[b]p1.[/b] Let $p > 5$ be a prime. It is known that the average of all of the prime numbers that are at least $5$ and at most $p$ is $12$. Find $p$.


[b]p2.[/b] The numbers $1, 2,..., n$ are written down in random order. What is the probability that $n-1$ and $n$ are written next to each other? (Give your answer in term of $n$.)


[b]p3.[/b] The Duke Blue Devils are playing a basketball game at home against the UNC Tar Heels. The Tar Heels score $N$ points and the Blue Devils score $M$ points, where $1 < M,N < 100$. The first digit of $N$ is $a$ and the second digit of $N$ is $b$. It is known that $N = a+b^2$. The first digit of $M$ is $b$ and the second digit of $M$ is $a$. By how many points do the Blue Devils win?


[b]p4.[/b] Let $P(x)$ be a polynomial with integer coefficients. It is known that $P(x)$ gives a remainder of $1$ upon polynomial division by $x + 1$ and a remainder of $2$ upon polynomial division by $x + 2$. Find the remainder when $P(x)$ is divided by $(x + 1)(x + 2)$.



[b]p5.[/b] Dracula starts at the point $(0,9)$ in the plane. Dracula has to pick up buckets of blood from three rivers, in the following order: the Red River, which is the line $y = 10$; the Maroon River, which is the line $y = 0$; and the Slightly Crimson River, which is the line $x = 10$. After visiting all three rivers, Dracula must then bring the buckets of blood to a castle located at $(8,5)$. What is the shortest distance that Dracula can walk to accomplish this goal?


[b]p6.[/b] Thirteen hungry zombies are sitting at a circular table at a restaurant. They have five identical plates of zombie food. Each plate is either in front of a zombie or between two zombies. If a plate is in front of a zombie, that zombie and both of its neighbors can reach the plate. If a plate is between two zombies, only those two zombies may reach it. In how many ways can we arrange the plates of food around the circle so that each zombie can reach exactly one plate of food? (All zombies are distinct.)


[b]p7.[/b] Let $R_I$ , $R_{II}$ ,$R_{III}$ ,$R_{IV}$ be areas of the elliptical region $$\frac{(x - 10)^2}{10}+ \frac{(y-31)^2}{31} \le 2009$$ that lie in the first, second, third, and fourth quadrants, respectively. Find $R_I -R_{II} +R_{III} -R_{IV}$ .


[b]p8.[/b] Let $r_1, r_2, r_3$ be the three (not necessarily distinct) solutions to the equation $x^3+4x^2-ax+1 = 0$. If $a$ can be any real number, find the minimum possible value of
 $$\left(r_1 +\frac{1}{r_1} \right)^2+ \left(r_2 +\frac{1}{r_2} \right)^2+   \left(r_3 +\frac{1}{r_3} \right)^2$$


[b]p9.[/b] Let $n$ be a positive integer. There exist positive integers $1 = a_1 < a_2 <... < a_n = 2009$ such that the average of any $n - 1$ of elements of $\{a_1, a_2,..., a_n\}$ is a positive integer. Find the maximum possible value of $n$.


[b]p10.[/b] Let $A(0) = (2, 7, 8)$ be an ordered triple. For each $n$, construct $A(n)$ from $A(n - 1)$ by replacing the $k$th position in $A(n - 1)$ by the average (arithmetic mean) of all entries in $A(n - 1)$, where $k \equiv n$ (mod $3$) and $1 \le k \le 3$. For example, $A(1) = \left( \frac{17}{3} , 7, 8 \right)$ and $A(2)  = \left( \frac{17}{3} , \frac{62}{9}, 8\right)$. It is known that all entries converge to the same number $N$. Find the value of $N$.


PS. You had better use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].
1. To find the probability that \( n-1 \) and \( n \) are written next to each other in a random permutation of the numbers \( 1, 2, \ldots, n \), we start by considering the total number of permutations of these \( n \) numbers, which is \( n! \).

2. Next, we need to count the number of permutations where \( n-1 \) and \( n \) are adjacent. We can treat the pair \( (n-1, n) \) as a single "block". This block can be placed in any of the \( n-1 \) positions in the sequence of \( n \) numbers. 

3. For each position of the block, the remaining \( n-2 \) numbers can be permuted in \( (n-2)! \) ways. Since the block \( (n-1, n) \) can appear as either \( (n-1, n) \) or \( (n, n-1) \), we have 2 choices for each position of the block.

4. Therefore, the total number of favorable permutations is \( 2 \cdot (n-1) \cdot (n-2)! \).

5. The probability is then given by the ratio of the number of favorable permutations to the total number of permutations:
\[
\frac{2 \cdot (n-1) \cdot (n-2)!}{n!}
\]

6. Simplifying this expression:
\[
\frac{2 \cdot (n-1) \cdot (n-2)!}{n \cdot (n-1) \cdot (n-2)!} = \frac{2}{n}
\]

The final answer is \(\boxed{\frac{2}{n}}\).