Consider a $2 \times n$ grid where each cell is either black or white, which we attempt to tile with $2 \times 1$ black or white tiles such that tiles have to match the colors of the cells they cover. We first randomly select a random positive integer $N$ where $N$ takes the value $n$ with probability $\frac{1}{2^n}$. We then take a $2 \times N$ grid and randomly color each cell black or white independently with equal probability. Compute the probability the resulting grid has a valid tiling.

Evin and Jerry are playing a game with a pile of marbles. On each players' turn, they can remove $2$, $3$, $7$, or $8$ marbles. If they can’t make a move, because there's $0$ or $1$ marble left, they lose the game. Given that Evin goes first and both players play optimally, for how many values of $n$ from $1$ to $1434$ does Evin lose the game? [i]Proposed by Evin Liang[/i] [hide=Solution][i]Solution.[/i] $\boxed{573}$ Observe that no matter how many marbles a one of them removes, the next player can always remove marbles such that the total number of marbles removed is $10$. Thus, when the number of marbles is a multiple of $10$, the first player loses the game. We analyse this game based on the number of marbles modulo $10$: If the number of marbles is $0$ modulo $10$, the first player loses the game If the number of marbles is $2$, $3$, $7$, or $8$ modulo $10$, the first player wins the game by moving to $0$ modulo 10 If the number of marbles is $5$ modulo $10$, the first player loses the game because every move leads to $2$, $3$, $7$, or $8$ modulo $10$ In summary, the first player loses if it is $0$ mod 5, and wins if it is $2$ or $3$ mod $5$. Now we solve the remaining cases by induction. The first player loses when it is $1$ modulo $5$ and wins when it is $4$ modulo $5$. The base case is when there is $1$ marble, where the first player loses because there is no move. When it is $4$ modulo $5$, then the first player can always remove $3$ marbles and win by the inductive hypothesis. When it is $1$ modulo $5$, every move results in $3$ or $4$ modulo $5$, which allows the other player to win by the inductive hypothesis. Thus, Evin loses the game if n is $0$ or $1$ modulo $5$. There are $\boxed{573}$ such values of $n$ from $1$ to $1434$.[/hide]

Let $m$ and $n$ be positive integers. In Philand, the Kingdom of Olymphics, with $m$ cities, and the Kingdom of Mathematicians for Fun, with $n$ cities, fight a battle in rounds. Some cities in the country are connected by roads, so that it is possible to travel through all the cities via the roads. In each round of the battle, if all cities neighboring, that is, connected directly by a road, a city in one of the kingdoms are from the other kingdom, that city is conquered in the next round and switches to the other kingdom. Knowing that between the first and second round, at least one city is not conquered, show that at some point the battle must end, i.e., no city can be captured by another kingdom.

An up-right path between two lattice points $P$ and $Q$ is a path from $P$ to $Q$ that takes steps of length $1$ unit either up or to the right. How many up-right paths from $(0, 0)$ to $(7, 7),$ when drawn in the plane with the line $y = x - 2.021$, enclose exactly one bounded region below that line?

Prair has a box with some combination of red and green balls. If she randomly draws two balls out of the box (without replacement), the probability of drawing two balls of the same color is equal to the probability of drawing two balls of different colors! How many possible values between $200$ and $1000$ are there for the total number of balls in the box? [i]Proposed by Andrew Wu[/i]

Let $S = \{1, 2, \dots, 9\}.$ Compute the number of functions $f : S \rightarrow S$ such that, for all $s \in S, f(f(f(s))) =s$ and $f(s) - s$ is not divisible by $3$.

A light pulse starts at a corner of a reflective square. It bounces around inside the square, reflecting off of the square’s perimeter $n$ times before ending in a different corner. The path of the light pulse, when traced, divides the square into exactly $2021$ regions. Compute the smallest possible value of $n$.

Evaluate $\dbinom{6}{0}+\dbinom{6}{1}+\dbinom{6}{4}+\dbinom{6}{3}+\dbinom{6}{4}+\dbinom{6}{5}+\dbinom{6}{6}$ [i]Proposed by Jonathan Liu[/i] [hide=Solution] [i]Solution.[/i] $\boxed{64}$ We have that $\dbinom{6}{4}=\dbinom{6}{2}$, so $\displaystyle\sum_{n=0}^{6} \dbinom{6}{n}=2^6=\boxed{64}.$ [/hide]

The numbers $1$, $2$, $3$, and $4$ are randomly arranged in a $2$ by $2$ grid with one number in each cell. Find the probability the sum of two numbers in the top row of the grid is even. [i]Proposed by Muztaba Syed and Derek Zhao[/i] [hide=Solution] [i]Solution. [/i]$\boxed{\dfrac{1}{3}}$ Pick a number for the top-left. There is one number that makes the sum even no matter what we pick. Therefore, the answer is $\boxed{\dfrac{1}{3}}$.[/hide]

The numbers $1$, $2$, $3$, and $4$ are randomly arranged in a $2$ by $2$ grid with one number in each cell. Find the probability the sum of two numbers in the top row of the grid is even. [i]Proposed by Muztaba Syed and Derek Zhao[/i] [hide=Solution] [i]Solution. [/i]$\boxed{\dfrac{1}{3}}$ Pick a number for the top-left. There is one number that makes the sum even no matter what we pick. Therefore, the answer is $\boxed{\dfrac{1}{3}}$.[/hide]

Find, with proof, all polynomials $f$ such that $f$ has nonnegative integer coefficients, $f$($1$) = $8$ and $f$($2$) = $2012$.

Teresa the bunny has a fair $8$-sided die. Seven of its sides have fixed labels $1, 2, \cdots , 7,$ and the label on the eighth side can be changed and begins as $1$. She rolls it several times, until each of $1, 2, \dots, 7$ appears at least once. After each roll, if $k$ is the smallest positive integer that she has not rolled so far, she relabels the eighth side with $k$. The probability that $7$ is the last number she rolls is $\tfrac ab,$ where $a$ and $b$ are relatively prime positive integers. Compute $100a + b$.

Leo the fox has a $5$ by $5$ checkerboard grid with alternating red and black squares. He fills in the grid with the numbers $1, 2, 3, \dots, 25$ such that any two consecutive numbers are in adjacent squares (sharing a side) and each number is used exactly once. He then computes the sum of the numbers in the $13$ squares that are the same color as the center square. Compute the maximum possible sum Leo can obtain.

Let $S = \{1, 2, \dots , 2021\}$, and let $\mathcal{F}$ denote the set of functions $f : S \rightarrow S$. For a function $f \in \mathcal{F},$ let \[T_f =\{f^{2021}(s) : s \in S\},\] where $f^{2021}(s)$ denotes $f(f(\cdots(f(s))\cdots))$ with $2021$ copies of $f$. Compute the remainder when \[\sum_{f \in \mathcal{F}} |T_f|\] is divided by the prime $2017$, where the sum is over all functions $f$ in $\mathcal{F}$.

Bamal, Halvan, and Zuca are playing [i]The Game[/i]. To start, they‘re placed at random distinct vertices on regular hexagon $ABCDEF$. Two or more players collide when they‘re on the same vertex. When this happens, all the colliding players lose and the game ends. Every second, Bamal and Halvan teleport to a random vertex adjacent to their current position (each with probability $\dfrac{1}{2}$), and Zuca teleports to a random vertex adjacent to his current position, or to the vertex directly opposite him (each with probability $\dfrac{1}{3}$). What is the probability that when [i]The Game[/i] ends Zuca hasn‘t lost? [i]Proposed by Edwin Zhao[/i] [hide=Solution][i]Solution.[/i] $\boxed{\dfrac{29}{90}}$ Color the vertices alternating black and white. By a parity argument if someone is on a different color than the other two they will always win. Zuca will be on opposite parity from the others with probability $\dfrac{3}{10}$. They will all be on the same parity with probability $\dfrac{1}{10}$. At this point there are $2 \cdot 2 \cdot 3$ possible moves. $3$ of these will lead to the same arrangement, so we disregard those. The other $9$ moves are all equally likely to end the game. Examining these, we see that Zuca will win in exactly $2$ cases (when Bamal and Halvan collide and Zuca goes to a neighboring vertex). Combining all of this, the answer is $$\dfrac{3}{10}+\dfrac{2}{9} \cdot \dfrac{1}{10}=\boxed{\dfrac{29}{90}}$$ [/hide]

Ava and Tiffany participate in a knockout tournament consisting of a total of $32$ players. In each of $5$ rounds, the remaining players are paired uniformly at random. In each pair, both players are equally likely to win, and the loser is knocked out of the tournament. The probability that Ava and Tiffany play each other during the tournament is $\tfrac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. Compute $100a + b.$

Evin and Jerry are playing a game with a pile of marbles. On each players' turn, they can remove $2$, $3$, $7$, or $8$ marbles. If they can’t make a move, because there's $0$ or $1$ marble left, they lose the game. Given that Evin goes first and both players play optimally, for how many values of $n$ from $1$ to $1434$ does Evin lose the game? [i]Proposed by Evin Liang[/i] [hide=Solution][i]Solution.[/i] $\boxed{573}$ Observe that no matter how many marbles a one of them removes, the next player can always remove marbles such that the total number of marbles removed is $10$. Thus, when the number of marbles is a multiple of $10$, the first player loses the game. We analyse this game based on the number of marbles modulo $10$: If the number of marbles is $0$ modulo $10$, the first player loses the game If the number of marbles is $2$, $3$, $7$, or $8$ modulo $10$, the first player wins the game by moving to $0$ modulo 10 If the number of marbles is $5$ modulo $10$, the first player loses the game because every move leads to $2$, $3$, $7$, or $8$ modulo $10$ In summary, the first player loses if it is $0$ mod 5, and wins if it is $2$ or $3$ mod $5$. Now we solve the remaining cases by induction. The first player loses when it is $1$ modulo $5$ and wins when it is $4$ modulo $5$. The base case is when there is $1$ marble, where the first player loses because there is no move. When it is $4$ modulo $5$, then the first player can always remove $3$ marbles and win by the inductive hypothesis. When it is $1$ modulo $5$, every move results in $3$ or $4$ modulo $5$, which allows the other player to win by the inductive hypothesis. Thus, Evin loses the game if n is $0$ or $1$ modulo $5$. There are $\boxed{573}$ such values of $n$ from $1$ to $1434$.[/hide]