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.

Let $N$ be a positive integer. Brothers Michael and Kylo each select a positive integer less than or equal to $N$, independently and uniformly at random. Let $p_N$ denote the probability that the product of these two integers has a units digit of $0$. The maximum possible value of $p_N$ over all possible choices of $N$ can be written as $\tfrac ab,$ where $a$ and $b$ are relatively prime positive integers. Compute $100a + b$.

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]

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.

Assume $\Omega(n),\omega(n)$ be the biggest and smallest prime factors of $n$ respectively . Alireza and Amin decided to play a game. First Alireza chooses $1400$ polynomials with integer coefficients. Now Amin chooses $700$ of them, the set of polynomials of Alireza and Amin are $B,A$ respectively . Amin wins if for all $n$ we have : $$\max_{P \in A}(\Omega(P(n))) \ge \min_{P \in B}(\omega(P(n)))$$ Who has the winning strategy. Proposed by [i]Alireza Haghi[/i]