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 $100a + b$.

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 $100a + b.$

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.

Blue rolls a fair $n$-sided die that has sides its numbered with the integers from $1$ to $n$, and then he flips a coin. Blue knows that the coin is weighted to land heads either $\dfrac{1}{3}$ or $\dfrac{2}{3}$ of the time. Given that the probability of both rolling a $7$ and flipping heads is $\dfrac{1}{15}$, find $n$. [i]Proposed by Jacob Xu[/i] [hide=Solution][i]Solution[/i]. $\boxed{10}$ The chance of getting any given number is $\dfrac{1}{n}$ , so the probability of getting $7$ and heads is either $\dfrac{1}{n} \cdot \dfrac{1}{3}=\dfrac{1}{3n}$ or $\dfrac{1}{n} \cdot \dfrac{2}{3}=\dfrac{2}{3n}$. We get that either $n = 5$ or $n = 10$, but since rolling a $7$ is possible, only $n = \boxed{10}$ is a solution.[/hide]

There are $n$ cities in a country, where $n \geq 100$ is an integer. Some pairs of cities are connected by direct (two-way) flights. For two cities $A$ and $B$ we define: $(i)$ A $\emph{path}$ between $A$ and $B$ as a sequence of distinct cities $A = C_0, C_1, \dots, C_k, C_{k+1} = B$, $k \geq 0$, such that there are direct flights between $C_i$ and $C_{i+1}$ for every $0 \leq i \leq k$; $(ii)$ A $\emph{long path}$ between $A$ and $B$ as a path between $A$ and $B$ such that no other path between $A$ and $B$ has more cities; $(iii)$ A $\emph{short path}$ between $A$ and $B$ as a path between $A$ and $B$ such that no other path between $A$ and $B$ has fewer cities. Assume that for any pair of cities $A$ and $B$ in the country, there exist a long path and a short path between them that have no cities in common (except $A$ and $B$). Let $F$ be the total number of pairs of cities in the country that are connected by direct flights. In terms of $n$, find all possible values $F$ Proposed by David-Andrei Anghel, Romania.

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]