also know $|z|=4$ or $a^2+b^2=16$. We want to find the line $81a-108b=k$ tangent to circle $a^2+b^2=16$.
Using $\frac{|ax+by+c|}{\sqrt{a^2+b^2}}=r$ we can substitute and get $\frac{|81(0)-108(0)-k|}{\sqrt{81^2+108^2}}=4$
\begin{align*} \frac{k}{\sqrt{18225}}&=4 \\\frac{k}{135}&=4 \\k&=\boxed{540} \end{align*}
~BH2019MV0
Follow Solution 1 to get $81a-108b$. We can let $a=4\cos\theta$ and $b=4\sin\theta$ as $|z|=4$, and thus we have $324\cos\theta-432\sin\theta$. Furthermore, we can ignore the negative sign in front of the second term as we are dealing with sine and cosine, so we finally wish to maximize $324\cos\theta+432\sin\theta$ for obviously positive $\cos\theta$ and $\sin\theta$.

Using the previous fact, we can use the [Cauchy-Schwarz Inequality](https://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality) to calculate the maximum. By the inequality, we have:
$(324^2+432^2)(\cos^2\theta+\sin^2\theta)\ge(324\cos\theta+432\sin\theta)^2$
$540^2\cdot1\ge(324\cos\theta+432\sin\theta)^2$
$\boxed{540}\ge324\cos\theta+432\sin\theta$
~eevee9406
Similar to the solutions above, we find that $Re((75+117i)z+\frac{96+144i}{z})=81a-108b=27(3a-4b)$, where $z=a+bi$. To maximize this expression, we must maximize $3a-4b$. Let this value be $x$. Solving for $a$ yields $a=\frac{x+4b}{3}$. From the given information we also know that $a^2+b^2=16$. Substituting $a$ in terms of $x$ and $b$ gives us $\frac{x^2+8bx+16b^2}{9}+b^2=16$. Combining fractions, multiplying, and rearranging, gives $25b^2+8xb+(x^2-144)=0$. This is useful because we want the maximum value of $x$ such that this quadratic has real roots which is easy to find using the discriminant. For the roots to be real, $(8x)^2-4(25)(x^2-144) \ge 0$. Now all that is left to do is to solve this inequality. Simplifying this expression, we get $-36x^2+14400 \ge 0$ which means $x^2 \le 400$ and $x \le 20$. Therefore the maximum value of $x$ is $20$ and $27 \cdot 20 = \boxed{540}$
~vsinghminhas
First, recognize the relationship between the reciprocal of a complex number $z$ with its conjugate $\overline{z}$, namely:
\[\frac{1}{z} \cdot \frac{\overline{z}}{\overline{z}} = \frac{\overline{z}}{|z|^2} = \frac{\overline{z}}{16}\]
Then, let $z = 4(\cos\theta + i\sin\theta)$ and $\overline{z} = 4(\cos\theta - i\sin\theta)$.
\begin{align*} Re \left ((75+117i)z+\frac{96+144i}{z} \right) &= Re\left ( (75+117i)z + (6+9i)\overline{z}    \right ) \\                                                &= 4 \cdot Re\left ( (75+117i)(\cos\theta + i\sin\theta) + (6+9i)(\cos\theta - i\sin\theta)    \right ) \\                                                &= 4 \cdot (75\cos\theta - 117\sin\theta + 6\cos\theta + 9\sin\theta) \\                                                &= 4 \cdot (81\cos\theta - 108\sin\theta) \\                                                &= 4\cdot 27 \cdot (3\cos\theta - 4\sin\theta) \end{align*}
Now, recognizing the 3 and 4 coefficients hinting at a 3-4-5 right triangle, we "complete the triangle" by rewriting our desired answer in terms of an angle of that triangle $\phi$ where $\cos\phi = \frac{3}{5}$ and $\sin\phi = \frac{4}{5}$
\begin{align*} 4\cdot 27 \cdot(3\cos\theta - 4\sin\theta) &= 4\cdot 27 \cdot 5 \cdot (\frac{3}{5}\cos\theta - \frac{4}{5}\sin\theta) \\                                                &= 540 \cdot (\cos\phi\cos\theta - \sin\phi\sin\theta) \\                                                &= 540 \cos(\theta + \phi) \end{align*}
Since the simple trig ratio is bounded above by 1, our answer is $\boxed{540}$
~ Cocoa @ [https://www.corgillogical.com/](https://artofproblemsolving.comhttps://www.corgillogical.com/)
(yes i am a corgi that does math)
Follow as solution 1 would to obtain $81a + 108\sqrt{16-a^2}.$
By the Cauchy-Schwarz Inequality, we have
\[(a^2 + (\sqrt{16-a^2})^2)(81^2 + 108^2) \geq (81a + 108\sqrt{16-a^2})^2,\]
so
\[4^2 \cdot 9^2 \cdot 15^2 \geq (81a + 108\sqrt{16-a^2})^2\]
and we obtain that $81a + 108\sqrt{16-a^2} \leq 4 \cdot 9 \cdot 15 = \boxed{540}.$

- [spectraldragon8](https://artofproblemsolving.comhttps://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8)
Follow solution 2 to get that we want to find the line $81a-108b=k$ tangent to circle $a^2+b^2=16$. The line turns into $a=\frac{k}{81}+\frac{4b}{3}$
Connect the center of the circle to the tangency point and the y-intercept of the line. Let the tangency point be $A$, the y-intercept be $C$, and the center be $B$. Drop the perpendicular from $A$ to $BC$ and call it $D$. Let $AD=3x$, $DC=4x$. Then, $BD=\sqrt{AB^2-AD^2}=\sqrt{16-9x^2}$. By similar triangles, get that $\frac{BD}{AD}=\frac{AD}{DC}$, so $\frac{\sqrt{16-9x^2}}{3x}=\frac{3x}{4x}$. Solve this to get that $x=\frac{16}{15}$, so $BC=\frac{20}{3}$ and $\frac{k}{81}=\frac{20}{3}$, so $k=\boxed{540}$
~ryanbear
Because $|z|=4$, we can let $z=4e^{i\theta}$. Then, substituting $i=e^{\frac{i\pi}{2}}$, we get that the complex number is
\begin{align*}
w&=4e^{i\theta}(75+117e^{\frac{i\pi}{2}})+\dfrac{1}{4}e^{-i\theta}(96+144e^{\frac{i\pi}{2}})\\
&=300e^{i\theta}+468e^{i(\frac{\pi}{2}+\theta)}+24e^{-i\theta}+36e^{i(\frac{\pi}{2}-\theta)}\\
\end{align*}
We know that the $\text{Re}(e^{i\alpha})=\cos(\alpha)$ from Euler's formula, so applying this and then applying trig identities yields
\begin{align*}
\text{Re}(w)&=300\cos{(\theta)}+468\cos{(\dfrac{\pi}{2}+\theta)}+24\cos{(-\theta)}+36\cos{(\dfrac{\pi}{2}-\theta)}\\
&=300\cos{(\theta)}-468sin{(\theta)}+24\cos{(\theta)}+36\sin{(\theta)}\\
&=324\cos{(\theta)}-432\sin{(\theta)}\\
\implies \dfrac{1}{108}\text{Re}(w)&=3\cos{(\theta)}-4\sin{(\theta)}\\
\end{align*}
We can see that the right-hand side looks an awful lot like the sum of angles formula for cosine, but 3 and 4 don't satisfy the pythagorean identity. To make them do so, we can divide everything by $\sqrt{3^2+4^2}=5$ and set $\cos{(\alpha)}::=\frac{3}{5}$ and $\sin{(\alpha)}::=\frac{4}{5}$. Now we have that
\[\dfrac{1}{540}\text{Re}(w)=\cos{(\theta+\alpha)}\]
Obviously the maximum value of the right hand side is 1, so the maximum value of the real part is $\boxed{540}$.
~Mooshiros
Let $c$ denote value of the above expression such that $\mathsf{Re} (c)$ is maximized. We write $z=4e^{i\theta}$ and multiply the second term in the expression by $\overline{z} = 4e^{-i\theta},$ turning the expression into
\[4e^{i\theta}(75+117i) + \frac{(96 + 144i)\cdot 4e^{-i\theta}}{4e^{i\theta}\cdot 4e^{-i\theta}} = 300e^{i\theta} + 468ie^{i\theta} + (24+ 36i)e^{-i\theta}.\]
Now, we write $e^{i\theta} = \cos\theta + i\sin\theta$. Since $\cos$ is even and $\sin$ is odd,
\begin{align*} &300(\cos\theta + i\sin\theta) +468i + (24+36i)(\cos\theta -i\sin\theta) \\ \iff & \mathsf{Re}(c) = 324\cos\theta -468\sin\theta \end{align*}
We want to maximize this expression, so we take its derivative and set it equal to $0$ (and quickly check the second derivative for inflection points):
\begin{align*}     &\mathsf{Re}(c) = 108\left(3\cos\theta - 4\sin\theta\right)\\     \frac{d}{d\theta} &\mathsf{Re}(c) = -324\sin\theta -468\cos\theta = 0, \end{align*}
so $\tan\theta = -\dfrac{468}{324} = -\dfrac{4}{3},$ which is reminiscent of a $3-4-5$ right triangle in the fourth quadrant (side lengths of $3, -4, 5$). Since $\tan\theta = -\frac{4}{3},$ we quickly see that $\sin\theta = -\dfrac{4}{5}$ and $\cos\theta = \dfrac{3}{5}.$ Therefore,
\begin{align*}  \mathsf{Re}(c) &= 108\left(3\cos\theta - 4\sin\theta \right) =  108\left(\frac{9}{5} + \frac{16}{5} \right)  = 108\cdot 5 = \boxed{\textbf{(540)}} \end{align*}
-Benedict T (countmath1)

answer: 540

question: Find the largest possible real part of \[(75+117i)z+\frac{96+144i}{z}\]where $z$ is a complex number with $|z|=4$.


problem: Among the 900 residents of Aimeville, there are 195 who own a diamond ring, 367 who own a set of golf clubs, and 562 who own a garden spade. In addition, each of the 900 residents owns a bag of candy hearts. There are 437 residents who own exactly two of these things, and 234 residents who own exactly three of these things. Find the number of residents of Aimeville who own all four of these things.

solution:
Let $w,x,y,z$ denote the number of residents who own 1,2,3 and 4 of these items, respectively. We know $w+x+y+z=900$, since there are 900 residents in total. This simplifies to
$w+z=229$, since we know $x=437$ and $y=234$.
Now, we set an equation of the total number of items. We know there are 195 rings, 367 clubs, 562 spades, and 900 candy hearts. Adding these up, there are 2024 (wow! the year!) items in total. Thus, $w+2x+3y+4z=2024$ since we are adding the number of items each group of people contributes, and this must be equal to the total number of items.
Plugging in x and y once more, we get $w+4z=448$. Solving $w+z=229$ and $w+4z=448$, we get $z=\boxed{073}$
-Westwoodmonster
Let $a,b,c$ denote the number of residents that own only a diamond ring and a bag of candy hearts, the number of residents that own only a golf club and a bag of candy hearts, and the number of residents that own only a garden spade and a bag of candy hearts, respectively. Let $x,y,z$ denote the number of residents that own only a diamond ring, a golf club, and a bag of candy hearts; the number of residents that own only a diamond ring, a garden spade, and a bag of candy hearts; and the number of residents that own only a golf club, a garden spade, and a bag of candy hearts. Let $n$ denote the number of people that own all $4$ items.
$a+x+y+n=195$ (the number of people that got diamond rings), $b+x+z+n=367$ (the number of people that got golf clubs), $c+y+z+n=562$ (the number of people that got garden spades). We also know $a+b+c=437$ (the number of people that own two objects), and $x+y+z=234$ (the number of people that own three objects). Adding the first three equations gives \[a+b+c+2(x+y+z)+3n=1124.\]
Substituting the second two equations gives $437+2\cdot 234+3n=1124$, so $n=\boxed{073}.$
~nezha33
We know that there are 195 diamond rings, 367 golf clubs, and 562 garden spades, so we can calculate that there are $195+367+562=1124$ items, with the exclusion of candy hearts which is irrelevant to the question. There are 437 people who owns 2 items, which means 1 item since candy hearts are irrelevant, and there are 234 people who own 2 items plus a bag of candy hearts, which means that the 234 people collectively own $234*2=468$ items. We can see that there are $1124-437-468=219$ items left, and since the question is asking us for the people who own 4 items, which means 3 items due to the irrelevance of candy hearts, we simply divide 219 by 3 and get $219/3=\boxed{073}$.
~Callisto531
Let $a$ be the number of people who have exactly one of these things and let $b$ be the number of people who have exactlty four of these objects. We have $a + 437 + 234 + d = 900,$ so $a + d = 229.$

Including those who have more than one object, we have
\[195 + 367 + 562 + 900 = a + 2\cdot 437 + 3\cdot 234 + 4d.\]
This is because we count those who own exactly $2$ objects twice, those who own $3$ thrice, and those who own $4$ four times. Solving gives $a + 4d = 448.$

Solving the system $a + 4d = 448, a + d = 229$ gives $3d = 219,$ so $d = \boxed{\textbf{(073)}}.$

-Benedict T (countmath1)

answer: 073

question: Among the 900 residents of Aimeville, there are 195 who own a diamond ring, 367 who own a set of golf clubs, and 562 who own a garden spade. In addition, each of the 900 residents owns a bag of candy hearts. There are 437 residents who own exactly two of these things, and 234 residents who own exactly three of these things. Find the number of residents of Aimeville who own all four of these things.


problem:
Let $\omega\neq 1$ be a 13th root of unity. Find the remainder when
\[\prod_{k=0}^{12}(2-2\omega^k+\omega^{2k})\]
is divided by 1000.

solution:
\[\prod_{k=0}^{12} \left(2- 2\omega^k + \omega^{2k}\right) = \prod_{k=0}^{12} \left((1 - \omega^k)^2 + 1\right) = \prod_{k=0}^{12} \left((1 + i) - \omega^k)((1 - i) - \omega^k\right)\]
Now, we consider the polynomial $x^{13} - 1$ whose roots are the 13th roots of unity. Taking our rewritten product from $0$ to $12$, we see that both instances of $\omega^k$ cycle through each of the 13th roots. Then, our answer is:
\[((1 + i)^{13} - 1)(1 - i)^{13} - 1)\]
\[= (-64(1 + i) - 1)(-64(1 - i) - 1)\]
\[= (65 + 64i)(65 - 64i)\]
\[= 65^2 + 64^2\]
\[= 8\boxed{\textbf{321}}\]
~Mqnic_
To find $\prod_{k=0}^{12} (2 - 2w^k + w^{2k})$, where $w\neq1$ and $w^{13}=1$, rewrite this is as
$(r-w)(s-w)(r-w^2)(s-w^2)...(r-w^{12})(s-w^{12})$ where $r$ and $s$ are the roots of the quadratic $x^2-2x+2=0$.
Grouping the $r$'s and $s$'s results in $\frac{r^{13}-1}{r-1} \cdot\frac{s^{13}-1}{s-1}$
the denomiator $(r-1)(s-1)=1$ by vietas.
the numerator $(rs)^{13} - (r^{13} + s^{13}) + 1 = 2^{13} - (-128) + 1= 8321$ by newtons sums
so the answer is $\boxed{321}$
-resources
Denote $r_j = e^{\frac{i 2 \pi j}{13}}$ for $j \in \left\{ 0, 1, \cdots , 12 \right\}$.
Thus, for $\omega \neq 1$, $\left( \omega^0, \omega^1, \cdots, \omega^{12} \right)$ is a permutation of $\left( r_0, r_1, \cdots, r_{12} \right)$.
We have
\begin{align*}\
\Pi_{k = 0}^{12} \left( 2 - 2 \omega^k + \omega^{2k} \right)
& = \Pi_{k=0}^{12} \left( 1 + i - \omega^k \right)
\left( 1 - i - \omega^k \right) \\
& = \Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - \omega^k \right)
\left( \sqrt{2} e^{-i \frac{\pi}{4}} - \omega^k \right) \\
& = \Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - r_k \right)
\left( \sqrt{2} e^{-i \frac{\pi}{4}} - r_k \right) \\
& = \left(
\Pi_{k=0}^{12} \left( \sqrt{2} e^{i \frac{\pi}{4}} - r_k \right)
\right)
\left(
\Pi_{k=0}^{12} \left( \sqrt{2} e^{-i \frac{\pi}{4}} - r_k \right)
\right) . \hspace{1cm} (1)
\end{align*}
The third equality follows from the above permutation property.
Note that $r_0, r_1, \cdots , r_{12}$ are all zeros of the polynomial $z^{13} - 1$.
Thus,
\[ z^{13} - 1 = \Pi_{k=0}^{12} \left( z - r_k \right) . \]
Plugging this into Equation (1), we get
\begin{align*}
(1)
& = \left( \left( \sqrt{2} e^{i \frac{\pi}{4}} \right)^{13} - 1 \right)
\left( \left( \sqrt{2} e^{-i \frac{\pi}{4}} \right)^{13} - 1 \right) \\
& = \left( - 2^{13/2} e^{i \frac{\pi}{4}} - 1 \right)
\left( - 2^{13/2} e^{-i \frac{\pi}{4}} - 1 \right) \\
& = 2^{13} + 1 + 2^{13/2} \cdot 2 \cos \frac{\pi}{4} \\
& = 2^{13} + 1 + 2^7 \\
& = 8321 .
\end{align*}
Therefore, the answer is $\boxed{\textbf{(321) }}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

answer: 321

question:
Let $\omega\neq 1$ be a 13th root of unity. Find the remainder when
\[\prod_{k=0}^{12}(2-2\omega^k+\omega^{2k})\]
is divided by 1000.


problem:
Find the number of ways to place a digit in each cell of a 2x3 grid so that the sum of the two numbers formed by reading left to right is $999$, and the sum of the three numbers formed by reading top to bottom is $99$. The grid below is an example of such an arrangement because $8+991=999$ and $9+9+81=99$.
\[\begin{array}{|c|c|c|} \hline 0 & 0 & 8 \\ \hline 9 & 9 & 1 \\ \hline \end{array}\]

solution:
Consider this table:
$\begin{array}{|c|c|c|} \hline a & b & c \\ \hline d & e & f\\ \hline \end{array}$
We note that $c+f = 9$, because $c+f \leq 18$, meaning it never achieves a unit's digit sum of $9$ otherwise. Since no values are carried onto the next digit, this implies $b+e=9$ and $a+d=9$. We can then simplify our table into this:
$\begin{array}{|c|c|c|} \hline a & b & c \\ \hline 9-a & 9-b & 9-c \\ \hline \end{array}$
We want $10(a+b+c) + (9-a+9-b+9-c) = 99$, or $9(a+b+c+3) = 99$, or $a+b+c=8$. Since zeroes are allowed, we just need to apply stars and bars on $a, b, c$, to get $\tbinom{8+3-1}{3-1} = \boxed{045}$. ~akliu

answer: 045

question:
Find the number of ways to place a digit in each cell of a 2x3 grid so that the sum of the two numbers formed by reading left to right is $999$, and the sum of the three numbers formed by reading top to bottom is $99$. The grid below is an example of such an arrangement because $8+991=999$ and $9+9+81=99$.
\[\begin{array}{|c|c|c|} \hline 0 & 0 & 8 \\ \hline 9 & 9 & 1 \\ \hline \end{array}\]


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

solution:
Denote $\log_2(x) = a$, $\log_2(y) = b$, and $\log_2(z) = c$.
Then, we have:
$a-b-c = \frac{1}{2}$
$-a+b-c = \frac{1}{3}$
$-a-b+c = \frac{1}{4}$
Now, we can solve to get $a = \frac{-7}{24}, b = \frac{-9}{24}, c = \frac{-5}{12}$. Plugging these values in, we obtain $|4a + 3b + 2c|  = \frac{25}{8} \implies \boxed{033}$. ~akliu
$\log_2(y/xz) + \log_2(z/xy) = \log_2(1/x^2) = -2\log_2(x) = \frac{7}{12}$
$\log_2(x/yz) + \log_2(z/xy) = \log_2(1/y^2) = -2\log_2(y) = \frac{3}{4}$
$\log_2(x/yz) + \log_2(y/xz) = \log_2(1/z^2) = -2\log_2(z) = \frac{5}{6}$
$\log_2(x) = -\frac{7}{24}$
$\log_2(y) = -\frac{3}{8}$
$\log_2(z) = -\frac{5}{12}$
$4\log_2(x) + 3\log_2(y) + 2\log_2(z) = -25/8$
$25 + 8 = \boxed{033}$
~Callisto531
Adding all three equations, $\log_2(\frac{1}{xyz}) = \frac{1}{2}+\frac{1}{3}+\frac{1}{4} = \frac{13}{12}$. Subtracting this from every equation, we have: \[2\log_2x = -\frac{7}{12},\] \[2\log_2y = -\frac{3}{4},\] \[2\log_2z = -\frac{5}{6}\] Our desired quantity is the absolute value of $4\log_2x+3\log_2y+2\log_2z = 2(\frac{7}{12})+3/2(\frac{3}{4})+\frac{5}{6} = \frac{25}{8}$, so our answer is $25+8 = \boxed{033}$.
~Spoirvfimidf

answer: 033

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