[u]Round 1[/u]

[b]p1.[/b] Solve the maze
[img]https://cdn.artofproblemsolving.com/attachments/8/c/6439816a52b5f32c3cb415e2058556edb77c80.png[/img]

[b]p2.[/b] Billiam can write a problem in $30$ minutes, Jerry can write a problem in $10$ minutes, and Evin can write a problem in $20$ minutes. Billiam begins writing problems alone at $3:00$ PM until Jerry joins himat $4:00$ PM, and Evin joins both of them at $4:30$ PM. Given that they write problems until the end of math team at $5:00$ PM, how many full problems have they written in total?


[b]p3.[/b] How many pairs of positive integers $(n,k)$ are there such that ${n \choose k}= 6$?


[u]Round 2 [/u]


[b]p4.[/b]  Find the sumof all integers $b > 1$ such that the expression of $143$ in base $b$ has an even number of digits and all digits are the same.


[b]p5.[/b]  Ιni thinks that $a \# b = a^2 - b$ and $a \& b = b^2 - a$, while Mimi thinks that $a \# b = b^2 - a$ and $a \& b = a^2 - b$. Both Ini and Mimi try to evaluate $6 \& (3 \# 4)$, each using what they think the operations $\&$ and $\#$ mean. What is the positive difference between their answers?


[b]p6.[/b] A unit square sheet of paper lies on an infinite grid of unit squares. What is the maximum number of grid squares that the sheet of paper can partially cover at once? A grid square is partially covered if the area of the grid square under the sheet of paper is nonzero - i.e., lying on the edge only does not count.


[u]Round 3[/u]


[b]p7.[/b] Maya wants to buy lots of burgers. A burger without toppings costs $\$4$, and every added topping increases the price by 50 cents. There are 5 different toppings for Maya to choose from, and she can put any combination of toppings on each burger. How much would it cost forMaya to buy $1$ burger for each distinct set of toppings? Assume that the order in which the toppings are stacked onto the burger does not matter.


[b]p8.[/b]  Consider square $ABCD$ and right triangle $PQR$ in the plane. Given that both shapes have area $1$, $PQ =QR$, $PA = RB$, and $P$, $A$, $B$ and $R$ are collinear, find the area of the region inside both square $ABCD$ and $\vartriangle PQR$, given that it is not $0$.


[b]p9.[/b] Find the sum of all $n$ such that $n$ is a $3$-digit perfect square that has the same tens digit as $\sqrt{n}$, but that has a different ones digit than $\sqrt{n}$.


[u]Round 4[/u] 


[b]p10.[/b]  Jeremy writes the string: $$LMTLMTLMTLMTLMTLMT$$ on a whiteboard (“$LMT$” written $6$ times). Find the number of ways to underline $3$ letters such that from left to right the underlined letters spell LMT.


[b]p11.[/b] Compute the remainder when $12^{2022}$ is divided by $1331$.


[b]p12.[/b] What is the greatest integer that cannot be expressed as the sum of $5$s, $23$s, and $29$s?


[u]Round  5 [/u]


[b]p13.[/b] Square $ABCD$ has point $E$ on side $BC$, and point $F$ on side $CD$, such that $\angle EAF = 45^o$. Let $BE = 3$, and $DF = 4$. Find the length of $FE$.


[b]p14.[/b]  Find the sum of all positive integers $k$ such that
$\bullet$ $k$ is the power of some prime.
$\bullet$ $k$ can be written as $5654_b$ for some $b > 6$.


[b]p15.[/b]  If $\sqrt[3]{x} + \sqrt[3]{y} = 2$ and $x + y = 20$, compute $\max \,(x, y)$.


PS. You should use hide for answers. Rounds 6-9 have been posted [url=https://artofproblemsolving.com/community/c3h3167372p28825861]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. Given the equations:
   \[
   \sqrt[3]{x} + \sqrt[3]{y} = 2
   \]
   and
   \[
   x + y = 20
   \]

2. We start by cubing both sides of the first equation:
   \[
   (\sqrt[3]{x} + \sqrt[3]{y})^3 = 2^3
   \]
   Expanding the left-hand side using the binomial theorem, we get:
   \[
   \sqrt[3]{x}^3 + 3\sqrt[3]{x}^2\sqrt[3]{y} + 3\sqrt[3]{x}\sqrt[3]{y}^2 + \sqrt[3]{y}^3 = 8
   \]
   Simplifying, we have:
   \[
   x + 3\sqrt[3]{x^2y} + 3\sqrt[3]{xy^2} + y = 8
   \]

3. We know from the second equation that \(x + y = 20\). Substituting this into the expanded equation, we get:
   \[
   20 + 3\sqrt[3]{xy}(\sqrt[3]{x} + \sqrt[3]{y}) = 8
   \]

4. Since \(\sqrt[3]{x} + \sqrt[3]{y} = 2\), we substitute this into the equation:
   \[
   20 + 3 \cdot 2 \cdot \sqrt[3]{xy} = 8
   \]
   Simplifying, we get:
   \[
   20 + 6\sqrt[3]{xy} = 8
   \]

5. Solving for \(\sqrt[3]{xy}\), we subtract 20 from both sides:
   \[
   6\sqrt[3]{xy} = 8 - 20
   \]
   \[
   6\sqrt[3]{xy} = -12
   \]
   \[
   \sqrt[3]{xy} = -2
   \]

6. Cubing both sides to solve for \(xy\), we get:
   \[
   (\sqrt[3]{xy})^3 = (-2)^3
   \]
   \[
   xy = -8
   \]

7. Now, we solve the system of equations:
   \[
   x + y = 20
   \]
   \[
   xy = -8
   \]

8. These are the roots of the quadratic equation:
   \[
   t^2 - (x+y)t + xy = 0
   \]
   Substituting the known values:
   \[
   t^2 - 20t - 8 = 0
   \]

9. Solving this quadratic equation using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
   \[
   t = \frac{20 \pm \sqrt{20^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1}
   \]
   \[
   t = \frac{20 \pm \sqrt{400 + 32}}{2}
   \]
   \[
   t = \frac{20 \pm \sqrt{432}}{2}
   \]
   \[
   t = \frac{20 \pm 12\sqrt{3}}{2}
   \]
   \[
   t = 10 \pm 6\sqrt{3}
   \]

10. Therefore, the solutions for \(x\) and \(y\) are:
    \[
    x = 10 + 6\sqrt{3}, \quad y = 10 - 6\sqrt{3}
    \]
    or
    \[
    x = 10 - 6\sqrt{3}, \quad y = 10 + 6\sqrt{3}
    \]

11. The maximum value of \(x\) or \(y\) is:
    \[
    \max(x, y) = 10 + 6\sqrt{3}
    \]

The final answer is \(\boxed{10 + 6\sqrt{3}}\)